aa r X i v : . [ m a t h . G R ] J a n Report on Freely Representable Groups
Wayne Aitken ∗ February 2, 2021
Abstract
This report is an account of freely representable groups, which are finitegroups admitting linear representations whose only fixed point for a noniden-tity element is the zero vector. The standard reference for such groups isWolf (1967) where such groups are used to classify spaces of constant positivecurvature. Such groups also arise in the theory of norm relations in algebraicnumber theory, as demonstrated recently by Biasse, Fieker, Hofmann, andPage (2020). This report aims to synthesize the information and results fromthese and other sources to give a continuous, self-contained development ofthe subject. I introduce new points of view, terminology, results, and proofs inan effort to give a coherent, detailed, self-contained, and accessible narrative.
This is an account of freely representable groups, which are finite groups admit-ting linear representations whose only fixed point for a nonidentity element is thezero vector. Such groups attracted my attention as being exactly the groups thatdo not have a general type of norm relation (as recently shown in [4]). Interestingly,such groups arose earlier as the key to the classification of Riemannian manifoldswith constant positive curvature (see [17]). Such groups were studied even earlierfrom a purely group theoretical point of view. For example, in 1905, W. Burn-side [5] proved necessary conditions for a group to be freely representable. In 1955,S. A. Amitsur [2] observed that all finite subgroups of a division ring are freelyrepresentable groups. Amitsur used this as the starting point of his classificationof such groups; the actual classification required class field theory. Finally, thesegroups are related to the subject of Frobenius complements in finite group theoryand groups with periodic cohomology. Freely representable groups include cyclic groups, and in general behave likecyclic groups in many interesting ways. Non-cyclic examples of freely representablegroups can be readily given. In fact, the quaternion group with 8 elements is a ∗ Thanks to Shahed Sharif and his student Antony Savage for showing me the usefulness ofnorm relations in the problem of finding units in algebraic number theory, and further thanks toShahed Sharif for discussions in 2019 that led me to wonder what groups have norm relations ofunity. Thanks to the authors of [4] for giving nice answers to such questions and for promptingmy interest in freely representable groups, an interest that led to this report. Finally, thanks tomy student Jason Martin for his help in reviewing various drafts of this report. See Wall [16] for periodic cohomology. This report does not consider Amitsur’s classification,Frobenius complements, or periodic cohomology. I hope to include some of these topics in sequelsor in a future version of the current report. My main sources for this material are [17], [4], and [14]. I have also incorporatedsome insights of Allcock [1] which makes several improvements to Wolf [17]. Ihave also consulted several standard works on finite group theory. In fact, a majorpurpose of this report is to synthesize the information and results from these varioussources to give a continuous, self-contained development of the subject. There isstill work to be done in this synthesis, but I think this report is a good start.I have tried to simplify and polish proofs whenever possible, and introduce newpoints of view, terminology, and results whenever they help the overall narrative.New proofs are given to several of the results of [17] with the goal of making theproofs of these results more accessible. For example Wolf uses Burnside’s theoremconcerning Sylow subgroups central in their normalizers, where my proofs avoid thistechnique. I also avoid transfer techniques more generally, and I avoid the Schur-Zassenhaus theorem (used in [1]). I explore Sylow-cyclic groups in more detail thanWolf and gives some results not found in Wolf [17] which might be new: for exampleTheorem 61 concerning the conjugacy of subgroups of the same order may be new.The characterization of freely representable groups in terms of the existence ofunique subgroups of each prime order (see Theorem 81) may be new. I make moreuse of the idea of a maximum cyclic conjugate (MCC) subgroup, which leads toa different perspective from Wolf, and the resulting theorem on the existence of aunique element of order 2 may be new (Corollary 67). The results from [4] cited hereon norm relations are given more elementary proofs which use less representationtheory (for example, the proof here does not use central primitive idempotents),and they are generalized a bit. In general the paper [4] has inspired me to use normrelations as a tool to study freely representable groups. For example, I have castethe proof of the pq -theorem (Theorem 18) using norm relations, which seems to benovel point of view, and have used norm relations in a few other places (Lemma 73and Theorem 74) to show that groups are not freely representable.I believe this report is reasonably self-contained at least for the solvable case;appendices have been given to help make the report more accessible. The non-solvable case requires me to draw from some high-level sources such as Suzuki [14].I hope to give complete proofs even in the non-solvable case in a sequel or futureedition of this report. Suppose a group G acts on a set S . We say that G acts freely if for all g = 1 in G and all s ∈ S we have gs = s . In other words, no nonidentity element has a fixed There are actually several types of “cycloidal groups” considered in this report besides freelyrepresentable groups. One interesting class is the collection of semiprime-cyclic groups. These aregroups with the property that any subgroup of order the product of two primes is cyclic. Theremarkable fact is that freely representable group and semiprime-cyclic groups correspond exactlyfor solvable groups. The semiprime-cyclic groups will be considered in more detail, from anotherperspective, in the sequel,
The Funakura Invariant and Norm Relations . G acts linearly on a vector space V . We say that the action is a free linear representation if V = 0 and if G acts freely on the set of nonzero vectorsof V . In other words, for all g ∈ G , the associated linear transformation V → V has eigenvalue 1 if and only if g = 1.Here are a few observations related to this definition: Lemma 1. If G is a free linear representation of V then any nonzero subrepresen-tation is also free. Lemma 2.
A free linear representation of V over F is faithful. In other words, ityields an injective homomorphism G ֒ → GL( V ) .Proof. Suppose g ∈ G is in the kernel of G → GL( V ). Let v = 0 be in V .Then gv = v since g is in the kernel. So g = 1 since G acts freely on nonzerovectors.A group G is freely representable over a field F if it possesses a free linearrepresentation of an F -vector space V . In this report we focus on finite groupsonly, so it is understood that all freely representable groups we consider are finite.This allows us to restrict our attention, when convenient, to finite dimensionalvector spaces: Lemma 3.
Let G be a finite group and let F be a field. If G is freely representableover F then there is a free linear representation of G on a finite dimensional vectorspace V . In fact we can choose V to be of dimension at most | G | .Proof. By assumption G has a free linear representation on some vector space W .By definition, W is not the zero space; let w ∈ W be a nonzero vector and let V be the span of { σw | σ ∈ G } . The result now follows from Lemma 1 Corollary 4.
Let G be a finite group and let F be a field. Then G is freely repre-sentable over F if and only if there is an irreducible free linear representation of G on a finite dimensional F -vector space V . Combining Lemma 3 with Lemma 2 yields the following:
Corollary 5.
Let G be a finite group and let F be a field. Then G is freely repre-sentable over F if and only if it is isomorphic to a subgroup Γ of GL n ( F ) , for somepositive integer n , such that the only element of Γ with eigenvalue is the identity. Note we can even take n be to at most | G | in the above corollary if we wish.If we do not specify F it is understood that F is C . This default is reason-able since C is in some sense the most basic field used in representation theory.But note that free representations over R are central to the theory of Riemannianmanifolds with constant positive curvature, so one might argue that R could havebeen a reasonable default. The following proposition shows this question is moot:there is no distinction between being freely representable over C and being freelyrepresentable over R . Lemma 6.
If a finite group G is freely representable over a field F then it is freelyrepresentable over all fields of the same characteristic at that of F . roof. Let F be a field and let F be its prime subfield. If we have a free repre-sentation of G on an F -vector space V then observe that it is a free representationof V regarded as an F -vector space (where we allow the dimension to increase).Conversely, suppose G is freely representable over F . By Corollary 5 we canassume that G is a finite subgroup of GL n ( F ) where n ≥ G with eigenvalue 1 is the identity. In other words, det( g − I ) = 0 forall g ∈ G not equal to the identity. Note that GL n ( F ) is a subgroup of GL n ( F ), andthe determinant of g − I with g ∈ GL n ( F ) is the same whether we take determinantsin GL n ( F ) or in GL n ( F ). Thus, by Corollary 5, G is freely representable over F .Now we consider some basic examples that will prove to be central in whatfollows: Example . Every cyclic group is freely representable. To see this let G be thegroup of N th roots of unity. This group G acts on V = C n by scalar multiplicationfor all n ≥
1. This construction gives a free representation over C of dimension n .Now if we view V = C n as a real vector space, we get instead a free representationover R of dimension 2 n . We can also find free representations of a cyclic group G over R directly by considering finite groups of rotations of R fixing the origin. Ofcourse this is equivalent, in the case n = 1, to the early construction. Example . We can extend the stock of examples in a very interesting way byreplacing C in Example 1 with the quaternions H . Let G be a finite multiplicativesubgroup of H × . Then G acts on H n by scalar multiplication. Now identify H n with R n in the usual way, and observe we get a 4 n -dimensional free representationover R . (See the relevant appendix for more information on the division algebra H .)We conclude that every finite subgroup of H × is a freely representable group.In particular, the quaternion group is a freely representable noncyclic group with 8elements.The classification of finite subgroups of H × is easy to describe in terms of finiterotations groups of R . First observe that any finite subgroup of H × sits in thegroup of elements H of norm 1, which is topologically the 3-sphere. There isa two-to-one surjective homomorphism H → SO(3) which sends h ∈ H to therotation v hvh − where v ∈ R and where R is identified with the R -span of the quaternions i , j , k .(See the appropriate appendix for more information). The kernel is {± } .This two-to-one mapping H → SO(3) can be used to classify finite subgroupsof H in terms of finite subgroups of SO(3). The fact that − H of order 2 helps us describe the correspondence (see Lemma 8).First consider any finite subgroup G of H of odd order . Then the restrictionof H → SO(3) to G has trivial kernel, so G is naturally isomorphic to a subgroupof SO(3) of odd order. All finite subgroups of SO(3) of odd order are cyclic, so G must be be cyclic in this case. Since H contains the unit circle subgroup of C × asa subgroup, we have cyclic subgroup of all odd orders in H . It turns out that the group H is isomorphic to the Lie group SU(2). We do not require thisfact in this report.
4e can form finite subgroups G of H of even order by taking preimages of finitesubgroups of SO(3). Conversely, every finite subgroup G of H of even order mustcontain − − H of order 2. This implies that sucha G is the preimage of its image (see Lemma 8 for details). So to classify the evenorder subgroups of H we can just look at preimages of finite subgroups of SO(3).See the appropriate appendix for a classification of subgroups of SO(3).We start the classification of even ordered subgroups of H by looking at thepreimage L of a cyclic subgroup C k of order k . The preimage of a group isomorphicto C k is isomorphic to C k (Lemma 8). Next we consider the preimage of noncyclicsubgroups of SO(3). Observe that the preimage of a noncyclic group cannot bea cyclic group. So the classification of noncyclic finite subgroups of SO(3) nowgives us a classification of noncyclic finite subgroup of H . Every noncyclic finitesubgroup of H is the preimage of one of the following: a dihedral group D m oforder 2 m where m ≥
2, the tetrahedral group T (which is isomorphic to A ),the octahedral group O (which is isomorphic to S ), or the icosahedral group I (which is isomorphic to A ). We call G a binary dihedral group, binary tetrahedralgroup, binary octahedral group , or binary iscosahedral group depending on its imagein SO(3). In this classification we include D , the dihedral group of order 4; thisis just the Klein four group C × C . The quaternion group of size 8 has such a D as its image. Remark.
With a little work, we can show that binary dihedral groups of equalorder are isomorphic, and up to isomorphism there is a unique binary tetrahedralgroup, a unique binary octahedral group, and a binary icosahedral group. (See thefollowing proposition). We denote these groups as 2 D n , T, O, and 2 I . Proposition 7.
Let G and G be subgroups of H of the same order and whoseimages under the standard map π : H → SO(3) are isomorphic. Then G and G are isomorphic. In fact, they are conjugate subgroups of H .Proof. We take it as known that isomorphic subgroups of SO(3) are conjugatesubgroups of SO(3). Thus π [ G ] = γ − π [ G ] γ for some γ ∈ SO(3). Now choose anelement h ∈ H mapping to γ and our goal is to show that G = h − G h .We start with the case where G and G have even order. Since G and h − G h have the same order it is enough to show the inclusion G ⊆ h − G h . So let g ∈ G .Then π ( g ) = γ − π ( g ) γ for some g ∈ G . Observe that g and h − g h have thesame image, so g − h − g h is in the kernel of π . But the kernel of π is {± } . So g = h − ( ± g ) h. But ± g ∈ G since − ∈ G (since G has even order). Thus G ⊆ h − G h asdesired.The remaining case is where G and G are odd of order k . Let G ′ i be thepreimage of π [ G i ]. In this case G i and π [ G i ] are cyclic of order k and so the G ′ i are cyclic of order 2 k . By the above argument, G ′ and G ′ are conjugate groups.Since G i is the unique subgroup of G ′ i of index 2, it follows that G and G mustbe conjugate as well. The binary tetrahedral group is isomorphic to SL ( F ), and the binary icosahedral group isisomorphic to SL ( F ). H and their images in SO(3) are part of a more general phenomenon: Lemma 8.
Let π : G → M be a surjective homomorphism between groups withkernel K of size . Then the following hold:1. The map π : G → M is a -to- map: the preimage of each t ∈ M has size .2. The preimage in G of any finite subgroup L of M is a finite subgroup of G oforder twice the order of L .3. Every finite subgroup H of G of odd order is isomorphic to its image in M .4. If g ∈ G has odd finite order then g and π ( g ) have the same order.Furthermore if G has a unique element of order then also5. Every finite subgroup H of G of even order is the preimage of its image π [ H ] .6. Every finite subgroup H of G of even order has order twice that of its im-age π [ H ] .7. If g ∈ G has even finite order then π ( g ) has order one-half the order of g .8. Let L be a finite cyclic subgroup of M of odd order k . Then its preimage in G is a cyclic subgroup of order k .9. Let L be a finite cyclic subgroup of M of odd order k . Then there is a uniquesubgroup of G of order k whose image in M is L .Proof. Suppose t ∈ M is given, and let g ∈ G map to t . Then g ′ maps to t if andonly if g ′ = hg for some h ∈ K . Since K has order 2, there are two such elements g ′ .So G → M is two-to-one. From this (1) and (2) follow.If H is a subgroup of G then the restriction of G → M to H has kernel H ∩ K .If H is finite of odd order, then H ∩ K must be the trivial group since it is asubgroup of K . Thus H is isomorphic to its image in M and (3) follows.Note that if g ∈ G then π ( g ) generates the image of h g i in M . If in addition g has finite odd order then, by (3), h g i and h π ( g ) i are isomorphic, so g and π ( g ) havethe same order. Thus (4) holds.If H is a finite subgroup of G of even order then it has an element of order 2 byCauchy’s theorem. Thus H contains K since there is a unique element of order 2.The kernel of the restriction of G → M to H has kernel H ∩ K , which in this caseis K itself. Since the image π [ H ] of H is isomorphic to H/K (first isomorphismtheorem) this implies that H has twice the order of π [ H ]. By (2) the preimage H ′ of π [ H ] also has order twice that of π [ H ]. Since H ⊆ H ′ , this implies that H = H ′ .Thus (5) and (6) hold.Note that if g ∈ G then π ( g ) generates the image of h g i in M . If in addition g has finite even order then, by (6), h g i and has twice the size of h π ( g ) i , so g hasorder twice that of π ( g ). Thus (7) holds.Let L be cyclic subgroup of M of odd order k . By (6) its preimage H hasorder 2 k . Let g ∈ H map to a generator of L . If g has even order then g hasorder 2 k by (7), so H is cyclic. If g has odd order then it has order k by (4). Let τ
6e the unique element of order 2 in G . Then gτ g − = τ by uniqueness, so gτ = τ g .Note that g and τ generate H , and so H is Abelian. This means that τ g hasorder 2 k . Thus H is cyclic. So (8) holds.Finally, let L be cyclic subgroup of M of odd order k . By (8), the preimage H of L in G is cyclic of order 2 k , and has a unique subgroup C of order k . The imageof C is L by (3). Since any subgroup of G with image L must be a subgroup of H ,this means that G has a unique subgroup of order k whose image is L . So (9)holds. Example . We can extend Example 2 from H to a general division ring D . Sup-pose G is a finite group of D × . The prime subfield F of D is either Q or F p forsome prime p . In the first case we say that D has characteristic 0. If the prime fieldis finite then we define the characteristic of D to be the size of the prime field. Notethat F is in the center of D in the sense that av = va for all a ∈ F and v ∈ D . Thisfollows from the observation that, for each nonzero v ∈ D , the map a v − av isa ring homomorphism F → D that maps 1 to 1. By definition of F and propertiesof ring homomorphisms this map must send any a ∈ F to itself.Observe that D is an F -vector space. Observe also that multiplication definesa linear representation of G on V = D . Suppose g ∈ G and v ∈ D . If g = 1 andif g has a fixed vector v ∈ D then gv = v . So ( g − v = 0. Since D is a divisionring and since g − = 0, we have v = 0. Thus we get a free linear representationof G on the F -vector space D . In other words, G is freely representable over F . Incharacteristic 0 we conclude that G is freely representable in the usual sense (forfields of characteristic zero).What about if F has finite prime characteristic? In that case the subset R offinite linear combinations P a i g i with a i ∈ F and g i ∈ G is closed under additionand multiplication. In fact R forms a subring of D . Observe that R is finite.Since R has no zero divisors, it must be a finite field by Wedderburn’s theorem.It is well-known that R × is a cyclic group if R is a finite field, and since G is asubgroup of R × it is also cyclic. So G is freely representable (over C ) in this caseas well. So all finite subgroups of division rings are freely representable , and in finiteprime characteristic they are actually cyclic. Norm relations of unity are of interest in algebraic number theory, both for theoreticreasons and for computational reasons. See [4] for a discussion on their history andapplications to algebraic number theory. One of my motivations for studying freelyrepresentable groups is the fact that such group are exactly the finite groups without The fact that D × is cyclic in prime characteristic was noticed by Herstein in 1953. Hersteinthen conjectured for in any characteristic that any subgroup of D × of odd order is cyclic, becausethis holds in cases such as D = H and for D of prime characteristic. In 1955, Amitsur [2] found acounter-example to the conjecture of size 63. Here we used the fact that (1) F × is cyclic if F is finite, and (2) Wedderburn’s theorem.We will give independent arguments for these facts later in the document as a consequence of thestructure theorem for Sylow-cyclic groups. See Section 8 below. Definition 1.
Let G be a finite group and let H be a subgroup. Let R be acommutative ring (with unity). The norm of H is defined to be the formal sum ofelements of H in R [ G ]: N H def = X σ ∈ H σ. If we fix a linear representation of a group G on an F -vector space V then wecan view V as an F [ G ]-module. If in addition the representation is a free linearrepresentation then norms of nontrivial subgroups have the interesting propertythat they annihilate V . Proposition 9.
Let G be a finite group and let F be a field. If G has a freelinear representation on an F -vector space V , and if H = { } is a subgroup of G ,then ( N H ) v = 0 for all v ∈ V .Proof. Let v ∈ V , and consider v ′ = ( N H ) v . Since H is nontrivial it has a non-identity element σ . Observe σv ′ = σ (( N H ) v ) = ( σ N H ) v = ( N H ) v = v ′ . Since G acts freely on nonzero vectors we have v ′ = 0. Thus ( N H ) v = 0 for allvectors v in V . Definition 2.
Let G be a finite group. A norm relation of unity for G is anexpression in Q [ G ] of the form = X H ∈H a H ( N H ) b H where H is the collection of nontrivial subgroups of G , where a H , b H ∈ Q [ G ], andwhere is the unit in Q [ G ] which can be viewed as the norm of the trivial group.If we replace Q in the above with a field F , then we call such a relation a normrelation of unity relative to F . Example . The first norm identity of unity that came to my attention was oneexploited by my colleague S. Sharif and heavily used by his student A. Savage in histhesis [12]. This norm relation involves the group G = C × C . In addition to { } and G itself, there are four subgroups H , H , H , H of G which are all cyclic oforder 3. In Z [ G ] we have the following (as in the proof of Theorem 18 below):3 · = N H + N H + N H + N H − N G giving us a simple norm relation of unity in Q [ G ]: = 13 N H + 13 N H + 13 N H + 13 N H − N G. Z [ G ] was then applied to bicubic Galois extensions K/ Q withGalois group identified with G . The above additive relation yields a multiplicativerelation for α ∈ K : α = N K/K ( α ) N K/K ( α ) N K/K ( α ) N K/K ( α ) N K/ Q ( α )where N K/L is the usual norm of field theory, and where K , . . . , K are the inter-mediate fixed fields associated, as in Galois theory, with the subgroups H , . . . , H .Sharif and Savage were interested in the case where α is a unit in the ring ofintegers O K . For such α the equation becomes α = ± N K/K ( α ) N K/K ( α ) N K/K ( α ) N K/K ( α ) . This identity helps one to identify units in O K given a predetermination of units ofeach O K i . See also [4] for generalization of this identity and applications to findingunits and other invariants of K for more general number fields K . Example . A earlier example, from 1966, can be found in Wada [15] for the casewhere G = C × C is the Klein four group. Let σ , σ , σ be the nontrivial elementsof G and let H i be the subgroup generated by σ i . Then we have2 · = N H + N H − σ N H which, when divided by 2, yields a norm relation of unity. In particular if K is a biquadratic extension of Q , say, with Galois group identified with G , andif K , K , K are the quadratic extensions associated with H , H , H , then we get α = N K/K ( α ) N K/K σ N K/K ( α ) . As in the previous example this was used to describe units in O K which was thenused to calculate class numbers of such biquadratic fields K . Example . A third example can be found in Parry [11] for G = C × C . As in theearlier example, let H , H , H , H be the distinct cyclic subgroups of order 3. Fixa generator σ of H and a generator τ of H . Switching H and H if necessary, wecan assume that H is generated by στ and that H is generated by στ . In Z [ G ]we have the following:3 · = N H + N H + N H − ( σ + στ ) N H which gives a norm identity of unity when we divide by 3. Parry used these tostudy bicubic extensions of Q . Lemma 10.
Let G be a finite group and let H be the collection of nontrivial sub-groups of G . Then the following are equivalent:1. G has a norm relation of unity relative to F .2. The two-sided ideal of F [ G ] generated by { N H | H ∈ H} is all of F [ G ] .3. The left ideal of F [ G ] generated by { N H | H ∈ H} is all of F [ G ] . . The right ideal of F [ G ] generated by { N H | H ∈ H} is all of F [ G ] .Proof. Clearly (1) ⇐⇒ (2), (3) = ⇒ (2), and (4) = ⇒ (2). Next we note thatthe implication (2) = ⇒ (3) is a consequence of the following fact: Let R be acommutative ring and let I be the left ideal of R [ G ] generated by { N H | H ∈ H} .Then I is a two-sided ideal of R [ G ] . To establish this claim it is enough to showthat ( N H ) g ∈ I for all H ∈ H and g ∈ G . This follows from the equation( N H ) g = gg − ( N H ) g = g N ( g − Hg ) . Similarly, (2) = ⇒ (4). Theorem 11.
Let G be a finite group and let F be a field of characteristic notdividing | G | . Let I be the two-sided ideal of F [ G ] generated by the norms N H ofnontrivial subgroups of G . Then G is freely representable over F if and only if I isa proper ideal of F [ G ] .Proof. Suppose that G has a free linear representation on the F -vector space V = 0.Let K be the kernel of the action of F [ G ]. In other words, K is the collection of allelements α ∈ F [ G ] such that αv = 0 for all v ∈ V . Observe that K is a two-sidedideal of F [ G ]. By the above proposition, N H ∈ K for all nontrivial subgroups H .Thus I ⊆ K . However, if v = 0 then v = 0. So is not in K . Thus K , andhence I , must be a proper ideal of F [ G ].Conversely, suppose that I is a proper ideal of F [ G ]. Let V = F [ G ] /I . Al-though V is an algebra over F , we will think of it just as an F [ G ]-module; in otherwords, V is an F -vector space with a linear action of G . Since I is a proper idealof F [ G ], the vector space V is not the zero space. For convenience let’s write [ α ]for the coset α + I in V .The goal is to show that the linear action of G on V is free. So suppose that g = 1in G and suppose that g [ α ] = [ α ] for some [ α ] ∈ V . Let C be the subgroup of G generated by g . So ( N C )[ α ] = k [ α ] where k is the size of C . We can rewrite thisas [ N C · α ] = k [ α ] where now we think of k is an element of F . Since N C · α ∈ I ,we have that k [ α ] = [0]. Since k is invertible in F we have [ α ] = [0]. This showthat the representation of G on V is a free linear representation.As a corollary we get the following: Theorem 12.
A finite group G has a norm relation of unity if and only if it is notfreely representable.Proof. Let H be the collection of nontrivial subgroups of G . Let I be the two-sidedideal of Q [ G ] generated by { N H | H ∈ H} .Suppose G has a norm relation of unity. By Lemma 10, I is all of Q [ G ]. ByTheorem 11, G is not freely representable over Q , hence is not freely representableover C (Lemma 6).Conversely, suppose G is not freely representable. In other words, suppose G isnot freely representable over C , and hence over Q (Lemma 6). By Theorem 11 theideal I must be all of Q [ G ]. So G has a norm relation of unity by Lemma 10.10 Basic Properties of Freely Representable Groups
We start with some important properties of freely representable groups that canbe proved without too much effort. We will see that the Abelian groups that arefreely representable are just the cyclic groups. And we will see various ways inwhich freely representable groups behave like cyclic groups, confirming the ideathat freely representable groups can play the same role for finite groups in generalthat the cyclic groups play for the finite Abelian groups. In other words, they canbe regarded as “cycloidal” groups.For example, every subgroup of a cyclic group is a cyclic group. The same holdsfor freely representable groups:
Proposition 13.
Every subgroup of a freely representable group is freely repre-sentable.Proof.
This follows from the definition.
Lemma 14. If A is a freely representable finite Abelian group then A is cyclic.Proof. (We will give here a proof that uses a bit of representation theory over C and the fact that finite subgroups of C × are cyclic. However, more algebraic argu-ments can be given that do not use C in any essential way. See the remark afterCorollary 19.)A freely representable group has an irreducible free linear representation over C (Corollary 4). Every irreducible linear representation of an Abelian group is one-dimensional and so gives a homomorphism into the circle group of C × (see the ap-pendix on group representations). Also, free linear representations are necessarilyfaithful (Lemma 2). So any freely representable finite Abelian group is isomomor-phic to a subgroup of the circle group in C × . Such groups are cyclic.Combining this lemma with Example 1 gives the following: Corollary 15.
A finite Abelian group A is freely representable if and only if it iscyclic. Combining the above lemma with Proposition 13 yields the following:
Corollary 16.
If a finite group G is freely representable then all its Abelian sub-groups are cyclic.Proof. Every subgroup of G is freely representable, so every Abelian subgroup iscyclic by the previous corollary. Remark.
In Example 3 we observed that if F is a division ring then any finitesubgroup of F × is freely representable. Thus all finite subgroups of F × are cyclicif F is a field by Lemma 14. If we use the above proof of Lemma 14 then theargument is somewhat circular since it uses the result for F = C . Also, in Example 3we used the result for finite F . However, the remark after Corollary 19 gives anargument for Lemma 14 that does not assume the result for F = C , and in Section 8we will give an independent argument for the result that F × is cyclic withoutassuming the result for finite F . 11ecall that finite subgroups of H × give examples of freely representable groups,and all such groups all have at most one element of order 2. This occurs moregenerally. (Note also that this is a basic property of cyclic groups as well). Proposition 17.
Suppose that G is a finite group of even order that is freelyrepresentable over a field F . Then G has a unique element of order two, and thiselement is in the center of G .Proof. By Corollary 5 we can think of G as a subgroup of GL n ( F ) such that theonly element of Γ with eigenvalue 1 is the identity. Since G has even order it hasat least one element of order 2 (Cauchy’s theorem). We will prove the result byshowing that if g ∈ GL n ( F ) has order 2 and does not have eigenvalue 1 then g = − I .To see this, observe that ( g − I )( g + I ) = 0 since g has order 2. Also observethat the null-space of g − I is { } since g fixes only the zero vector. Thus g − I isinvertible, and so g + I = 0. In other words, g = − I as desired. Remark.
The above proof can be adapted to prove the following: if F has charac-teristic 2 then every freely representable group over F is of odd order. Remark.
By the above proposition, every freely representable group of even orderhas a normal subgroup of order 2. The Feit-Thompson theorem states that everynon-Abelian simple group has even order. So the above proposition implies thata simple group is freely representable if and only if it is cyclic (and necessarily ofprime order).
Remark.
The above proposition implies that many familiar non-Abelian groups,such as the A n if n ≥ S n if n ≥ Remark.
The quaternion group with 8 elements is the classic example of a non-Abelian group with a single element of order two. Later we will see that every2-group with only one such element is freely representable, and will form a well-known class of groups called the generalized quaternion groups .We give another class of groups where freely representable and cyclic coincide.
Theorem 18.
Let G be a group of order equal to the product of at most two primeswhere the two primes in question can be equal. Then G is freely representable ifand only if it is cyclic.Proof. One direction is clear from previous results, thus it is enough to show thatif G not cyclic then it is not freely representable. So suppose that G is not cyclic.This means that all the nontrivial cyclic subgroups of G have prime order, andevery nonidentity element of G is in a unique cyclic group. Let C be the collectionof nontrivial cyclic subgroups of G . Then every nonidentity element of G is inexactly one C ∈ C . So X C ∈C N C = ( k − + N G where k is the size of C . This yields a norm relation of unity since k >
1. So G isnot freely representable by Theorem 12 12 emark. The above theorem cannot be extended to all groups whose size is aproduct of three primes: we have seen that the quaternion group with 8 = 2 elements is freely representable, but not cyclic. Remark.
The proof of the above theorem generalizes as follows: suppose G is theunion of two or more nontrivial proper subgroups such that when you remove theidentity element from each you get a partition of G − { e } . Then G has a normrelation of unity, and so is not freely representable. For example, dihederal groupsmust have such norm relations and so we see (in a second way) that they are notfreely representable. This idea will arise later (See Lemma 73). Corollary 19. If G is a finite freely representable group then every subgroup of G of order pq , where p and q are primes, is cyclic.Remark. We can use this result to give another proof of Lemma 14 (all finite Abelianfreely representable groups are cyclic) using the structure theorem for finite Abeliangroups.We can even an even more elementary algebraic proof of Lemma 14 based on theabove result without appealing to the structure theorem for finite Abelian groups.
We assume we have a Abelian subgroup A with the property that all subgroups ofsize p are cyclic where p is any prime dividing | A | and we show that all suchgroups are cyclic. Start by fixing a prime p dividing | A | (if | A | = 1 we are doneof course) and consider the endomorphism x x p . Let K be the kernel and let I be the image. By Cauchy’s theorem (in the easy case of Abelian groups), thegroup K is a nontrivial p -group. Suppose K contains two distinct subgroups C and C of order p . Then C C is a group of order p , and so is cyclic. This meansthat C = C since a cyclic group of order p has a unique subgroup of order p .This is a contradiction, thus K is a group of order p . This means that I hasindex p in G , so by induction we can assume I is cyclic. If I has order prime to p then G = KI ∼ = K × I is cyclic. So we assume p divides the order of I . Let g ∈ G be an element mapping to a generator of I . So p divides the order of g , whichmeans that g has order p times the order of I . Thus g has order | I | · | K | = | G | .We now consider the Cartesian products of groups of relatively prime order. Proposition 20.
Suppose A and B are finite groups of relatively prime order, andlet F be a field. If A has a free linear representation on a finite dimensional F -vector space V A , and B has a free representation on a finite dimensional F -vectorspace on V B , then the associated representation of A × B on V A ⊗ V B is a free linearrepresentation, where A × B acts on V A ⊗ V B according to the rule ( a, b )( v ⊗ v ) = a ( v ) ⊗ b ( v ) . Proof.
Suppose gv = v where g is a nonidentity element of A × B . Then g k v = v for all powers of g . In particular, if p is a prime divisor of the order of g then thereis an element σ of order p such that σv = v . Since A and B have relatively primeorders, σ must be in A or B (where A and B are regarded as subgroups of A × B ).Suppose σ ∈ A . Let e , . . . , e n be a basis for V B . Then v = v ⊗ e + . . . + v n ⊗ e n v i ∈ V A . Since σv = v ,( σv ) ⊗ e + . . . + ( σv n ) ⊗ e n = v ⊗ e + . . . + v n ⊗ e n . Thus σv i = v i for each i . So v i = 0 since the representation on V A is a free linearrepresentation. Thus v = 0 as desired. Similarly, v = 0 if σ ∈ B .This yields a generalization of a result about cyclic groups: if A and B haverelatively prime orders then A × B is cyclic if and only if both A and B are cyclic. Corollary 21.
Suppose A and B are finite groups of relatively prime order.Then A × B is freely representable if and only if A and B are both freely repre-sentable. Next we investigate freely representable p -groups. If p is odd these turn out tocyclic, as we will soon see. In the case where p = 2 we can get generalizations ofquaternion groups in addition to cyclic groups, but the details will have to wait fora later section.We start with a few lemmas. The first is motivated by the following question:if a prime p divides the order of a freely representable group G , is there a uniquesubgroup of order p ? This is a natural question since we always have uniquenesswhen G is cyclic, and we have established uniqueness for general freely representablegroups G in the case p = 2. The following shows that p dividing the order of thecenter is a sufficient condition. (We can even weaken the hypothesis a bit: insteadof requiring that G be freely representable, we just assume G has the property thatall subgroups of order p are cyclic.) Lemma 22.
Suppose p is a prime and G is a finite group with the property thatevery subgroup of G of order p is cyclic. If p divides the order of the center Z of G then G has exactly one subgroup of order p , and that group is a subgroup of Z .Proof. By Cauchy’s theorem Z has a subgroup Z p of order p . Let C p be a subgroupof order p . Then H = C p Z p is a subgroup with at most p elements. Since everyelement of H has order 1 or p , the order of H is p or p . Thus H is cyclic, andso H has a unique subgroup of order p . Thus C p = Z p . Lemma 23.
Suppose G is a group of order p k where p is a prime. Suppose that H and H are distinct Abelian subgroups of G of index p . If p = 2 then the map x x p is a homomorphism from G to H ∩ H . If p = 2 then the map x x is ahomomorphism from G to H ∩ H .Proof. Since H and H have index p in G , they are normal subgroups of G (Proposition 184). Thus Z = H ∩ H is also a normal subgroup of G . Observethat G = H H and that Z is in the center of G since H and H are Abelian. Since H and H both have index p , if g ∈ G then g p ∈ Z .Given x, y ∈ G we form the commutator [ x, y ] = x − y − xy . So [ x, y ] is definedby xy = yx [ x, y ] . In the interesting case where G is non-Abelian Z is all of the center. Hence the notation Z for this subgroup. (Why? If Z ′ is the center then Z ′ H i is Abelian, so is H i .)
14n the case where a ∈ H and b ∈ H we have a − b − ab ∈ H ∩ H since H , H are both normal. So [ a, b ] ∈ Z . This means that G/Z is Abelian since G = H H and H , H are Abelian. In particular, [ x, y ] ∈ Z for all x, y ∈ G .If x, y ∈ G then x p y = yx p since x p ∈ Z and Z is in the center. However, x n y = x · · · xy = x · · · xyx [ x, y ] = . . . = yx n [ x, y ] n since [ x, y ] is in the center. Comparing these expressions when n = p gives[ x, y ] p = 1 . Another consequence of the fact that [ x, y ] is in the center of G is that( xy ) p = xyxy · · · xyxy = x p y p [ y, x ] m where m = 1 + . . . + ( p −
1) = p ( p − /
2. When p is odd, we have that m is amultiple of p , so [ y, x ] m = 1 for all x, y ∈ G and so( xy ) p = x p y p . If p = 2 then ( xy ) = x y [ y, x ] which doesn’t necessarily give a homomorphism.But since [ y, x ] = 1 and since [ y, x ] is in the center,( xy ) = xyxyxyxy = x y [ y, x ] = x y . Here is an interesting group theoretical consequence (interesting beyond justthe theory of freely representable groups):
Proposition 24. If p is an odd prime then every p -group G with only one subgroupof order p is cyclic.Proof. We proceed by induction on k ≥ p k is the order of G . The basecase k = 1 is clear so assume k ≥
2. So suppose G has a unique subgroup C oforder p . Let H be a subgroup of index p in G (which exists by Proposition 185).Let g ∈ G be any element of G not in H . If g generates G we are done, so assume g generates a proper subgroup. By Proposition 185 there is a subgroup H of index p in G containing g .Observe that H and H can have only one subgroup of order p since thatholds of G . So H and H are cyclic by the induction hypothesis. By the abovelemma, x x p is a homomorphism from G to H ∩ H . Observe that the kernelis C , the unique subgroup of G of order p . Thus the image of x x p has p k − elements. However H ∩ H has order bounded by p k − , a contradiction.These results can be collected together to give another family of groups wherefreely representable coincides with cyclic: Theorem 25.
Suppose G is a p -group where p is an odd prime. Then the followingare equivalent.1. G is freely representable. . Every subgroup of G of order p is cyclic.3. G has a unique subgroup of order p .4. G is cyclic.Proof. We have (1) = ⇒ (2) by Theorem 18. We have (2) = ⇒ (3) by Lemma 22(the center is nontrivial by Proposition 182). We have (3) = ⇒ (4) by Proposi-tion 24. Finally (4) = ⇒ (1) as in Example 1. Remark.
The quaternion group on 8 elements shows that this theorem does nothold for p = 2. We will investigate 2-groups in a later section. Remark.
It follows from the above theorem that if G is freely representable thenfor every odd prime p the p -sylow subgroup is cyclic. An important case is whereevery Sylow subgroup is cyclic (even for p = 2). Such groups have a long history.Frobenius and Burnside proved such groups are solvable (and even metacyclic). Infact, such groups constitute one of the three classes of groups studied by Burnsidein his classification of groups that could be freely representable ([6] and [5]). Wewill review the theory of such groups in Section 7. Now that we have explored the more accessible properties of freely representablegroups and have some feel for them, we will take a break from group theory anddiscuss their central role in Riemannian geometry in order to further motivatetheir study. After this we will return to group theoretic matters and survey theclassification of such groups.Freely representable groups are of importance in Riemannian geometry becauseof the following (where S n is the unit sphere in R n +1 ): Theorem 26.
Suppose Γ is a finite group of isometries of S n acting freely on S n .Then Γ is freely representable. Conversely if G is a freely representable group, thenfor some n there is a finite group of isometries Γ of S n isomorphic to G actingfreely on S n .Proof. We appeal to the following standard result in differential geometry: theisometry group of S n can be identified with the orthogonal group O ( n + 1). In par-ticular, an orthogonal linear tranformation R n +1 → R n +1 restricts to an isometryof any sphere S n ⊆ R n +1 centered at the origin.Assume Γ is a finite subgroup of O ( n + 1) acting freely on a S n ⊆ R n +1 .This means that the inclusion map gives a free representation of Γ, so Γ is freelyrepresentable.Conversely, if G is a freely representable, then G is freely representable over R by Lemma 6. So there is a free representation of G on a real vector space V . We canform a positive definite inner product on V that is G -invariant (using the standardaveraging technique). In particular, every element of G acts as an orthogonaltransformation with respect to such an inner product. Fix an orthonormal basisand identify V with R n +1 for some n ≥
0. Since the representation is free, it is16aithful. So G is isomorphic its image Γ in O ( n + 1). Since the representation isfree, Γ acts freely on the unit sphere S n . Remark. If n is 0 or 1 then any finite group of isometries acting freely of S n is cyclic.However, in Example 1 we showed how to construct free linear representation over R of cyclic groups for any even dimension. So we can safely remove the cases n = 0and n = 1 in the above theorem, and thus assume S n is simply connected: if G is afreely representable group then for some n ≥ there is a finite group of isometries Γ of S n isomorphic to G acting freely on the simply connect sphere S n .Remark. If n + 1 is odd then any element of O ( n + 1) must have a real eigenvalue,and that eigenvalue must equal to ±
1. Thus the square of every element must haveeigenvalue +1. This means that for any finite subgroup Γ of O ( n + 1) acting freelyon S n , the square of every element is the identity. So all complex eigenvalues ofelements of Γ are ±
1, but we cannot have nonidentity elements of Γ with eigen-value +1 since Γ acts freely. So Γ is a subgroup of {± I } . In other words, if n iseven then the only finite subgroup of O ( n + 1) acting freely on S n are {± I } andthe trivial group { I } .This means we can restrict our attention to S n for odd n : if G is a freelyrepresentable group, then for some odd n ≥ there is a finite group of isometries Γ of S n isomorphic to G acting freely on the simply connect sphere S n . Observe that Example 2 gives interesting non-Abelian freely representablegroups action on S .The groups Γ of the above theorem are of importance in the classification ofspaces of positive curvature because of the following theorem: Theorem 27.
Suppose M is a complete, connected Riemannian manifold of con-stant positive sectional curvature K and of dimension n ≥ . Then M is isomet-ric to S n / Γ where Γ is a finite subgroup of isometries of S n acting freely on thesphere S n of curvature K .If both Γ and Γ ′ are finite subgroup of isometries of S n acting freely on S n ,then S n / Γ is isometric to S n / Γ ′ if and only if Γ and Γ ′ are conjugate subgroupsof O ( n + 1) . Corollary 28.
A finite group G is a freely representable group if and only if itoccurs as the fundamental group of a complete, connected Riemannian manifold ofconstant positive sectional curvature of dimension n ≥ .Remark. As pointed out in an earlier remark, if n ≥ {± I } or the trivial group { I } . This means that there areonly two complete, connected Riemannian manifold of constant positive sectionalcurvature K of dimension n : S n itself and real projective space. Remark.
Since cyclic groups are freely representable, they occurs as the fundamen-tal group of such manifolds. Such manifolds are lens spaces of constant curvature.
The Riemannian manifolds of constant curvature play a central role in geometryas a whole. The study of such manifolds is geometry par excellence . It arose out of17he hyperbolic geometry of Lobatchevsky-Bolyai-Gauss. In fact, it predates hyper-bolic geometry since Euclidean and spherical geometry are special cases. In 1854Riemann gave his account of elliptic geometry, and began the development of Rie-mann geometry (as a generalization of Gauss’s differential geometry of surfaces).Beltrami played an essential role by providing models for non-Euclidean geometryin 1868.In the 1870s, William Clifford began exploring interesting topologies that canoccur with surfaces of constant curvature. He was an early proponent of Riemann’sformulation of differential geometry, which was revolutionary at the time. Cliffordwas a remarkable figure, who tragically died at age 33 in 1879. Clifford algebras arenamed for him, and some of Clifford’s ideas about curvature anticipate Einstein’stheory of gravity.Starting in the early 1870s Felix Klein became interested non-Euclidean geom-etry and in classifying surfaces of constant curvature. He pioneered the techniquesof using covering spaces for studying such surfaces in 1891. Klein coined the term“hyperbolic” and “elliptic” geometry (and even “parabolic” geometry for Euclideangeometry, but that term has not caught on).In 1891, Killing posed the classification question for spaces of positive curvaturein n -dimensions and called this the Clifford-Klein spherical space form problem(The Clifford-Kleinschen Raumproblem ).The term “space form” is a term for complete, connected manifold of constantcurvature or equivalently (the Killing-Hopf theorem) manifolds of the form M/ Γwhere M is the (simply connected) standard hyperbolic, Euclidian, or spheric spacesand where Γ is a group of isogenies acting freely and properly discontinuously on M .This last condition means that for each x ∈ M there is a neighborhood U suchthat γU and U are disjoint if γ ∈ Γ is not the identity. In the case where M isa sphere, this just means that Γ is finite since the sphere is compact. The term“space form” can also refer to this analogous situation in differential topology orthe topology of manifolds more generally, where Γ is not restricted to isometries.H. Hopf (1926) modernized Killing’s spherical space form problem, and high-lighted the use of group theoretical methods. He studied the examples where Γ is abinary dihedral group, thus establishing an infinite family of topologically distinctthree-dimensional spaces of positive curvature.Threfall and Seifert classified three dimensional spherical spaces in 1930 whichuses the interesting subgroups of H × . Georges Vincent made significant progresstoward the general classification in 1947. Vincent classifed all the solvable freelyrepresentable groups and their free representations over R . This lead to the classi-fication of spherical space forms in all dimensions not congruent to 3 mod 4. Wolfin the 1960s completed the classification of freely representable groups, classifiedtheir real free representations, and so classified all complete, connected Rieman-nian manifolds of constant curvature [17]. He simplied Vincent’s approach by usingwork by Zassenhaus, and was thus able to expand the results to non-solvable freelyrepresentable groups.As mentioned above Burnside had studied examples of freely representablegroups as early as the 1905, but outside the context of geometry. More specificallyhe studied groups whose Sylow subgroups are cyclic except for 2-Sylow subgroups As far as I know, Vincent only published one paper. -Groups For odd primes p the only freely representable p -groups are cyclic. However, inExample 2 we saw that any finite subgroup of H × is freely representable and manyof these are non-Abelian 2-groups. We will see that all freely representable 2-groupsare in fact isomorphic to a subgroup of H × . Definition 3. A generalized quaternion group is a group isomorphic to a non-cyclic finite group of H × whose order is a power of 2. For example, the standardquaternion group {± , ± i , ± j , ± k } qualifies as a generalized quaternion group.It is immediate from this definition that generalized quaternion groups arefreely-representable groups. As discussed in Example 2, every noncyclic finite sub-groups of H × is the preimage of a finite noncyclic subgroup of SO(3) under the stan-dard two-to-one homomorphism H → SO(3). So a generalized quaternion groupmust be the preimage of a noncyclic subgroup of SO(3) whose order is a power oftwo. The only such subgroups of SO(3) are dihedral groups D n where n ≥ Proposition 29.
A group is a generalized quaternion group if and only if it isisomorphic to D n where n ≥ is some power of . Thus every generalized quater-nion group has order k for some k ≥ , and for every k ≥ there is a uniquegeneralized quaternion group of order k up to isomorphism.Proof. Most of this is clear from the above discussion. The uniqueness is a conse-quence of the fact that dihedral subgroups of SO(3) of a fixed order are isomorphism(see appendix on finite subgroups of SO(3)), together with Proposition 7.Let Q be a generalized quaternion group of order 2 n with n ≥
3. We identify Q with a subgroup of H × . As mentioned in Example 2 (and Lemma 8), Q has aunique element of order 2, namely −
1. Also Q is a subgroup of H , and restrictingthe map H → SO(3) to Q gives a two-to-one homomorphism Q → D where D is a dihedral subgroup of SO(3) of order 2 n − . The kernel of this homomorphismis {± } .As in our discussion in Example 2, the image of any nontrivial cyclic subgroup C of Q of order k is a cyclic subgroup of D of order k/
2. In other words, an elementof Q of order k > k/
2. Since D is dihedral oforder 2 n − , it has generators ρ and τ such that ρ has order 2 n − and τ has order 2,and such that τ ρτ = ρ − . Let R ∈ Q map to ρ and let T ∈ Q map to τ . So R has order 2 n − and T has order 4. Note that R generates a cyclic subgroup of Q of index 2. Since R n − and T are elements of order 2, they are both equalto −
1, and so T = R n − . The subgroup h R, T i of Q generated by R and T mapsonto D since ρ and τ generate D , so h R, T i has order 2 n (Lemma 8), meaningthat h R, T i = Q . 19n D we have that τ ρτ = ρ − so that ( τ ρ ) = 1. This means that τ ρ hasorder 2 (it cannot have order 1 since τ and ρ are not inverses; otherwise D wouldbe generated by τ and would have order 2). Thus T R has order 4, and (
T R ) mustbe -1. Note also that T − = − T since T = −
1. Thus
T RT − R = T R ( − T ) R = − ( T R ) = 1 , so T RT − = R − . In summary, we have the following relations between the generators R and T : R n − = 1 , T = R n − , T RT − = R − . These relations are sufficient to characterize Q : Proposition 30.
The quaternion group group Q with n elements is isomorphicto the following abstract group G described by two generators R and T and threerelations R n − = 1 , T = R n − , T RT − = R − . Proof.
Using the given relations we see that we
T R = R − T , and so every elementof G can be written as R i T j where 0 ≤ i ≤ n − − ≤ j ≤
1. So G has at most 2 n elements.Next observe that the specific choice of elements we also call R and T in Q givenabove satisfies the desired relations, giving us a homomorphism G → Q (where weview G as a quotient of a free group generated by two elements). Since R and T in Q generates Q , the homomorphism G → Q is surjective. Since Q has order 2 n and G has order at most 2 n , this map is an isomorphism and G has order exactly 2 n .Our next goal is to show that every freely representable 2-group is a generalizedquaternion group. It turns out that we can do so by exploiting the fact that freelyrepresentable groups of even order have a unique element of order 2. To this endwe will require a few lemmas about 2-groups with a unique element of order 2. Lemma 31. If A is an Abelian group of order k with a unique element of order then A is cyclic.Proof. This follows from the structure theorem for finite Abelian groups.A direct proof can be given. We proceed by induction. The base case k = 1is clear, so suppose A is Abelian with order 2 k where k ≥
2, and assume A hasa unique element of order 2. Consider the homomorphism A → A defined bythe rule x x . The kernel of this map has two elements, so the image C hasorder 2 k − . Since C is a subgroup of A of order 2 k − , it has a unique element oforder 2. So, by induction, C is cyclic. Let h ∈ A map to a generator of C , so h has order 2 k − . Since the order of h is even, the order of h is half the order of h .So h has order 2 k , and A must be cyclic with generator h . Lemma 32.
Suppose that G is a -group with a unique element of order .If | G | ≤ then G is cyclic or is the quaternion group with elements. roof. Suppose that G is not cyclic. Then | G | = 4 or | G | = 8, and all elementsof G have order 1, 2, or 4. Note that G has one element of order 1 and one elementof order 2, so G has | G | − | G | = 4, then any of theelements of order 4 generate G so G is in fact a cyclic group. So. we can assume G is a noncyclic group of order 8 with 6 elements of order 4.Let R be any element of order 4. Note that the subgroup generated by R has 2elements of order 4, so there are 4 additional elements of order 4. Let T be one ofthese additional elements of order 4. Note that the group generated by R and T has more than 4 elements, so must be all of G .Finally we work out enough relations between R and T to determine G . Since R has order 4, we have the relation R = 1. Since R and T have order 2, they mustbe equal. So we get a second relation R = T . The group generated by R hasindex two in G so must be a normal subgroup of G . Thus the conjugate T RT − of R is a power of R . It must have order 4 and so this conjugate is R or R = R − .If T RT − = R then R and T commute and so G is Abelian. This can’t happenby the previous lemma. So we must have T RT − = R − as a third relation. ByProposition 30, these three relations insure that the group G is a quotient of aquaternion group of order 8. Since G has size 8 this means that G is isomorphic tothe quaternion group with 8 elements.Next we see that with one exception 2-groups with a unique element of order 2must have a unique cyclic subgroup of index 2. Lemma 33.
Let G be a group of order n with a unique element of order andwith at least two distinct cyclic subgroups of index . Then G is isomorphic to thequaternion group with elements.Proof. Observe that G is non-Abelian: otherwise G would be cyclic by Lemma 31and would only have one subgroup of index 2. Let H and H be distinct cyclicsubgroups of index 2. Because H and H have index 2 in G , they are normalsubgroups of G . Thus the intersection H ∩ H is also normal. Observe that G must have order at least 8 since otherwise it could not have two distinct subgroupsof index 2.Let Z be the center of G . Observe that ZH i is Abelian, and so cannot be allof G . This means ZH i = H i and so Z ⊆ H i . Hence Z ⊆ H ∩ H . Observethat G = H H , and that H ∩ H ⊆ Z is in the center of G = H H since H and H are cyclic. We conclude that Z = H ∩ H .From this we see that the inclusion H ֒ → G = H H induces an isomor-phism H /Z → G/H (since Z = H ∩ H ). So [ H : Z ] = 2. Since H hasorder 2 n − we have that Z has order 2 n − . Since Z is Abelian, it is cyclic byLemma 31. Let C be the unique subgroup of Z of index 2 in Z . So C is a cyclicgroup of order 2 n − .Since G/H i has order 2, if x ∈ G then x ∈ H i . So x ∈ Z = H ∩ H .Furthermore, since Z/C has order 2, we have x ∈ C for all x ∈ G . In otherwords, x x is a map G → C . By Lemma 23 the map x x is actuallya homomorphism G → C . Let K be the kernel. Observe that K consists of allelements of G of order 1, 2, or 4. Since C has 2 n − elements, the image of G → C has at most 2 n − elements, so K has at least 8 elements.21s noted above G has at least 8 elements. Suppose G has more than 8 ele-ments, so n ≥
4. Since Z has order 2 n − this means Z contains a subgroup Z oforder 4. Then Z ⊆ K by definition of K . Since K has order at least 8, there isan element b ∈ K not in Z . Since Z is in the center, Z h b i would be an Abeliansubgroup of K of order at least 8. By Lemma 31, this means Z h b i would be cyclicof order 8 or more. But every element of K has order at most 4, a contradiction.We conclude that G has 8 elements. Since it is not cyclic, it must be the quaterniongroup by the previous lemma.We know that a generalized quaternion group Q has a cyclic subgroup of index 2and a unique element of order 2. Now we prove the converse for noncyclic groups: Lemma 34.
Suppose G is a noncyclic group of order n ≥ with a cyclic sub-group C of index and a unique element of order . Then G is a generalizedquaternion group.Proof. Because G has order 2 n , it is enough to find generators in G satisfying therelations of Proposition 30. We simply choose R to be any generator of C and T to be any element outside of C . These elements R and T generate G since C hasindex 2 in G . We just need to show that R and T satisfy the desired relations. Thefirst relation, R n − = 1, is immediate.Observe that C is normal in G since its index in G is 2, and so T ∈ C since G/C has order 2. Let C ′ be the cyclic subgroup of C generated by T . If C ′ = C then T generates G , so G is cyclic, a contradiction. So C ′ is a proper subgroup of C .Let H be the unique subgroup of C containing C ′ with [ H : C ′ ] = 2. Let H bethe subgroup of G generated by T . So C ′ also has index 2 in H and | H | = | H | .Every subgroup of a normal cyclic subgroup of G must be normal in G (sincethere is at most one subgroup of a cyclic group of any given order). Thus H isa normal subgroup of G . This implies that Q = H H is a subgroup of G . Everyelement of Q can be written as h or hT with h ∈ H since T ∈ H . Thus Q hasorder 2 | H | . Since H and H have equal order, they both have index 2 in Q . Bythe previous lemma Q is the quaternion group of 8 elements.Since Q is the quaternion group with 8 elements, T has order 4. Now T and R n − have order 2, and so are equal: T = R n − . Also H has order 4 so C has order at least 4.Note that T acts via conjugation on C since C is normal in G . The square T is in C so acts trivially on C . However, T cannot act trivially on C since G is not Abelian (otherwise it would be cyclic by Lemma 31). Thus T acts as anautomorphism of C of order 2. One such automorphism is x x − . If | C | ≥ x x n − +1 and x x n − − .Assume for now that C has order 8 or more. Then H (which has four elements)is a proper subgroup of C . Observe that x x n − +1 maps R to R , so restrictsto the identity on any proper subgroup of C , including H . But T acts nontriviallyon H (since H H = Q is non-Abelian), so this gives a contradiction. Supposeinstead that T acts as x x n − − . Then( T R ) = T ( T − RT ) R = T R n − = 122ince T and R n − are each the unique element of order 2 in G . Thus T R hasorder 2. But
T R is not in C , so cannot be the unique element of order 2.Thus in any case T acts as x x − on C . So we get the third and last relation T RT − = R − . As before, let G be a 2-group with a unique element of order 2. The above tellsus that if G has a cyclic subgroup of index 2 it is either cyclic or a generalizedquaternion group. The following tells us that if G has a generalized quaternionsubgroup of index 2 then G is itself a generalized quaternion group. Since G musthave a subgroup of index 2, we can combine these two cases to construct a simpleinduction to argue that G is either cyclic or is a generalized quaternion group. Lemma 35.
Let G be a -group with a unique element of order . If G has a gen-eralized quaternion subgroup Q of index then G is itself a generalized quaterniongroup.Proof. The subgroup Q has a unique cyclic subgroup C of index 2 in Q in the casewhere | Q | >
8. If | Q | = 8 then Q has three such cyclic subgroups C , C , C . Notethat Q is normal in G since it has index 2. By uniqueness of C for | Q | > C is also a normal subgroup of G . Now consider the case | Q | = 8. Any τ ∈ G permutes the set { C , C , C } via conjugation, but Q acts trivially on this set sinceeach C i has index 2 and so is normal in Q . In other words, the 2-element group G/Q acts on the three element set { C , C , C } . At least one C i must be fixed by thisaction, and so must be normal in G . So in any case Q has a cyclic subgroup C ofindex 2 that is a normal subgroup of G .Since C is a normal subgroup of G , if τ ∈ G then τ acts on C by conjugation.It must act as x x k for some odd k , since all automorphisms of C are of thisform. So τ acts as x x k . But τ ∈ Q since Q has index 2 in G . So τ acts as either x x − or x x since Q is a generalized quaternion group. Notethat k
6≡ − j if 2 j ≥ τ acts as the trivial map x x . This implies that τ ∈ C .If τ C then H τ = C h τ i is a subgroup of G (since C is normal in G ). Asabove we have τ ∈ C , so every element of H τ is of the form h or hτ with h ∈ C .So [ H τ : C ] = 2 and hence [ G : H τ ] = 2. If H τ is cyclic then G is a generalizedquaternion group by the previous lemma, and we are done ( G is not cyclic sinceit contains Q as a subgroup). If H τ is not cyclic, then it must be a generalizedquaternion group by the previous lemma. In particular, τ acts on C by the auto-morphism x x − .Let τ ∈ G − Q and σ ∈ Q − C . If either H τ or H τσ is cyclic, then as pointed outabove, G is a general quaternion group. If neither are cyclic, then τ and τ σ bothact on C by the automorphism x x − . But this means σ ∈ Q − C acts triviallyon C , a contradiction.We are now ready to assert the main results about 2-groups. Theorem 36.
Let G be a -group with a unique element of order . Then G iscyclic or is isomorphic to a general quaternion group. roof. We proceed by induction on k ≥ | G | = 2 k . The base case k = 1 isclear so assume k ≥
2. Let H be a subgroup of index 2 (Proposition 185). By theinduction hypothesis, H is cyclic or isomorphic to a generalized quaternion group.If H is cyclic, the result follows from Lemma 34. If H is a generalized quaterniongroup, the result follows from Lemma 35. Corollary 37.
Let G be a nontrivial -group. Then the following are equivalent:1. G is freely representable.2. Every subgroup of G of order is cyclic.3. G has a unique element of order .4. G is a cyclic or generalized quaternion group.Proof. The implication (1) ⇒ (2) is covered by Theorem 18. Note that 2 dividesthe order of the center of G (Proposition 182), so the implication (2) ⇒ (3) iscovered by Lemma 22. The implication (3) ⇒ (4) is covered by the above theorem.The implication (4) ⇒ (1) is covered by the discussion of Example 2.It will be convenient to give the groups of the above corollary a special name: Definition 4.
A 2 -cycloidal group is a 2-group that is either cyclic or a generalizedquaternion group. Equivalently a 2-cyloidal group is a freely representable 2-group.Here is another way in which 2-cycloidal groups behave like cyclic groups:
Proposition 38.
Let G be a -group. Then G is a -cycloidal group if and only ifevery Abelian subgroup A of G is cyclic.Proof. We assume | G | ≥ | G | = 1 is clear. Suppose G is a 2-cycloidalgroup. Then G and hence every nontrivial subgroup A of G has a unique elementof order 2. By Lemma 31, any Abelian subgroup A of G is cyclic.Conversely, suppose every Abelian subgroup A of G is cyclic. Since every sub-group of order 4 is Abelian, every subgroup of order 4 is cyclic. Thus G is a2-cycloidal group by Corollary 37.Here are some additional properties of 2-cycloidal groups Proposition 39.
Every subgroup H of a -cycloidal group G is a -cycloidal group.Proof. Every nontrivial subgroup of G must contain a unique element of order 2. Proposition 40.
Let G be a -cycloidal group. Then G contains a cyclic subgroupof index . If G is not the quaternion group with 8 elements then the cyclic subgroupof index is unique.Proof. We can identify G with a subgroup of H . The image of G under the doublecover H → SO(3) has image G ′ which is either dihedral or cyclic. So G ′ has acyclic subgroup H ′ of index 2, and its preimage H has index 2 in G (see Lemma 8).Lemma 33 covers the uniqueness claim for generalized quaternion groups.24 roposition 41. Let G be a generalized quaternion group. For all ≤ k ≤ | G | there is a generalized quaternion subgroup H of G of order k .Proof. We can think of G as a subgroup of H . As in Example 2, there is a naturalhomomorphism H → SO(3), and the preimage of any subgroup of order n of SO(3)is a subgroup of order 2 n of H . The image of G is a dihedral group D of order | G | / D is G . A standard property of a dihedral group of order 2 n is that it contains a dihedral subgroup of order 2 m for each m > n .So let E be a dihedral subgroup of D of order 2 k − , and let H be the preimageof E . Since H is the preimage of a dihedral 2-group, H is a generalized quaterniongroup.What is required to force a 2-group to be cyclic? The following gives an answer: Proposition 42.
Let G be a -group of order at least . Then G is cyclic if andonly if G has a unique subgroup of order and that subgroup is cyclic.Proof. If G is cyclic, then it has a unique subgroup of any order dividing | G | , andthat subgroup is unique. So the result follows.Now suppose G has a unique subgroup H of order 4, and that H is cyclic.Note that 2 divides the order of the center of G (Proposition 182), so G containsa unique element of order 2 by Lemma 22. By Theorem 36, G is either cyclic ora generalized quaternion group. We just need to eliminate the second possibility.If G is a generalized quaternion group, then it has a subgroup isomorphic to thequaternion subgroup with 8 elements, and that subgroup contains three subgroupsof order 4. Now that we have classified the p -groups that are freely representable, we turn tothe general case. This document gives a self-contained treatment in the solvablecase, but when we get to the non-solvable case some results will be given withoutproof, with citations to the group-theoretic literature instead. A necessary but far from sufficient condition for a group G to be freely repre-sentable is that every Sylow subgroup of G is freely representable. We have by ourprevious results that a p -Sylow subgroup G with p odd will be freely representableif and only if G is cyclic. A 2-Sylow subgroup of G is freely representable if andonly if it is cyclic or a generalized quaternion group. We have an informal notionof cycloidal where we want this adjective to apply to families of groups that be-have in important ways like cyclic groups. We won’t commit to a final definitionof “cycloidal” here, but will keep the notion a bit open. Let’s agree to (1) admitall finite subgroups of H × as cycloidal. Let’s also agree that (2) every subgroup ofa cycloidal group is cycloidal, and (3) an Abelian group is cycloidal if and only ifit is actually cyclic. These stipulations are enough to force a definition of cycloidalin the case of p -groups (see Theorem 25, Proposition 183, and Corollary 37): At least in this version of this document. The hope is to give a unified account that incor-porates the non-solvable case in a sequel, or a future version of this report. efinition 5. Let p be an odd prime. Then a p -group is cycloidal if it is cyclic.As above, a 2-group is cycloidal if it is cyclic or a generalized quaternion group. Agroup G is Sylow-cycloidal if every Sylow subgroup is cycloidal in the above sense.A group G of even order is Sylow-cycloidal-quaternion if every Sylow subgroup iscycloidal and if the 2-Sylow subgroups are quaternion. A group G is Sylow-cyclic if every Sylow subgroup is cyclic. So when we classify freely representable groups we can limit our attention toSylow-cycloidal groups. We will divide our classification of Sylow-cycloidal groups,and hence freely representable groups, into three categories:1.
Sylow-cyclic groups . These turn out to be solvable groups.2.
Solvable Sylow-cyclic-quaternion groups .3.
Non-Solvable Sylow-cyclic-quaternion groups . Remark.
According to the Sylow theorems, all p -Sylow subgroups are conjugatehence isomorphic. So to determine if G is Sylow-cycloidal or Sylow-cyclic, it isenough to look at one p -Sylow subgroup for each prime p . Remark.
It is interesting to note that every freely representable group has theproperty that all its Sylow subgroups are isomorphic to subgroups of H × . Alsonote that Sylow-cyclic groups are those whose Sylow subgroups are isomorphic tosubgroups of C × . Remark.
Although not all Sylow-cycloidal groups are freely representable, the col-lection of Sylow-cycloidal groups is of great interest since such groups are exactlythe groups whose Abelian subgroups are all cyclic. A proof of this fact is providedbelow (Theorem 48). Such groups arise in the classification of groups of periodiccohomology (see Wall [16]). Now we consider some basic observations concerning Sylow-cycloidal groups.
Proposition 43.
Let G be a Sylow-cycloidal group. Then every subgroup H of G is a Sylow-cycloidal group.Proof. Let H p be a p -Sylow subgroup of H for a prime p dividing | G | . By a Sylowtheorem (Theorem 180) H p is a subgroup of a p -Sylow subgroup G p of G . When p is odd, G p is cyclic, and so the subgroup H p of G p must also be cyclic. Similarly,if p = 2 then H p must be a 2-cycloidal group (Proposition 39). Proposition 44.
Let A be an Abelian Sylow-cycloidal group. Then A is cyclic,hence A is freely representable. As far as I know, the terminology introduced in this definition is new. Unfortunately I have not explored periodic cohomology in this document, but I hope to coverthis topic either in a sequel or in a future version of this report. roof. Every Abelian group is the product of its Sylow subgroups. Since A isAbelian, all its Sylow subgroups must be Abelian, and hence cyclic. The productof cyclic subgroups of pairwise relatively prime orders is cyclic. Remark.
This gives another proof of Lemma 15 that is less dependent on represen-tation theory or the fact that all finite subgroups of C × is cyclic. Proposition 45. If G is a Sylow-cycloidal group, and N is a normal subgroup ofodd order, then G/N is a Sylow-cycloidal group.Proof. If G p is a p -Sylow subgroup, then its image G ′ p in G/N is a p -Sylow subgroupof G/N , assuming p divides the order of G/N . If p is odd then G ′ p is the imageof a cyclic group, so G ′ p is itself cyclic. If p = 2 then the restricted canonicalmap G p → G/N has trivial kernel G p ∩ N since | N | is odd, so G ′ p is isomorphicto G p , and so is a 2-cycloidal group. Proposition 46.
Suppose A and B are subgroups of G such that | A | and | B | arerelatively prime and such that | A || B | = | G | . Then G is Sylow-cycloidal if and onlyif both A and B are Sylow-cycloidal.Proof. One direction follows just because A and B are subgroups of G . So supposethat A and B are Sylow-cycloidal. Let p be a prime dividing | A | . Every p -Sylowsubgroup P of A is in fact a p -Sylow subgroup of G (since p cannot divide | B | ), andis either cyclic or is a generalized quaternion group. The same holds for any primedividing | B | . Since | G | = | A || B | , for any prime p dividing the order of G there is ais p -Sylow subgroup that is either cyclic or is a generalized quaternion group. Thisimplies that G is Sylow-cycloidal. Corollary 47.
The product or even the semi-direct product of two Sylow-cycloidalgroups of relatively prime orders is itself Sylow-cycloidal.
Theorem 48.
Let G be a finite group. Then G is Sylow-cycloidal if and only ifevery Abelian subgroup of G is cyclic.Proof. Suppose that G is Sylow-cycloidal with Abelian subgroup A . Then A isSylow-cycloidal (Proposition 43). Hence A is cyclic (Proposition 44).Now suppose every Abelian subgroup of G is cyclic. Let P be a p -Sylow subgroupof G . If H is a subgroup of P of order p then H is Abelian (Proposition 183) andso is cyclic. If p is odd then P must itself be cyclic (Theorem 25). If p = 2then P must be cyclic or a generalized quaternion group (Corollary 37). Thus G isSylow-cycloidal. Suppose A and B are subgroups of an Abelian group with A ∩ B = { } , and consider thenatural map A × B → AB . This must be an isomorphism. This is a nice general fact. To see it, note that the restricted canonical map G p → G/N haskernel G p ∩ N . The largest power of p dividing | G/N | times the largest power of p dividing | N | is equal to the order of G p . Similarly, | G ′ p | times | G p ∩ N | is also equal to the order of G p .An inequality forces equality: | G ′ p | is equal to the largest power of p dividing | G/N | . Note alsothat G p ∩ N must be the p -Sylow subgroup of N .
27e saw above that every even-ordered freely representable group has a uniqueelement of order 2. This generalizes to many, but not all, Sylow-cycloidal groups.The following lemma is useful in this respect:
Lemma 49.
Let G be a finite group. Suppose G has a p -Sylow subgroup having aunique subgroup order p . Then G has a unique subgroup of order p if and only if G has a normal subgroup of order p .In particular, if G has a -Sylow subgroup that has a unique element of order then G has a unique element of order if and only if G has a normal subgroup oforder .Proof. One direction is clear: a unique subgroup of order p is characteristic hencenormal in G .Conversely, suppose A is a normal subgroup of G of order p , and let B beany subgroup of G order p . By a Sylow theorem (Theorem 180), A is containedin a p -Sylow subgroup P , and B is contained in a p -Sylow subgroup P . By aSylow theorem (Theorem 181), P and P are conjugate. We also know that A is the unique subgroup of P of order p , and B is the unique subgroup of P oforder p (uniqueness for all p -Sylow subgroups follows from the fact that all p -Sylowsubgroups are conjugate). This uniqueness property implies that the conjugationmap that carries P to P must carry A to B . But A is normal and so conjugationcarries A to itself. Thus A = B as desired. The term “Sylow-cyclic” might be new, but these groups have a long history withcontributions by early group theory pioneers such as H¨older, Frobenius, Burnside,and later in the 1930s by Zassenahaus. They have been classified since the early20th century, and have a relatively simple structure: they are the semi-direct prod-ucts of cyclic groups of relatively prime order. Some, but not all, of these groupsare freely representable:
Example . Any group of square-free order is Sylow-cyclic. This includes dihedralgroups with 2 n elements where n is square free and odd. Such dihedral groupscannot be freely representable since they do not have a unique element of order 2. Thus there are an infinite number of orders such that there exists Sylow-cyclicgroups of that order that are not freely representable.
Later we will see that if anon-Abelian group has square-free order then it cannot be freely representable.
Example . Let G be a binary dihedral group 2 D n where n is a square free odd inte-ger. Such groups occur as subgroups of H × so are freely representable. They mustthen be Sylow-cyclic (since the 2-Sylow subgroups have four elements). So thereare infinite number of orders of Sylow-cyclic groups that are freely representable.Later we will see there are an infinite number of odd orders of non-Abelian Sylow-cyclic groups that are freely representable, and none of these can be isomorphic to Suzuki [14] and a few other authors call these groups “ Z -groups”, perhaps as a pun on“Zassenhaus” and “Zyklische Gruppe”. I started using “Sylow-cyclic” before learning of the term“ Z -group”. I have a slight preference for the more descriptive term “Sylow-cyclic”.
28 subgroup of H × . These will be our first examples of freely representable groupsthat are not isomorphic to subgroups of H × .We continue with some basic observations about Sylow-cyclic groups: Proposition 50.
Any Abelian Sylow-cyclic group is cyclic, and hence is freelyrepresentable.Proof.
This is a special case of Proposition 44.
Proposition 51.
Any subgroup H of a Sylow-cyclic group G is also a Sylow-cyclicgroup.Proof. Let H p be a p -Sylow subgroup of H for a prime p dividing | G | . By a Sylowtheorem (Theorem 180), H p is a subgroup of a p -Sylow subgroup G p of G . Since G p is cyclic, the subgroup H p of G p must also be cyclic. Proposition 52.
Any quotient group
G/N of a Sylow-cyclic group G is also aSylow-cyclic group.Proof. The proof is similar to the proof of Proposition 45. The basic idea is thatany Sylow subgroup of G maps to a Sylow subgroup of G/N (or a trivial group).
Proposition 53.
Suppose A and B are subgroups of G such that | A | and | B | arerelatively prime and such that | A || B | = | G | . Then G is Sylow-cylic if and only ifboth A and B are Sylow-cyclic.Proof. The proof is similar to that of Proposition 46.
Corollary 54.
The product or semi-direct product of two Sylow-cyclic groups ofrelatively prime orders is itself Sylow-cyclic.Example . Let p be an odd prime and let q be any prime dividing p −
1. Thecyclic group C p of size p has F × p for its automorphism group, which is cyclic andcontains a unique subgroup of order q . Let C q be a cyclic group of size q , and fix anisomorphism with the subgroup of F × p of size q . This gives a nontrivial action of C q on C p . Thus the semidirect product G = C p ⋊ C q is a non-Abelian Sylow-cyclicgroup whose order is a multiple of p . By Theorem 18, G is not freely representable.For example, we can construct a Sylow-cyclic group of order 21 that is not freelyrepresentable; here p = 7 and q = 3.The converse of Corollary 54 true, and is a deeper result (proved by Burnsideby 1905). Toward this end, we start with the following interesting result: Theorem 55.
Suppose that G is Sylow-cyclic and that q is the largest prime di-viding | G | . Then G has a unique q -Sylow subgroup and this subgroup is normal. The proof is adapted from M. Hall [9], proof of Theorem 9.4.3. roof. We start by defining a large-prime divisor of | G | to be a divisor d of | G | suchthat the primes dividing d are as large as or larger than any prime dividing | G | /d .Then we establish the following claim: for every large-prime divisor d of | G | , theset E d = { x ∈ G | x d = 1 } has exactly d elements.To establish the claim, let d be the largest large-prime divisor that is a counter-example. Note that d < | G | since the claim holds for d = | G | . Let p be thelargest prime divisor of the complementary divisor | G | /d . Then pd is a large-primedivisor, and we get to assume that E pd has pd elements. Let p k be the largestpower of p dividing d . Any p -Sylow subgroup of G is cyclic, and so has an elementof order p k +1 . Such an element is in E pd but not in E d . So E pd is strictly largerthan E d .We partition the nonempty set E pd − E d by declaring two elements equivalentif they generate the same cyclic subgroup. If g ∈ E pd − E d has order t , then theset of elements that generate h g i has φ ( t ) elements. Now p − φ ( t ) sincethe order of t is divisible by p . So E pd − E d has size m ( p −
1) for some positive m .Next, we note that E d has order nd for some positive n by Frobenius’s theorem(Theorem 186). Thus pd = nd + m ( p − , so d | m ( p − . By choice of p , there is no prime divisor of d strictly smaller than p . Thus p − d . So d actually divides m . We write m = m d for m >
0, and pd = nd + m d ( p − , so p = n + m ( p − . The only solution to this last equation with positive n and m is n = m = 1. Thisshows that E d has d elements, so d cannot be a counter-example.Now let p be the largest prime divisor of | G | and let p l be the largest primepower of p dividing | G | . Every p -Sylow subgroup of G is contained in E p l . Since p l is a large-prime divisor of | G | , we must have that every p -Sylow subgroup of G isthe set E p l since they have the same size. So there is a unique p -Sylow subgroup,and it is normal since E p l is invariant under conjugation.By repeated application of the above result we get the following: Corollary 56.
Suppose that G is Sylow-cyclic and that p is the smallest primedividing the order of G . Then G has a quotient of order p . In particular, G has anontrivial Abelian quotient. We also get the following result by repeatedly applying the above theorem. Corollary 57.
Every Sylow-cyclic group is solvable.
An important step in the classification is the following:
Proposition 58.
Let G ′ be the commutator subgroup of a Sylow-cyclic group G .Then G ′ and G/G ′ are cyclic. I have seen this result attributed to H¨older (1859–1877). roof. Consider the derived series. In other words, recursively define G ( n +1) tobe the commutator subgroup of G ( n ) starting with G (1) = G ′ . By Corollary 56,if G ( n ) is nontrivial then G ( n +1) is a proper subgroup of G ( n ) . Thus G ( n ) = 1 forsufficiently large n .The result is clear for Abelian G , so assume G is non-Abelian. Let m be thelargest integer such that G ( m ) is nontrivial. Since G ( m ) is an Abelian Sylow-cyclicgroup, it is cyclic. Observe that G acts on G ( m ) by conjugation. Thus
G/A is isomorphic to a subgroup of the automorphism group of G ( m ) where A is thecentralizer of G ( m ) . Since G ( m ) is cyclic, its automorphism group is ( Z /t Z ) × where t is the order of G ( m ) . Thus G/A is Abelian, which means that G ′ ⊆ A . In otherwords, if g ∈ G ′ and h ∈ G ( m ) then g and h commute.Suppose m >
1. Then G ( m − /G ( m ) is an Abelian Sylow-cyclic group, so iscyclic. Let g ∈ G ( m − be such that its image in G ( m − /G ( m ) is a generator.Since g ∈ G ′ , it commutes with every element of G ( m ) . So G ( m − is Abeliansince G ( m − = h g i G ( m ) . This implies G ( m ) is trivial, a contradiction. So m = 1,as desired.Since m = 1, G ′ = G ( m ) is cyclic. Since G/G ′ is Abelian and Sylow-cyclic, it iscyclic as well.Now we are ready for Burnside’s result on the structure of Sylow-cyclic groups: Theorem 59.
Let G be a Sylow-cyclic group and let A be the commutator subgroupof G . Then, as above, both A and G/A are cyclic. Let m be the order of A andlet n be the order of G/A , so G has order mn . Let b ∈ G be any element mappingto a generator of G/A , and let B be the subgroups of G generated by b . Then thefollowing hold: • The subgroup B has order n , and so its generator b has order n . • The subgroup B is a complement of A in G , so G = AB = A ⋊ B . • The subgroups A and B have relatively prime orders m and n . • The order m of A is odd. • The normalizer of B and centralizer of B in G are both equal to B . • There are m subgroups of G of order n , and they are all conjugate in G to B . • There there is a pair of subgroups of order n that together generates all of G .Proof. By the previous proposition, A and G/A are cyclic. Choose b ∈ G so thatits image in G/A is a generator, and choose a to be a generator of A . Let B = h b i .Observe that a and b together generate G and G = AB . Consider the commutator a ′ def = [ a, b ] = a − b − ab ∈ A. Every subgroup of A is normal in G (since A is normal in G , and has at mostone subgroup of any given order), so the group A ′ generated by a ′ is normal in G . Any automorphism of a group restricts to an automorphism of its commutator subgroup.Thus, by induction, conjugation automorphisms (inner automorphisms) on G restricts to auto-morphisms of G ( m ) . G/A ′ is generated by the images of a and b , and these images commute.So G/A ′ is Abelian, which means that A ′ contains the commutator subgroup A . Inother words, A = A ′ .Since a and a ′ commute, we have [ a t , b ] = ( a ′ ) t for all positive integers t . Supposenow that a t commutes with b . Then1 = [ a t , b ] = ( a ′ ) t . So t must be a multiple of the order of A since a ′ is a generator of A . Hence a t = 1.So 1 is the only element of A that centralizes B . In other words A ∩ Z ( B ) = { } .Of course the center Z ( G ) is a subgroup of Z ( B ), and B is also a subgroup of Z ( B )since B is cyclic. So A ∩ Z ( G ) = { } and A ∩ B = { } . This last equation showsthat B is a complement to A and so B has size n . In particular, the restrictionof G → G/A to B is an isomorphism B → G/A .Now suppose g = a s b t normalizes B . Since B is cyclic, this means that a s normalizes B . Note that a s ba − s ∈ B and b ∈ B map to the same element underthe isomorphism B → G/A so a s ba − s = b . Thus a s ∈ A ∩ Z ( B ) = { } . Weconclude that the normalizer of B is just B itself: N ( B ) = Z ( B ) = B . By thestabilizer-orbit theorem, there are exactly m distinct conjugates of B in G .Next we show that the orders of A and B are relatively prime. Suppose otherwisethat a prime p divides the order of both A and B . Then A and B each havea subgroup of order p , call them A ′ and B ′ . Note that A ′ and B ′ are distinctsince A ∩ B = { } . Since A ′ is the unique subgroup of the normal subgroup A of order p , the subgroup A ′ must be normal in G . This means A ′ B ′ is a group oforder p . Let C be a p -Sylow subgroup of G containing A ′ B ′ . Then C contains atleast two subgroups ( A ′ and B ′ ) of order p , contradicting the fact that C is cyclic.Next we show that A has odd order. This is clear if G has odd order so supposethat G has even order. By Corollary 56, G has a quotient G/K isomorphic tothe 2-Sylow subgroup of G , so K has odd order. The result follows from the factthat G/K is Abelian and that A is the commutator subgroup so A ⊆ K .Now we show that every subgroup of G of order n is conjugate to B . Since B has exactly m conjugates, it is enough to show that there are at most m subgroupsof G of order n . Suppose H is a subgroup of order n . Then its image under themap G → G/A must be all of
G/A since A ∩ H = { } , and the restriction H → G/A is an isomorphism. The image of b in G/A generates
G/A , so H has an elementof the form a t b which generates H . There are only m elements of the form a t b , sothere can be at most m subgroups of order n .Finally, note that the above arguments apply to any other choice of b . Inparticular, if b is the original choice, then ab also has order n . The group generatedby h b i and h ab i contains b and a , so is all of G .This yields an interesting corollary: Corollary 60.
Let G be a Sylow-cyclic subgroup with commutator subgroup A .Then the center Z ( G ) of G has the property that A ∩ Z ( G ) = { } . In fact, Z ( G ) isthe intersection of all subgroups B of G of order n = | G/A | .Proof. By the above theorem there is a cyclic complement B of A suchthat Z ( B ) = B . Thus Z ( G ) ⊆ Z ( B ) = B . All subgroups of G of size n B , so Z ( G ) is contained in the intersection Z of all subgroupsof G of order n . Also A ∩ Z ( G ) = { } since A ∩ B = { } .Note that Z commutes with each subgroup of order n since such groups arecyclic. Since groups of order n generate G , we conclude that Z = Z ( G ).Here is another application of the above theorem, one that shows a strongparallelism between Sylow-cyclic groups and cyclic groups: Theorem 61.
Let G be a Sylow-cyclic group of order N . For any divisors d of N ,there is a subgroup of G of order d , and all subgroups of order d are conjugate in G .Proof. By the above theorem, we can write G as A ⋊ B where A and B are cyclicof relatively prime orders and where A is the commutator subgroup of G .Let d be a divisor of | G | and write d = ef where e divides | A | and f divides | B | .Let A e be the unique subgroup of A of order e . Since A e is the unique subgroupof A of order e , and since A is normal in G , it follows that A e is normal in G . Let B f be the unique subgroup of B of order f . Since A e ∩ B f = { } , the subgroup A e B f has order d = ef .Now suppose that H and H are subgroups of order d . Replacing G if necessary,we can assume that G is generated by H and H . (And we choose A and B forthe new G ). As before, write d = ef where e divides | A | and f divides | B | . Notethat under the map G → G/A , both H and H have images of size f , and therestriction H i → G/A has kernel A ∩ H i of size e . Since A and G/A are cyclic, thismeans that H and H have the same image in G/A and that A ∩ H = A ∩ H .Let A = A ∩ H = A ∩ H and let G /A be the image of H i in G/A , where G is a subgroup of G containing A . Since G is generated by H and H , we musthave G = G , and so f = n = | G/A | . Let B i a subgroup of H i of order n (whichexists since H i → G/A is surjective). Note that A B = H and A B = H .By Theorem 59, B and B are conjugate subgroups of G . Thus H and H areconjugate. Remark.
Let G be a Sylow-cyclic group of order N . In light of the above theorem,we observe that for any divisor d of N the following are equivalent: (1) G has aunique subgroup of order d , (2) G has a characteristic subgroup of order d , and(3) G has a normal subgroup of order d .We can classify divisors d of | G | based on whether or not there is a uniquesubgroup of order d . Alternatively we can classify divisors d based on whetheror not the subgroups of order d are cyclic. We are particularly interested in thedivisors d for which both conditions hold: there is a normal cyclic subgroup oforder d . Since every subgroup of a cyclic normal subgroup is cyclic normal, weconclude that if there is a normal cyclic subgroup of of order d , then the same istrue for each divisor of d . The following proposition and corollaries give us furtherresult about such subgroups that are normal and cyclic. Lemma 62.
Let G be a finite group with the property that any two subgroups of thesame prime power order are conjugate. If A and B are cyclic normal subgroups of G then AB is a cyclic normal subgroup of order equal to the least common multipleof | A | and | B | . roof. Note that AB is a normal subgroup of G since A and B are normal. So wejust need to show AB is cyclic of the specified order.First we consider the case where A and B are of relatively prime order. Since A and B are normal, and since A ∩ B = { } , we have that AB is isomorphic to A × B .Here A × B is cyclic of order equal to | A || B | and so the result holds.Next we consider the case where A and B are p -groups for the same prime p .Either | A | divides | B | or | B | divides | A | . Suppose, say, that | A | divides | B | . Then A has the same order as a subgroup of B (Theorem 179), but there is a uniquesubgroup of order | A | in G since A is normal. So A ⊆ B and so AB = B . Theresult holds in this case as well.In the general case, we have AB = P · · · P k Q · · · Q l where each P i is a Sylow subgroup of A and each Q j is a Sylow subgroup of B .If X and Y are two Sylow subgroups in the above product then XY = Y X since X and Y are normal subgroups of G (all subgroups of a cyclic normal subgroup arecyclic normal subgroups). Thus we can rearrange the terms and reduce to thespecial cases listed above Corollary 63.
Let G be a Sylow-cyclic subgroup of order N . Then there is amaximum cyclic characteristic (MCC) subgroup µ ( G ) of G that contains all normalcyclic subgroups. Every subgroup of µ ( G ) is normal cyclic, and is characteristic.Proof. By Theorem 61, a subgroup of G is normal if and only if it is characteristic.Define µ ( G ) to be the product of all normal cyclic subgroups of G . Theorem 61allows us to use the above lemma to conclude that µ ( G ) is a normal cyclic sub-group, so is characteristic. By construction it contains all normal cyclic subgroupsof G , so is the maximum (under the inclusion relation) among cyclic characteristicsubgroups. Note that every subgroup of a normal cyclic group is a normal cyclicgroup.The following shows that a necessarily condition for a Sylow-cyclic group G ofeven order to be freely representable is that µ ( G ) have even order. Corollary 64.
Let G be a Sylow-cyclic subgroup of order N > and let D be theorder of the maximal cyclic characteristic subgroup µ ( G ) of G . Then the maximumprime divisor of N divides D to the same order as it divides N , so D > . Also G has a unique cyclic subgroup of order d if and only if d divides D . Thus G has aunique element of order two if and only if D is even.Proof. The first claim follows from Theorem 55 and the previous corollary. Theother claims follow directly from the previous corollary.We can give other useful characterizations of µ ( G ). We start with the followinglemma: Lemma 65.
Let G be a Sylow-cyclic group with commutator subgroup G ′ . Thenthe centralizer Z ( G ′ ) of G ′ in G is equal to G ′ · Z ( G ) where Z ( G ) is the centerof G . Furthermore G ′ and Z ( G ) have relatively prime orders and G ′ · Z ( G ) is acharacteristic cyclic subgroup isomorphic to G ′ × Z ( G ) . roof. Let B be a cyclic complement of G ′ . Suppose ab ∈ Z ( G ′ ) where a ∈ G ′ and b ∈ B . Of course a ∈ Z ( G ′ ) since G ′ is cyclic, so b ∈ Z ( G ′ ). Also b ∈ Z ( B )since B is cyclic, so b ∈ Z ( G ′ B ) = Z ( G ). We conclude that Z ( G ′ ) ⊆ G ′ · Z ( G ).The other inclusion is clear.Since Z ( G ) ⊆ B and so G ′ ∩ Z ( G ) = { } (Corollary 60) we have that Z ( G ) isnormal of relatively prime order to | G ′ | and so G ′ · Z ( G ) must be isomorphic to thecyclic group G ′ × Z ( G ). Proposition 66.
Let G be a Sylow-cyclic group with commutator subgroup G ′ andmaximum cyclic characteristic subgroup µ ( G ) . Then µ ( G ) = Z ( G ′ ) = G ′ · Z ( G ) where Z ( G ′ ) is the centralizer of G ′ in G , and where Z ( G ) is the center of G .Proof. Since µ ( G ) is an Abelian group containing G ′ , we have µ ( G ) ⊆ Z ( G ′ ).By the above lemma, Z ( G ′ ) is a characteristic cyclic subgroup of G , which forcesequality.A necessary condition for a Sylow-cyclic group of even order to be freely repre-sentable is that its center have even order: Corollary 67.
Let G be a Sylow-cyclic group with maximal cyclic characteristicsubgroup µ ( G ) and center Z ( G ) . Then G has a unique element of order if andonly if the center Z ( G ) has even order.Proof. This follows from the fact that µ ( G ) = G ′ · Z ( G ) (as above) and the factthat the commutator subgroup G ′ has odd order (Theorem 59).Above we considered subgroups that are normal and cyclic. Now we add a thirdrequirement that the quotient be cyclic as well: Definition 6.
Let G be a finite group. A metacyclic kernel C is any normalsubgroup of G such that both C and G/C are cyclic. For example, we haveshown that the commutator subgroup of a Sylow-cyclic group is a metacyclic kernel.Observe that any metacyclic kernel is a characteristic cyclic subgroup of G and sois contained in the MCC subgroup µ ( G ). Lemma 68.
Let G be a finite group with a metacyclic kernel K . If C is a cyclicsubgroup of G containing K then C is also a metacyclic kernel.Proof. There is a natural correspondence between subgroups of
G/K and subgroupsof G containing K , and that this correspondence is well-behaved for normal sub-groups. Let C be the image of C in G/K . Since
G/K is cyclic, C is a normalsubgroup of G/K . It follows that C is a normal subgroup of G . The quotient G/K by C is cyclic, since G/K is cyclic. But this quotient is isomorphic to
G/C . This is not a standard term in the literature, as far as I know. I came up with this terminologybased on the term metacyclic group which is a fairly standard term for a group G with a normalsubgroup C such that C and G/C are both cyclic. emma 69. Let G be a Sylow-cyclic subgroup. Let G ′ be the commutator subgroupand let Z ( G ′ ) be the centralizer of G ′ . Then a subgroup C of G is a metacyclickernel if and only if G ′ ⊆ C ⊆ Z ( G ′ ) . Proof.
First suppose that C is a metacyclic kernel. Since G/C is cyclic, henceAbelian, it follows that G ′ ⊆ C . Since C is cyclic, hence Abelian, C ⊆ Z ( G ′ ). Soone implication is established.Suppose G ′ ⊆ C ⊆ Z ( G ′ ). Since Z ( G ′ ) is cyclic (Proposition 66), we havethat C is cyclic. So C is a metacyclic kernel by the above lemma. Remark.
Let G be a Sylow-cyclic group and let Z ( G ′ ) be the centralizer of thecommutator group G ′ in G , which is also equal to the MCC subgroup of G (Propo-sition 66). Then Z ( G ′ ) = µ ( G ) is the maximum metacyclic kernel , and G ′ the minimum metacyclic kernel . These groups are important invariants of G , and µ ( G )will be used to identify whether or not G is freely representable. Corollary 70.
Let G be a Sylow-cyclic group and let µ ( G ) be the MCC subgroupof G . Then µ ( G ) is a maximal cyclic subgroup of G .Proof. Since µ ( G ) is the maximum among metacyclic kernels, it is maximal amongall cyclic groups by Lemma 68.In practice µ ( G ) can be calculated using automorphisms from any semidirectdecomposition of G into cyclic groups of relatively prime order: Lemma 71.
Suppose G is the semidirect product A ⋊ B of two cyclic groups ofrelatively prime order. As usual we identify A and B with subgroup of G . Thenthe MCC subgroup µ ( G ) of G is AK where K is the kernel of the associated actionmap B → Aut( A ) .Proof. Note that G is Sylow-cyclic and that A is a metacyclic kernel. Since K acts trivially on A via conjugation, the group AK is Abelian. Since AK is Sylow-cyclic, AK is cyclic. Thus AK is a metacyclic kernel (Lemma 68).By Lemma 69, AK is a subgroup of Z ( G ′ ) = µ ( G ). Let g ∈ µ ( G ). Write g as ab where a ∈ A and b ∈ B . Observe that b ∈ µ ( G ) since a ∈ AK ⊆ µ ( G ). Since µ ( G )is an Abelian group containing A , we see that b acts trivially on A and so b ∈ K .Thus g = ab is in AK . We conclude that AK = µ ( G ).We will need the following later in our proof of Wedderburn’s theorem. Corollary 72.
Let G be a Sylow-cyclic group with MCC subgroup µ ( G ) . Then | µ ( G ) | > [ G : µ ( G )] . Proof.
Write G as the semi-direct product A ⋊ B of two cyclic groups of relativelyprime order m = | A | and n = | B | . Let K be the kernel of the associated actionmap B → Aut( A ), and let I be the image. The automorphism group of A isisomorphic to ( Z /m Z ) × where m is the order of A , so | I | < m and n = | B | = | K || I | < | K | m.
36y the above lemma, | µ ( G ) | = | A || K | , so | µ ( G ) | = m | K | > n. However,
G/AK is a quotient of
G/A ∼ = B . So | G/µ ( G ) | ≤ n . Remark.
Let A be a cyclic group of order m and let B be a group of order n , where m and n are relatively prime. Let a be a generator of A and let b be a generator of B .We construct a semi-direct product G of A with B by choosing an automorphism σ of A associated with conjugation by b . Such σ is of the form x x r where r isany r ∈ ( Z /m Z ) × of order dividing n . So r n = 1 in Z /m Z . In G we have b − ab = a r , [ a, b ] = a − b − ab = a r − . Clearly the commutator subgroup G ′ will be contained in A since the quotientis Abelian (isomorphic to B ). Since [ a, b ] = a r − we have (cid:10) a r − (cid:11) ⊆ G ′ , andsince G/ h [ a, b ] i is Abelian, we have G ′ ⊆ h [ a, b ] i . Thus G ′ = (cid:10) a r − (cid:11) . If we want G ′ = A we will also need r − m . Example . As an illustration we classify groups G of order 210 = 2 · · · A = G ′ . Since A is a proper subgroup of G of odd order, its possible ordersare 1 , , , , , , , or 105. We will use the notation of the above remark.If A has order 1 then G is cyclic and µ ( G ) = G . This gives us the only Abelianexample.If A has order 3 then we choose r ∈ ( Z / Z ) × so that r − r = −
1. Here B has order 70 and the kernel K of the map B → Aut( A )has order 35. So µ ( G ) has order 105 and so all subgroups of odd order are cyclicand normal. Note that in this case there is a subgroup isomorphic to the dihedralgroup D .If A has order 5, then B has order 42. We choose r ∈ ( Z / Z ) × so that r − r has order dividing 42. This means that r is − K of the map B → Aut( A ) has order 21. So µ ( G ) has order 105 and soall subgroups of odd order are cyclic and normal. Note that in this case there is asubgroup isomorphic to the dihedral group D .If A has order 7, then B has order 30. Every element of r ∈ ( Z / Z ) × has orderdividing 30, so we just need r − r = 2 , , , , − • If r = − K of the map B → Aut( A ) has order 15. So µ ( G )has order 105 and all subgroups of odd order are cyclic and normal. Note inthis case there is a subgroup isomorphic to D . • If r = 2 , K of the map B → Aut( A ) has order 10. So µ ( G )has order 70. • If r = 3 , K of the map B → Aut( A ) has order 5.So µ ( G ) has order 35. Note in this case there is a subgroup isomorphic to D .37ote that if we replace our generator b of B with its inverse the corresponding r changes to its multiplicative inverse. So r = 3 , r = 2 , A hasorder 7.If A has order 15 then B has order 14. Then we choose r ∈ ( Z / Z ) × whichis isomorphic to C × C . This means that r should have order 2, which meansthat r = 4 , , − r − r = − K of the map B → Aut( A ) has order 7. So µ ( G ) hasorder 105 and so all subgroups of odd order are cyclic and normal. Note that inthis case there is a subgroup isomorphic to the dihedral group D .If A has order 21 then B has order 10. Then we choose r ∈ ( Z / Z ) × whichis isomorphic to C × C . This means that r should have order 2, which meansthat r = 8 , , −
1. But we also want r − r = − K of the map B → Aut( A ) has order 5. So µ ( G ) hasorder 105 and so all subgroups of odd order are cyclic and normal. Note that inthis case there is a subgroup isomorphic to the dihedral group D .If A has has order 35 then B has order 6. Then we choose r ∈ ( Z / Z ) × where r − r has order dividing 6. This means that r ≡ − , r ≡ , , , , − r = 4 , , , , − . We can divide this into two subtypes depending on the order of r ∈ ( Z / Z ) × : • If r = − r has order 2 and the kernel K of B → Aut( A ) has order 3.So µ ( G ) has order 105, and all subgroups of odd order are cyclic and normal.Here G contains D as a subgroup. • Otherwise (for r = 4 , , ,
24) we have r of order 6 and the kernel K of themap B → Aut( A ) has order 1. So µ ( G ) = A . Here G also contains D asa subgroup. Note that r = 4 and r = 9 are inverses and so yield isomorphicsemidirect products. Similarly, 19 and 24 give similar results.Finally if A has order 105, then B has order 2. In order to satisfy the require-ments that r have order at most 2 and that r − , ,
7, wemust have r ≡ − , r = − µ ( G ) = A has 105 elements and G is D .All in all we have 12 groups of order 210 up to isomorphism. Later we will see thatnone of them are freely representable except the cyclic group. In fact, only one non-Abelian example (where A has order 7) satisfies the necessary condition that µ ( G )has even order and so there is a unique element of order two (Corollary 67). Now we take-up the question of which Sylow-cyclic groups are freely representable.38 emma 73.
Suppose G is the semi-direct product A ⋊ B of two cyclic groupswhere B has prime order p and where A has order q k for a prime q not equal to p .Then if G is freely representable, G must be cyclic.Proof. By Corollary 54, G is Sylow-cyclic. So by Theorem 59 the commutatorsubgroup G ′ of G is cyclic, and G ′ and G/G ′ have relatively prime orders. Since G/A is isomorphic to B and so is Abelian, G ′ ⊆ A . So either G ′ = 1 or G ′ = A .Start by assuming G ′ = A . By Theorem 59, any element of g ∈ G whose imagein G/A generates
G/A must generate a group h g i of order p . Since every nontrivialelement of G/A generates this group, every element g ∈ G − A is contained in aunique subgroup of G of order p .Let C be the collection of subgroups of G consisting of A together with allsubgroups of order p . Then every nonidentity element of G is in exactly one C ∈ C .So X C ∈C N C = ( k − + N G where k is the size of C . This contradicts Theorem 12 since k > A is not G ′ . This means that G ′ = 1. So G is Abelian.Since G is Sylow-cyclic and Abelian, G must be cyclic. Theorem 74.
Suppose G is the semidirect product A ⋊ B of two cyclic groupswhere B has prime order p and where p does not divide the order of A . Then G isfreely representable if and only if G is cyclic.Proof. One implication is clear, so suppose G is freely representable. Since A isnormal and cyclic in G , all the subgroups of A are normal and cyclic in G (sincethere is at most one subgroup of A of any given order). So if q is a prime dividing theorder of A then the q -Sylow subgroup A q of A will be a normal q -Sylow subgroupof G . So viewing B as a subgroup of G , we have that H q = A q B = A q ⋊ B isa subgroup of G . Since H q is a subgroup of a freely representable group, H q is afreely representable group. By the above lemma H q is cyclic.In particular B acts trivially by conjugation on A q for each Sylow subgroup of A .Since A is cyclic, it is generated by its Sylow subgroups. Thus B acts trivially onall of A . This implies that G is Abelian. Since G is Sylow-cyclic this means that G is cyclic.We can leverage this to get a necessary condition for a Sylow-cyclic group G tobe freely representable. Basically the condition is that G is a semi-direct productof cyclic groups using a relatively weak action: Lemma 75.
Let A and B be cyclic subgroups of relatively prime orders of a group G such that G = AB = A ⋊ B . If G is freely representable then the kernel K of theassociated action homomorphism B → Aut( A ) must contain all subgroups of B of prime order. roof. Suppose B p is a subgroup of B of prime order p . Observe that AB p isa subgroup of G of order | A | p (since A is normal in G ) and is the semi-directproduct A ⋊ B p where the action of B p on A is just the restriction of the actionof B on A .Since G is freely representable, AB p is as well. So by the above theorem, AB p is cyclic, hence Abelian. Thus B p acts trivially on A . In other words, B p is in thekernel K of the action of B on A . Corollary 76.
Suppose G is the semi-direct product A ⋊ B of two cyclic groupsof relatively prime order. Suppose B has square free order. Then G is freely repre-sentable if and only if G is cyclic.In particular, if G is a group of square free order then G is freely representableif and only if G is cyclic.Proof. One direction is clear, so we assume G is freely representable. So by theabove theorem, the order of the kernel K of the action homomorphism B → Aut( A )must be divisible by exactly the primes that divide the order of B . This meansthat K = B , so G = AB = A × B is cyclic.We can restate the above lemma in terms µ ( G ): Corollary 77.
Let G be a freely representable Sylow-cyclic group, and let µ ( G ) bethe MCC subgroup of G . Then every element of G of prime order is in µ ( G ) . Inparticular, every prime dividing the order of G must divide the order of µ ( G ) .Proof. Write G as AB = A ⋊ B where A and B are cyclic subgroups of G of relativelyprime orders. Let g = ab be an element of prime order where a ∈ A and b ∈ B .Under the projection A ⋊ B → B the element g maps to b , so b is of prime orderor is the trivial element. In either case, by the above lemma, b ∈ K where K isthe kernel of the action map B → Aut( A ). So g = ab ∈ AK . But AK = µ ( G )(Lemma 71) so g ∈ µ ( G ).This gives a necessary condition for a Sylow-cyclic group to be freely repre-sentable. In order to show it is sufficient we use induced representations, but in avery basic manner. The following, which we take as given, is all that we need toknow about induced representations here: Proposition 78.
Let G be a finite group with subgroup H , and let F be a field.Suppose that H acts linearly on an F -vector space W . Then there is a linear actionof G on an F -vector space V containing W such that (1) the action of G on V restricts to the given action of H on W , (2) if g H, . . . , g k H are the distinct leftcosets in G/H then the vector space W is the direct sum of the spaces g i W : V = M g i W. This representation, called the induced representation, is unique up to a F [ G ] -module isomorphism fixing W . roposition 79. Let G be a finite group with a subgroup H that contains allelements of G of prime order. Suppose W is an F -vector space with a linear rep-resentation of H on W , and suppose V is a F -vector space containing W with arepresentation of G induced by the representation of H on W . If the linear repre-sentation of H on W is a free linear representation, then the linear representationof G on V is also a free linear representation.Proof. Let g H, . . . , g k H be the distinct left cosets in G/H . Suppose σ ∈ G is notthe identity and that σ ( v ) = v where v ∈ V is equal to v = g w + . . . + g k w k where w i ∈ W . Let m > σ and let p be a prime dividing m .Then τ = σ m/p has order p and also fixes v . By assumption τ ∈ H . Observe thatthe conjugate τ i = g − i τ g i also has order p so is in H . Since τ g i = g i τ i τ v = τ g w + . . . + τ g k w k = g ( τ w ) + . . . + g k ( τ k w k ) . Since τ v = v , g ( τ w ) + . . . + g k ( τ k w k ) = g w + . . . + g k w k . By the direct sum property of induced representations, g i ( τ i w i ) = g i w i for each 1 ≤ i ≤ k . Multiplying by the inverse of g i gives τ i w i = w i for the inducedrepresentation. Since τ i ∈ H and w i ∈ W , we have τ i w i = w i in the originalrepresentation of H on W . This representation is a free linear representation byassumption, so w i = 0 for each i . This implies v = 0, showing that the inducedrepresentation is a free linear representation. Corollary 80.
Let G be a finite group with freely representable subgroup H . If H contains all elements of G of prime order, then G is also freely representable. Now we are ready for the main theorem:
Theorem 81.
Let G be a Sylow-cyclic group and let µ ( G ) be its maximal charac-teristic cyclic subgroup. The following are equivalent:1. G is freely representable.2. Every prime dividing the order of G also divides the order of µ ( G ) .3. For every prime p dividing the order of G there is a unique subgroup of G of order p .Proof. Suppose G is freely representable. Then by Corollary 77 every prime divid-ing the order of G must divide the order of µ ( G ). So (1) = ⇒ (2).Now assume (2). If p divides | G | then, by assumption, p divides | µ ( G ) | . Thusthere is a unique subgroup of G of order p by Corollary 64. So (2) = ⇒ (3) holds.Finally assume (3). Let g ∈ G be an element of prime p order. By assumptionthe subgroup h g i is the unique subgroup of G of order p . This implies that h g i is acharacteristic cyclic subgroup, and so h g i is contained in µ ( G ). Since µ ( G ) is cyclic,it is freely representable. Thus G is freely representable by Corollary 80.41e can now strengthen Lemma 75. Corollary 82.
Let A and B be cyclic subgroups of G of relatively prime orderssuch that G = A ⋊ B (so A is normal in G ). Let K be the kernel of the actionhomomorphism B → Aut( A ) associated to the semidirect product. Then G is freelyrepresentable if and only if every prime dividing | B | divides | K | .Proof. Recall that µ ( G ) is AK (Lemma 71).If every prime dividing | B | divides | K | then every prime dividing the order of G must divide | A | or | K | . Hence every prime dividing the order of G divides the orderof µ ( G ), and so G is freely representable by the above theorem.Conversely, if G is freely representable, then every prime p dividing the orderof B must divide the order of µ ( G ) = AK ∼ = A × K by the above theorem. But p does not divide the order of A , so p divides the order of K as desired. Example . Suppose A a cyclic group of odd prime order p . Let q be any primedividing p −
1. Let B be a cyclic group of order q k with k >
1. We can iden-tify Aut( A ) with F × p which is cyclic of order p −
1. Let B ′ be the unique quotientof B of size q , and fix an injective homomorphism B ′ → Aut( A ). Now have B acton A by the composition B → B ′ → Aut( A )and let G be the associated semidirect product A ⋊ B . The kernel K of this actionhomomorphism has order q k − . By Lemma 71 we have that µ ( G ) ∼ = A × K , andthis has pq k − elements. By the above theorem, G is a freely representable non-Abelian group of order pq k . In the case of order p · such groups arose alreadyas binary dihedral groups 2 D p in H × . But if we take q = 2 we can conclude thefollowing: There are an infinite number of odd orders such that there exists non-Abelian Sylow-cyclic groups of that order that are freely representable (for example,for a fixed q take an infinite sequence of primes p ≡ q ). Note that such Sylow-cyclic groups cannot be isomorphic to subgroups of H × since all finite non-Abeliansubgroups of H × have order divisible by 4.The smallest such order of this type of group of odd order is 7 · = 63. Notethat if G is a noncyclic freely representable group of order 5 · then G is A ⋊ B where A is cyclic subgroup of G of order 5 and B is a cyclic subgroup of G oforder 9. Furthermore, µ ( G ) has order 15 or 45. The second case cannot happensince G is not cyclic. The first case cannot happen either: the automorphism groupof A has order 4, so the kernel K of the action of B on A must be all of K ,so µ ( G ) = AK = G . We conclude that 63 is the smallest odd order possible for anoncyclic freely representable group.Here is another interesting application of Theorem 81. Proposition 83.
Let G be a freely representable Sylow-cyclic group and let N bea normal subgroup of G of index p . If p does not divide the order of G then G = N C p ∼ = N × C p where C p is a subgroup of G of order p (and is the unique subgroup of order p ). roof. Since G is freely representable, there is a unique subgroup C p of order p ,and so C p must be normal in G . Since N ∩ C p = { } we have N C p ∼ = N × C p .Finally, G = N C p since [ G : N ] = p .We cannot hope to generalize Theorem 81 to all freely representable groups.For example, the binary tetrahedral group 2 T is freely representable (as a subgroupof H × ), but does not have a unique subgroup of order 3. However, one implicationholds in general: Proposition 84.
Let G be a finite group with the property that for each prime p dividing the order of G there is a unique subgroup of order p . Then G is freelyrepresentable.Proof. Let p , . . . , p k be the primes dividing the order of G . Let C p i be the be theunique subgroup of order p i . Then each C p i is normal and H def = C p · · · C p k ∼ = C p × · · · × C p k is a cyclic subgroup of G . So H is freely representable. Now use Corollary 80. Remark.
The classification of Sylow-cyclic groups is enough to yield significantapplications to differential geometry. In fact, by a theorem of Vincent (1947),every complete connected Riemannian manifold of constant positive curvature ofdimension not congruent to 3 modulo 4 has a fundamental group that is Sylow-cyclic. From this Vincent was able to give a full classification of such manifoldswhen the dimension is not congruent to 3 modulo 4. Wolf [17] completed theclassification to all dimensions by classifying freely representable groups beyondthe Sylow-cyclic groups.
Consider the automorphism group Aut( G ) where G is a Sylow-cyclic group of oddorder, and let O (Aut( G )) be the maximal odd normal subgroup of Aut( G ). Thenwe can use the above results to show that Aut( G ) /O (Aut( G )) is an Abelian 2-group. This is clear if G is cyclic since Aut( G ) is an Abelian group. The key togeneralizing this is to relate this quotient to the corresponding quotient for thecyclic subgroup µ ( G ). We start with a lemma. Lemma 85.
Let G be a Sylow-cyclic group of odd order and let φ be an automor-phism of G such that φ is the identity map. If φ fixes the MCC subgroup µ ( G ) then φ fixes all of G .Proof. Let P be a nontrivial Sylow subgroup of G . We will show that φ acts triviallyon P . Since the Sylow subgroups of G generate G , this gives the result. Let A bethe commutator subgroup of G . Since A and G/A have relatively prime orders,either P ⊆ A or P ∩ A = { } . In the first case φ acts trivially on P since A ⊆ µ ( G ).So from now on we assume that P ∩ A = { } .Observe that the image P of P in G/A is isomorphic to P . Since P is a cyclicgroup of odd order, it has a unique automorphism of order 2. So φ acts on P eitheras x x or as x x − . 43irst suppose that φ acts on P as x x . So if b ∈ P then φ ( b ) = ab forsome a ∈ A . Observe then that b = φ ( b ) = φ ( ab ) = φ ( a ) φ ( b ) = a ( ab ) = a b. Thus a = 1. Since A is a cyclic group of odd order a = 1, and φ ( b ) = b . Weconclude that φ acts trivially on P .Finally suppose that φ acts on P as x x − and let c ∈ P be a generator.Thus φ ( c ) = c − a for some a ∈ A . Let B be a complement of A and observethat P is conjugate to a Sylow subgroup of B . So replacing B with a complementof B if necessary, we can assume P is a subgroup of B . Note that φ ( c ) = c since c − and c have distinct images in P (and P has odd order greater than 1). Thus c is notin the center Z ( G ) since Z ( G ) ⊆ µ ( G ). Since c centralizes B , it cannot centralize A .Let a ∈ A be such that cac − = a . Note that cac − ∈ A so cac − = φ ( cac − ) = φ ( c ) aφ ( c ) − = c − a aa − c = c − ac. So c ac − = a . However c has odd order, so this implies that cac − = a , acontradiction. So φ cannot act on P as x x − . Proposition 86.
Let G be a Sylow-cyclic group of odd order. Let µ ( G ) be the MCCsubgroup of G , and let O (Aut( G )) be the maximal odd normal subgroup of theautomorphism group Aut( G ) . Then the quotient Aut( G ) /O (Aut( G )) . is isomorphic to a -group inside Aut( µ ( G )) . In particular, this quotient is anAbelian -group.Proof. Observe that Aut( µ ( G )) is Abelian since µ ( G ) is a cyclic group. Recall thatAbelian groups are the products of their Sylow subgroups, and soAut( µ ( G )) = A A = A × A where A is the subgroup of Aut( µ ( G )) consisting of elements of odd order, andwhere A is the 2-Sylow subgroup of Aut( µ ( G )). Since µ ( G ) is characteristic in G we have a homomorphism Aut( G ) → Aut( µ ( G )). We also have the projectionhomomorphism Aut( µ ( G )) = A A → A . Let K be the kernel of the compositionAut( G ) → Aut( µ ( G )) → A . Observe that K contains O (Aut( G )).Claim: K contains only elements of odd order. Suppose otherwise that ψ ∈ K has order 2 k , and let φ = ψ k . Then φ has order 2 and is in K . The imageof φ in Aut( µ ( G )) is just the restriction φ | µ ( G ) and it is in the kernel of of theprojection Aut( µ ( G )) = A A → A . In other words, φ | µ ( G ) ∈ A and so has oddorder. Since φ has order 2, we conclude that φ | µ ( G ) has order 1. By the previouslemma φ is the identity, a contradiction.So the claim has been established. This means K = O (Aut( G )) and we havean injection Aut( G ) /O (Aut( G )) ֒ → A . The result follows. 44
Applications to Division Rings
The classification of Sylow-cyclic fields can be used to prove Wedderburn’s theorem.Along the way we will see an argument that every finite subgroup of a field is cyclic.This section is independent of Example 3 where we used (1) Wedderburn’s theoremand (2) the fact that F × is cyclic for any finite field. We start with the following. Lemma 87.
Let D be a division ring and let F be its prime subfield. Then everyfinite subgroup G of D × is freely representable over F .Proof. We let G act on V = D by left multiplication. Note that V is an F -vectorspace. This action is a free linear action since D has no zero divisors. Remark.
In particular, if F has characteristic zero then G is freely representable(and so is Sylow-cycloidal). This result is the starting point for Amitsur’s classi-fication of finite subgroups of D × where D is a division ring (1955 [2]). Amitsurused class field theory to complete the classification. Lemma 88.
Let G be a group of order p where p is a prime. Suppose F is a fieldof characteristic not equal to p . If G is freely representable over F then G is cyclic.Proof. Suppose G is freely representable but not cyclic. This means that everynonidentity element of G is in a unique cyclic group of order p . Observe that thereare k = ( p − / ( p −
1) = p + 1 such cyclic groups. Let C be the collection of cyclicsubgroups of G of order p . Then X C ∈C N C = ( k − + N G = p + N G. Since p is nonzero in F , this contradicts Theorem 11.Suppose D is a division ring and that G is a finite subgroup of D × . If F is theprime subfield of D , then let F ( G ) be the F -span of G in D . (Warning: F ( G ) isanalogous to the group ring F [ G ], but they are not the same since G might not belinearly independent.) Lemma 89.
Let
D, G, F, F ( G ) be as above, and suppose F is F p for some prime p .Then F ( G ) is a finite division ring of order a power of p .Proof. Let k be the size of a basis of F ( G ) for scalar field F , and observe that F ( G )has finite size p k . Observe that F ( G ) is closed under multiplication, so we concludethat F ( G ) is a subring of D . Next suppose a ∈ F ( G ) is nonzero. Then themap x ax is an injective map F ( G ) → F ( G ) since F ( G ) is contained in a divisionring. Since F ( G ) is finite, this map is surjective, and ab = 1 for some b ∈ F ( G ). Thisimplies that F ( G ) is a division ring. Lemma 90.
Let D be a division ring whose prime field F has prime characteris-tic p . Then D × has no elements of order p .Proof. Suppose g ∈ D × has order p , and let G be the group generated by g . Observethat G is a subgroup of F ( G ) × , and F ( G ) × has order p k − k ≥
1. Sothe order of g ∈ F ( G ) × fails to divide the order of F ( G ) × , a contradiction.45 orollary 91. Let G be a subgroup of D × where D is a division ring. Then everysubgroup of G of order p , where p is a prime, is cyclic.Proof. Let H be a subgroup of G of order p where p is a prime. By the abovelemma, we can assume that p is not the characteristic of the prime field F of D .By Lemma 87, H is freely representable over F . So H is cyclic by Lemma 88. Corollary 92.
Suppose G is a finite subgroup of F × where F is a field. Then G is cyclic.Proof. By the previous corollary, every subgroup of G of order p is cyclic, for anyprime p . Since G is Abelian, G is cyclic by the finite structure theorem of Abeliangroups (or you can use the more elementary argument given in the remark afterCorollary 19). Proposition 93.
Let D be a division algebra and let G be a finite subgroup of D × .Then G is a Sylow-Cycloidal groupProof. If p is odd, then any q -Sylow subgroup of G is cyclic by Theorem 25. If p = 2then any q -Sylow subgroup is either cyclic or a generalized quaternion group byCorollary 37.Now suppose F has prime characteristic p . In this case G is a subgroupof F ( G ) × , and F ( G ) × has order p k − k ≥
1. So p cannot divide theorder of G , so there are no p -Sylow subgroups of G that we need to worry about.We conclude that all Sylow-subgroups of G have the desired form, and that G is a Sylow-cycloidal group. Lemma 94.
Let D be a division algebra of prime characteristic p , and let G be afinite subgroup of D × . Then G is a Sylow-cyclic groupProof. Let q be any odd prime dividing the order of G . By Corollary 91 andTheorem 25, the q -Sylow subgroups of G are cyclic.Suppose the 2-Sylow subgroup of G are not also cyclic. By Corollary 91 andProposition 37 the 2-Sylow subgroups of G must be generalized quaternion groups.By Proposition 41, G must then contain a subgroup Q that we can identify with thequaternion group of size 8. By Lemma 89 we have the division algebra F p ( Q ). Theidea of the remainder of the proof is to argue that there cannot be a “quaternionring” over F p .To proceed we solve 1 + x + y = 0 over F p . If p = 2 then x = 1 , y = 0 is asolution. Otherwise, observe that there are ( p + 1) / F p . So as x variesin F p , the expression − − x takes on ( p + 1) / p − / F p . Sowe choose x so that − − x is a square, and we choose y so that y is − − x . Wecan exchange x and y if necessary, and assume y = 0. Then in F p ( Q ) we have(1 + x i + y j )(1 − x i − y j ) = 1 + x + y = 0 . Since F p ( Q ) has no zero divisors, we have 1 + x i + y j = 0 or 1 − x i − y j = 0.Since y = 0 this means that j ∈ F p ( h i i ). But F p ( h i i ) is a field, and so i and j commute, a contradiction. 46he following is a result of Herstein. He proved it as a corollary of Wedderburn’stheorem, but we will prove Wedderburn’s theorem as a corollary of this result. Theorem 95 (Herstein) . Let D be a division ring of prime characteristic p , andlet G be a finite subgroup of D × . Then G is cyclic.Proof. First we consider the case where D is finite and G = D × . By the abovelemma G is a Sylow-cyclic group. Let C be the maximum cyclic characteristic(MCC) subgroup of G . Such a subgroup C exists by Corollary 63 and, by Corol-lary 70, C is maximal among cyclic subgroups of G . Observe that F p ( C ) is a field,so F p ( C ) × is cyclic. By the maximality of C , this means that C = F p ( C ) × .Let q be the number of elements of F p ( C ), and let q k be the number of elementsof D . (Here we use the fact that D is a vector space over any subfield). If k > | G/C | = q k − q − q k − + . . . + q + 1 ≥ q + 1 > q − | C | which contradicts Corollary 72. Thus k = 1 and so G = C , and G is cyclic.In general, we consider F p ( G ). Since F p ( G ) × is cyclic, as we have just shown,and since G is a subgroup of F p ( G ) × , we conclude that G is cyclic as well. Corollary 96 (Wedderburn) . Every finite division ring D is a field.Proof. By the above theorem D × is cyclic. So D must be a commutative ring. Suppose G is a solvable Sylow-cycloidal group and O ( G ) is the maximal normalsubgroup of G of odd order. Then our first important result will be to describe thepossible quotients G/O ( G ). We will show that G/O ( G ) is isomorphic to either acyclic 2-group, a generalized quaternion group, the binary tetrahedral group 2 T orthe binary octahedral group 2 O . In particular, G/O ( G ) is isomorphic to a solvablesubgroup of H × . This result divides Sylow-Cycloidal groups into four mutuallyexclusive types. We then focus on each type individually. G/O ( G ) We start more generally than with Sylow-cycloidal groups. We say a finite group G satisfies the (2 , condition if every 2-Sylow subgroup of G is cyclic or is the quotientof a generalized quaternion group and every 3-Sylow subgroup is cyclic. I learned the technique of classifying freely representable groups G by their quotients G/O ( G )from a recent paper by Daniel Allcock [1]. My approach generalizes the scope of Allcock a bit.Although my proof is different and more elementary that that in [17], I also lean on Wolf [17],and thus indirectly on Zassenhaus (1936), for guidance. Zassenhaus adopts a more general scopethan mine by only restricting the 2-Sylow subgroup (see Lemma 6.1.9 of Wolf [17] attributed toZassenhaus). This means that the 2-Sylow subgroups of G are cyclic, dihedral, or generalized quaternion,but we will not need this fact. emma 97. Let G be a group that satisfies the (2 , condition. Then every sub-group and quotient of G satisifies the (2 , condition.Proof. Observe that every p -Sylow subgroup of a quotient G/N is a quotient ofa p -Sylow subgroup of G . Also every p -Sylow subgroup of a subgroup H of G isa p -group and so is a subgroup of a p -Sylow subgroup of G .The class of cyclic p -groups is closed under the processes of quotient and sub-group. Since the class of cyclic and generalized quaternion 2-groups is closed undersubgroup, the class of quotients of such groups is closed under quotient and sub-groups. Definition 7.
Let G be a finite solvable group and let G, G ′ , G ′′ , . . . , G ( k ) = { } be the derived series of commutator subgroups. Then the characteristic Abeliansubgroup of G , which we denote as A ( G ), is defined to be the first Abelian term ofthe series. Observe that A ( G ) is Abelian and characteristic, and if G is nontrivialthen A ( G ) is also nontrivial. Lemma 98.
Let G be a solvable group satisfying the (2 , condition. Then either G has order of the form m n or there is a prime p ≥ such that there is a p -subgroup K of G that is a characteristic subgroup of G .Proof. Let q be a prime dividing A ( G ) (if no such q exists then G has order 2 and we are done). If q ≥ p = q and choose K to be the p -Sylow subgroup of A ( G ), and we are done. Otherwise we define an elementarycharacteristic subgroup N as the solutions of x q = 1 in A ( G ).Since G satisfies the (2 ,
3) condition, the same is true of N and G/N . If q = 3this means that N is cyclic of order 3. If q = 2 then N is generated by one ortwo elements, so is either cyclic of order 2 or is the Klein four group. So anyautomorphism of N has order 1 , , or 3.If G/N has order of the form 2 m n we are done, so we will now assume that G/N is not of that form. By induction we can assume that
G/N has a p -subgroup L/N where p ≥ L is a subgroup of G containing N such that L/N is a characteristic subgroup of
G/N . Since N is a characteristic subgroup, thisimplies that L must also be a characteristic subgroup of G . Let K be a p -Sylowsubgroup of L and observe that K is a complement of N in L , so L ∼ = N ⋊ K . Notethat K acts trivially on N since all automorphism of N have order prime to p . Thismeans that L is isomorphic to N × K , and so K is the unique p -Sylow subgroupof L . Since L is a characteristic subgroup of G , its unique p -Sylow subgroup K isalso characteristic subgroup of G . Corollary 99.
Let G be a solvable group satisfying the (2 , condition. Let O ( G ) be the maximal normal subgroup of G of order relatively prime to . Then G/O ( G ) has order of the form m n . Corollary 100.
Let G be a solvable group satisfying the (2 , condition. Let O ( G ) be the maximal normal subgroup of G of odd order. Then G/O ( G ) has order of theform m n , and every nontrivial normal subgroup of G/O ( G ) has even order. m n . Of course we know what such group are when m = 0 orwhen n = 0, so we focus on the case where m and n are positive. Lemma 101.
Let G be a solvable Sylow-cycloidal group of order m n with m and n positive. Assume also that every normal subgroup of G is of even order.Then the following hold • G contains a normal subgroup N isomorphic to the quaternion group Q . • Every element of G of order acts nontrivially on N . • G contains four subgroups of order , and the action of G on these subgroupsgives a homomorphism G → S with kernel Z ( G ) . • The center Z ( G ) has two elements. • G/Z ( G ) is isomorphic to either A or S , and so G has order 24 or 48.Proof. Let N be a maximal normal 2-subgroup (i.e., the intersection of the 2-Sylowsubgroups of G ). Observe that characteristic Abelian subgroup A ( G/N ) of
G/N must be a nontrivial 3-group since if it had order divisible by 2 then we could violatethe maximality of N via the 2-Sylow subgroup of A ( G/N ). Let
H/N be the uniquecyclic group of order 3 in A ( G/N ). Here H is a subgroup of G containing N , andsince H/N is characteristic in A ( G/N ), we see that
H/N is characteristic in
G/N .Since N is characteristic in G we then see that H is characteristic in G . Let C bea cyclic subgroup of H of order 3. Then we have that H = N C is a semidirectproduct N ⋊ C . If C acts trivially on N then H would be isomorphic to N × C andwould contain a characteristic subgroup of order 3. This violates our assumptionon normal subgroups of G . So C acts nontrivially on N .Up to isomorphism the only 2-Cycloidal group with an automorphism of order 3is the quaternion group Q , so N is isomorphic to Q . Thus H has 24 elements.Denote by − N of order 2, which must be the unique elementof H of order 2. Note that − C , and so must be in thecenter of H = N C . In fact, the action of C on the quaternions only fixes ± Z ( H ) is {± } .Let L be the normalizer of C in H . Observe that L ∩ N and C are normalin L , and so L is isomorphic ( L ∩ N ) × C . This means C is characteristic in L .So L cannot be normal in H , otherwise C would be normal in H , and hence in G ,a contradiction. So L must have index at least 4 in H . However, L contains C and Z ( H ). Thus L has order 6. Since L is isomorphic to ( L ∩ N ) × C = Z ( H ) × C ,it is cyclic of order 6. By the orbit-stabilizer theorem (and the fact that p -Sylowgroups are conjugate) we get that H has exactly four 3-Sylow subgroups. Thisaction gives a homomorphism H → S . The kernel of this action is the intersectionof the normalizers. Our description of L applies to all the normalizers, and theintersection is seen to be Z ( H ). So Z ( H ) is the kernel of the action H → S .Observe that the image of L under H → S is isomorphic to L/Z ( H ), so is cyclicof order 3, and it fixes the element corresponding to C . This observation appliesnot just to C but to all 3-Sylow subgroups of H . So the image of H → S contains49ll subgroup of order 3. This means that the image contains all of A , and must infact be A since H/ {± } is a group of size 12.If G = H , then we are done. So for the remainder of the proof we assume H isnot all of G . We claim that G has has order 2 m m >
3. Otherwise everyelement h of order 3 in H is of the form g for some element of g of order 9. Sincesuch an g acts on N as an automorphism of order 1 or 3, this forces h to act triviallyon N , a contradiction to a previous conclusion.The 2-Sylow subgroups of G cannot be normal in G by maximality of N . Sothere are three 2-Sylow subgroups S , S , S since the number of such groups mustdivide | G | / m . Since G acts transitively on { S , S , S } by conjugation, we geta homomorphism G → S whose image is S or A . The kernel is a normal 2-subgroup of G of index 3 or 6 in G . By maximality of N , this kernel is containedin N . So the index of N in G is a divisor of 6, and since it has size at least 6it is equal to 6. Hence (1) G has size 48, (2) each S i is a generalized quaterniongroup of size 16, (3) H is the kernel of G → S , (4) the homomorphism G → S issurjective with each S i mapping to a different subgroup of order 2 in S , and (5) H has index 2 in G and its image under G → S is A .Observe that every 3-Sylow subgroup of G is contained in H since [ G : H ] = 2.Thus the set of 3-Sylow subgroups of G has size 4, giving a homomorphism G → S extending the earlier surjection H → A . This means that the kernel K of G → S has order 2 or 4. The image of K under G → S is a normal 2-subgroup of S ,so K has trivial image in S . This means that K ⊆ H . But we already know thatthe kernel of H → S is Z ( H ). Hence K = Z ( H ). This implies that G/Z ( H ) isisomorphic to S .We conclude by showing that Z ( H ) = Z ( G ). First observe that under themap G → S the center Z ( G ) maps to the center of S which is trivial. So Z ( G )is contained in the kernel Z ( H ). Conversely, since S ∩ S ∩ S is the maximalnormal 2-subgroup of G , this intersection is H . Thus Z ( H ) ⊆ S i , and so Z ( H )is the unique subgroup of S i of size 2. So Z ( H ) is contained in Z ( S i ) for each i .Since S , S , S generates G we have that Z ( H ) ⊆ Z ( G ). Remark.
By a theorem of Burnside, all groups of order p m q n are solvable for distinctprimes p and q . So accepting this result allows us to drop the solvability assumptionfor groups of order 2 m n . I will keep this assumption just to make the proofs a bitmore accessible.There are two cases in the above lemma. We now link these cases to specificsubgroups of H × . Proposition 102. If G is a solvable Sylow-cycloidal group of order 24 with nonormal subgroup of order 3, then G is isomorphic to the binary tetrahedral group T .Proof. By the above lemma, we can identify Q with a normal subgroup of G .Also, G contains four subgroups of order 3, each which acts nontrivially on Q , andthe center of G has two elements. Note that we have an embedding of G/Z ( G )into Aut( Q ). This implies that if α and β are distinct elements of order 3, theiractions on Q are distinct (otherwise α = ǫβ where ǫ ∈ Z ( G ), and ǫ must be 1since α has order 3). 50ote that any automorphism φ of order 3 of Q must permute its three sub-groups of order 4, and cannot fix any such subgroup (otherwise it would have tofix all three, and then φ would fix all of Q ). So there are only 4 possible valuesof φ ( i ), and they are contained in {± j , ± k } . Once φ ( i ) is known, there are onlytwo possibilities for φ ( j ). Since i and j generate Q , this gives 8 possibilities. Allof these possibilities must occur as the automorphism associated with elements oforder 3, since there are 8 such elements.Thus we can find an element g ∈ G of order 3 such that it acts on Q by sending i to j and sending j to k . Let C be the subgroup generated by g . Then G ∼ = Q ⋊ C where C acts on G as specified. Thus G is unique up to isomorphism.Since 2 T is freely representable, it is a Sylow-cycloidal group. Also 2 T / {± } isisomorphic to A . Since A is solvable and has no normal subgroups of order 3,the same is true of 2 T . So 2 T satisfies our assumptions for G , and so must beisomorphic to any such G .Here is a variant of the above: Corollary 103.
Let G be a finite group of order 24 that has a unique element oforder but does not have a unique subgroup of order . Then G is isomorphic tothe binary tetrahedral group T .Proof. Let P be a 2-Sylow subgroup of G . Then P has a unique element oforder 2, so must be isomorphic to Q or a cyclic group of order 8. In particular, G is a Sylow-cycloidal group. Note that the subgroups of order 3 of G are conjugatesince they are Sylow subgroups, so they cannot be normal since there is more thanone. The result now follows from Proposition 98 if we grant that G is solvable.It is well-known that all subgroups of order 24 are solvable, but to see thisdirectly for G here let C be the unique subgroup of order 2 and let C be anysubgroup of order 3. Then the subgroup C C = C ⋊ C must be cyclic since C has no automorphism. So the normalizer of C has at least 6 elements. Thismeans that there are at most 4 subgroups of G of order 3. Since the numberof such Sylow subgroups is congruent to 1 modulo 3, we see there are exactly 4subgroups of order 3 and C C is the normalizer of C . The action of G on the setof subgroups of order 4 gives a homomorphism G → S . The kernel of the action isthe intersection of the normalizers for subgroups of order 3. Our description of thenormalizer shows this intersection is C . Since S is solvable, this implies that G is solvable. Proposition 104.
Let G be a solvable Sylow-cycloidal group of order 48 suchthat G/Z ( G ) is isomorphic to S . Then G is isomorphic to the binary octahedralgroup O . Furthermore G has elements of order 3, and subgroups of order 3,and these elements generate the unique subgroup of G of index , and this index 2subgroup is a binary tetrahedral group.Proof. We start by checking that G = 2 O satisfies the hypothesis. Recall that 2 O has 48 elements. We note that 2 O has center containing the elements ±
1, and51ince 2 O/ {± } ∼ = O ∼ = S the center is exactly {± } (since S has trivial center).So 2 O/Z (2 O ) is isomorphic to S .Now let G be any subgroup of size 48 such that G/Z ( G ) ∼ = S . Observethat Z ( G ) has order 2 since G has order 48 and S has order 24. So the gen-erator of Z ( G ) is the unique element of G of order 2 (Lemma 49). For now, we fixa particular isomorphism G/Z ( G ) → S . Since Z ( G ) has order 2, the order of theimage of g ∈ G in S is either the same as the order of g or has half the order of g .For example if g maps to an element of order 2 then g must have order 2 or 4. Butin this case g is not the unique element of order 2 since that maps to the identityelement, so g has order 4 and g is the unique element of order 2.In contrast if g has order 3, then g must map to an element of order 3 since 3is odd. In particular if g has order 3 then g maps to a 3 cycle in S . So if S isthe set of elements of order 3 in G then we have a map S → T where T is theset of three cycles. Now let t ∈ T be given. The subgroup h t i of S correspondsto a subgroup H t of G of order 6 containing Z ( G ) as a normal subgroup. Infact, H t = Z ( G ) C = Z ( G ) ⋊ C where C is a 3-Sylow subgroup of H t . Since Z ( G )has no nontrivial automorphisms, H t is just Z ( G ) × C , so C is the unique subgroupof G order 3 whose image in S is h t i . Thus there is exactly one element of G that maps to t . This means that S → T is a bijection. In particular, there are 8elements of order 3 in G , and 4 subgroups of order 3 in G .Let H be the subgroup of G generated by the set S of elements of order 3. Theimage in S is A (since T generates A ). This implies that H is also of even orderand so must contain the unique subgroup Z ( G ) of G of order 2. Thus H/Z ( G ) ∼ = A ,and so H has order 24. Also if H has a normal subgroup of size 3, then its imagein A would also have a normal subgroup of size 3. But this is not the case. Sowe have that H is isomorphic to the binary tetrahedral group 2 T (see previousproposition). We also note that H is the only subgroup of G of order 24. So seethis note that any other such group H ′ would have to contain the unique elementof G order 2 and so would contain Z ( G ). Thus its image in S would have size 12.But A is the only subgroup of S of index two and so H ′ and H would have thesame image in S and so would be equal.This gives us enough information to describe G in terms of relations. The finalspecification of G will not depend on a particular isomorphism G/Z ( G ) ∼ = S butonly on the fact that such an isomorphism exists. Note that the subgroup H (theunique subgroup of G of index 2) and the subset S do not depend on the map.Choose an element g ∈ G of order 3. It can be any such element. Next choose g to be any element of order 3 such that the product g g has order 4. To show thiscan be done it is useful to choose a map G/Z ( G ) ∼ = S (by permuting the number ifnecessary from a given map) so that g corresponds to (1 2 3). Then the element g corresponding to (1 2 4) will work since g g maps to (1 3)(2 4) of order 2 in S andso g g has order 4 in G . In fact, once we have chosen a suitable g and g , we canpermute the numbering of the four elements permuted by S so that g correspondsto (1 2 3) and g corresponds to (1 2 4). Note that (34)(1 2 3)(34) = (1 2 4) so ifwe choose τ ∈ G − H mapping to (34) then τ g τ − is an element of order 3 A subgroup N of S of index 2 is normal, and contains all three cycles since G/N has order 2.Since three cycles generate A any such N would have to be A . S → T is a bijection, this means that τ g τ − = g . We also have τ g − τ − = g − , τ g τ − = g , τ g − τ − = g − . Here we made use of τ − gτ = τ gτ − for all g ∈ G since τ has order 2 in Z ( G ).By conjugating other 3-cycles in S by (3 4) we can see that τ gτ − = g − for the other four elements g ∈ S . This gives us eight relations, one for eachelement of S . We verified they held by using a particular map G/Z ( G ) ∼ = S , butthe actual relations themselves do not depend on the map. In addition we have aninth relation τ = − − H .Now consider the free product H ∗ C where C is an abstract cyclic group oforder 4 with generator called τ . Let K be the normal subgroup generated by the 9relations discussed above. Note that every element of H ∗ C /K can be writtenas a or aτ where a is in the image of H and τ is the image of τ . This follows fromthe fact that S generates H . Thus the group H ∗ C /K has at most 48 elements.However, G satisfies these relations so there is a homomorphism H ∗ C /K → G ,and this is a surjection since G is generated by τ and H . Thus G is isomorphicto H ∗ C /K .So if G and G are solvable Sylow-cycloidal groups of order 48 that contain acommon subgroup H of size 24, and if G /Z ( G ) and G /Z ( G ) are both isomorphicto S , then G must be isomorphic to G since both are isomorphic to H ∗ C /K (and K does not depend on G i but only on H ). More generally, if G and G areare solvable Sylow-cycloidal groups of order 48 such that G /Z ( G ) and G /Z ( G )are isomorphic to S , then as observed above each G i has a subgroup isomorphicto 2 T . By identifying these subgroups we reduce to the situation where G and G share a subgroup of order 24. We conclude that G and G are isomorphic underthese conditions.Here are some useful observations linking the 2-Sylow subgroup of G to thequotient G/O ( G ). Proposition 105.
Let G be finite group and let O ( G ) be its maximal normal sub-group of odd order. Then G and G/O ( G ) have isomorphic -Sylow subgroups. Corollary 106.
Let G be a solvable Sylow-cycloidal group whose -Sylow sub-group S is not quaternionic of order 8 or 16. Then G/O ( G ) is isomorphic to S . Corollary 107.
Let G be a Sylow-cycloidal group. Then G is a Sylow-cyclic groupif and only if G/O ( G ) is a cyclic -group. So we divide the solvable Sylow-cycloidal groups G into four mutually exclusivetypes:1. Sylow-cyclic groups. These are the Sylow-cycloidal groups where G/O ( G ) iscyclic. 53. Quaternion type. These are defined to encompass the Sylow-cycloidal groupswhere G/O ( G ) is a generalized quaternion group.3. Binary tetrahedral type. These are defined to encompass the Sylow-cycloidalgroups where G/O ( G ) ∼ = 2 T .4. Binary octahedral type. These are defined to encompass the Sylow-cycloidalgroups where G/O ( G ) ∼ = 2 O .In addition, there are non-solvable Sylow-cycloidal groups that we will considerlater. These are Sylow-cycloidal groups that contain a perfect Sylow-cycloidal sub-group. As noted out above, a Sylow-cycloidal group G is of this type if and only if G/O ( G )is a cyclic 2-group. Such groups were treated in Section 7. In particular, this typeof group G is freely representable if and only if it has a unique subgroup of order p for each prime p dividing the order of G .In order to compare with later results it is convenient to break out the odd partfrom the even part: Proposition 108.
Let G be a Sylow-cyclic group of order k n where n is oddand where k ≥ . Then G has a normal Sylow-cyclic subgroup M of order n .For such M , the group G is freely representable if and only if (1) M is freelyrepresentable and (2) G has a unique element of order .Proof. Let M = O ( G ), so G/M is a cyclic 2-group (Corollary 107), and O ( G ) hasodd order by definition, so M has order n .If G is freely representable then (1) and (2) hold by earlier results. So assume (1)and (2). To show G is freely representable, it is enough to show that G has aunique subgroup of order p for each p dividing 2 k n . For p = 2 we are covered byassumption (2). For an odd prime p we note that every subgroup of G of order p isa subgroup of M since G/M is a 2-group. By (1) there is a unique subgroup of M (and hence of G ) of order p . As noted in Proposition 105, these groups have 2-Sylow subgroups that are gener-alized quaternion groups, and if conversely if G is a solvable Sylow-cycloidal groupwith 2-Sylow subgroups that are generalized quaternion groups of order 32 or morethen G must be of this type. (If G is a solvable Sylow-cycloidal group with 2-Sylowsubgroups isomorphic to the quaternion group Q of order 8, it can either be of thistype or of binary tetrahedral type. If G is a solvable Sylow-cycloidal group with 2-Sylow subgroups isomorphic to the generalized quaternion group Q of order 16,it can either be of this type or of binary octahedral type.)Observe that for Sylow-cycloidal groups G of quaternion type, any 2-Sylowsubgroup Q of G functions as a complement for the normal subgroup O ( G ) of G .54hus G = O ( G ) Q = O ( G ) ⋊ Q. In particular, up to isomorphism G is determined by O ( G ) and the action of Q on O ( G ). Proposition 109.
Let G be a Sylow-cycloidal group of quaternion type. Then G has a unique element of order 2. This element is in the center of G .Proof. Let Q be a 2-Sylow subgroup of G , which is a generalized quaternion group(Proposition 105). We start by considering the action of Q on G . By Proposition 86,we have that A = Aut( O ( G )) /O (Aut( O ( G ))) is an Abelian 2-group. The actionof Q gives a map into A : Q → Aut( O ( G )) → Aut( O ( G )) /O (Aut( O ( G ))) = A. Since A is Abelian, this map has a nontrivial kernel. Thus if C is the uniquesubgroup of Q of order 2, then C is in this Kernel. In particular, the imageof C in Aut( O ( G )) must land in O (Aut( O ( G ))). But O (Aut( O ( G ))) has oddorder, so the image of C in Aut( O ( G )) is trivial. Thus C acts trivially on O ( G ).Since G = O ( G ) Q we have that C is in the center of G .Observe that the subgroup O ( G ) C of G corresponds to the unique subgroupof G/O ( G ) of order two. If g ∈ G has order 2, then its image in G/O ( G ) is theunique element of order 2, and so g ∈ O ( G ) C . Since C is in the center of G , wehave O ( G ) C = O ( G ) × C , and so every element of O ( G ) C of order 2 must bein C . We conclude that every element of order 2 is in C . In other words, there isa unique element of order 2 in G .Let G be a Sylow-cycloidal group of quaternion type and let C be its unique sub-group of order 2. Let M = O ( G ) C . Observe that every element of odd prime ordermust be in O ( G ), so every element of G of prime order is in M . By Corollary 80, G is freely representable if and only if M is freely representable. Since M ∼ = O ( G ) × C we have the M is freely representable if and only if O ( G ) and C are freely repre-sentable (Corollary 21), but of course C is freely representable. Thus we get thefollowing: Theorem 110.
Let G be a Sylow-cycloidal group of quaternion type. Then thefollowing are equivalent:1. G is freely representable.2. O ( G ) is freely representable.3. For each prime p dividing the order of G , there is exactly one subgroup of G of order p .Proof. The equivalence (1) ⇐⇒ (2) was addressed in the discussion proceeding thestatement of the theorem. The implication (3) = ⇒ (1) follows from Proposition 84.So we just need to verify that (2) implies (3). We know that (3) holds for p = 2by the previous proposition (independent of whether (2) is true or not). So supposethat (2) holds and that p is an odd prime dividing the order of G . Observe that55very subgroup of order p of G must actually be a subgroup of O ( G ) since G/O ( G )has even order. Finally, (2) implies O ( G ) has exactly one subgroup of order p byTheorem 81.Now we consider other characterizations of this type of group. Proposition 111.
Let G be a finite group. Then G is a Sylow-cycloidal groupof quaternion type if and only if it is a semidirect product M ⋊ Q where M is aSylow-cyclic group of odd order and Q is a generalized quaternion group.If G is a semidirect product M ⋊ Q where M is a Sylow-cyclic group of oddorder and Q is a generalized quaternion group, then G is freely representable if andonly if M is freely representable.Proof. We mentioned earlier that if G is a Sylow-cycloidal group of quaterniontype then G = O ( G ) ⋊ Q where Q is any 2-Sylow subgroup of G and where Q is a generalized quaternion group. Conversely suppose G is of the form M ⋊ Q .Then M is isomorphic to a normal subgroup of G of odd order, and the 2-group Q is isomorphic to the corresponding quotient of G . Thus O ( G ) must be isomorphicto M , and G/O ( G ) is isomorphic to Q . So by the above theorem, G is freelyrepresentable if and only if M is freely representable. Proposition 112.
Let G be a finite group. Then G is a Sylow-cycloidal group ofquaternion type if and only (1) the -Sylow subgroups of G are generalized quater-nion groups, and (2) G has a Sylow-cyclic subgroup M of index . In this case G is freely representable if and only if M is freely representable.Proof. Suppose is a Sylow-cycloidal group of quaternion type, so
G/O ( G ) is ageneralized quaternion group. This quotient contains a cyclic subgroup of index 2which we can write as M/O ( G ) where M is a subgroup of G containing O ( G ). Notethat M has index 2 in G . Since O ( G ) has odd order, the 2-Sylow subgroups of M are isomorphic to the cyclic group M/O ( G ). Thus M is a Sylow-cyclic subgroupof G . We also note that G/O ( G ) is isomorphic to the 2-Sylow subgroups of G , sothese 2-Sylow subgroups of G are generalized quaternion groups.Conversely, suppose (1) and (2) hold. Since M is a Sylow-cyclic group, thequotient M/O ( M ) is a cyclic 2-group. Since G/M has order 2, the subgroup O ( G )must be a subgroup of M , so O ( G ) ⊆ O ( M ) by the maximality of O ( M ). Since M is normal in G and since O ( M ) is characteristic in M it follows that O ( M ) is normalin G . Thus O ( G ) = O ( M ). Since M/O ( M ) = M/O ( G ) is a 2-group and since G/M is a 2-group, it follows that
G/O ( G ) is a 2-group. Any Sylow 2-group of G is thusisomorphic to G/O ( G ), and so G/O ( G ) is a generalized quaternion group. Alsonote that since G/M has order 2, all odd order Sylow-subgroups of G are containedin M , and so are cyclic. Thus G is a Sylow-cycloidal group of quaternion type.If G is freely representable, then so is the subgroup M . Conversely, if M isfreely representable then so is O ( G ) since, as noted above, O ( G ) is a subgroupof M . Thus G is freely representable by Theorem 110. Proposition 113.
Let G be a finite group and let S be a -Sylow subgroup of G .Then G is a Sylow-cyclic group if and only if S is cyclic and G has a normalSylow-cyclic subgroup M of index | S | . Similarly, G is a Sylow-cycloidal group of uaternion type if and only if S is a generalized quaternion group and G has anormal Sylow-cyclic subgroup M of index | S | .Proof. If G is Sylow-cyclic group or a Sylow-cycloidal group of quaternion type,then let M = O ( G ) which is normal. Conversely, suppose there is a normal Sylow-cyclic subgroup M of index | S | . This group M has odd order, and is maximal withthis property, so M = O ( G ). It follows that G/O ( G ) ∼ = S . Note also that everyodd ordered Sylow subgroup of G is actually in M since G/M has even order. Thusevery odd ordered Sylow subgroup of G is cyclic. Let G be a Sylow-cycloidal group of binary tetrahedral type and let O ( G ) be itsmaximal normal subgroup of odd order. Then G/O ( G ) is isomorphic to the binarytetrahedral group 2 T , so the 2-Sylow subgroups of G are all isomorphic to thequaternion group of order 8 (Proposition 105). What is interesting about this caseis that the 2-Sylow subgroup of G is unique, and so is characteristic: Proposition 114.
Let G be a Sylow-cycloidal group of binary tetrahedral type.Then G has a unique -Sylow subgroup of order , and this -Sylow subgroup isisomorphic to the quaternion group with 8 elements. Furthermore, the -Sylowsubgroup of G centralizes O ( G ) .Proof. We start with the fact that 2 T has a cyclic quotient C of order 3. Let H bethe kernel of the composition G → G/O ( G ) ∼ = 2 T → C . Note that every 2-Sylowsubgroup of G must be in the kernel H . So we just need to show that H has aunique 2-Sylow subgroup. Let Q be a 2-Sylow subgroup of G , and observe that Q is a complement for O ( G ) in H . So H = O ( G ) Q = O ( G ) ⋊ Q. To prove the uniqueness result for H , and hence for G , it is enough to show that Q acts trivially on O ( G ).By Proposition 86, A = Aut( O ( G )) /O (Aut( O ( G ))) is an Abelian 2-group.Let G act on O ( G ) by conjugation. Then we have the composition G → Aut( O ( G )) → Aut( O ( G )) /O (Aut( O ( G ))) = A. Since the codomain is a 2-group, we have O ( G ) is in the Kernel. So we get a map2 T ∼ = G/O ( G ) → A. The kernel must contain the commutator subgroup of 2 T . The commutator sub-group of 2 T contains the commutator subgroup of Q , so contains the unique sub-group C of 2 T of order 2. But 2 T /C is isomorphic to T = A , whose commutatorsubgroup is the normal subgroup of order 4. Thus the commutator subgroup (2 T ) ′ of 2 T has index 3 in 2 T . Thus we get a homomorphism(2 T ) / (2 T ) ′ → A. Since (2 T ) / (2 T ) ′ has order 3, and A is a 2-group, we get that the image of G in A istrivial. In other words, the image of G in Aut( O ( G )) has odd order. In particular,the image of Q in Aut( O ( G )) must be trivial. So Q acts trivially on O ( G ).57 orollary 115. Let G be a Sylow-cycloidal group of binary tetrahedral type.Then G has a unique element of order . Let
M/O ( G ) be the subgroup of G/O ( G ) of order 8. Here M is a subgroupof G containing O ( G ), and since M/O ( G ) is normal in G/O ( G ) we have that M isa normal subgroup of G . Let Q be the 2-Sylow subgroup of G . Observe that Q is in M since G/M has order 3. In fact Q is a complement for O ( G ) in M , so M = O ( G ) Q = O ( G ) × Q since Q acts trivially on O ( G ). We have that Q is freely representable, so M isfreely representable if and only if O ( G ) is freely representable (Corollary 21). Thefurther analysis of freely representable depends on whether or not 9 divides theorder of G . Theorem 116.
Let G be a Sylow-cycloidal group of binary tetrahedral type. If divides the order of G then the following are equivalent.1. G is freely representable.2. O ( G ) is freely representable.3. For each prime p dividing the order of G , there is exactly one subgroup of G of order p .Proof. The property of being freely representable is inherited by subgroupsso (1) = ⇒ (2).Now suppose that (2) holds. Note that (3) holds for p = 2 (independent ofwhether (2) is true or not) by the above Corollary. Suppose that C is a subgroupof G of order p where p is an odd prime. If p = 3 then C is in O ( G ) since G/O ( G )has order prime to p . If p = 3, then C is contained in a 3-Sylow subgroup P of orderat least 9. The map P to G/O ( G ) has image of size 3, so it has a nontrivial kernel.Since P is cyclic, all nontrivial subgroups of P contain C . So C is in the kernelof P → G/O ( G ). In other words, C is in O ( G ). Since all subgroups of odd primeorder of G are in O ( G ), we have uniqueness for each prime order by Theorem 81.So (3) holds.Finally the implication (3) = ⇒ (1) follows from Proposition 84. Theorem 117.
Let G be a Sylow-cycloidal group of binary tetrahedral type. If does not divide the order of G then G has a subgroup H isomorphic to T and G = O ( G ) H ∼ = O ( G ) ⋊ T. Furthermore, the following are equivalent.1. G is freely representable.2. O ( G ) is freely representable and G ∼ = O ( G ) × T .3. For each prime p = 3 dividing the order of G there is exactly one subgroupof G of order p , and there are subgroups of order in G . roof. Let C be a 3-Sylow subgroup of G , and let Q be the unique 2-Sylowsubgroup of G . Then Q is normal in G , so H = Q C is a subgroup of G oforder 24. Note that O ( G ) has order prime to 24, so H maps isomorphically ontothe quotient G/O ( G ). Hence H is isomorphic to 2 T . Also observe that H is acomplement to O ( G ) so G = O ( G ) H ∼ = O ( G ) ⋊ T. Next observe that O ( G ) C = O ( G ) ⋊ C is a Sylow-cyclic subgroup of G . Soif G is freely representable, then same is true of L = O ( G ) C . If L = O ( G ) C isfreely representable then, by Theorem 81, C is the only subgroup of L of order 3.Thus C is normal in L , and so L = O ( G ) × C . In particular C acts triviallyon O ( G ). Since H = Q C and since Q acts trivially on O ( G ) then H acts triviallyon O ( G ). So (1) = ⇒ (2) holds.Note that (2) = ⇒ (1) by Corollary 21.Next observe that if G ∼ = O ( G ) × T then every subgroup of prime order p = 2 , G must be in O ( G ) and the subgroups of prime order p = 2 or p = 3 correspondto the subgroup of 2 T of prime order. If O ( G ) is freely representable then there isone subgroup of order p for each p = 2 , G (and hence theorder of O ( G )) (Theorem 81). Observe that 2 T has one subgroup of order 2 andfour subgroups of order 3. So (2) = ⇒ (3).Finally suppose (3) holds. We can conclude that O ( G ) is freely representableby Proposition 84. Since H itself has four subgroups of order 3, we have at leastfour subgroups of order 3. Note the four subgroups of H of order 3 map to distinctsubgroups of G/O ( G ). As before, let L = O ( G ) C where C is a cyclic subgroupof H . Any subgroup of order 3 in L maps to the same subgroup of G/O ( G ) as C .Since there are only 4 subgroups of order 3 in G we conclude that C is the uniquesubgroup of order 3 in L . So C is normal in L , and L = O ( G ) × C . Thus C acts trivially on O ( G ). Since H = Q C and since Q acts trivially on O ( G ), weconclude that G = O ( G ) × H . So (3) implies (2).For convenience we have divided into cases depending on whether or not 9divides the order of G . Now we will see another more unified approach. As beforelet Q be the unique 2-Sylow subgroup. Let S be a 3-Sylow subgroup of G . Wenote that Q S maps onto G/O ( G ), which implies that S is not in the centralizerof Q . So S must act on Q by sending a generator to an automorphism of Q oforder 3.Recall that O ( G ) acts trivially on Q . Consider M = O ( G ) S = O ( G ) ⋊ S .Note that the image of M in Aut( Q ) has order 3. Also note that G = Q M = Q ⋊ M. Proposition 118.
Let G be a Sylow-cycloidal group of binary tetrahedral type, andlet Q be its unique -Sylow subgroup of of G , which is isomorphic to the quaterniongroup with 8 elements. Then there is a complement M to Q in G which is a Sylow-cycloidal group of odd order: G = QM = Q ⋊ M. Here M → Aut Q has image of order . Moreover, G is freely representable if andonly if M is freely representable. roof. The only thing left to show is that if M is freely representable, then G isfreely representable. Since M is freely representable, then the same is true of O ( G )since it is isomorphic to a subgroup of M . If 9 divides the order of G then G mustbe freely representable by Theorem 116. So from now on we assume that 9 doesnot divide the order of G .Since M is freely representable, it has a unique subgroup C of order 3 (Theo-rem 81). Since O ( G ) and C are normal in M we have M = O ( G ) × C and C actstrivially on O ( G ). Note that Q acts trivially on O ( G ), so H = QC acts triviallyon O ( G ). This means that G = O ( G ) × H . However, H is isomorphic to G/O ( G ).Thus G ∼ = O ( G ) × T . So G is freely representable by Theorem 117.We can strengthen the above: Proposition 119.
Let G be a finite group and let Q be a -Sylow subgroup of G .Then G is a Sylow-cycloidal group of binary tetrahedral type if and only if Q isa normal quaternionic subgroup of G and there exists a non-normal Sylow-cyclicsubgroup M of index | S | in G . In this case G ∼ = Q ⋊ M where Q is a quater-nion group with 8 elements and where the action map M → Aut Q has image ofsize 3. Additionally, in this case G is freely representable if and only if M is freelyrepresentable.Proof. In light of the previous proposition we just need to show that G is a Sylow-cycloidal group of binary tetrahedral type under the assumption that Q is a normalquaternionic group in G and there exists a non-normal Sylow-cyclic subgroup M of index | S | in G .Under these assumption, every Sylow-subgroup of M of odd order is actuallya Sylow-subgroup of G since [ G : M ] is even, and every Sylow-subgroup of G isconjugate to a Sylow-subgroup of M by the Sylow theorems. Thus every Sylow-subgroup of G of odd order is cyclic. Note also that M is isomorphic to G/Q since M is of odd order. Thus G/Q and Q are solvable, and so G is solvable.Thus G is a solvable Sylow-cycloidal group. In addition O ( G ) is a proper sub-group of M since O ( G ) is normal. So G/O ( G ) is not a 2-group. Thus G is notSylow-cyclic, and is not of quaternion type. So G is either of binary tetrahedral orbinary octahedral type. Note that the image of Q in G/O ( G ) is a 2-Sylow subgroupof G/O ( G ) that is normal. This rules out the binary octahedral type (since theexistence of such a normal 2-Sylow subgroup in 2 O gives a unique 2-Sylow sub-group in its quotient S , contradicting the fact that two-cycles of S generates S ).Thus G is a Sylow-cycloidal group of binary tetrahedral type. We start with some basic observations about key subgroups of this type of Sylow-cycloidal group:
Proposition 120.
Let G be a Sylow-cycloidal group of binary octahedral type.Then every -Sylow subgroup of G is a generalized quaternion group of order 16,and these -Sylow subgroups are not normal in G . In addition G contains a uniquesubgroup H of index , and this subgroup is of binary tetrahedral type. Moreover, G contains a unique quaternion subgroup Q of order 8, and this group Q is contained n H . Finally, G contains exactly four subgroups of index ; these subgroups areconjugate subgroups of H hence are conjugate in G ; these subgroups are not normalin H hence are not normal in G ; these subgroups each contain O ( G ) as a subgroupof index 3; these subgroups are Sylow-cyclic groups of odd order; and these subgroupsare maximal among subgroups of odd order.Proof. Every 2-Sylow subgroup S of G is isomorphic to a 2-Sylow subgroupof G/O ( G ) ∼ = 2 O , so is a generalized quaternion group of order 16 (Proposi-tion 105). The image S in G/O ( G ) of a 2-Sylow subgroup S contains the uniqueelement of G/O ( G ) of order 2, so these Sylow subgroups S correspond to Sylowsubgroups of the quotient O ∼ = S . But Sylow subgroups of S are not normal (anormal 2-Sylow subgroup N in S would have to contain all two cycles since S /N has order 3, but the collection of two cycles generate S ). Thus such an S is notnormal in G/O ( G ), and so S cannot be normal in G .Let H be the subgroup of G containing O ( G ) such that H/O ( G ) corresponds tothe binary tetrahedral subgroup of G/O ( G ). Clearly O ( G ) ⊆ O ( H ), but since H isnormal in G and since O ( H ) is characteristic in H , it follows that O ( H ) is normalin G , and so O ( G ) = O ( H ). Hence H/O ( H ) is a binary tetrahedral group.Suppose L is any subgroup of G of index 2. Thus L is normal in G . Sincethe quotient G/L has two elements and O ( G ) has odd order, O ( G ) is containedin L . Note that L/O ( G ) has index 2 in G/O ( G ), so L/O ( G ) = H/O ( G ) (Propo-sition 104). This gives us L = H , and so H is the unique subgroup of index 2 in G . By Proposition 114, H has a unique subgroup Q H of order 8 and this groupis a quaternion group. This group is characteristic in H and so is normal in G .By the Sylow theorems, Q H is contained in some 2-Sylow subgroup of G , andhence in all since Q H is normal (and all 2-Sylow subgroups are conjugate). Everyquaternion subgroup Q of G of order 8 is contained in some 2-Sylow subgroup S of G by the Sylow theorems, so both Q H and Q are subgroups of S . Thus Q = Q H since every general quaternion group contains a unique subgroup isomorphic to thequaternion group with 8 elements. So Q H is the unique subgroup Q H isomorphicto the quaternion group with 8 elements.Each subgroup of order 3 in G/O ( G ) is uniquely of the form M/O ( G ) where M is a subgroup of G containing O ( G ). Each such M is of index 16 in G and con-tains O ( G ) as a subgroup of index 3. Since each such M is of odd order, M is aSylow-cyclic group. Having index 16 in G , each such M must be maximal amongsubgroups of odd order in M . Since M = O ( G ) this means M cannot be normalin G since O ( G ) is the maximal subgroup of odd order.Next we argue that each subgroup of index 16 in G arises in this way. Suppose M has index 16 in G . Then O ( G ) M/O ( G ) is isomorphic to M/ ( M ∩ O ( G )) which hasodd order. Thus O ( G ) M has odd order. But since M has index 16 in G , there is nostrictly larger subgroup of odd order. So M = O ( G ) M , and so O ( G ) is containedin M . In particular, such a group corresponds to a subgroup M/O ( G ) of G/O ( G )of index 16 and order 3.By Proposition 104, there are 4 subgroups of G/O ( G ) of order 3, and theyare all contained in H/O ( G ). They constitute the 3-Sylow subgroups of H/O ( G )hence are conjugate in H/O ( G ) by the Sylow theorems, and thus cannot be normal61n H/O ( G ). This means that there are 4 subgroups of G of index 16, they are allcontained in H , they are conjugate in H , and cannot be normal in H .In particular, there is a unique subgroup of order 2 in any group of this type: Proposition 121.
Let G be a Sylow-cycloidal group of binary octahedral type.Then G has a unique element of order .Proof. By the above proposition, we have a normal subgroup of order 8 in G , andthis group has a unique subgroup of order 2. Thus we have a normal subgroup oforder 2 in G. The result follows from Lemma 49.The main theorem about freely representable groups of this type is as follows: Theorem 122.
Let G be a Sylow-cycloidal group of binary octahedral type. Let H be the unique subgroup of G of index , which is of binary tetrahedral type. Let M be any of the four subgroups of G of index 16, which is Sylow-cyclic subgroup of H of odd order. Then the following are equivalent1. G is freely representable.2. H is freely representable.3. M is freely representable.Furthermore, if divides the order of G then if O ( G ) is freely representable, thenso is G .Proof. The implication (1) = ⇒ (2) = ⇒ (3) is clear since a subgroup of a freelyrepresentable group is freely representable (Proposition 13). So we just need toshow that (3) = ⇒ (1). In fact we will proceed by showing (3) = ⇒ (2) = ⇒ (1).Suppose that M is freely representable. By Proposition 118, H is also freelyrepresentable. Since G/H has order 2, any subgroup of G odd prime order p isa subgroup of H . Also, by the previous proposition, G has a unique subgroup oforder 2 and this is a subgroup of H since H has even order. By Corollary 80 weconclude that G is freely representable.Now suppose 9 divides the order of G and that O ( G ) is freely representable.Note that O ( H ) is characteristic is H , and H is normal in G , so O ( H ) is normalin G . This implies that O ( H ) ⊆ O ( G ) so that O ( H ) is freely representable. Thus H is freely representable by Theorem 116, which as we have seen implies that G isfreely representable. We make some observations about solvable Sylow-cycloidal groups in general.
Proposition 123.
Let G be a solvable Sylow-cycloidal group. If G is not a Sylow-cyclic group, then G always has a unique element of order 2. If G is a Sylow-cyclicgroup then G has a unique element of order 2 if and only if G has a normal subgroupof order 2. roof. This was proved for each type individually. See Lemma 49, Proposition 109,Corollary 115, and Proposition 121.
Proposition 124.
Let G be a solvable Sylow-cycloidal group of order k n where n is odd and where k ≥ . Then G has a Sylow-cyclic subgroup M of order n withthe following property: G is freely representable if and only if (1) M is freely repre-sentable and (2) G has a unique element of order . In particular, if G is not itselfSylow-cyclic then G is freely representable if and only if M is freely representable.Proof. This was proved for each type individually. See Proposition 108, Proposi-tion 111, Proposition 118, and Theorem 122.
10 The Non-Solvable Case: the Group SL ( F p ) . A very important family of Sylow-cycloidal groups is SL ( F p ). The first goal here isto show that SL ( F p ) is Sylow-cycloidal for all odd primes p . In some sense these,together with the Sylow-cycloidal groups we have considered up to now, are all thatare needed to form the most general Sylow-cycloidal groups. Along the way we seesome very striking results about the cyclic (or equivalently the Abelian) subgroupsof SL ( F p ). These results leads naturally to the classification of normal subgroupsof SL ( F p ) and nonsolvability results. We will also consider an amusing necessarycondition for these groups to be freely-representable: p is a Fermat prime. It turnsout that p = 3 or 5 is a necessary and sufficient condition, but we will not prove thishere (at least not in this version of the document). Interestingly, p = 3 and p = 5are the two cases that occur as subgroups of H × .A few of the initial results can be proved for any finite field F , but we will needto specialize to F of odd prime order if we want Sylow-cycloidal groups. Proposition 125.
Let F be a finite field of order q . Then SL ( F ) is a group oforder ( q − q ( q +1) . If q is odd then there is a unique element of order in SL ( F ) .Proof. The first row of an invertible 2-by-2 matrix is nonzero, so we can limit ourattention to q − q − q choices for the second row for each giventop row. So there are ( q − q − q ) = ( q − q ( q + 1)elements of GL ( F ). The determinant homomorphism GL ( F ) → F × is surjectivesince even the invertible diagonal matrices map onto F × . So the kernel of this map,which is SL ( F ), has order ( q − q ( q + 1).Now we assume q is odd. Every element α of order 2 in SL ( F p ) has hasan eigenvalue 1 or − α − I )( α + I ) = 0. Any element of SL ( F p ) witheigenvalue ± α = ± (cid:18) a (cid:19) , α = (cid:18) a (cid:19) . We conclude that the only element of order 2 in SL ( F p ) is − I .63ext we classify elements of order greater than 2 by the number of eigenvaluesin F , starting with two and working down to zero eigenvalues. Lemma 126.
Let F be a finite field of order q and let α ∈ SL ( F ) be an elementwith two distinct eigenvalues in F . Then α is contained in a cyclic subgroup C of SL ( F ) of order q − where C has the property that that there is a basis suchthat every element of C is a diagonal matrix. Furthermore, the existence of suchan α implies that q > .Proof. Fix a basis of eigenvectors for A . Then A is contained in the subgroup C consisting of matrices of the form (cid:18) a a − (cid:19) with respect to the chosen basis, where a ∈ F × . Since F × is cyclic, this group is acyclic group of order q − α is 1 and since they are distinct, wemust have q > Lemma 127.
Let F be a finite field of order q and characteristic p . Let v ∈ F be anonzero vector. Then the set of elements D v of SL ( F ) with exactly one eigenvalueand with eigenvector v is a subgroup of SL ( F ) isomorphic to {± } × F where F isthe additive group of F (of size q ). In particular D v is Abelian and each elementof F has order divisible by p if q is odd, and divisible by p = 2 if q is even.Let α ∈ SL ( F ) be an element with exactly one eigenvalue in F . If q is oddthen α is a contained in a cyclic subgroup C of SL ( F ) of order p . If q is eventhen α is contained in a cyclic subgroup C of SL ( F ) of order p = 2 .Proof. Choose a basis whose first element is v . Then D v consists of the matriceswhose representation with respect to this basis is of the form ± (cid:18) a (cid:19) for some a ∈ F . Observe that (cid:18) a (cid:19) (cid:18) b (cid:19) = (cid:18) a + b (cid:19) , and in fact that D v is isomorphic to {± } × F . In particular, every element of D v iscontained in a cyclic subgroup of order 2 p if q is odd, and a subgroup of order p = 2if q is even. Lemma 128.
Let F be a finite field of order q and let α ∈ SL ( F ) be an elementwith no eigenvalues in F . Then α is contained in a cyclic subgroup of SL ( F ) of order q + 1 . More specifically, let M ( F ) be the ring of -by- matrices withentries in F , where we view F as a subring via the diagonal embedding. Then thesubring E = F [ α ] of M ( F ) generated by F and α is a field. This field E has thefollowing properties: E has size q . • The group K = SL ( F ) ∩ E × is a cyclic group of order q + 1 that contains α . • The determinant homomorphism E × → F × is the map x x q +1 , and thekernel of this map is K = SL ( F ) ∩ E × . • The Galois group of E over F has two elements. Its nontrivial element σ isthe automorphism x x q . For all x ∈ K = SL ( F ) ∩ E × we have σx = x − .Proof. Using the ring homomorphism F [ X ] → M ( F ) sending X to α , we seethat F [ α ] is isomorphic to F [ X ] / h f i where F [ X ] is the polynomial ring in one-variable and f is the minimal polynomial of α in F [ X ]. By the Cayley-Hamiltontheorem, f divides the characteristic polynomial of α , so in this case f must bean irreducible quadratic polynomial since the characteristic polynomial of α hasno roots in F . This implies that F [ X ] / h f i is a quadratic field extension of F . Inparticular, E = F [ α ] is a field of size q . Note that E × is cyclic of order q − F × is the subgroup of E × of size q −
1, we have that that β ∈ E × isin F × if and only if β q − = 1. This implies that β ∈ E is in F if and only if β q = β .Now let β be a generator of the cyclic group E × . In E [ X ] we have the polynomial( X − β )( X − β q ) = X − ( β + β q ) X + β q +1 Observe that ( β + β q ) q = β q + β q = β q + β and that ( β q +1 ) q = β q β q = β q . These follow since β q − = 1, so β q = β . Also q is a power of the characteristic p of E and in fields of characteristic p we have the identity ( a + b ) p = a p + b p ,so ( a + b ) q = a q + b q for all a, b ∈ E . We conclude that X − ( β + β q ) X + β q +1 lies in F [ X ] and so must be the minimal polynomial of β in F [ X ] (since β is notin F ). By the Cauchy-Hamilton theorem, it is the characteristic polynomial of β .In particular the determinant of β is β q +1 . Note also that we have establishedthat x x q is an automorphism σ of the field E , and that it fixes F .Consider the determinant homomorphism E × → F × . This sends the genera-tor β to β q +1 , so sends any element in E × to its q + 1 power. In particular thekernel K must be the cyclic subgroup of E × of size q + 1 since q + 1 divides q − K = E × ∩ SL ( F ) since it is in the kernel of the determinant map.So α ∈ K .Also note that any automorphism of E fixing F is determined by its action onthe generator β of E × , and that β must map to a root of ( X − β )( X − β q ). Thusthere are only two elements of the Galois group of E over F (i.e., the automorphismsof E fixing F ): the identity x x and x x q . Let σ be the map x x q . Notethat if x ∈ K then x q +1 = 1 so x q = x − . Thus the restriction of σ to K is themap x x − .We can combine these three lemmas:65 emma 129. Let F be a finite field of order q and characteristic p . Let α ∈ SL ( F ) be an element not equal to or − . Then the following hold: • The element α has two distinct eigenvalues in F if and only if α has orderdividing q − . • The element α has exactly one eigenvalue in F if and only if α has orderdividing p . • The element α has no eigenvalues in F if and only if α has order dividing q +1 .Furthermore α can only have two distinct eigenvalues if q > .Proof. One direction of each implication follows directly from Lemmas 126, 127,and 128. To see the converse, observe that the GCD of any two distinct elementsof { q − , p, q + 1 } is 2 is q is odd, and is 1 is q is even. Now if q is odd then − α is not 1 or −
1, we concludethat the order of α cannot divide two of { q − , p, q + 1 } . From this the conversefollows.A maximal cyclic subgroup of a finite group G is defined to be a cyclic subgroupof G that is not contained in a cyclic group of G of larger order. Of course everycyclic subgroup is contained in an maximal cyclic group (since G is finite), buta given cyclic subgroup might be contained in several maximal cyclic groups in ageneral group G . For G = SL ( F ) we can establish uniqueness. Corollary 130.
Let F be a finite field of order q and characteristic p and let C bea cyclic subgroup of SL ( F ) . • If q > is odd then C is a maximal cyclic subgroup if and only if it hasorder q − , p , or q + 1 . • If q = 3 then C is a maximal cyclic subgroup if and only if it has order p = 6 or q + 1 = 4 . • If q > is even then C is a maximal cyclic subgroup if and only if it hasorder q − , , or q + 1 . • If q = 2 then C is a maximal cyclic subgroup if and only if it has order or .Proof. Let α be a generator of C . Since C is a maximal cyclic group, it has thelisted orders by Lemmas 126, 127, and 128 depending on the number of eigenvaluesof α in F .Conversely, suppose q > C has order exactly q −
1, 2 p ,or q + 1. Let D be a maximal cyclic subgroup containing C . Then D also hasorder q −
1, 2 p , or q + 1 by the above. Note that (1) the order of C divides theorder of D , (2) the order of C is at least 3 (since q > { q − , p, q + 1 } is 2. So it is impossible for C and D to have differentorders. So C = D . A similar but modified argument works for q = 3, or q > q = 2.We are now ready for the first main theorem.66 heorem 131. Let p be an odd prime. Then SL ( F p ) is a Sylow-cycloidal groupwhose -Sylow subgroups are not cyclic.Proof. Let q be an odd prime dividing the order ( p − p ( p +1) of the group SL ( F p )and let q k be the maximal power that q divides ( p − p ( p +1). By Cauchy’s theorem,there is an element α of order q . Let C be a maximal cyclic subgroup SL ( F p )containing α . By the above corollary, C has order p − p or p + 1. Since q ≥ q and hence q k divides exactly one of p − p or p + 1. Since q divides | C | this means that if | C | = p − p + 1 then q k divides | C | . If | C | = 2 p then q = p (since q is odd) and so q k divides | C | as well (and k = 1 in this case).In any case, C has a cyclic subgroup of order q k . Since all q -Sylow subgroups areconjugate, we have established that all q -Sylow subgroups are cyclic.Since SL ( F p ) has a unique element of order 2 (Proposition 125), each 2-Sylowsubgroup of SL ( F p ) has a unique element of order 2 (by Cauchy’s theorem). Thuseach 2-Sylow subgroup S of SL ( F p ) is either cyclic or quaternionic (Corollary 37).So SL ( F p ) is a Sylow-cycloidal group.Suppose S is a cyclic 2-subgroup of SL ( F p ), and let C be a maximal cyclicsubgroup containing S . By the above corollary, S has order dividing | C | which iseither p − p or p −
1. Since 2 divides both p − p + 1, the largest powerof 2 diving | C | is less than the largest power of 2 dividing the order ( p − p ( p + 1)of SL ( F p ). Thus S is not a 2-Sylow subgroup.Here is a partial converse. Proposition 132.
Suppose that F is a finite field. If SL ( F ) is a Sylow-cycloidalgroup then F = F p for some prime p .Proof. By Lemma 127 there is a subgroup D of SL ( F ) isomorphic to the additivegroup F . If SL ( F ) is a Sylow-cycloidal group then the Abelian group D ∼ = F mustbe cyclic. This can only happen if F = F p where p is the characteristic of F . Remark.
The case SL ( F ) is special. It has order 6, and so is Sylow-cyclic sinceit is of prime free order. As we have seen, all its cyclic subgroups are of order 2or 3, so it is not cyclic and so must be dihedral. Note that SL ( F ) fails to have aunique element of order 2.The next major result is that all cyclic subgroups of SL ( F p ) of the same orderare conjugate when p is a prime. It is a bit easier to show that such groups areconjugate in GL p ( F p ), but the following two lemmas will give us tools to achieveconjugacy in SL ( F p ) instead of GL p ( F p ). Lemma 133.
Let α ∈ GL ( F ) where F is a finite field of order q and let d ∈ F × .If α has two distinct eigenvalues in F or no eigenvalues in F then there is anelement β ∈ GL ( F ) such that βαβ − = α and det β = d .Proof. Suppose α has two eigenvalues in F and let v , v ∈ F be a basis of eigen-vectors. Then let β be such that its associated linear tranformation maps v dv and v v . Then β clearly works.Suppose that α has no eigenvalues in F . By Lemma 128, α ∈ E × where E × isa cyclic subgroup of GL ( F ) of order q − E × is the multiplicative groupof a field); furthermore, the determinant map on E × → F × is given by x x q +1 F × has order q −
1. Now just let β ∈ E × be an elementof determinant d . Lemma 134.
Let p be a prime and let C be a cyclic subgroup of SL( F p ) .Let d ∈ F × p . Then there is an there is an element β ∈ GL ( F ) such that βCβ − = C and det β = d .Proof. Let α be a generator of C . If α has two distinct eigenvalues or no eigenvaluesthen the result follows from the previous lemma. So suppose that α has exactlyone eigenvalue in F and let v be an eigenvector. Let v be such that v , v form abasis for F . The representation for α in this basis is of the form (cid:18) e a e (cid:19) where e is 1 or −
1. So let β be an element with the following matrix representation(for this same basis v , v ): (cid:18) d
00 1 (cid:19)
Note that (cid:18) d
00 1 (cid:19) (cid:18) e a e (cid:19) (cid:18) d −
00 1 (cid:19) = (cid:18) e da e (cid:19) = (cid:18) e a e (cid:19) m where m is an odd positive integer such that m ≡ d (mod p ).As mentioned above, our next goal is to establish that cyclic subgroups (equiv-alently, Abelian subgroups) of the same order of SL ( F p ) are conjugate. We startwith cyclic groups of order p −
1, followed by order p , then order p + 1, ending withgeneral order. Lemma 135.
Let F be a finite field of order q . The vector space F has q + 1 distinct one-dimensional subspaces. Let L and L be two distinct one-dimensionalsubspaces of F . Then the set elements of SL ( F ) with a basis of eigenvectorsin L ∪ L forms a cyclic subgroup of SL ( F ) of order q − .Now assume q > . Then all cyclic subgroups of SL ( F ) of order q − arise inthis way. There are q ( q + 1) cyclic subgroups of order q − and they are conjugate.Two distinct cyclic subgroups of order q − have intersection {± } .Proof. Counting the number of one-dimensional subspaces is straightforward.Let v , v be a basis for F such that v ∈ L and v ∈ L . Then α ∈ SL ( F )has a basis of eigenvectors in L ∪ L if and only if it has form α = (cid:18) a a − (cid:19) with respect to this basis, where a ∈ F × . So the set of such matrices is a cyclicsubgroup of order q − q > C be a cyclic subgroup of SL ( F ) of order q −
1, and let α be a generator of C .So α is not 1 or − q − >
2. By Lemma 129, α has two distinct eigenvalues,68nd its eigenvectors determine two distinct one-dimensional subspaces L and L of F . So α is in a cyclic group of order q − C has the desired form.Suppose C and C ′ are cyclic subgroup of SL ( F ) of order q −
1. Suppose C isdefined using L , L and C ′ is define using L ′ , L ′ . Suppose g ∈ C ∩ C ′ is not ± g has distinct eigenvalues (Lemma 129). Let v be an eigenvector of g .Then v ∈ L i and v ∈ L ′ j for some i, j , so L i = L ′ j since these are one-dimensional.After renumbering we can assume L = L ′ . Let v be an eigenvector of g not inthe span of v . Then v is in L and L ′ so L = L ′ . Thus C = C ′ . In otherwords, distinct cyclic subgroup of SL ( F ) of order q − {± } . (Notethat − ∈ C and − ∈ C ′ by Proposition 125 if q is odd, and trivially if q is even).So different choices of { L , L } will produce difference cyclic subgroups of or-der q − q − >
2. So there are q ( q + 1) such cyclic subgroups.Suppose C and C ′ are cyclic subgroup of SL ( F ) of order q −
1. Suppose C is defined using L , L and C ′ is define using L ′ , L ′ . Let α be a generator for C .Let β ∈ GL ( F ) give a linear transformation mapping L to L ′ and L to L ′ .Let β ∈ GL ( F ) be such that β αβ − = α and det( β β ) = 1 (see Lemma 133).Let β = β β . Observe that C ′ = βCβ − . So C and C ′ are conjugate. Lemma 136.
Let p be an odd prime. The vector space F p has p + 1 distinct one-dimensional subspaces. Let L be a one-dimensional subspaces of F p . Then the setelements of SL ( F p ) with exactly one eigenvalue, and with an eigenvector in L ,forms a cyclic subgroup of SL ( F p ) of order p .All cyclic subgroups of SL ( F p ) of order p arise in this way. There are p + 1 cyclic subgroups of order p and they are conjugate. Two distinct cyclic subgroupsof order p have intersection {± I } .Proof. Counting the number of one-dimensional subspaces is straightforward.Let v , v be a basis for F p such that v ∈ L . Then α ∈ SL ( F p ) has exactly oneeigenvalue and has an eigenvector in L if and only if can be written as α = ± (cid:18) a (cid:19) with respect to this basis, where a ∈ F p . Observe that the set of such matricesforms a cyclic group isomorphic to {± } × F p where F p here is the additive group.Let C be a cyclic subgroup of SL ( F p ) of order 2 p , and let α be a generatorof C . Then α has exactly one eigenvalue by Lemma 129. If L is an eigenspacefor α , then as above α is contained in a group of order 2 p of the given form. So C is a group of the desired form.Suppose C and C ′ are cyclic subgroup of SL ( F p ) of order 2 p where C is definedusing L and C ′ is define using L ′ . Suppose g ∈ C ∩ C ′ is not ±
1. Then g has aunique eigenvalue, and the eigenspace associated to that eigenvalue has dimensionone (since g = ± L = L ′ and so C = C ′ . In other words, distinct cyclicsubgroup of SL ( F p ) of order 2 p intersect in {± } .So different choices of L will produce difference cyclic subgroups of order 2 p . Sothere are p + 1 such cyclic subgroups.Suppose C and C ′ are cyclic subgroup of SL ( F p ) of order 2 p . Suppose C isdefined using L and C ′ is define using L ′ . Let β ∈ GL ( F p ) represent a linear69ransformation sending L to L ′ . Let β ∈ GL ( F ) be such that β Cβ − = C andso that det( β β ) = 1 (see Lemma 134). Let β = β β . Observe that C ′ = βCβ − .So C and C ′ are conjugate. Lemma 137.
Let F be a finite field of order q and let M ( F ) be the ring of -by- matrices with entries in F . Then the following hold: • If C is a cyclic subgroup of SL ( F ) of order q + 1 then there is a uniquequadratic field extension E of F in M ( F ) containing C , and C is the uniquesubgroup of E × of order q + 1 . • Two distinct cyclic subgroups of SL ( F ) of order q + 1 are conjugate in SL( F ) and have intersection {± } . • If C is a cyclic subgroup of SL ( F ) of order q + 1 then there is a γ ∈ SL ( F ) such that x γxγ − is an automorphism of the group C sending any x ∈ C to x − .Proof. Let α be a generator of C . By Lemma 129, α cannot have eigenvalues in F .So by Lemma 128, the subring E = F [ α ] of of M ( F ) is a field of order q , and α iscontained in K = SL ( F ) ∩ E × , which is the unique subgroup of E × of order q + 1.So in fact C = K . Since E = F [ α ] any quadratic field extension E of F in M ( F )containing C would contain α and so F [ α ], and hence be equal to E . So E is theunique such field.Suppose C and C ′ are two subgroups of SL ( F ) size q + 1 and that E and E ′ arethe respective quadratic extensions of F containing C and C ′ . Suppose β ∈ C ∩ C ′ isnot ±
1. Then β cannot be of the form c c ∈ F since the determinant of β is 1. Thus β generates a proper extension of F contained in the intersection E ∩ E ′ .So E = E ′ = F [ β ] since [ E : F ] = [ E : F ] = 2 and [ E ′ : F ] ≥
2. We concludethat C = C ′ . In other words, if C and C ′ are distinct subgroups of SL ( F ) ofsize q + 1 then their intersection is {± } .Next fix a generator g of F × . In other words, g has order q −
1. Suppose C isa cyclic subgroup of SL ( F ) of order q + 1 and let E be the field extension of F inside M ( F ) of size q containing C . Since E × has order q −
1, and since q + 1is even, there is a subgroup of E × of order 2( q − E × hasan element β such that β = g . Fix a nonzero vector v ∈ F and let v = βv (viewing v as a column vector). By the Cayley-Hamilton theorem, the characteristicpolynomial of β is X − g , so β has no eigenvalues in F . Thus v , v must be abasis of F . Note also that βv = gv .Suppose C ′ is another cyclic subgroup of SL ( F ) of order q + 1 and let E ′ bethe field extension of F inside M ( F ) of size q containing C ′ . Let β ′ ∈ ( E ′ ) × besuch that v ′ = v and v ′ = β ′ v ′ forms a basis with β ′ v ′ = gv ′ .Let γ ∈ GL ( F ) be chosen so that γ maps v = v ′ to itself, and maps v to v ′ .Note that γ βγ − = β ′ since both sides of this equation map v ′ v ′ and v ′ gv ′ .Since F [ β ] = E and F [ β ′ ] = E ′ this means that the map x γ xγ − is an isomorphism E → E ′ between fields. Hence it sends the unique subgroup C of E × of order q + 1 to the unique subgroup C ′ of ( E ′ ) × of order q + 1.70n other words, γ Cγ − = C ′ . Let γ ∈ GL ( F ) be chosen so that γ γ hasdeterminant 1 and so that γ Cγ − = C (Lemma 133 applied to a generator α of C and to d = det γ − ). If γ = γ γ then γCγ − = C ′ and γ ∈ SL ( F ). So C and C ′ are conjugate in SL ( F ).A special case of this construction is where C = C ′ and where β ′ = − β sothat γ βγ − = − β . Observe that x γ xγ − must then be a field automorphismfixing F (thought of as diagonal matrices). Since F × has C for its unique subgroupof order q +1, this automorphism acts on C . Lemma 128 implies that it sends x ∈ C to x − . So if γ = γ γ then γxγ − = x − for all x ∈ C .These lemmas combine to yield the following: Lemma 138.
Let p be an odd prime, and let C and C ′ be two distinct maximalcyclic subgroups of SL ( F p ) . Then C ∩ C ′ = {± } . Furthermore, if C and C ′ havethe same order then C and C ′ are conjugate subgroups of SL ( F p ) .Proof. By Corollary 130, the orders of C and C ′ are in the set { p − , p, p + 1 } .If C and C ′ happen to have different orders then C ∩ C ′ has order dividing 2 sincethe GCD of any two of { p − , p, p + 1 } is 2. Since C and C ′ are of even order theyboth contained − − C ∩ C ′ = {± } .Now we consider the case where C and C ′ have the same order in { p − , p, p +1 } .Then C and C ′ are conjugate and C ∩ C ′ = {± } by Lemma 135, Lemma 136, orLemma 136 (depending on the order of C ).The above lemma yields the conjugacy results we want but only for maximal cyclic subgroups of SL ( F p ). The following lemma and proposition make it possibleto establish the result for cyclic subgroups more generally. Lemma 139.
Let C be a cyclic subgroup of SL ( F p ) where p is an odd prime. If theorder of C is at least then there is a unique maximal cyclic subgroup of SL ( F p ) containing C .Proof. Suppose D and D are distinct maximal cyclic subgroups of SL ( F p ) con-taining C . By the above lemma D ∩ D = {± } , contradicting the assumption onthe size of C . Proposition 140.
Let C be a cyclic subgroup of SL ( F p ) where p is a odd prime.If C has order at least then the centralizer Z ( C ) is the unique maximal cyclicsubgroup of SL ( F p ) containing C . So if C is a maximal cyclic subgroup of SL ( F p ) then C = Z ( C ) .Proof. Let D be the unique maximal cyclic subgroup of SL ( F p ) containing C (seeprevious lemma). Suppose that h ∈ Z ( C ) and let H be the subgroup generatedby h . Observe that HC is an Abelian subgroup of SL ( F p ). Since SL ( F p ) isSylow-cycloidal (Theorem 131) the group HC must be cyclic and so is containedin a maximal cyclic subgroup D ′ . By the previous lemma D = D ′ since bothcontain C . Thus h ∈ D . We have established that Z ( C ) ⊆ D . The other inclusionis clear.We are ready for the second main theorem.71 heorem 141. Let p be an odd prime, and let C and C ′ be two cyclic subgroupsof SL ( F p ) of the same order. Then C and C ′ are conjugate subgroups of SL ( F p ) .If, in addition, C = C ′ then C ∩ C ′ ⊆ {± } .Proof. If C and C ′ have order 1 or 2 then the result is clear since − ( F p ) of order 2. So from now on assume that C and C ′ have equalorder at least 3. By the above proposition, Z ( C ) and Z ( C ′ ) are maximal cyclicsubgroups. By Corollary 130, the orders of Z ( C ) and Z ( C ′ ) are restricted to theset { p − , p, p + 1 } . So the order of C and C ′ must divide an element of theset { p − , p, p + 1 } , and in fact divides a unique element of { p − , p, p + 1 } sincethe GCD of any two distinct elements is 2. This means that Z ( C ) and Z ( C ′ ) havethe same order, namely the unique element of { p − , p, p + 1 } that is a multipleof the order of C .Thus Z ( C ) and Z ( C ′ ) are conjugate subgroups by Lemma 138. Since C is theunique subgroup of Z ( C ) of its order, and the same is true of C ′ in Z ( C ′ ), weconclude that C and C ′ are conjugate as well.If, in addition, C = C ′ then Z ( C ) cannot equal Z ( C ′ ) since C is the uniquesubgroup of Z ( C ) of its order, and the same is true of C ′ in Z ( C ′ ). So the inter-section Z ( C ) ∩ Z ( C ′ ) is {± } by Lemma 138. Thus C ∩ C ′ ⊆ {± } .Next we wish to count the number of cyclic subgroups of SL ( F p ) of each order.We have already counted such groups when the order is p − p , so we startwith the order p + 1 case. Lemma 142.
Let p be an odd prime. The number of cyclic subgroup of SL ( F p ) of order p + 1 is equal to ( p − p .Proof. We consider two approaches. One is to note that each element g of SL ( F p )outside of {± } is contained in a unique maximal cyclic subgroup of SL ( F p ),namely the centralizer Z ( g ) (see Lemma 140). Also a cyclic subgroup of SL ( F p )is maximal if and only if its order is p − p = 3), 2 p , or p + 1. So we canpartition the elements of SL ( F p ) outside of {± } : if c n is the number of cyclicsubgroups of order n we have | SL ( F p ) | − c p − (( p − −
2) + c p (2 p −
2) + c p +1 (( p + 1) − p = 3 since ( p − − p − p ( p + 1) − p ( p + 1)( p −
3) + ( p + 1)(2 p −
2) + c p +1 ( p − . Now we solve for c p +1 .This first approach is legitmate, but there is also a more group theoretical /Galois theoretical approach. Let C be any cyclic group of order p + 1, and let E be the field F [ C ] inside of M ( F p ) generated by the elements of C . Then thenormalizer N ( C ) acts via conjugation not just on C but on all of E = F [ C ]. Infact we get a homomorphism from N ( C ) to the Galois group G of E over F , andkernel of this homomorphism is just Z ( C ). Since E is a quadratic extension of F p ,the Galois group G has order 2 (see Lemma 128). So Z ( C ) has index 1 or 2in N ( C ). Lemma 137 shows that N ( C ) is not Z ( C ) so the index is actually 2,72nd Proposition 140 show that Z ( C ) = C . Thus [ N ( C ) : C ] = 2 and so N ( C )has order 2( p + 1). By the orbit-stabilizer theorem we have that the number ofconjugates of C is ( p − p ( p + 1)2( p + 1) = 12 ( p − p which counts the number of cyclic subgroups of order p + 1 since all such groupsare conjugate (Theorem 141).Our calculations culminate with a full census of the cyclic subgroups of SL ( F p ).Since SL ( F p ) is Sylow-cyloidal, this gives a full census of Abelian subgroups as well. Proposition 143.
Let p be an odd prime and let c m be the number of cyclic sub-groups of SL ( F p ) of order m . Then c = c = 1 . If m > divides p − then c m = 12 p ( p + 1) . If m > divides q then c m = p + 1 . If m > divides p + 1 then c m = 12 p ( p − . Otherwise c m = 0 .Proof. Since there is a unique element of order 2 (Proposition 125) and order 1, wehave c = c = 2. So from now on we assume m > m divides p − p > m > C of order p − Z ( C ) for the maximum cyclic subgroup containing C (Proposi-tion 140). By Corollary 130, Z ( C ) has order in the set { p − , p, p + 1 } , and m divides the order of Z ( C ) since C is a subgroup of Z ( C ). Since m divides p −
1, itcannot divide 2 p or p + 1 as well (since the GCD of p − m > Z ( C ) is a cyclic group of order p −
1. Conversely each cyclic group of order p − m . Thus there is a one-to-one correspondences be-tween cyclic subgroups of order m and subgroups of order p −
1. By Lemma 135there are p ( p + 1) cyclic subgroups of order p −
1. Thus c m = p ( p + 1).The case where m divides 2 p is similar: there is a one-to-one correspon-dence C Z ( C ) between cyclic subgroups of order m and cyclic subgroups oforder 2 p . We can use Lemma 136 to conclude that there are c m = p + 1 cyclicsubgroups of order m .The case where m divides p + 1 is also similar: there is a one-to-one correspon-dence C Z ( C ) between cyclic subgroups of order m and cyclic subgroups oforder p + 1. We can use Lemma 142 to conclude that there are c m = ( p − p cyclic subgroups of order m . Corollary 144.
Suppose that p is an odd prime. The only normal cyclic subgroupsof SL ( F p ) are { } and {± } . roof. Suppose that C is a normal subgroup of order m . By Theorem 141 all cyclicsubgroups of order m are conjugate. Since C is normal, this means that there isonly one subgroup of order C . According to the previous proposition, this canonly happen if m = 1 ,
2. The result follows now from the fact that − Corollary 145.
Suppose that p is an odd prime. Then the only normal Sylow-cyclicsubgroups of SL ( F p ) are { } and {± } .Proof. Suppose that N is a normal Sylow-cyclic subgroup of SL ( F p ). By The-orem 59 the commutator subgroup N ′ of N is cyclic of odd order. Since N ′ ischaracteristic in N it must be normal in SL ( F p ). By the above corollary, N ′ isthe trivial group. So N is Abelian, hence cyclic since N is Sylow-cyclic. So by theabove corollary N is { } or {± } . Lemma 146.
Suppose that p is an odd prime and that N is a normal subgroupof SL ( F p ) not equal to { } or {± } . Then N has odd index in SL ( F p ) .Proof. Let N be a 2-Sylow subgroup of N . By the above corollary, N cannot becyclic, so N must be a general quaternion group. In particular, N contains a cyclicsubgroup of order 4. All cyclic subgroups of order 4 are conjugate in SL ( F p ) byTheorem 141. Since N is normal, it must contain all cyclic subgroups of order 4.Let P be a 2-Sylow subgroup of SL ( F p ). Note that P is generated by its cyclicsubgroups of order 4. So P must be contained in N . The result follows. Lemma 147.
Let p be an odd prime. Suppose SL ( F p ) has a normal subgroupnot equal to { } or {± } or SL ( F p ) . Then SL ( F p ) has a normal subgroup of oddprime index.Proof. Let N be a normal subgroup not equal to { } or {± } or SL ( F p ). Then N has odd index in SL ( F p ) by the previous lemma. Observe that G = SL ( F p ) /N is a Sylow-cyclic group of odd order. By Corollary 56, G has a nontrivial cyclicquotient. Hence G has a quotient of odd prime order. This means that SL ( F p )has a quotient of odd prime order as well, and the result follows.Now we are ready for the third main result: Theorem 148.
Let p ≥ be an odd prime. Then the normal subgroups of SL ( F p ) are { } , {± } , and SL ( F p ) itself.Proof. Suppose there are normal subgroups that differ from { } , {± } , and SL ( F p )itself. By the previous lemma, there is a normal subgroup N such that N has primeindex q in SL ( F p ). Since the order of SL ( F p ) is ( p − p ( p + 1) we have that q divides an element of the set { p − , p, p + 1 } . It cannot divide two elements of thisset since q ≥
5. So there are three cases based on which element is divisible by q .The next step is to partition SL ( F p ) as follows: • Let Γ = {± } . • Let Γ be all elements g of order m ≥ Z ( h g i ) of h g i has order p −
1. 74
Let Γ be all elements g of order m ≥ Z ( h g i ) of h g i has order 2 p . • Let Γ be all elements g of order m ≥ Z ( h g i ) of h g i has order p + 1.These sets are clearly disjoint. They form partition of SL ( F p ) since every elementoutside Γ has order at least 3 (Proposition 125), and Z ( h g i ) has order in theset { p − , p, p + 1 } (see Proposition 140 and Corollary 130).Next we further partition each Γ i . We start with Γ . Every element g ∈ Γ is in exactly one cyclic subgroup of order p −
1, namely Z ( h g i ) (it cannot be in asecond cyclic group of order p − can be partitioned intosets of the form C ∩ Γ where C is a cyclic subgroup of order p −
1. What is thesize of C ∩ Γ ? By Corollary 130, if C is a cyclic subgroup of order p − C equal to Z ( h g i ) for all g ∈ C of order m ≥ C not in C ∩ Γ are the elements of order 1or 2, namely the elements ±
1. So each partition C ∩ Γ has size ( p − − p − c p − = p ( p + 1).So Γ has c p − ( p −
3) = p ( p + 1)( p −
3) elements. A similar calculation can bemade for Γ and Γ .In fact, we can list the proportion of elements of SL ( F p ) in Γ , Γ , Γ as follows: • The proportion of elements of SL ( F p ) in Γ is p ( p + 1)( p − − p − p ( p + 1) = 12 p − p − < . • The proportion of elements of SL ( F p ) in Γ is( p + 1)(2 p − p − p ( p + 1) = 2 p . • The proportion of elements of SL ( F p ) in Γ is ( p − p ( p + 1 − p − p ( p + 1) = 12 p − p + 1 < . Claim: the complement of N is contained in Γ i for some i ∈ { , , } . Forinstance, suppose q divides p −
1. Then clearly Γ ⊆ N . Note that q cannot dividethe order of any g ∈ Γ since q does not divide 2 p . So each g ∈ Γ must have trivialimage under SL ( F p ) → SL ( F p ) /N . In other words, Γ ⊆ N . Similarly q cannotdivide the order of any g ∈ Γ since q does not divide p + 1. So each g ∈ Γ musthave trivial image under SL ( F p ) → SL ( F p ) /N . In other words, Γ ⊆ N . So thecomplement of N in SL ( F p ) must be contained in Γ . The other cases are similar.In fact, if N c is the complement SL ( F p ) − N then • If q divides p − N c ⊆ Γ . • If q divides p then N c ⊆ Γ . 75 If q divides p + 1 then N c ⊆ Γ .This leads to a contradiction when p ≥
5. Since N is a proper subgroupof SL ( F p ), the complement N c must contain at least one-half of the elementsof SL ( F p ). But the formulas for proportion of the elements of Γ , Γ , Γ in SL ( F p )shows that these proportions are each strictly smaller than one-half. So no suchnormal subgroup N of index q exists. Corollary 149.
Let p ≥ be an odd prime. Then SL ( F p ) is a non-solvableSylow-cycloidal group, and the quotient PSL ( F p ) def = SL ( F p ) / {± } is a simple group.Remark. The simple groups PSL ( F p ) were studied even by Galois, and were theearliest known non-Abelian simple groups outside of the alternating groups.The above takes care of the case p ≥
5. The case p = 3 is of interest as well: Proposition 150.
The group SL ( F ) is a solvable Sylow-cycloidal group isomor-phic to the binary tetrahedral group T .Proof. The order of SL ( F ) is 2 · · ( F ) is a Sylow-cycloidal group whose 2-Sylow subgroups are notcyclic. Thus the 2-Sylow subgroups of SL ( F ) must be isomorphic to the quater-nion group Q with 8 elements.According to Proposition 143, there are 3 cyclic subgroups of SL ( F ) of order 4which is exactly the number of such cyclic subgroups of Q . Thus there is a unique 2-Sylow subgroup of SL ( F ), and so it is normal. This explains why SL ( F ) issolvable. Proposition 143 also asserts that there are 4 cyclic subgroups of SL ( F )of order 3, and these subgroups are conjugate (Theorem 141). Thus SL ( F ) hasno normal subgroup of order 3. By Proposition 102, SL ( F ) is isomorphic to thebinary tetrahedral group 2 T .We have considered cyclic subgroups and normal subgroups of SL ( F p ). Nowwe will consider subgroups of odd order. We will do so using the following resulton normalizers of cyclic subgroups. Lemma 151.
Let C be a cyclic subgroup of SL ( F p ) where p is an odd prime.Then the normalizer N ( C ) of C is the normalizer N ( Z ( C )) of the maximal cyclicsubgroup Z ( C ) containing C . Furthermore, • If Z ( C ) has order p − then N ( Z ( C )) is a non-Abelian subgroup of SL ( F p ) of order p − . • If Z ( C ) has order p then N ( Z ( C )) is a non-Abelian subgroup of SL ( F p ) oforder ( p − p . • If Z ( C ) has order p + 1 then N ( Z ( C )) is a non-Abelian subgroup of SL ( F p ) of order p + 1)) . roof. Since Z ( C ) is cyclic, C is a characteristic subgroup of Z ( C ). Thus C isnormal in N ( Z ( C )) since Z ( C ) is normal in N ( Z ( C )). Hence N ( Z ( C )) ⊆ N ( C ) . We show equality by showing that both subgroups in this inclusion have the sameorder.We focus on the case where Z ( C ) has order p −
1. The other cases are simi-lar. By Theorem 141 the group SL ( F p ) acts transitively on the collection of cyclicsubgroups of order | C | . By Proposition 143 the orbit of C under this action hassize p ( p + 1). So by the orbit-stabilizer theorem, the stabilizer N ( C ) under con-jugation has the following size: | N ( C ) | = ( p − p ( p + 1) p ( p + 1) = 2( p − . This calculation is valid for N ( Z ( C )) as well since Z ( C ) is cyclic of size p + 1.So N ( Z ( C )) has order 2( p −
1) as well. Equality follows.
Corollary 152.
Suppose H is a noncyclic subgroup of SL ( F p ) of odd order,where p is an odd prime. Then H has order dividing ( p − p . Furthermore H con-tains a normal subgroup C of order p , and H is subgroup of the normalizer N ( C ) of C in SL ( F p ) .Proof. Since SL ( F p ) is a Sylow-cycloidal group, the subgroup H of odd order is aSylow-cyclic group. By Theorem 59, H has a nontrivial normal cyclic subgroup C ,and so H ⊆ N ( C ). If C has order dividing p − p − N ( C ) /Z ( C ) is even,and the image of H in N ( C ) /Z ( C ) is trivial. Thus H ⊆ Z ( C ) and so H is cyclic.So we are left with the case where C has order dividing 2 p . In other words, C hasorder p . By the above proposition N ( C ) has order ( p − p .Given the above corollary, the normalizer N ( C ) of a cyclic subgroup C of order p warrants our further attention. Lemma 153.
Let C p be a subgroup of SL ( F p ) of order p , where p is an odd prime.Then C p is conjugate to the subgroup (cid:26)(cid:18) b (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) b ∈ F p (cid:27) ∼ = F p . The normalizer N ( C p ) is conjugate to the subgroup (cid:26)(cid:18) a b a − (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) a ∈ F × p , b ∈ F p (cid:27) . Furthermore, N ( C p ) is a semidirect product C p ⋊ C p − where C p − is the cyclicgroup (cid:26)(cid:18) a a − (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) a ∈ F × p (cid:27) ∼ = F × p . Viewing C p − as F × p and C p as F p , the action of a ∈ C p − on C p is given by b a b. roof. Let α be a generator of C p . By Lemma 129, α has exactly one eigenvaluein F p . If we work in a basis v , v where v is an eigenvector of α , then α has theform (cid:18) b (cid:19) for some nonzero b ∈ F p . It follows that C p has the desired form (up to conjugation).From now on we assume C p is this particular subgroup of SL ( F p ).Next consider the subgroup H def = (cid:26)(cid:18) a b a − (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) a ∈ F × p , b ∈ F p (cid:27) . Observe that (cid:18) a b a − (cid:19) a is a surjective homomorphism H → F × p with kernel C p . So C p is a normal subgroupof H , and so H ⊆ N ( C p ). By the above lemma N ( C p ) has size ( p − p , which isalso the size of H . So H = N ( C p ). Note that C p − (as defined in the statementof the lemma) is a complement to C p in H = N ( C p ) and so N ( C p ) is a semidirectproduct C p ⋊ C p − . The last assertion follows from the calculation: (cid:18) a a − (cid:19) (cid:18) b (cid:19) (cid:18) a a − (cid:19) − = (cid:18) a a − (cid:19) (cid:18) b (cid:19) (cid:18) a − a (cid:19) = (cid:18) b (cid:19) Here is the next major result:
Theorem 154.
Let H be a noncyclic subgroup of SL ( F p ) of odd order, where p is an odd prime. Then H contains a cyclic subgroup of order p and H is containedin the normalizer N ( C ) of C in SL ( F p ) . Each such H is isomorphic to F p ⋊ T where T is a nontrivial subgroup of F × p of odd order. Here the action of a ∈ T isgiven by b a b .Furthermore, given such a semidirect product, F p ⋊ T there is a non-Abeliansubgroup of SL ( F p ) isomorphic to F p ⋊ T .Proof. The existence of C follows from Corollary 152. By the above lemma, anysuch H is isomorphic to a subgroup of F p ⋊ F × p where F p corresponds to C . Observethat H is actually isomorphism to F p ⋊ T where T is the image of H in F × p . Notethat T is nontrivial since H is noncyclic.Conversely, given a nontrivial odd order subgroup T of F × p , we have a subgroupof N ( C ) isomorphic F p ⋊ T where here C is any subgroup of SL ( F p ) of order p .(This follows from the above lemma).This leads to some important corollaries: Corollary 155.
Suppose p is an odd prime. Then SL ( F p ) contains a noncyclicsubgroup of order the product of two primes if and only if p is not a Fermat prime. roof. First suppose SL ( F p ) has a noncyclic subgroup H of order | H | = q q where q ≤ q are primes. Since SL ( F p ) is a Sylow-cycloidal group, any subgroupof prime squared order is cyclic. Thus q = q .First suppose that q = 2. By Theorem 59 we have that H is isomorphic to asemidirect product C q ⋊ C where C q and C are cyclic subgroups of order q = q and 2 respectively. The only nontrivial action of C is the one where the nontrivialelement of C acts on C q by x x − . Thus H is a dihedral group of order 2 q ,and so contains q elements of order 2. However, SL ( F p ) has a unique element oforder 2. Thus we can assume that q and q are odd.By the above theorem H is of order pr where r is an odd prime dividing p − p is not a Fermat prime.Conversely, suppose p is not a Fermat prime, and let r be an odd prime divid-ing p −
1. Let T be a cyclic subgroup of F × p of order r . Then SL ( F p ) contains anon-Abelian subgroup of order pr (isomorphic to a semidirect product F p ⋊ T ) bythe above theorem. Corollary 156.
Suppose p is an odd prime. If p is not a Fermat prime then SL ( F p ) is not freely representable.Proof. Use the previous corollary, Theorem 18, and the fact that every subgroupof a freely representable group must be freely representable.There are similar limits of subgroups of SL ( F p ) even of even order, but we willnot cover this here.
11 Classification of Non-Solvable Sylow-Cycloidal Groups
We take as given a substantial result of Suzuki from 1955 [14].
Theorem 157.
Suppose G is a non-solvable Sylow-cycloidal Group. Then G has asubgroup isomorphic to SL ( F p ) for some prime p ≥ . In fact, G has a subgroup H of index or such that H is isomorphic to SL ( F p ) × M where M is a Sylow-cyclicsubgroup of G of order prime to ( p − p ( p + 1) .Remark. Note that 2 T ∼ = SL ( F ) played an important role in the above classifica-tion of solvable Sylow-cycloidal groups, so it is interesting that SL ( F p ) with p ≥ Remark.
The group SL ( F ) can be shown to be isomorphic to the binary icosahe-dral subgroup of H × , so is freely representable. Suzuki’s theorem allows us to focus our attention to SL ( F p ): Proposition 158.
Suppose G is a non-solvable Sylow-cycloidal group. Let H , M ,and SL ( F p ) be as in Suzuki’s theorem. Then G has a unique element of order .Moreover, G is freely representable if and only if (1) M and SL ( F p ) are freelyrepresentable. I hope to include this in a sequel, or future version of this document. We will use previ-ous results on solvable Sylow-cycloidal groups. For the non-solvable case we will need Suzuki’stheorem. I hope to cover this in a sequel, or in a future edition of this document. roof. Since M is of odd order and SL ( F p ) has a unique subgroup of order 2(Proposition 125), the group H ∼ = M × SL ( F p ) has a unique element of order 2. Soif G = H then we are done. In general, H is normal in G , and H has a characteristicsubgroup C of order 2. Thus C must be normal in G . By Lemma 49, G has aunique element of order 2.Next suppose G is freely representable. Then M and SL ( F p ) must be freelyrepresentable since they are isomorphic to subgroups of G .Finally suppose that M and SL ( F p ) are freely representable. Then the prod-uct H ∼ = M × SL ( F p ) is freely representable by Corollary 21. Since G/H hasorder 1 or 2, any subgroup of odd prime order is in the kernel of G → G/H and sois a subgroup of H . Since G and H have a unique subgroup of order 2, we concludethat every subgroup of G of prime order is a subgroup of H . Since H is freelyrepresentable, the same is true of G by the technique of induced representations(Proposition 79). Definition 8.
A finite group G is perfect if its commutator subgroup G ′ is all of G .In other words, if G has no nontrivial Abelian quotient. Theorem 159. If p ≥ then SL ( F p ) is a perfect group, and is a Sylow-cycloidalgroup.Proof. The fact that G = SL ( F p ) is a perfect group follows from Theorem 148 andits corollary. In fact, by these earlier results, G ′ = { } or G ′ = {± } or G ′ = G .However, SL ( F p ) / {± } is of non-prime order and is simple, so cannot be Abelian.This means that G ′ cannot be {± } or { } . Thus G = G ′ .The group SL ( F p ) is Sylow-cycloidal by Theorem 131. Lemma 160.
The group SL ( F ) is freely representable.Proof. Recall that the binary icosahedral group 2 I is the preimage of the icosahedralgroup under the standard two-to-one map H → SO(3). Since 2 I is a subgroupof H , the group 2 I is freely representable. Finally 2 I , as mentioned above, 2 I isisomorphic to SL ( F ).Here is a theorem that Wolf attributes to Zassenhaus (1935). See Wolf [17],Section 6.2 for a proof. The proof is long: running to 14 pages, and I have notstudied it yet. Recently shorter proofs have appeared (which I also need to study):see Mazurov [10] and Allcock [1]. Theorem 161 (Zassenhaus) . Suppose G is a perfect freely representable group thatis not trivial. Then G is isomorphic to SL ( F ) .Remark. This gives another argument that a simple group is freely representable ifand only if it is cyclic.
Corollary 162. If p > then SL ( F p ) is a non-solvable Sylow-cycloidal group thatis not freely representable. If we combine Zassenhaus and Suzuki’s results we get the following:80 heorem 163.
Let G be a non-solvable group with a unique element of order 2.Then G is freely representable if and only G has a subgroup H of index or anda freely representable subgroup M of order prime to such that H is isomorphicto SL ( F ) × M .Proof. One direction is straightforward based on Suzuki and Zassenhaus’s results(because every freely representable group is Sylow-cycloidal, see Section 6, andevery subgroup of a freely representable group is freely representable).In the other direction we have that H is freely representable by Lemma 160 andCorollary 21. Every element of prime order of G must be in H : for odd primesjust note that all such elements must be in the kernel of G → G/H , and for theprime 2 use the uniqueness. Now use the technique of induced representations(Proposition 79)If we combine Proposition 123 and Proposition 158 we get the following:
Proposition 164.
Suppose G is a Sylow-cycloidal group that is not a Sylow-cyclicgroup. Then G has a unique element of order two.
12 Semiprime-Cyclic Groups
An important necessary condition for G to be freely representable is the following:if H is a subgroup of order pq where p and q are primes then H is cyclic. The classof such groups forms an interesting class of “cycloidal” groups, and is surprisinglyclose to being the same as the class of freely representable groups. There aredifferences only in the nonsolvable case. We call such groups “semiprime-cyclic”groups:
Definition 9.
A finite group G is said to be semiprime-cyclic if every sub-group H ⊆ G whose order is a semiprime (the product of two primes) is cyclic.We restate Corollary 19 using this new terminology: Proposition 165.
Every freely representable group is semiprime-cyclic.
Now we consider some basic properties of semiprime-cyclic groups:
Proposition 166.
The subgroups of a semiprime-cyclic group are semiprimecyclic.Proof.
This follows from the definition.
Proposition 167.
Suppose that A and B are finite groups of relatively prime order.Then A × B is semiprime-cyclic if and only if both A and B are. See Corollary 19 above. Wolf [17] calls this the pq-condition . Note that we allow p = q inthis condition. roof. One direction is clear since A and B can be identified with subgroups of theproduct A × B . Conversely, suppose that A and B are both semiprime-cyclic. Let D be a subgroup of A × B of order pq where p, q are prime. Suppose pq also dividesthe order of A . Then the image of D under A × B → B is trivial. Thus D ⊆ A ,and hence D is cyclic. Similarly if pq divides the order of B , then D is cyclic.So we can focus on the case where p divides the order of A and q divides theorder of B . Then the image of D under A × B → B must have order q and thekernel D ∩ A must have order p . Similarly, D ∩ B has order q . Thus both terms ofthe inclusion ( D ∩ A )( D ∩ B ) ⊆ D have order pq , and so this is an equality. Thismeans that D is isomorphic to ( D ∩ A ) × ( D ∩ B ), and D must must be cyclic sinceit is isomorphic to the product of cyclic groups of relatively prime order. Proposition 168.
Let G be a finite group with subgroup H . If H is semiprime-cyclic and if H contains every element of G of prime order, then G is alsosemiprime-cyclic.Proof. Let D be a subgroup of G of order pq where p and q are prime. Let a, b ∈ D where a has order p and b has order q . By assumption a, b ∈ H , so D is a subgroupof H . Thus D is cyclic. Proposition 169.
A finite Abelian group A is semiprime-cyclic if and only if it iscyclic.Proof. If A is cyclic then it is semiprime-cyclic since all subgroups of A are cyclic.If A is semiprime-cyclic then it follows that A is cyclic from the decomposition ofAbelian groups into cyclic subgroups of prime power order. (See the remarks afterCorollary 19 for a more elementary argument). Proposition 170.
Let G be a p -group for some prime p . Then G is semiprime-cyclic if and only if G is cyclic or (if p = 2 ) is isomorphic to a general quaterniongroup.Proof. A p -group G is semiprime-cyclic if and only if all subgroups of G of order p are cyclic. So the result follows from Theorem 25 and Corollary 37. Corollary 171. If G is semiprime-cyclic then G is a Sylow-cycloidal group Now we determine which Sylow-cyclic groups are semiprime-cyclic. We startwith a special case:
Lemma 172.
Let C p be cyclic of order p and let C q k be cyclic of order q k where p and q are distinct primes. If a semidirect product G = C q k ⋊ C p is semiprime-cyclicthen G is cyclic.Proof. If k = 1 the result holds by definition. So assume k ≥
2, and we proceed byinduction. Consider the homomorphism C q k → C q k defined by x x q ; let A beits image and let K be its kernel. Observe that A is cyclic of order q k − since agenerator of C q k maps to an element of order q k − . Thus K has order q .The action of C p on C q k restricts to an action on A since A is the only subgroupof C q k of order q k − . By induction we can assume A ⋊ C p is cyclic, so C p actstrivially on A . Let a ∈ A be a generator, and let S be the elements of C q k mapping82o a under x x q . Since K has q elements, S also has q elements. Note that C p acts on S with orbits of size p or 1. Since S has q elements, there is an orbit ofsize 1. In other words, C p fixes an element of S . But the elements of S generate C q k .So C p acts trivially on C q k . Thus C q k ⋊ C p is Abelian. So C q k ⋊ C p is cyclic sinceit is semiprime-cyclic and Abelian.Next we extend the special case a bit: Lemma 173.
Let A be a finite cyclic group, and let C p be cyclic of order p where p is a prime not dividing | A | . If a semidirect product G = A ⋊ C p is semiprime-cyclicthen G is cyclic.Proof. Suppose G = A ⋊ C p is semiprime-cyclic but not cyclic. Then it is non-Abelian (all Abelian semiprime-cyclic groups are cyclic), and so C p acts nontriviallyon A . The action of C p on A restricts to an action on any subgroup of A (since A has at most one subgroup of any given order). The groups C p cannot act triviallyon all Sylow subgroups of A since A is generated by such groups. So C p actsnontrivially on some Sylow subgroup P of A . However, by the previous lemma, thegroup P ⋊ C p is cyclic, a contradiction. Proposition 174.
Let G be a Sylow-cyclic group. Then G is semiprime-cyclic ifand only if G is freely representable.Proof. One implication has been established (Proposition 165), so we can assumethat G is semiprime-cyclic. Since G is Sylow-cyclic, we can write G as A ⋊ B where A and B are cyclic groups of relatively prime order. Consider the kernel K associated homomorphism B → Aut( A ). If C is a cyclic subgroup of B of primeorder, then A ⋊ C is cyclic by the above lemma. So C is contained in K . ByCorollary 82 we conclude that G is freely representable.We can strengthen the above to include Sylow-cyclic-quaternion groups: Theorem 175.
Let G be a solvable finite group. Then G is semiprime-cyclic ifand only if G is freely representable.Proof. One implication has been established (Proposition 165), so we can assumethat G is semiprime-cyclic. If G is Sylow-cyclic, we appeal to the previous propo-sition. If G is not Sylow-cyclic then, by Proposition 124, there is a Sylow-cyclicsubgroup M of G with the property that G is freely representable if and only if M is freely representable. Since M is semiprime-cyclic, M is freely representable bythe previous proposition. Thus G is freely representable.The above theorem is remarkable given that it does not hold for non-solvablegroups: if p > p = 17 then SL ( F p ) is a non-solvablesemiprime-cyclic group, but is not freely representable (Corollary 162 and Corol-lary 155).We have the following version of Suzuki’s theorem (Theorem 157):83 heorem 176. Suppose G is a finite group with a unique element of order .Then G is a non-solvable semiprime-cyclic group if and only if G has a subgroup H of index or such that H is isomorphic to SL ( F p ) × M where p is a Fermatprime and where M is a freely representable group of order prime to ( p − p ( p + 1) .Proof. Suppose G is a non-solvable semiprime-cyclic group. By Corollary 171, G is Sylow-cycloidal. So by Suzuki’s theorem (Theorem 157), G has a subgroup H ofindex 1 or 2 such that H is isomorphic to SL ( F p ) × M where p ≥ M is a Sylow-cyclic subgroup of order prime to ( p − p ( p + 1). Since SL ( F p )and M are isomorphic to subgroups of G they are both semiprime-cyclic. Thus p is a Fermat prime (Corollary 155). By Proposition 174, M is freely representable.Conversely, assume the existence of such an H . By Corollary 155, SL ( F p )is semiprime-cyclic. Also SL ( F p ) is non-solvable (Corollary 149) so the same istrue of G . Also M is semiprime-cyclic (Proposition 174). Thus the product H issemiprime-cyclic (Proposition 167).Suppose g is an element of G of prime order. If g has odd order, then its imageunder G → G/H is trivial so g ∈ H . If g has order 2, then g ∈ H simply because H has an element of order 2, and the element of order 2 in G is unique. Sinceevery element of prime order in H is in G then G must also be semiprime-cyclic(Proposition 168).Note that semiprime-cyclic groups have unique elements of order 2: Proposition 177. If G is a semiprime-cyclic group then G has a unique elementof order .Proof. If G is solvable, then it is freely-generated by Theorem 175, so G has aunique element of order 2. If G is non-solvable, then it is at least Sylow-cycloidal(Corollary 171), and so has a unique element of order 2 by Proposition 164. Remark.
The above proof seems like a very high-powered way to prove such a basicresult. Is there a direct proof?We can use the close connection between freely representability and thesemiprime-cyclic conditions to characterize which groups are not freely repre-sentable. Proposition 178.
Let G be a finite group. Then G is not freely representable ifand only if G has a subgroup B such that either (1) B is a non-cyclic group of orderthe product of two primes, or (2) B is is isomorphic to SL ( F p ) for some Fermatprime p ≥ .Proof. Suppose G is not freely representable. If G is solvable then G cannot besemiprime-cyclic (semiprime-cyclic implies freely representable for solvable groupsby Theorem 175), so G has a subgroup B satisfying condition (1). Next suppose G isnon-solvable and does not contain a subgroup satisfying (1). Then G is semiprime-cyclic and so has a subgroup H and a Fermat prime p describe by Theorem 176.Now p = 5 or else G would be freely representable (Theorem 163). Thus H , andhence G , has a subgroup B isomorphic to SL ( F p ) for some Fermat prime p ≥ This is the condition highlighted in [4]. I think of this as the “poison pill” proposition. G has such a subgroup B . In either case, such a B is notfreely representable, so G is not freely representable. Appendix: Sylow Theorems
We take these as given.
Theorem 179 (First Sylow Theorem) . For every prime power p k dividing theorder of G there is a subgroup of G of order p k . Theorem 180 (Second Sylow Theorem) . Let G be a finite group of order divisibleby a prime p . Every subgroup H of G that is a p -group is contained in a p -Sylowsubgroup P of G . Theorem 181 (Third Sylow Theorem) . Let G is a finite group of order mp k where m is not divisible by p , where p is a prime, and where k ≥ . All p -Sylowsubgroups of G are conjugate. The number t of p -Sylow subgroups of G divides m and t ≡ p ) . Appendix: Some facts about p -groups. Proposition 182.
Every nontrivial p -group has nontrivial center.Proof. We let G act on G by conjugation. Each orbit has size a power of p , andan element is in the center Z if and only if it is in an orbit of size 1. Since | G | isdivisible by p , and each orbit involving a g Z has size divisible by p , it followsthat | Z | is also divisible by p . Proposition 183.
Let G be a group of order p . Then G is either cyclic or isisomorphic to the product of two cyclic group of order p .Proof. Let Z be the center of G . By the previous proposition Z has order p or p .Suppose Z has order p , and let g ∈ G be an element outside of Z . We let G acton G by conjugation. The stabilizer of g is a group containing g and every elementof Z . Thus the stabilizer of g is all of G . This means that g is in the center, acontradiction.Thus Z has order p and so G = Z must be Abelian. If G is cyclic, we aredone, so assume A is any nontrivial cyclic subgroup. Let B be the cyclic subgroupgenerated by any b A . Observe that A ∩ B = { } . Thus the map A × B → AB is an isomorphism. The result follows from the fact that G = AB . Proposition 184.
Suppose G is a p -group for prime p and H is a subgroup ofindex p in G . Then H is a normal subgroup of G . roof. We let G act on the collection of subgroups by conjugation. The stabilizerof H , which is just the normalizer of H , must be G or H . Suppose the normalizeris H , so the orbit H of H has size p . Note that the normalizer of any g − Hg in H must necessarily be g − Hg .Now we restrict the action of G on H to an action of H on H . Clearly H ∈ H is fixed under this action. If g − Hg is not H , then the stabilizer of g − Hg is theintersection of H with g − Hg , which is a proper subgroup of H . So the orbitof such g − Hg under the action of H must have size greater than 1. The orbitof H has size 1 and the orbit of g − Hg has size at least p . However, both ofthese orbits (under H ) are contained in H , which has only p elements. This is acontradiction. Second proof (induction).
Suppose that H does not contain the center Z and ob-serve that H is normal in G = ZH . If H contains the center Z then by the inductionhypothesis H/Z is normal in
G/Z and so H is normal in G . Third proof.
This is actually a consequence of the following.
Proposition 185.
Every finite p -group G is solvable. In fact if H is a subgroupof G then there is a composition series { e } = G ( G ( G . . . ( G k = G such that H = G i for some i . (Here G j is a normal subgroup of G j +1 of index p .)Proof. Every p -group has a nontrivial center (Proposition 182), and so has a normalsubgroup of order p . This is enough to show that any p -group is solvable. Inparticular, if H is a subgroup of G then H has a composition series. Now we needto show that we can extend the series. In other words if we need establish theclaim that if G j is a proper subgroup of G then we can find a subgroup G j +1 suchthat G j is a normal subgroup of G j +1 of index p .We prove this claim by induction on k . There are two cases. If G j containsthe center Z of G then consider G j /Z inside G/Z and argue (by the inductionhypotheis) that there is a subgroup G j +1 containing Z such that G j /Z is normalof index p in G j +1 /Z . This implies G j +1 is as desired. If G j does not contain thecenter Z , then observe that G j is a proper normal subgroup of G j Z , so the desiredgroup G j +1 corresponds to any subgroup of order p in G j Z/G j . Appendix: Frobenius’s Theorem
Theorem 186 (Frobenius) . Suppose n divides the order of a finite group G . Thenthe number of elements in { x ∈ G | x n = 1 } is a multiple of n . Actually we will view Frobenius’s theorem as a special case of Theorem 189,which seems to be easier to prove than trying to prove Frobenius’s theorem directly(see [9] and Zassenhaus). We will use the following notation:86 efinition 10.
Let G be a group and let n be a positive integer. If a ∈ G then a /n is the set of x ∈ G whose n th power is n . We extend this notation as follows:if C is a subset of G then C /n def = { x ∈ G | x n ∈ C } Remark.
We will need the following easy identities: C = C ,( C ∪ C ) /n = C /n ∪ C /n , ( a /n ) /n = a n n , and g − a /n g = ( g − ag ) /n . We start with a few lemmas.
Lemma 187.
Let G be a finite group and let p be a prime dividing the order of G .If a ∈ G has order divisible by p then | a /p | is also divisible by p .Proof. If x ∈ a /p has order m then x p = a has order m/ gcd( p, m ) = n . Inparticular, p | m since p | n , so x has order m = pn . We partition a /p usingthe equivalence relation that defines x and y to be equivalent if and only if x and y generate the same cyclic subgroup of G . For any x ∈ a /p consider thehomomorphism h x i → h a i defined by x x p . The kernel has p -elements, and so itis a p -to-one map. The equivalence class of x is the elements mapping to a underthis map, so it has p elements. This shows that | a /p | is kp where k is the numberof such equivalence classes (and k is the number of cyclic subgroup of order pn containing a ).In the next lemma, recall that the centralizer Z a of a ∈ G is the subgroup ofelements of G that commute with a . Lemma 188.
Let G be a finite group. If p is a prime dividing the order of G andif a is in the center of G then p also divides | a /p | .Proof. Our proof will be by induction on the order of G , the case | G | = 1 beingtrivial. So suppose | G | > p be a prime dividing the order of G .Let A be the subgroup of the center of G consisting of elements of order primeto p . By the previous lemma, it is enough to show that p divides | a /p | for all a ∈ A .Let a , a ∈ A . Consider the map A → A defined by x x p . Its kernelis trivial, so it is a bijection. Thus there is a u ∈ A such that u p = a a − .Consider the map a /p → a /p defined by the rule x ux . Observe that it iswell-defined and injective, so is a bijection. Thus | a /p | is the same for all a ∈ A .In particular, | A /p | = | A || a /p | for any a ∈ A .Let B be the set of elements b of G whose centralizer Z b is a proper subgroupof G of order divisible by p . Observe that Z b contains b /p for any b ∈ B sinceif x p = b then x − bx = x − x p x = x p = b (in other words b /p relative to Z b isthe same as b /p relative to G ). Also observe that b is in the center of Z b . So bythe inductive hypothesis applied to Z b , the prime p divides | b /p | for each b ∈ B .Thus p divides | B /p | . 87et C be the set of elements of G not in A or B . In other words, C is theelements whose centralizer Z b has order prime to p .For each x ∈ G let C x be the set of conjugates of x in G . So G acts on C x transitively with stabilizer Z x . The orbit size | C x | is | G | / | Z x | . Also if x ∈ C then forany y ∈ C x we have Z y and Z x have the same order. Since g − x /p g = ( g − xg ) /p ,we have that | y /p | = | x /p | if y ∈ C x . In particular, | C /px | = | C x || x /p | . In particular, if x ∈ C then the orbit C x has size a multiple of p . For x ∈ C and y ∈ C x , we have that Z y has the same order as Z x , so Z y has order prime to p .In other words, if x ∈ C then C x ⊆ C . We conclude that | C /px | = | C x || x /p | isdivisible by p for each x ∈ C , and | C /x | is divisible by p since C is the disjointunion of such C x .Since | G | = | ( A ∪ B ∪ C ) /p | = | A /p | + | B /p | + | C /p | = | A || a /p | + | B /p | + | C /p | and since | G | , | B /p | , and | C /p | are divisible by p , we see that | A || a /p | is divisibleby p . However | A | is not divisible by p since A is an Abelian group whose elementshave their orders prime to p . So | a /p | is divisible by p as desired. Theorem 189.
Let G be a finite group and let n be a positive integer. If C is asubset of G closed under conjugation then | C /n | is a multiple of gcd( n | C | , | G | ) .In particular, if a ∈ G is in the center of G , and if n divides | G | , then | a /n | isa multiple of n .Proof. We can restate the result as asserting that | C /n | is a Z -linear combinationof n | C | and | G | .Observe that the result holds trivially for G = { } regardless of n . So we caninductively assume the result is true for all subgroups of G (regardless of n ). Notealso that this result holds in G itself for n = 1 so we can inductively assume theresult holds for all divisors of n (for our given G ).Suppose C = { a } and n = p is a prime dividing | G | . Then a must be in thecenter of G , and the result holds by the previous lemma. Also if C = { a } and n = p is a prime not dividing | G | , the result holds trivially.Next consider the case C = { a } where a ∈ G in the center of G and where n iscomposite. Write n as n n where n , n are proper divisors of n . Since n < n wehave by induction that | a /n | = u ( n ·
1) + v | G | for some u, v ∈ Z . Similarly, (cid:12)(cid:12)(cid:12)(cid:12)(cid:16) a /n (cid:17) /n (cid:12)(cid:12)(cid:12)(cid:12) = u ′ n ( un + v | G | ) + v ′ | G | for some u ′ , v ′ ∈ Z . In particular, (cid:12)(cid:12)(cid:12)(cid:12)(cid:16) a /n (cid:17) /n (cid:12)(cid:12)(cid:12)(cid:12) = ( uu ′ ) n + v ′′ | G | for some v ′′ ∈ Z . Thus the result holds for C = { a } and composite n .88e have established the result for C = { a } with a in the center. In general,by the identity ( C ∪ C ) /n = C /n ∪ C /n it is enough to prove the result forsets C which are in a single orbit under the action of G by conjugation. We canalso assume | C | >
1. For any a ∈ C and g ∈ G we have g − a /n g = ( g − ag ) /n .So a /n has the same size for all a ∈ C . Thus for any particular a ∈ C , | C /n | = | C || a /n | . Let Z a be the centralizer of a particular a ∈ C . In other words, Z a is the subgroupstabilizing a under the conjugation action. Note that | Z a || C | = | G | by the orbit-stabilizer principle. So Z a is a proper subgroup of G since | C | >
1. Also a /n is asubset of Z a since if x n = a then x − ax = x − x n x = x n = a , and a is in the centerof Z a . So by the induction hypothesis applied to Z a and { a }| a /n | = u ( n ·
1) + v | Z a | = un + v | G | / | C | . for some u, v ∈ Z . Thus | C /n | = | C || a /n | = | C | ( un + v | G | / | C | ) = un | C | + v | G | as desired. This establishes the result for general C closed under conjugation. Appendix: Group Representations
We assume that the reader is familiar with the construction of the group ring R [ G ]where R is a commutative ring with unity and where G is a group (see Dummitand Foote [7], Section 7.2). Observe that R [ G ] is a commutative ring if and onlyif G is an Abelian group.A linear representation of G on a vector space V over a field F is a homomor-phism Φ : G → GL( V ) from G to the group GL( V ) of invertible linear transfor-mations of V . If V is F n then GL( V ) can be identified with GL n ( F ), the groupof n -by- n invertible matrices with coefficients in F . If we fix an ordered basis of V of size n then we can identify V with F n , and so identify GL( V ) with GL n ( F ).If Φ : G → GL( V ) is a linear representation of G then, for g ∈ G and v ∈ V ,we often write gv for Φ( g ) v if there is no chance of confusion. This is the usualconvention for group actions in general.We can also think of a linear representation of G as a way to give an F -vectorspace an F [ G ]-module structure. In other words, each F [ G ]-module V supplies,canonically, a linear representation of G on the F -vector space V .Suppose V is a representation, thought of as an F [ G ]-module, then a subrep-resentation of V is a submodule. This can be thought of as a subspace W ⊆ V invariant under the action of G . We will use the term nontrivial for representa-tions where V has at least dimension one. A representation is irreducible if its onlysubrepresentations are itself and the zero subspace.In representation theory it is often convenient to use the field F = C . In thiscase all irreducible representations of finite Abelian groups A are one-dimensional,and the image of A under A → GL( V ) = C × is a cyclic group.89his report uses representation theory, but not in a deep way; certainly muchless representation theory than mentioned in Wolf [17]. Inducted representationsand the tensor product of representations are used. Such ideas are reviewed in thedocument as needed. Appendix: Orthogonal Transformations
Recall that R n has an inner product, and this inner product can be used to definedistances and angles. We write h x, y i for the inner product between vectors. Whenwe say “orthogonal” or “orthonormal” it will be with respect to this standard innerproduct. We write e , . . . , e n for the standard orthonormal basis of R n . Definition 11. An orthogonal matrix is a square matrix with real entries suchthat the columns are orthonormal with respect to the standard inner product. Proposition 190.
Let L be a linear transformation from R n → R n . Then thefollowing are equivalent.(i) L preserve the inner product. In other words, for all x, y ∈ R n h Lx, Ly i = h x, y i . (ii) L maps any orthonormal basis to an orthonormal basis.(iii) L maps some orthonormal basis to an orthonormal basis.(iv) The matrix representation of L with respect to any orthonormal basis of R n is an orthogonal matrix.(v) The matrix representation of L with respect to some orthonormal basis of R n is an orthogonal matrix. The above motivates the following definition:
Definition 12. An orthogonal transformation R n → R n is a linear map such that h Lx, Ly i = h x, y i for all x, y ∈ R n . Since distances and angles are defined in terms of this innerproduct, orthogonal transformations preserve angles and lengths.The set of orthogonal transformation of R n is seen to be a group. We call thisgroup the orthogonal group and write it as O( n ). If we fix an orthonormal basis, wewill also identify O( n ) with the group of n -by- n orthogonal matrices. The subgroupof matrices of determinant 1 will be written SO( n ). Proposition 191.
The group O ( n ) consists of all isometries of R n fixing the origin.Example . The rotation of R counter-clockwise by θ sends e to (cos θ, sin θ ),and e to (cos( θ + π/ , sin( θ + π/ (cid:18) cos θ − sin θ sin θ cos θ (cid:19) . Note this is orthogonal with determinant 1. Clearly composition of such rotationscorresponds to addition of angles (mod 2 π ).90 roposition 192. Let A be an n -by- n real matrix. The matrix A is orthogonal ifand only if A T A = I. Corollary 193.
So every orthogonal matrix A has inverse A T . Every orthogonalmatrix has determinant or − . Corollary 194.
The transpose of an orthogonal matrix is orthogonal matrix withthe same determinant. An n -by- n real matrix if orthogonal if and only if its rowsare orthonormal. The group
O(2)
Now we investigate O(2) in more detail. Every unit vector of R is of theform (cos θ, sin θ ). So any orthogonal transformation maps e to some such unitvector. Now e will have to map to an orthogonal vector(cos( θ ± π/ , sin( θ ± π/ ∓ sin θ, ± cos θ )(where we use the standard trig identities for angle addition). So the associatedmatrix has two possibilities (cid:18) cos θ − sin θ sin θ + cos θ (cid:19) or (cid:18) cos θ + sin θ sin θ − cos θ (cid:19) If the matrix is of determinant 1 then we have the rotation by θ we consideredabove: (cid:18) cos θ − sin θ sin θ cos θ (cid:19) . We conclude that SO(2) consists of rotation vectors. Note that SO(2) acts freelyon the set R − { (0 , } . We see that SO(2) is isomorphic to the circle group, whichcan be expressed in terms of the angle group R / π Z . Theorem 195.
The group
SO(2) is equal to the rotation group, and so is isomor-phic to the circle group R / π Z . This group acts freely on R − { (0 , } . Now matrices of O(2) outside of SO(2) have form (cid:18) cos θ sin θ sin θ − cos θ (cid:19) The trace is 0 and the norm is 1. This means that the characteristic polynomialmust be X − trace( M ) X + det M = X − . So it has eigenvalues 1 and −
1. This means there is a basis of eigenvectors (recallthat every real eigenvalue of a real matrix has a real eigenvector); such a basis isorthogonal:
Proposition 196.
All real eigenvalues of an orthogonal matrix are or − . If x is an eigenvector of eigenvalue and y is an eigenvector of eigenvalue − , then x, y are orthogonal. roof. Suppose
M x = λx where M is an orthogonal matrix and where x = 0. Then h x, x i = h M x, M x i = h λx, λx i = λ h x, x i . Thus λ = 1. So λ = ± M x = x and M y = − y . Then h x, y i = h M x, M y i = h x, − y i = − h x, y i . So 2 h x, y i = 0, and so h x, y i = 0.So if you have a matrix of O(2) outside of SO(2), then it has characteristicpolynomial X −
1. So has a fixed unit vector x , and an orthogonal eigenvector y mapping y to − y . If fixes the span of x , and send an orthogonal vector to itsinverse, so represents a reflection. (And does not act freely on R − { (0 , } ). The group
O(3)
We start with the following:
Lemma 197. If M ∈ O(3) then M has at least one eigenvalue in the set {± } .Proof. The characteristic polynomial is cubic in R [ X ] and so has a root.Let L be an orthogonal transformation of R . If we choose an orthonormalvector whose first term is an eigenvector then we have representation ± ∗ ∗ ∗ ∗ ∗ ∗ Since this must be an orthogonal matrix, it really has form ± ∗ ∗ ∗ ∗ And the 2-by-2 lower right submatrix is orthogonal.
Lemma 198. If M ∈ SO(3) then M has eigenvalue +1 .Proof. From the previous lemma, M has an eigenvalue of 1 or −
1. Suppose it haseigenvalue −
1, and choose an orthonormal basis whose first term is an eigenvectorwith eigenvalue −
1. In terms of this basis, the orthogonal transformation becomes − ∗ ∗ ∗ ∗ . The submatrix must be in O(2) and must have determinant −
1. This means ithas eigenvalue 1, which gives us an eigenvector of the form (0 , ∗ , ∗ ) in R witheigenvalue 1. 92 emma 199. If M ∈ O(3) has determinant − then M has eigenvalue − .Proof. Apply the above lemma to − M . Proposition 200. If M ∈ SO(3) there is an orthonormal basis for which thematrix M has the form θ − sin θ θ cos θ Proof.
Choose the first term of the basis to have eigenvalue 1. So we get ∗ ∗ ∗ ∗ . The lower right submatrix has determinant 1 and is in SO(2), so it has the desiredform.
Remark.
This shows that if M ∈ SO(3) is not the identity then it is a rotation ofspace. It fixes a one-dimension subspace called the “axis of rotation”.
Proposition 201. If M ∈ O(3) has determinant − has there is an orthonormalbasis for which the matrix M has the form − θ − sin θ θ cos θ = − θ − sin θ θ cos θ Proof.
The proof is similar to that of the previous proposition.
Remark.
This shows that any such transformation is a rotation followed by a re-flection across the plane perpendicular to the axis. (Or is just a reflection if θ = 0). Theorem 202.
The group
SO(3) is the group of rotation matrices for R (and theidentity matrix). Finally, we will need the following:
Lemma 203.
Suppose A ∈ O(3) . If det A = 1 then the fixed space of A is dimen-sion 1 (for a rotation) or dimension 3 (for the identity). If det A = − then thefixed space of A has dimension 2 (for a reflection of planes) or dimension 0. Inparticular A is a rotation if and only if the fixed space is dimension 1.Proof. This is fairly clear from the geometric description of the associated operators.We will discuss the case of det A = − x , x , x such that Ax = − Ax , and that A fixes the span of x and x , and is a rotation of angle θ on that subspace. Soif y = c x + c x + c x is fixed then c = − c , so c = 0. Thus y is in the spanof x and x . Thus y is fixed only if either θ is a multiple of 2 π , or if y = 0. If θ isa multiple of 2 π then A is a refection with fixed space spanned by x and x . If θ is not a multiple of 2 π then the fixed space is the zero space.93 emark. Although we do not much about the topology of O(3), we note rotationsare parameterized by three parameters: a point on the sphere for the axis and anangle of rotation. In fact O(3) is a 3-dimensional Lie group. The subgroup SO(3)and its other coset in O(3) are the connected components of O(3).
Appendix: Finite subgroups of
SO(3)
The collection of rotations of R fixing the origin can be identified with SO(3),the group of orthogonal 3-by-3 matrices with determinant +1. Subgroups of SO(3)include cyclic and dihedral groups, but the other finite subgroups are related in aninteresting way to the Platonic solids (tetrahedron, cube, octahedron, dodecahe-dron, icosohedron). In fact, the finite subgroups of SO(3) are as follows: • Cyclic subgroups of all orders. We use the denotation C n for any such groupof order n • Dihedral subgroups of all even orders ≥
4. Any dihedral subgroup of or-der 2 n > n -gon centered at theorigin. We also consider here the rotational symmetry group of a nonsquarerectangle centered at the origin. It has order 4, and is isomormorphic to theKlein 4-group. We use the denotation D n for any such group of order 2 n . • Tetrahedral subgroups of order 12. Each of these is the subgroup of rotationalsymmetries of a regular tetrahedron centered at the origin. This group isisomorphic to the alternating group A . This can be seen by looking at theaction on the four vertices. We use the denotation T for any such group. • Octahedral subgroups of order 24. Each of these is the subgroup of rotationalsymmetries of a regular octahedron centered at the origin. Each such groupis also the subgroup of rotational symmetries of a cube centered at the originThis group is isomorphic to the alternating group S . This can be seen bylooking at the action on the set of four pairs of opposite faces. We use thedenotation O for any such group. • Icosahedral subgroups of order 60. Each of these is the subgroup of rotationalsymmetries of a regular icosahedron centered at the origin. Each such group isalso the subgroup of rotational symmetries of a regular dodecahedron centeredat the origin. This group is isomorphic to the alternating group A . We usethe denotation I for any such group.Another interesting fact is that two finite subgroups of SO(3) are isomorphic asabstract groups if and only if they are conjugate subgroups of SO(3).See Artin [3] (Sections 4.5 and 5.9), Goodman [8] (Chapter 4 and Section 11.3),and Sternberg [13] (Section 1.8) for proofs of these facts. Appendix: The Quaternion Division Ring H We assume some familiarity with the ring of quaternions H (see Chapter 7 of Dum-mit and Foote [7] and Exercise 1 on page 306 of Artin [3]). Here we highlight the94onnection between H × and SO(3) which plays an important role in this document.The ring H is a non-commutative ring, and it is also a four-dimensional R -vectorspace. In other words it is an algebra over R . Every non-zero element is invertible,so it is a division ring, in fact a division algebra over R . If α = a + b i + c j + d k isin H then we define its conjugate α to be a − b i − c j − d k . Observe that conjugationsatisfies the law αβ = β α (use R -linearity to reduce to the case where α, β , andhence αβ , are are in {± , ± i , ± j , ± k } ).We consider H to contain R as a subring (we can even think of C as a subring asthe span of 1 , i ). Note that R commutes with any element of H . For α ∈ H , observethat α ∈ R if and only if α = α . Let H be the R -span of the quaternions i , j , k . Wecan and will identify H with R . An element α ∈ H is in H if and only if α = − α .Every element of H can be written uniquely as c + v where c ∈ R and v ∈ H ; wecall c the real part of α = c + v , and v the imaginary part of α .Observe that if u, v ∈ R , then the standard inner product is related to theproduct uv or uv in H . In fact, h u, v i R = − Real ( uv ) = Real ( uv ) . (Check using R -linearity to reduce to the case where u, v ∈ { i , j , k } ). This relationis even more direct if u = v ; observe that if u ∈ H then u = u = ( − u ) = u , so u is real. This means we can drop the symbol Real and write | u | R = h u, u i R = − u = uu. We can extend the definition of | α | to all α ∈ H by defining | α | = αα. (Warning, this does not equal to − α in general). Observe that if we write α as c + v in terms of its real and imaginary parts, then | α | = ( c + v )( c + v ) = ( c + v )( c − v ) = c − v = c + | v | ≥ . Note that this is the value of the standard norm when we think of H as R .Observe that this norm gives a homomorphism H × → R × + defined by α αα where R × + is the multiplicative group of positive real numbers. So the elements H of norm one is forms a subgroup of H × . Note that geometrically, H is the 3-spherein R . (As a group it is a compact Lie group isomorphic to SU .)When we define H we seem to put special emphasis on i , j , k . However, thefollowing shows we can really have chosen other orthonormal vectors to play therole of i , j , k . More precisely, all unit length vectors in H behave like i , j , k inhaving a square of −
1, and any orthonormal pair behaves like any two of i , j , k : Proposition 204. If x ∈ H has unit length then x = − . If x, y ∈ H areorthonormal then xy = − yx .Now suppose x, y ∈ H are orthonormal and define z to be xy . Then x, y, z forman orthonormal basis of H which satisfy the laws xy = z, yz = x, zx = y. roof. Above we observed that if x ∈ H then − x is the norm of x . So, due tothe assumption of unit length, x must be − x, y are unit length and orthogonal. Then x = y = − z = xy is zero since h x, y i = 0. So z is in H . Finally z haslength 1 · z = −
1. Next we show that z is orthogonalto x and y . Observe that xz = xxy = − y , which has real part zero, so h x, z i = 0.Similarly, zy = xyy = − x has real part zero and so h z, y i = 0. So x, y, z are areorthonormal, and they must be a basis since H has dimension three over R .Next we verify that xy = − yx . Since x, y, and z = xy are in H , yx = ( − y )( − x ) = y x = xy = z = − z = − xy. This result was based only on the assumption that u, v are orthonormal, so it appliesto y, z or z, x as well: zy = − yz and xz = − zx .We have xy = z by definition. Above we noted that zy = − x so yz = x . Wealso noted that xz = − y so zx = y .There is an action of H × on H by conjugation. More specifically, if Φ( h ) isdefined as the map α hαh − , then Φ( h ) is an R -linear map H → H (in fact, it isan algebra automorphism). What is interesting about this action is that if | h | = 1then Φ( h ) is an R -linear automorphism of H : Lemma 205.
Let h ∈ H × and let Φ( h ) be as above. Then1. If c ∈ R then Φ( h ) c = c for any h ∈ H × .2. If h ∈ H and α ∈ H then Φ( h ) α = Φ( h ) α.
3. If h ∈ H and v ∈ H then Φ( h ) v ∈ H .4. If h ∈ H and α ∈ H then the real part of Φ( h ) α is equal to the real part of α .Proof. The statement (1) is clear. For (2) observe that Φ( h ) α = hαh . SoΦ( h ) α = hαh = h α h = h α h = Φ( h ) α. For (3), recall that a vector v is in H if and only if v = − v . Observe thatΦ( h ) v = Φ( h ) v = Φ( h )( − v ) = − Φ( h ) v so Φ( h ) v ∈ H .For (4), write α = c + v . ThenΦ( h ) α = Φ( h ) c + Φ( h ) v = c + Φ( h ) v. Since Φ( h ) v ∈ H the result follows.In particular, H acts on H by conjugation. This action preserves the inner-product: 96 emma 206. Suppose Φ is as above. If h ∈ H and if x, y ∈ H then h Φ( h ) x, Φ( h ) y i = h x, y i . Proof.
Observe that − Real ((Φ( h ) x ) (Φ( h ) y )) = − Real (Φ( h )( xy )) = − Real ( xy ) . In particular, if h ∈ H then Φ( h ) acts on H as an orthogonal transformation.For convenience we identify H with R using the standard orthonormal basis i , j , k . Proposition 207.
The map h Φ( h ) gives a homomorphism Φ : H → O(3) . We will now identify the kernel and the image of this homomorphism. For thekernel we need the following:
Proposition 208.
The elements of H commuting with a nonzero x ∈ H are theelements in the R -span of , x . The elements of H commuting with all of H is R .In particular, the center of the ring H is R .Proof. Suppose x ∈ H . Normalize so x has unit length. Let y ∈ H be suchthat x, y are orthonormal. Let z = xy . As above, x, y, z is an orthonormal basisof H . Observe that for a, b, c, d ∈ R x ( a + bx + cy + dz ) = ax − b + cz − dy, ( a + bx + cy + dz ) x = ax − b − cz + dy. The first result follows by comparing coefficients. In particular, R is the set ofelements commuting with both i and j . The remaining claims are now clear. Proposition 209.
The kernel of
Φ : H → SO ( R ) is {± } .Proof. It is clear from the definition of Φ that 1 and − h is inthe kernel, then hu = uh for all u ∈ H . So h ∈ R from the previous result. Since h has unit length, h = ± Proposition 210.
The group H × , and hence H , has − as its a unique elementof order .Proof. Suppose that x = 1 with x ∈ H . Then ( x − x + 1) = 0. Since H is adivision ring, x − x + 1 = 0. So x = ±
1. The result follows.
Remark.
As we saw earlier, x = 1 for all x ∈ H ∩ H . Note that H ∩ H is theunit sphere in H ∼ = R . So x − H .Finally we present an argument that the image of our homomorphism is thesubgroup SO(3). We use the following (see the earlier appendix): Proposition 211.
Suppose that A ∈ O(3) is not the identity matrix. Then thefollowing are equivalent: A is a rotation of R . • A fixes exactly a one-dimensional subspace of R (called the axis of rotation). • det A = +1 , so A ∈ SO(3) . Lemma 212.
Let v ∈ H be of unit length. Let h = r + sv where r, s ∈ R where r + s = 1 and such that − < r < so that h ∈ H . Then the orthogonaltransformation Φ( h ) fixes v . Furthermore if u is a unit vector orthogonal to v thenthe cosine of the angle formed by u and Φ( h ) u is r − . In particular, Φ( h ) is arotation with axis of rotation spanned by v .Proof. It is clear that h has unit length since r + s = 1. By Proposition 208, thegiven vector v commutes with h = r + sv . SoΦ( h )( v ) = h − vh = h − hv = v. Let u be orthogonal to v and of unit length, and let w = uv . ThenΦ( h ) u = ( r + sv ) u ( r − sv ) = r u − rsuv + rsvu − s vuv = r u − rsw − s u. Thus(Φ( h ) u ) u = ( r u − rsw − s u ) u = − r − rsv + s = ( s − r ) + 2 rsv. So h Φ( h ) u, u i = − Real ((Φ( h ) u ) u ) = r − s = r − (1 − r ) = 2 r − . Since u , and hence Φ( h ) u , have unit length, the above inner product is just thecosine of the angle formed by u and Φ( h ) u .Note this cosine is not 1 since r <
1. Thus u is not fixed by Φ( h ). A generalvector of H can be written as av + bu for some u of unit length orthogonal to v .The image of av + bu under Φ( h ) is av + bu ′ where u ′ = u . Thus if b = 0, the vector av + bu is not fixed by Φ( h ). This means that the space of vectors fixed by Φ( h ) isone-dimensional. By the above proposition, Φ( h ) must be a rotation.Now we are ready for the identification of the image. Proposition 213.
The image of
Φ : H → O(3) is SO(3) .Proof.
Every element not equal to ± H can be written as r + sv for some unitvector v ∈ H and some r, s ∈ R with r + s = 1 and − < r <
1. By the aboveresult, its image is a rotation, and so has determinant 1.Conversely, suppose A is in SO(3). Then we need to find an element h of H with Φ( h ) = A . If A is the identity, then h = ± A is a rotation. Let v be a unit vector in H in the axis of rotation of A , and let θ be the angle of rotation around this axis. Choose r so that cos θ = 2 r −
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