Lifting in Frattini Covers and A Characterization of Finite Solvable Groups
aa r X i v : . [ m a t h . G R ] D ec LIFTING IN FRATTINI COVERS AND A CHARACTERIZATION OFFINITE SOLVABLE GROUPS
ROBERT M. GURALNICK AND PHAM HUU TIEP
Abstract.
We prove a lifting theorem for odd Frattini covers of finite groups. Usingthis, we characterize finite solvable groups as those finite groups which do not containnontrivial elements x i , i = 1 , ,
3, with x x x = 1 and x i a p i -element for distinct primes p i . Introduction
Let G be a finite group. Suppose that X is a finite group such that X/F = G . We saythat X is a Frattini cover of G if F is contained in the Frattini subgroup Φ( X ) of X .Frattini covers have been studied considerably with respect to coverings of curves (andinfinite towers of coverings of curves).Fried [F1, F2] introduced the modular tower problem and has made several interestingconjectures regarding them which generalize the fact that for an elliptic curve definedover a number field, the torsion subgroup defined over the number field is bounded. See[F2, BF, FK, D] for much more about this problem and its motivation and interestingexamples. Theorems 1.1 and 2.5 have some interesting consequences for Hurwitz spaces.In particular, it implies the existence of certain Hurwitz spaces for certain Frattini coversrelated to the modular tower program.Our first result gives a lifting critetion in Frattini covers. Theorem 1.1.
Let p be an odd prime. Let X be a Frattini cover of G = X/F with F a p -group. Assume that p does not divide the order of the Schur multiplier of G . Let g , . . . , g r ∈ G satisfy (i) G = h g , . . . , g r i , (ii) g · · · g r = 1 , and (iii) the order of each g i is coprime to p .Then, for any f ∈ F , there exist x i ∈ X , with x i F = g i , | x i | = | g i | , such that x · · · x r = f . In particular, we can take f = 1. This shows that the lifting invariant defined byFried vanishes in this setting. Furthermore, by a standard compactness argument, wecan take X to be the p -universal Frattini cover (see [F2]). More generally, if we drop the Mathematics Subject Classification.
Primary 20D10, 20D06, 20D05 ; Secondary 14G32, 14H30.
Key words and phrases.
Characterizations of finite solvable groups, lifting in Frattini covers.The authors gratefully acknowledge the support of the NSF (grants DMS-1001962 and DMS-0901241).They are also grateful to M. Fried for helpful comments on the topic of the paper. assumption that p does not divide the order of the Schur multiplier, we see that the onlyobstruction to lifting is lifting to the maximal central Frattini cover (and this is a trueobstruction – see the examples in Section 2 and Section 5). See Theorem 2.5 for a moregeneral result which does not assume the condition on the Schur multiplier.Of course, we can replace p by any set π of odd primes and let F be a (necessarilynilpotent) π -group (see Theorem 2.5). We also construct examples to show that Theorem1.1 fails if p = 2, see Proposition 6.1.Using this result and the Thompson classification of the finite simple groups in whichevery proper subgroup is solvable [T], we obtain the following characterization of finitesolvable groups (Barry [B] asked whether this was true). Theorem 1.2.
Let G be a finite group. Then G is solvable if and only if x x x = 1 forall nontrivial p i -elements x i of G for distinct primes p i , i = 1 , , . The forward implication is trivial. The reverse implication crucially depends on resultsof Thompson [T], and Proposition 2.1 which essentially follows from a result of Isaacs [I1].Thompson proved this result if one considers all triples of nontrivial elements of corpimeorder using his result on simple groups with all proper subgroups solvable. Kaplan andLevy [KaL] (see also [GL]) proved a variant of the previous result – in their result, x is a2-element, x is a p -element for some odd prime p and x is any nontrivial element whoseorder is coprime to 2 p . We actually prove a somewhat stronger result by showing that inTheorem 1.2 it suffices to assume that p = 2 and p ∈ { , } (see Theorem 3.4). An evenstronger result, Theorem 1.4 characterizing p -solvable groups, is also obtained (but usingthe full classification of finite simple groups).There have been many characterizations of finite solvable groups. We mention a few:(1) Every 2-generated subgroup is solvable [T] (see also [Fl]);(2) Every pair of conjugate elements generate a solvable group [G1];(3) The proportion of pairs of elements which generate a solvable group is greaterthan 11 /
30 [GW];(4) If x, y ∈ G , then x, y g generate a solvable group for some g ∈ G [DG].See [DG, GL, G2] for more characterizations and other references.We obtain another characterization of solvable groups by combining our Theorem 3.4with the proof of [B, Theorem 2]: Corollary 1.3.
Let G be a finite group. The following are equivalent: (i) G is solvable; (ii) For all distinct primes p i and for all Sylow p i -subgroups P i of G with ≤ i ≤ , | P P P | = | P || P || P | . (iii) For all distinct primes p i and for all Sylow p i -subgroups P i of G , where ≤ i ≤ , p = 2 , and p ∈ { , } , | P P P | = | P || P || P | . Note that Theorem 1.2 (and Theorem 3.4) depends on Thompson’s results but not onthe full classification of finite simple groups. In fact, the only results in this paper whichdepend on the full classification are the ones in § IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 3
In the next section, we prove Theorem 1.1 and the more general Theorem 2.5. In section3 we prove Theorem 3.4 (which includes Theorem 1.2). In section 4, we elaborate on thefact that one cannot just take elements of prime order in Theorem 1.2. In section 5, usingthe full classification of finite simple groups, we characterize the p -solvable finite groups(for p ≥ Theorem 1.4.
Let p be an odd prime and G be a finite group. Then G is p -solvable ifand only if G does not admit a triple of nontrivial elements x, y, z with xyz = 1 , x a -element, y a p -element, and z a q -element for any odd prime q = p . In section 6, we give examples to show that Theorem 1.1 fails for p = 2. In the finalsection, using Proposition 2.1, we give a short proof of a theorem of Feit and Tits [FT]about the minimal dimension of a representation of a group which has a section isomorphicto a given simple group.We use the notation of [Atlas] for various finite simple groups (in particular, L n ( q ),U n ( q ), S n ( q ), and O ± n ( q ) stands for PSL n ( q ), PSU n ( q ), PSp n ( q ), and P Ω ± n ( q ), respec-tively). 2. Lifting
The following statement is a key ingredient in our further considerations. It is essentiallya consequence of a result of Isaacs [I1, Theorem 9.1], and it is probably also known toDade. For the sake of completeness, we give an independent proof of it.
Proposition 2.1.
Let G be a finite group and N a non-abelian normal p -subgroup of G for a prime p . Assume that N is minimal among noncentral normal subgroups of G . (i) If ϕ ∈ Irr( Z ( N )) is nontrivial on [ N, N ] , then ϕ is fully ramified with respect to N/Z ( N ) , i.e. ϕ N = eθ for some θ ∈ Irr( N ) and e = | N/Z ( N ) | . (ii) Assume p > . Then G/Z ( N ) splits over N/Z ( N ) . In particular, N is not containedin Φ( G ) .Proof.
1) Observe that since N is non-abelian, N is noncentral in G . Since 1 < Z ( N ) < N ,the minimality of N implies that Z ( N ) ≤ Z ( G ), and so Z ( N ) = Z ( G ) ∩ N . Similarly,[ N, N ] ≤ Z ( G ), whence [ N, N ] ≤ Z ( N ) and N is nilpotent of class 2. It follows thatfor all x, y, z ∈ N , [ x, z ][ y, z ] = [ xy, z ]. In particular, for all x ∈ N , the set [ x, N ] := { [ x, n ] | n ∈ N } is a subgroup of [ N, N ]. Next we claim that [ x, N ] = [
N, N ] for all x ∈ N \ Z ( N ). Assume the contrary: X := [ x, N ] < [ N, N ]. Since N has class 2, Y := { y ∈ N | [ y, N ] ≤ X } is a subgroup of N containing x and Z ( N ). Next, X ≤ Z ( G )is normal in G , hence Y ⊳ G . If Y = N , then [ N, N ] ≤ X , a contradiction. Therefore Y < N , and so Y ≤ Z ( G ) ∩ N = Z ( N ) by minimality of N , again a contradiction.2) Now we prove (i). By the assumption, ϕ ( z ) = 1 for some z ∈ [ N, N ]. Let θ ∈ Irr( N )be any irreducible constituent of ϕ N . We claim that θ ( x ) = 0 for all x ∈ N \ Z ( N ).(Indeed, let Θ be a complex representation affording the character θ . By the previousparagraph, there is some n ∈ N such that z = x − n − xn . Since Θ( z ) = ϕ ( z ) I , it followsthat Θ( n ) − Θ( x )Θ( n ) = ϕ ( z )Θ( x ). Taking traces, we see that θ ( x )( ϕ ( z ) −
1) = 0 and so
ROBERT M. GURALNICK AND PHAM HUU TIEP θ ( x ) = 0 by the choice of z .) It is well known that in this case ϕ is fully ramified withrespect to N/Z ( N ) (cf. [I1, Lemma 2.6]).3) From now on, we assume that p >
2. Let M ≥ Z ( N ) be chosen such that M/Z ( N ) =Ω ( N/Z ( N )). Then Z ( G ) ∩ N = Z ( N ) < M ≤ N and M ⊳ G . The minimality of N again implies that M = N , i.e. N/Z ( N ) is elementary abelian. Now for any x, y ∈ N wehave x p ∈ Z ( N ) and so [ x, y ] p = [ x p , y ] = 1. Since [ N, N ] is abelian, it follows that [
N, N ]is also elementary abelian. As p >
2, this also implies that ( xy ) p = x p y p for all x, y ∈ N .4) Now we pick any nontrivial λ ∈ Irr([
N, N ]). Since Z ( N ) ≥ [ N, N ] is abelian, λ extends to ϕ ∈ Irr( Z ( N )), which is G -invariant because Z ( N ) ≤ Z ( G ). Let K := Ker( ϕ )and let P := N/K ; in particular, K ≤ Z ( G ). We will now show that P ∼ = p n + , anextraspecial p -group of exponent p of order p n for some n ≥ K ∩ [ N, N ] = Ker( λ ) has index p in [ N, N ] as [
N, N ] is elementary abelian.Hence [
P, P ] = K [ N, N ] /K is cyclic of order p ; in particular, P is non-abelian. Choose N ≥ Z ( N ) such that N /K = Z ( N/K ). Then N < N and N ⊳ G . The minimality of N implies that N = Z ( N ). Also, ϕ is a faithful linear character of Z ( N ) /K = N /K .We have shown that Z ( P ) = Z ( N ) /K is cyclic.Next we claim that P contains noncentral elements of order p . Indeed, fix x ∈ P such that | x | = exp( P ) = p s . Since P is non-abelian, we can find y / ∈ h x, Z ( P ) i . Then | y | = p t with 1 ≤ t ≤ s by the choice of x . Also, since P/Z ( P ) ∼ = N/Z ( N ) is elementaryabelian, x p , y p ∈ Z ( P ). Now x p , respectively y p , is an element in the cyclic group Z ( P ),of order p s − , respectively p t − , and t ≤ s . It follows that there is some integer k suchthat y p = x kp . As shown in 3), we now have ( x − k y ) p = x − kp y p = 1 and x − k y / ∈ Z ( P ), i.e. x − k y is a noncentral element of order p in P , as desired.Let N := { x ∈ N | x p ∈ K } . If x, y ∈ N , then by 3) we have ( xy ) p = x p y p ∈ K and so xy ∈ N . Hence N ⊳ G . By the previous claim, N is not contained in Z ( N ) = Z ( G ) ∩ N .The minimality of N now implies that N = N . We have shown that exp( P ) = p .Since Z ( P ) is cyclic, we also have Z ( P ) ∼ = C p ∼ = [ P, P ], and so Z ( P ) = [ P, P ]. But
P/Z ( P ) ∼ = N/Z ( N ) is elementary abelian, hence Φ( P ) = Z ( P ). Thus P is an extraspecial p -group of exponent p , as stated.5) It is well known that Aut ( P ), the group of all automorphisms of P which acttrivially on Z ( P ), is a semidirect product IS , where I = Inn( P ) ∼ = P/Z ( P ) is of order p n and S ∼ = Sp n ( p ). Now set C := C G ( N/K ) so that H := G/C embeds in Aut ( P ). Since[ N, N ] K , we have Z ( N ) ≤ N ∩ C < N , and so N ∩ C = Z ( N ) by the minimality of N .Thus N C/C ∼ = N/Z ( N ) ∼ = P/Z ( P ) ∼ = I . Now we can certainly write H as a semidirectproduct of N C/C and H ∩ S . Let U ≥ C be such that U/C = H ∩ S ; in particular U ∩ N C = C since S ∩ I = 1. Then G = ( N C ) U = U N , and U ∩ N = ( U ∩ N C ) ∩ N = C ∩ N = Z ( N ), i.e. G/Z ( N ) splits over N/Z ( N ).Suppose now that N ≤ Φ( G ). Then G = Φ( G ) U implies that G = U , contradictingthe fact that U ∩ N = Z ( N ). (cid:3) IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 5
If we consider groups of even order, the previous result fails. For example, there isa non-split extension of an extraspecial 2-group of order 2 a by an orthogonal groupO a (2) (if a ≥ Lemma 2.2.
Let p be a prime and let G be a finite group generated by elements g , . . . , g r .Assume that either G = O p ( G ) , or all g i are p ′ -elements. Let N be an abelian p -subgroupof G that is a minimal noncentral normal subgroup of G . Let C i := [ g i , N ] . Then ( g C )( g C ) · · · ( g r C r ) = ( g . . . g r ) N . Furthermore, every element of g i C i is N -conjugateto g i and is contained in the coset g i N .Proof. Note that, since N is normal and abelian, C i = { [ g i , n ] | n ∈ N } and it is asubgroup of N . Clearly, any element of g i C i is a conjugate of g i (by an element of N ). Itis straightforward to see that( g C )( g C ) · · · ( g r C r ) = ( g · · · g r ) D · · · D r , where D r = C r , D r − = C g r r − , . . . , D = C g ··· g r .Suppose that [ G, N ] = N . Then [ G, N ] is central in G by the minimality of N . If h isany p ′ -element of G , this implies that [ h p a , n ] = [ h, n ] p a = 1 for all n ∈ N , if exp( N ) = p a .It follows that all p ′ -elements of G centralize N . By the hypothesis, G is generated by p ′ -elements, whence G centralizes N , a contradiction. Hence [ G, N ] = N .Set M := D · · · D r . Note that [ g r , N ] ≤ M and so g r acts trivially on N/M . Also,[ g r , M ] ≤ [ g r , N ] = D r ≤ M , and so M g r = M . Hence C r − = D g − r r − ≤ M g − r = M ,whence g r − acts trivially on N/M . Continuing in this manner, we see that each g i actstrivially on N/M . Since G = h g , . . . , g r i , it follows that M ≥ [ G, N ] = N . The laststatement is obvious. (cid:3) We can now prove Theorem 1.1.
Proof.
Let G = X/F as in the statement. We induct on | F | . If F = 1, there is nothingto prove. So assume that this is not the case. Observe that O p ( X ) = X and so p doesnot divide | X/ [ X, X ] | . Indeed, X/O p ( X ) F ≥ O p ( G ) = G , whence O p ( X ) F = X and so O p ( X ) = X since F = Φ( X ).First assume that F ≤ Z ( G ). Since p is coprime to | X/ [ X, X ] | , F ≤ [ X, X ]. Thus X isa central extension of G with kernel F contained in [ X, X ]. Hence
F ֒ → Mult( G ) by [I2,Corollary 11.20]. In particular, p divides | Mult( G ) | , a contradiction.Thus F is not central and so we can take N ≤ F to be a minimal normal noncentralsubgroup of X . By Proposition 2.1(ii), N is abelian.By the induction hypothesis applied to X/N , we can choose x i ∈ X of order coprimeto p such that g i = x i F and x . . . x r = f n for some n ∈ N . Note that h x , . . . x n i = X .(Otherwise Y := h x , . . . x n i = X is contained in a maximal subgroup M of X . But Y F = X and F = Φ( X ) ≤ M , so M = X , a contradiction.) By Lemma 2.2 applied to X ,there exist y i ∈ x i N with y i conjugate to x i such that y · · · y r = ( x · · · x r ) n − = f . (cid:3) Note that Theorem 1.1 can be viewed as a result about branched coverings of Riemannsurfaces. Suppose that f : Y → P is a Galois branched covering of a Riemann surface Y ROBERT M. GURALNICK AND PHAM HUU TIEP with Galois group G and all ramification coprime to a given prime p >
2. Let G = X/F be a Frattini cover with F a p -group and assume that F = [ X, F ]. Then there exists anunramified F -cover Z → Y with Z → P Galois.Bailey and Fried [BF] have shown that if p = 2 then lifting to a central Frattini extensionis not always possible.More generally, we cannot remove the condition that the prime p is coprime to | Mult( G ) | in Theorem 1.1. We give families of such examples for any prime p . Example 2.3.
Let G = L ( q ) with q > q not a Fermat prime.Let C be a conjugacy class of elements of odd prime power order dividing ( q − / G = h x, y i with x, y ∈ C and z − = xy ∈ C (it is a straightforward computationto see that these exist – see [GM, Lemma 3.14] or [M]). Let X = SL ( q ), and let D be the conjugacy of elements of odd order which is the lift of C to X . Then we cannotfind u, v ∈ D and w ∈ D − with uvw = 1 and X = h u, v, w i . (Indeed, there is some α ∈ F × q such that αu , αv , and α − w all have a one-dimensional fixed point subspace onthe natural X -module F q . Now Scott’s Lemma [S] implies that h αu, αv, α − w i cannot actirreducibly on F q and so h u, v, w i 6 = X . See also [GM]). Example 2.4.
Let G = A , the alternating group on 7 letters. Using the character tableof G as given in [Atlas], one can check that there are elements x, y, z ∈ G of order 2, 5,and 7, respectively, with xyz = 1 and G = h x, y i . However, if we denote by ˆ x , ˆ y , ˆ z thelifts of the same order of these elements in the central cover X = 3 · A , then there are no( u, v, w ) ∈ ˆ x X × ˆ y X × ˆ z X such that uvw = 1, as one can check using the character tableof X in [Atlas].The previous two examples are given for p = 2 and p = 3. In §
5, using the classificationof irreducible groups generated by pseudoreflections, we will give a family of examples forany odd prime p , cf. Example 5.2.We can prove a version of Theorem 1.1 where we do allow p to divide the order ofthe Schur multiplier. Of course, this includes Theorem 1.1 as a special case (since then J = F ). Theorem 2.5.
Let X be a Frattini cover of G = X/F with F of odd order. Set J = [ F, X ] .Let g , . . . , g r ∈ G satisfy (i) G = h g , . . . , g r i , (ii) g · · · g r = 1 , and (iii) the order of each g i is coprime to | F | .Let X i := { x i ∈ X | x i F = g i , | x i | = | g i |} . Then X · · · X r is a coset of J in F .Proof. First assume that J = 1, i.e. F ≤ Z ( X ). It follows that there is a unique lift x i ∈ X of g i with | x i | = | g i | and x i F = g i . Then x . . . x r = f ∈ F .Now go back to the general case. Then F/J ≤ Z ( X/J ) and so, as before, g i has aunique lift to X/J of the same order. It follows that X · · · X r is contained in some coset f J of J in F . It remains to show that each element in f J is a product of elements in X i . IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 7
We will induct on | J | . Suppose first that J is central in X . Then for any h ∈ X of order m coprime to | F | and any w ∈ F , we have 1 = [ h m , w ] = [ h, w ] m , whence [ h, w ] = 1 and so h centralizes F . In particular, x i ∈ X i centralizes F for i = 1 , . . . , r . But h x , . . . , x r i = X since F ≤ Φ( X ) and G = h g , . . . , g r i . It follows that F ≤ Z ( X ), J = 1, and so we aredone by the previous case.Now we may assume that J is not central in X . Let N be a minimal normal noncentralsubgroup of X contained in J ; in particular, N is a p -group for some prime p dividing | F | and so p >
2. By Proposition 2.1, N is abelian. Observe that, if x i ∈ X i and n ∈ N ,then | n − x i n | = | x i | = | g i | and n − x i nF = x i F = g i , i.e. n − x i n ∈ X i . By the inductionhypothesis, the statement holds for X/N . Applying Lemma 2.2, we see that the statementholds in X as well. (cid:3) The coset of J in the previous result can be thought of as the lifting invariant (orobstruction). As Fried has observed, the different lifting invariants give rise to differentorbits for the Hurwitz braid group acting on the corresponding Nielsen classes of G , i.e.on { ( h , . . . , h r ) | h h , . . . , h r i = G, h · · · h r = 1 , h i ∈ g Gi } . See [BF, F1] for more details. 3.
Solvable Groups
We first need Thompson’s result on minimal simple groups [T].
Theorem 3.1.
Let G be a finite simple group such that every proper subgroup is solvable.Then G is one of the following groups: (a) L ( p ) with p ≥ an odd prime and p ≡ ± ; (b) L (2 p ) with p a prime; (c) L (3 p ) with p an odd prime; (d) Sz(2 p ) with p an odd prime; (e) SL (3) . We need one more preliminary result.
Lemma 3.2.
Let G be a finite group that is not solvable but has the property that everyproper subgroup is solvable. Then the solvable radical R ( G ) of G is Φ( G ) , the Frattinisubgroup of G ; in particular, R ( G ) is nilpotent. Moreover, G/R ( G ) is a non-abeliansimple group with all proper subgroups being solvable, and every prime divisor of | R ( G ) | divides | G/R ( G ) | .Proof. Suppose that R := R ( G ) is not contained in Φ( G ). Then G = RM for somemaximal subgroup M of G . Then M is solvable, whence G/R is solvable and so is G . Ofcourse, Φ( G ) is nilpotent, so R = Φ( G ) is nilpotent. Now G/R has no nontrivial solvablenormal subgroups; also, by the hypothesis, it has no proper non-solvable subgroups. Hence
G/R is simple non-abelian. Finally, suppose that a prime divisor p of | R | is coprime to | G/R | . Since R is nilpotent, O p ( R ) ∼ = R/O p ′ ( R ). By the Schur-Zassenhaus Theorem, ROBERT M. GURALNICK AND PHAM HUU TIEP
R/O p ′ ( R ) has a complement H/O p ′ ( R ) in G/O p ′ ( R ). In this case, H is a proper non-solvable subgroup of G , a contradiction. (cid:3) We next prove Theorem 1.2 in the case that G is quasi-simple. Lemma 3.3. (i)
Let G = SL ( q ) with q ≥ , respectively G = Sz( q ) with q ≥ , and let p be any odd prime divisor of | G | . Then there exist nontrivial elements x i ∈ G , ≤ i ≤ ,such that x x x = 1 , x is a -element, x is a p -element, and x is an s -element forsome prime s = 2 , p . (ii) Let G be a finite quasi-simple group with all proper subgroups being solvable. Thenthere exist nontrivial elements x i ∈ G , ≤ i ≤ , such that x x x = 1 , x is a -element, x and x are p i -elements, where p < p are odd primes and p ∈ { , } .Proof. (i) Recall that the number of triples ( y , y , y ) ∈ G × G × G with y y y = 1 and y i ∈ x Gi is equal to(1) | x G | · | x G | · | x G || G | · X χ ∈ Irr( G ) χ ( x ) χ ( x ) χ ( x ) χ (1) . The character tables for G = SL( q ) and Sz( q ) are well known (and are for example in Chevie [Ch]). First we consider the case G = Sz( q ) with q ≥
8. Then choose x of order2, x of order p , and x of some prime order s , where s | ( q −
1) if p | ( q + 1), and s | ( q + 1)otherwise.Next suppose that G = SL ( q ) with q = r f ≥ r a prime. • Assume r = 2 and let ǫ = ± p | ( q − ǫ ). Then we can choose x of order 2, x of order p , and x of some prime order s dividing q + ǫ . • Next assume that p = r . Choose ǫ = ± q ≡ ǫ (mod 4); in particular,( q + ǫ ) / s = 2 , p . Now we can choose x of order 4 (in a maximaltorus C q − ǫ ), x of order p , and x of order s (in a maximal torus C q + ǫ ). • Now let r = 2 , p but q
6≡ ±
1( mod 4 p ). Then there is some ǫ = ± | ( q − ǫ )and p | ( q + ǫ ), and we choose x ∈ C q − ǫ of order 4, x ∈ C q + ǫ of order p , and x of order s = r .In all the above cases, the principal character 1 G of G is the only irreducible characterthat is nonzero at x , x , and x altogether, and so we are done by (1).Suppose now that G = SL ( q ) with q = r f ≥ r = 2 , p and q ≡ ǫ (mod 4 p ) for some ǫ = ±
1. Choose x ∈ C q − ǫ of order 4, x ∈ C q − ǫ of order p , and x of some prime order s dividing ( q + ǫ ) / s = 2 , p ). Then there are precisely two irreducible characters of G which are nonzero at all x i : the principal character 1 G , and the Steinberg characters Stof degree q , with | χ ( x ) χ ( x ) χ ( x ) | = 1. Hence we are done in this case as well.(ii) It suffices to prove the statement for G , the Schur cover of G/Z ( G ). Hence we mayassume that G = SL ( q ), SL (3), Sz( q ), or 2 · Sz(8) by Theorem 3.1. We will choose p = 3 in the first two cases, and p = 5 in the last two cases. If G = SL ( q ) or Sz( q ), thenwe are done by (i). For G = SL (3), we can choose x of order 2, x of order 3, and x IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 9 of order 13, and again 1 G is the only irreducible character that is nonzero at x , x , and x . For G = 2 · Sz(8), we can choose x a noncentral 2-element, x of order 5, and x ofprime order 13, and check our statement by using (1) and [Atlas]. (cid:3) We can now prove a slightly stronger version of Theorem 1.2.
Theorem 3.4.
Let G be a finite non-solvable group. There exist nontrivial elements x i ∈ G , ≤ i ≤ , such that x x x = 1 , x is a -element, x and x are p i -elements,where p < p are odd primes and p ∈ { , } .Proof. Suppose the result is false. Let G be a minimal counterexample. Thus, everyproper subgroup of G is solvable, and so G is perfect and G/ Φ( G ) is non-abelian simpleby Lemma 3.2.1) We first claim that O ( G ) = 1. If not, then the result holds for G/O ( G ). Thus,we can choose nontrivial x i , ≤ i ≤
3, where x i are p i -elements with p = 2, p < p odd primes, p ∈ { , } , and x x x = y ∈ O ( G ). Then ( y − x ) x x = 1 and y − x is a2-element (and is nontrivial since x and x are). This proves the claim.2) By Lemma 3.3(ii), Φ( G ) > Z ( G ). Let N be a subgroup of Φ( G ) that is normal in G but is not central. Moreover, take N to be a minimal such subgroup. Since N ≤ Φ( G )is nilpotent, the minimality of N implies that N is a p -group. Observe that p > O ( G ) = 1. By Proposition 2.1(ii), N is abelian. Choose x i ∈ G , 1 ≤ i ≤ x i nontrivial p i -elements, p = 2, p < p odd primes, p ∈ { , } , and x x x = n ∈ N (thisis possible since G/N satisfies the theorem). Since every proper subgroup of G is solvable, G/N = h x N, x N, x N i by Thompson’s theorem [T]. Furthermore, since N ≤ Φ( G ), wesee that G = h x , x , x i . Recall that G is perfect, and so G = O p ( G ). Hence by Lemma2.2, there are conjugates y i of x i for i = 1 , , y y y = ( x x x ) n − = 1, acontradiction to the fact that G was a counterexample. (cid:3) The examples of Sz(8) and SL (2) show that Theorem 3.4 is best possible, in the sensethat one cannot always demand one of the primes to be 3, respectively 5. In §§
4, 5 wewill address possible refinements of Theorem 3.4 in some other directions.4.
Prime Order Elements
It is quite easy to see that Theorems 1.2 and 3.4 fail if we insist that the elements haveprime order. For example, if G ∼ = SL ( q ) with q odd and has order divisible by only 3primes, then since every involution in G is central, no product of 3 elements of distinctprime order (with one of the primes equal to 2) can be trivial (otherwise we would havetwo elements of prime order in G/Z ( G ) with product being trivial). For q = 5 this wasobserved (with a more complicated proof) in [B].More generally, we point out the following: Theorem 4.1.
Let G be a finite group. Then there exists a finite group X with X/F ∼ = G such that F is abelian, F ≤ Φ( X ) , and F contains all elements of prime order in X . Proof.
Let g ∈ G be of prime order p , and let D be a one-dimensional trivial h g i -moduleover F p . Since dim H ( h g i , D ) = 1, there exists a finite F p G -module W g with an element β g ∈ H ( G, W g ) such that the restriction of β g to h g i is nonzero in H ( h g i , W g ). (Justtake the induced module for example). By taking direct sums, we see that there exists afinite G -module W and an element β ∈ H ( G, W ) such that the restriction of β to eachsubgroup of prime order in G is nonzero. This allows us to construct an exact sequence1 → W → E → G → , where W contains all elements of prime order in E (since β | C is nonzero in H ( C, W ) forevery subgroup of prime order C of G ).Now choose X to be a minimal subgroup of E that surjects onto G . Clearly F := X ∩ W is abelian and contains all elements of prime order in X . It only remains to show that F ≤ Φ( X ). If not, then we could choose a proper subgroup X of X with X = X F andso X surjects onto G , contradicting the choice of X . (cid:3) In particular, given any non-solvable G , the subgroup generated by all elements ofprime order in the extension X specified in Theorem 4.1 is abelian, and so any productof elements in X of distinct prime orders p i has order equal to the product of the primes.We close by noting that this example (especially for the prime 2) has a nice consequencefor the inverse Galois problem. As far as we know, this was first observed by Serre. Corollary 4.2.
If every finite group occurs as a Galois group of a Galois extension K of Q , then every finite group occurs as a Galois group over a totally real Galois extension K ′ of Q .Proof. By Theorem 4.1, there is a finite group X such that G = X/F and F contains allinvolutions of X . Assume that X = Gal( K / Q ) for some Galois extension K of Q . Then G = Gal( K F / Q ). It remains to show that K F is totally real. Since K F / Q is Galois, itsuffices to show that K F is real. Note that the complex conjugation σ acts on K . Thus, σ is an involution in X and so contained in F . Hence σ is trivial on K F , whence K F isreal. (cid:3) Finite Groups with (2 , p, q ) -triples First we recall a result essentially proved by Gow in [Gow]:
Lemma 5.1.
Let G be a quasisimple Lie-type group of simply connected type, and let x, y ∈ G be any two regular semisimple elements. Then x G · y G contains every noncentralsemisimple element of G .Proof. The proof is essentially given in [Gow]. Let p be the defining characteristic of G . Observe that Z := Z ( G ) is a p ′ -group, so P Z/Z ∈ Syl p ( G/Z ) when P ∈ Syl p ( G );also, C G/Z ( P Z/Z ) contains no nontrivial p ′ -element (cf. [Gow]). Hence C G ( P ) containsno noncentral semisimple element. It follows that if g ∈ G \ Z ( G ) is semisimple, then p divides | g G | . Next, G has exactly s := | Z | p -blocks of maximal defect B , . . . , B s , and a IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 11 single p -block of defect zero consisting of the Steinberg character St. For any i = 1 , . . . , s ,there is an irreducible character χ i ∈ B i of p ′ -degree, and so the algebraic integer ω χ i ( g ) :=( χ i ( g ) /χ i (1)) ·| g G | is divisible by p . This implies that ω χ ( g ) := ( χ ( g ) /χ (1)) ·| g G | is divisibleby p for every irreducible character χ ∈ B i . On the other hand, | St( x ) | = | St( y ) | = 1since x, y are regular semisimple. Thus X χ ∈ Irr( G ) χ ( x ) χ ( y ) χ ( g ) χ (1) · | g G | ≡ St( x )St( y )St( g )St(1) · | g G | = ± [ G : C G ( g )] p ′ p ) , whence g − ∈ x G · y G by (1). (cid:3) Example 5.2.
Let p > m > p . Assume moreoverthat m does not divide p −
1. Let q be a prime with mp | ( q − G = SL p ( q ). Forbrevity, we will outline the arguments only for p > x i of G as follows:(a) x is irreducible on a hyperplane (of the natural F q G -module V = F pq ) and has aneigenvalue 1;(b) x = diag ( a, b, . . . , b ) where ab p − = 1 with b ∈ F × q of order m ;(c) x is irreducible on a subspace of codimension 2 of V , and has two eigenvalues, 1and b − , in F q ; furthermore it has order coprime to | x | .Note that the orders of x i , i = 1 , ,
3, are all coprime to p and x and x are regularsemisimple elements. It is easy to arrange so that | x | and | x | are coprime (by taking x to have prime order ℓ , a primitive prime divisor of q p − −
1, cf. [Zs]).2) Observe that if y i is conjugate to x i with y y y = 1, then h y , y i acts reducibly on V . This follows by Scott’s Lemma [S] applied to the elements y , b − y , by .3) Let 1 = d ∈ F ∗ q be of order p . Then d / ∈ { , a, b, ab − } since b , a = b − p , and ab − = b − p all have order dividing m . By Lemma 5.1 applied to the classes x G , x G ,( d − x ) G , there exist u i conjugate to x i such that u u u = dI . Let U = h u , u , u i . Weclaim that U = G for any such choice of u i .First we see that U acts irreducibly. Since u acts irreducibly on a hyperplane, the onlyother possibility would be that U preserves a hyperplane or a 1-space. But then therewould be a choice of eigenvalues e i of the u i with e e e = d . Since e = 1, e ∈ { a, b } and e ∈ { , b − } , e e e is contained in { , a, b, ab − } and so cannot be equal to d .Now U is an irreducible group containing pseudoreflections (up to scalar – namely, u ).Note also that, modulo scalars, U is the normal closure of u in U . (For if N = h u U i , then,modulo scalars, U/N is generated by u N and also by u N , and so has order dividingboth | u | and | u | , which are coprime). Thus, U is generated by pseudoreflections of oddorder modulo scalars. This implies that U is primitive and tensor-indecomposable, andthis also excludes most of the “obvious” examples of pseudoreflection groups. Now theclassification of finite pseudoreflection groups (cf. [GS, 7.1]) implies that U = G .4) Let ¯ u i ∈ L p ( q ) be the image of u i . Then¯ u ¯ u ¯ u = 1 , h ¯ u , ¯ u , ¯ u i = L p ( q ) by the result of 3). But ¯ u , ¯ u , ¯ u do not lift to elements in G of order coprime to p withproduct equal to 1 according to 2).Glauberman’s classification of S -free simple groups [Gl], together with results of Gold-schmidt [Gol], implies that Suzuki groups Sz(2 a +1 ), a = 1 , , . . . , are the only finitenon-abelian simple 3 ′ -groups. We will need the full classification of finite simple groupsto prove the following statement. Lemma 5.3.
Let S be a finite non-abelian simple group. Suppose that every propersubgroup of S is -solvable. Then S is either a Suzuki group Sz( q ) , or one of the groupslisted in Theorem 3.1.Proof. Certainly, S = A n for any n ≥ A is not 3-solvable. Similarly, one can checkusing [Atlas] that each of the 26 sporadic simple groups has a section isomorphic to A .Next suppose that S is a finite simple group of Lie type over F q with q ≥
4. If thetwisted Lie rank of S is at least 2, then a proper section of S is isomorphic to L ( q ) whichis again not 3-solvable. Otherwise S ∼ = L ( q ), Sz( q ), U ( q ), or G ( q ). In the first case,it is easy to check that S must then be one of the groups listed in Theorem 3.1. Thesecond case is included in the lemma’s conclusion. The last two cases cannot happen asotherwise S has a section ∼ = L ( q ). Also observe that SL ( q ) embeds in D ( q ).Finally, suppose that S is a finite simple group of Lie type over F q with q = 2 , D ( q ), L (7), or SL (3). Considering Levi subgroups or subsystemsubgroups of S , one readily checks that S has a section isomorphic to A or L (7). (cid:3) Let p, q, r be primes. By a ( p, q, r ) -triple in a finite group G we mean a triple ( x, y, z )of nontrivial elements in G such that x is a p -element, y is a q -element, z is an r -element,and xyz = 1. (Note that in fact the order of p, q, r does not matter, since if xyz = 1, then y ( x z − ) z = 1, etc.) Theorem 3.4 then states that any finite non-solvable group admitseither a (2 , , p )-triple (for a prime p ≥ , , p )-triple for a prime p ≥
7. Nowwe can characterize finite groups G for which only the latter can happen. In fact, we cancharacterize the finite p -solvable groups (with p >
2) as precisely the ones which do notadmit any (2 , p, q )-triple for any prime q = 2 , p .To this end, first we use the classification of finite simple groups to describe the minimalnon- p -solvable simple groups. Lemma 5.4.
Let S be a finite non-abelian simple group and p ≥ a prime with p dividing | G | . If every proper subgroup of S is p -solvable, then either S = L ( p ) , A p , or one of thefollowing holds. (i) S = L ( q ) with p | ( q − . (ii) S = L n ( q ) , n ≥ is odd, and p divides q n − but not Q n − i =1 ( q i − . (iii) S = U n ( q ) n ≥ is odd, and p divides q n − ( − n but not Q n − i =1 ( q i − ( − i ) . (iv) S = Sz( q ) . (v) S = G ( q ) , and p divides q − q + 1 but not q − . (vi) S = F ( q ) , q ≥ , and p | ( q − q + 1) . IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 13 (vii) S = D ( q ) , and p divides q − q + 1 but not q − . (viii) S = E ( q ) , and p divides q − but not Q i =8 , , , , ( q i − . (ix) ( S, p ) is one of: ( M , , ( J , or , ( Ly, or , ( J , or , ( F i ′ , , ( BM, , ( M, or or .Proof. By the assumption, every proper section of S is p -solvable. If S = A n , then n = p since A p − is a p ′ -group and A p is not p -solvable. The sporadic simple groups are treatedusing [Atlas].Suppose now that S is a simple group of Lie-type in characteristic r . If r = p ≥
5, then S has a section isomorphic to L ( p ) which is not p -solvable, whence S = L ( p ). Now weconsider the cases with r = p type-by-type. The cases S = L ( q ) and S = Sz( q ) are listedin (i) and (iv). Suppose S = L n ( q ) with n ≥
3, but p | ( q i −
1) for some 2 ≤ i ≤ n − n = 3 then q ≥ p ≥
5. Hence, a proper simple section L n − ( q )of S is not p -solvable, a contradiction. Also, the case p | ( q n −
1) with 2 | n is excluded byconsidering a section S n ( q ) of S . Thus we arrive at (ii). Similarly, we arrive at (iii) if S = U n ( q ) with n ≥
3. Next suppose that S = S n ( q ) with n ≥
2. If p | ( q n −
1) then aproper section L ( q n ) of S is not p -solvable. Otherwise a proper section S n − ( q ) of S isnot p -solvable. Similarly, if S = O n +1 ( q ) with n ≥
3, then either O +2 n ( q ) or O − n ( q ) is not p -solvable. If S = O +2 n ( q ) with n ≥
4, then either O +2 n − ( q ), or O − n − ( q ), or L n ( q ) is not p -solvable. Suppose S = O − n ( q ) with n ≥
4. Then S has two proper sections isomorphicto O +2 n − ( q ) and O − n − ( q ), and another proper section isomorphic to U n ( q ) if n is odd andL ( q n ) if n is even. At least one of these three sections is not p -solvable, a contradiction.If G = G ( q ), then p cannot divide q − ( q ), so we arrive at (v). If G = G ( q ), then either L ( q ) or U ( q ) is not p -solvable. If G = F ( q ), then by consideringsections S ( q ) and U ( q ) of S (and SL (3) for q = 2) we see that p | ( q − q + 1) and q ≥ G = D ( q ) but p | ( q − ( q ) is not p -solvable. If G = F ( q ), then a proper section O ( q ) or D ( q ) is not p -solvable. If G = E ( q ), thena proper section O +10 ( q ), L ( q ), or D ( q ) is not p -solvable. If G = E ( q ), then a propersection O − ( q ), U ( q ), or D ( q ) is not p -solvable. If G = E ( q ), then a proper section E ( q ), E ( q ), or L ( q ) is not p -solvable. If G = E ( q ) and p | Q i =8 , , , , ( q i − E ( q ), U ( q ), or D ( q ) is not p -solvable. (The aforementioned sectionsof exceptional groups of Lie type come from subgroups of maximal rank described in[LSS].) (cid:3) Lemma 5.5.
Let q < p be odd primes. Then there exist (2 , q, p ) -triples in A p .Proof. Write p = sq + t with s ≥ < t < q . Let y ∈ A p be an element of order q with s nontrivial cycles. Let O , . . . , O s be the nontrivial orbits of y and O the set of allfixed points of y (acting on { , , . . . , p } ). Take an involution x ∈ S p which is a productof s + t − x moves exactly one point a i − of O i to a i ∈ O i +1 for i = 1 , . . . , s −
1, and for each j ∈ O , xj is not in O and disjoint from the points a , . . . , a s − (this is possible since sq − s − ≥ q ≥ t ). For any u ∈ A p , let ind( u ) bethe difference between p and the number of cycles of u , i.e. the codimension of the fixed point subspace of u on the natural permutation module V = C p . Then ind( x ) = s + t − y ) = s ( q − s + t − x ∈ A p . By construction h x, y i is atransitive subgroup of A p . Applying Scott’s Lemma [S] to the action of h x, y i on V , wesee that ind( x ) + ind( y ) + ind( xy ) ≥ p − . Hence, ind( xy ) ≥ p −
1, which forces xy to be a p -cycle. Thus, ( x, y, ( xy ) − ) is a (2 , q, p )-triple. (cid:3) Proposition 5.6.
Let G be a finite quasisimple group and p ≥ a prime. Supposethat every proper subgroup of G is p -solvable but G is not p -solvable. Then G admits a (2 , p, s ) -triple for some odd prime s = p .Proof. We apply Lemma 5.4 to S = G/Z ( G ) and may assume that G is a Schur cover of S .1) If S = L ( p ), then G = SL ( p ) since p ≥
5, and so we are done by Lemma 3.3(i).Suppose S = A p . If p = 7, then by Lemma 5.5, S admits a (2 , p, q )-triple, which then liftsto a (2 , p, q )-triple in G = 2 A p by p. 1) of the proof of Theorem 3.4. Similarly, if p = 7,then by Lemma 5.5, S admits a (2 , , , , A and then lifts to a (2 , , G = 6 A . The same argument shows that 6 · L (9)admits a (2 , , · Sz(8) admits a (2 , p, s )-triple for any p ∈ { , , } .The cases S = L ( q ) and Sz( q ) now follow from Lemma 3.3(i).The sporadic groups listed in Lemma 5.4 can certainly by checked using (1) and [Atlas].But we point out a slightly easier way to check it as follows. Consider the case S = M (sothat G = S ) and let p ∈ { , , } . Pick a prime s ∈ { , , } \ { p } , x ∈ S of order p , y ∈ S of order s , and z ∈ S of order 32. Then | C G ( x ) | = p , | C G ( y ) | = s , | C G ( z ) | = 128;also, if χ ∈ Irr( G ) is non-principal then χ (1) ≥ | Irr( G ) | = 194, it followsthat | X χ ∈ Irr( G ) χ ( x ) χ ( y ) ¯ χ ( z ) χ (1) | ≥ − · √ · · > , and so a (2 , p, s )-triple exists. As another example, consider the case S = F i ′ (so wemay assume G = 3 S and p = 29). Choosing x ∈ G of order p , y ∈ G of order s = 17, and z ∈ G of order 16, we have | C G ( x ) | = 3 p , | C G ( y ) | = 3 s , | C G ( z ) | = 96. Now if χ ∈ Irr( G ) isnon-principal and χ ( x ) χ ( y ) ¯ χ ( z ) = 0, then χ (1) ≥ | Irr( G ) | = 260, it followsthat | X χ ∈ Irr( G ) χ ( x ) χ ( y ) ¯ χ ( z ) χ (1) | ≥ − · √ · · > . The same argument works for S = M with ( p, s ) = (23 , S = J with ( p, s ) = (7 , S = Ly with ( p, s ) = (37 , S = J with ( p, s ) = (29 , S = BM with ( p, s ) =(31 , S is a simple group of Lie-type in characteristic r = p (and not isomorphic to any of the aforementioned simple groups). The last assumptionimplies that G is a Lie-type group of simply connected type corresponding to S . IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 15
2) Assume in addition r = 2, so that we are in the cases (ii), (iii), (v), (vii), or (viii) ofLemma 5.4. In all these cases, the conditions on p imply that a p -element x ∈ G is regularsemisimple (see e.g. [MT, Lemma 2.3] for exceptional groups). Clearly, Sylow subgroupsof G cannot be central (since | S | and | G | have same set of prime divisors). Suppose thatwe can find a regular semisimple s -element y ∈ G , for a suitable prime s = 2 , p . Then wecan apply Lemma 5.1 to get a noncentral 2-element z ∈ x G · y G , yielding a (2 , p, s )-triple. • Suppose S = L n ( q ). If n ≥
5, or n = 3 but q is not a Mersenne prime, then wecan choose s to be a primitive prime divisor of q n − −
1. If n = q = 3, then p = 13 andwe are done by Lemma 3.3(i). In the remaining case, G = SL ( q ) and q = 2 t − ≥ G contains a regular semisimple element y of order 2 t (witheigenvalues α , α − , 1, for some α ∈ F × q of order 2 t ). Also choose s to be an odd primedivisor of q − t −
2. By Lemma 5.1, x G · y G contains a noncentral semisimple s -element z , giving rise to a (2 , p, s )-triple. • Suppose S = U n ( q ). If n ≥
5, or n = 3 but q is not a Fermat prime, then we canchoose s to be a primitive prime divisor of ( − q ) n − −
1. If n = q = 3, then p = 7. In thiscase, picking x of class 7 A , y in class 2 A , and z in class 3 B in the notation of [Atlas], wesee that 1 G is the only irreducible character χ of G such that χ ( x ) χ ( y ) ¯ χ ( z ) = 0, whence G admits a (2 , , G = SU ( q ) and q = 2 t + 1 ≥ G contains a regular semisimple element y of order 2 t (with eigenvalues α , α − , 1, for some α ∈ F × q of order 2 t ). Also choose s to be an odd prime divisor of q + 1 = 2 t + 2. By Lemma 5.1, x G · y G contains a noncentral semisimple s -element z ,giving rise to a (2 , p, s )-triple. • Suppose S = G ( q ). Then there exist some ǫ = ± p | ( q + ǫ √ q + 1), andsome odd prime s | ( q − ǫ √ q + 1). Now G contains a regular semisimple s -element, andso we are done. • Suppose S = D ( q ), and let s be a primitive prime divisor of q −
1. Then G containsa regular semisimple s -element (of type s as listed in [DM]), and so we are done. • Suppose S = E ( q ), and let s be a primitive prime divisor of q −
1. By [MT, Lemma2.3], G contains a regular semisimple s -element, and so we are done again.3) Assume now that r = 2, so that we are in the cases (ii),(iii), (vi)–(viii) of Lemma5.4. In all these cases, the conditions on p again imply that a p -element x ∈ G is regularsemisimple. • Suppose S = L n ( q ) or S = U n ( q ). Set ǫ = 1 in the SL-case and ǫ = − S = SL (2), then we can choose an odd prime s dividing q n − − ǫ n − but not Q n − i =1 ( q i − ǫ i ). By the choice of s , G contains a regular semisimple s -element y , which iscontained in a (unique) maximal torus of type T ,n − of G , in the notation of [LST]. Thesame is true for S = SL (2) if we choose s = 3 and y ∈ S of order 9. Also, x is containedin a (unique) maximal torus of type T n . Let χ ∈ Irr( G ) be any character of G that isnonzero at both x and y . By Propositions 2.3.1 and 2.3.2 of [LST], the tori T n and T ,n − are weakly orthogonal. Hence [LST, Proposition 2.2.2] implies that χ ∈ Irr( G ) must beunipotent. Now the proofs of Theorems 2.1 and 2.2 of [MSW] imply that such a unipotent character χ is either trivial or the Steinberg character St. Since x, y are regular, we alsohave | St( x ) | = | St( y ) | = 1. It follows by (1) that x G · y G ⊇ G \ Z ( G ). In particular, weget a noncentral 2-element z ∈ x G · y G , yielding a (2 , p, s )-triple. • Suppose S = D ( q ), and let s be a primitive prime divisor of q −
1. Then G containsa regular semisimple s -element y (of type s as listed in [DM]). Let T , respectively T , be the unique maximal torus containing x , respectively y . If Φ m ( q ) denotes the m th cyclotomic polynomial in q , then | T | = Φ ( q ) and | T | = Φ ( q ) . The order of thecentralizer of any semisimple element in the dual group G ∗ ∼ = G is listed in [DM, Tables1.1, 2.2]. Using this, it is easy to see that the centralizer of no nontrivial semisimpleelement of G ∗ can have order divisible by both | T | and | T | . Thus the dual tori T ∗ and T ∗ in G ∗ intersect trivially, and so T and T are weakly orthogonal. By [LST,Proposition 2.2.2], any irreducible character χ ∈ Irr( G ) that is nonzero on both x and y must be unipotent. Note that the p -parts of Φ ( q ) and of | G | are the same. Hence, ifΦ ( q ) divides χ (1), then χ has p -defect 0 and so χ ( x ) = 0. Similarly, if Φ ( q ) divides χ (1), then χ has s -defect 0 and so χ ( y ) = 0. Inspecting the list of unipotent charactersas given in [Sp], we see that χ = 1 G , St, or the unique unipotent character ρ of degree q ( q + 1) /
2. Choosing z to be a unipotent element of class D ( a ) of [Sp], we see that ρ ( z ) = St( z ) = 0. It now follows by (1) that z ∈ x G · y G , giving rise to a (2 , p, s )-triple. • Suppose S = F ( q ) with q >
2. Then there exist some ǫ = ± p | ( q + q +1 + ǫ √ q ( q + 1)), and some primitive prime divisor s of q −
1, and G contains a regularsemisimple s -element by [MT, Lemma 2.3]. In particular, | C G ( x ) | = q + q + 1 + ǫ p q ( q + 1) , | C G ( y ) | = q − q + 1 . Next, take z to be a regular unipotent element, so that | C G ( z ) | ≤ q [LiS]. In particular, | χ ( x ) χ ( y ) ¯ χ ( z ) | < p | C G ( x ) | · | C G ( y ) | · | C G ( z ) | < (2 . q for any χ ∈ Irr( G ). Now G has exactly two irreducible characters of degree ( q − q +1) p q/
2, and all other nontrivial irreducible characters have degree at least q ( q − q +1)( q − q + 1), cf. [Lu]. Also, | Irr( G ) | ≤ q + 4 q + 17 < (1 . q [FG, Table 1]. It nowfollows by (1) that | X χ ∈ Irr( G ) χ ( x ) χ ( y ) ¯ χ ( z ) χ (1) | > − (1 . q · (2 . q q ( q − q + 1)( q − q + 1) − (2 . q ( q − q + 1) p q/ > , whence z ∈ x G · y G , giving rise to a (2 , p, s )-triple. • Suppose S = E ( q ), and let s be a primitive prime divisor of q −
1. By [MT,Lemma 2.3], G contains a regular semisimple s -element y . Then | C G ( x ) | = Φ m ( q ) with m ∈ { , } , and | C G ( y ) | = Φ ( q ). Next we choose z to be a regular unipotent element,so that | C G ( z ) | ≤ q [LiS]. Now for any nontrivial χ ∈ Irr( G ) we have that χ (1) >q ( q −
1) by the Landazuri-Seitz-Zalesskii bound [LS], and | χ ( x ) χ ( y ) ¯ χ ( z ) | < p | C G ( x ) | · | C G ( y ) | · | C G ( z ) | < q . IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 17
On the other hand, | Irr( G ) | < (5 . q [FG, Table 1]. It now follows by (1) that | X χ ∈ Irr( G ) χ ( x ) χ ( y ) ¯ χ ( z ) χ (1) | > − (5 . q · q q ( q − > , whence z ∈ x G · y G , giving rise to a (2 , p, s )-triple. (cid:3) We can now prove Theorem 1.4. We restate the result.
Theorem 5.7.
Let G be a finite group. Then the following statements are equivalent: (i) G admits no (2 , p, q ) -triple for any odd prime q = p ; (ii) G is p -solvable. (iii) If T is a Sylow -subgroup of G , P a Sylow p -subgroup of G and Q a Sylow q -subgroup of G (for any prime q = p ), then T P ∩ Q = 1 .Proof.
1) Suppose that G is p -solvable but admits a (2 , p, q )-triple for some prime q = 2 , p .Choose such a G of minimal order. Since p divides | G | and G is p -solvable, G cannotbe simple. Let 1 < N < G be a normal subgroup of G . If x, y, z all belong to N , then( x, y, z ) is a (2 , p, q )-triple in N , and so by minimality of G , N cannot be p -solvable,contradicting the p -solvability of G . So at least one of x, y, z is not contained in N . Inthis case, since xyz = 1, all of them are outside of N by order consideration. It followsthat ( xN, yN, zN ) is a (2 , p, q )-triple in G/N , and so
G/N is not p -solvable by minimality,again a contradiction.2) From now on we will assume that G is a not p -solvable and aim to show that G admits a (2 , p, q )-triple for some q = 2 , p . Consider a minimal counterexample G to thisclaim, so that G has a composition factor S which is not p -solvable but G admits no(2 , p, q )-triple with q = 2 , p . The minimality of G implies that any proper subgroup of G is p -solvable and that G is perfect.Let N be any proper normal subgroup of G , in particular, N is p -solvable. Supposein addition that N is not contained in Φ( G ). Then G = M N for a maximal subgroup
M < G . Then S is also a composition factor of M , whence M is not p -solvable, acontradiction. Thus every proper normal subgroup of G is contained in Φ( G ). It followsthat G/ Φ( G ) is simple and so it is isomorphic to S . Since every proper subgroup of G is p -solvable, the same holds for S , whence S is one of the groups listed in Lemma 5.3 if p = 3 and in Lemma 5.4 if p ≥ G ) ≤ Z ( G ). Then G is a quasisimple group with all proper subgroupsbeing p -solvable.Assume p = 3. Then S = Sz( q ) as S is not 3-solvable, so S is one of the groups listedin Theorem 3.1. This in turn implies that all proper subgroups of S are solvable. Thisis also true for any proper subgroup H < G . (Indeed, in this case H Φ( G ) < G , whence H Φ( G ) / Φ( G ) < S is solvable and Φ( G ) is nilpotent.) Now Lemma 3.3 (and its proof)implies that G admits a (2 , , q )-triple for some prime q = 2 ,
3, a contradiction.
On the other hand, if p ≥
5, then according to Proposition 5.6, G also admits a (2 , p, q )-triple for some q = 2 , p , again a contradiction.We have shown that Φ( G ) > Z ( G ). Furthermore, if O ( G ) = 1, then G/O ( G ) admitsa (2 , p, q )-triple for some prime q = 2 , p by minimality of G , which can then be lifted toa (2 , p, q )-triple in G (see p.1) of the proof of Theorem 3.4). Thus O ( G ) = 1.Let N be a subgroup of Φ( G ) that is normal in G but is not central. Moreover, take N to be a minimal such subgroup. Then the minimality implies that N is an r -group forsome prime r , and r > O ( G ) = 1. This in turn implies by Proposition 2.1(ii)that N is abelian. By the minimality of G , there are some nontrivial elements x i ∈ G ,1 ≤ i ≤
3, such that x is a 2-element, x is a p -element, x is a q -element for some prime q = 2 , p , and x x x = n ∈ N . In particular, the subgroup L/N = h x N, x N, x N i of G/N admits a (2 , p, q )-triple. According to 1),
L/N , and so L , is not p -solvable. Hence L = G . This implies that G = h x , x , x i since N ≤ Φ( G ). Now arguing as in p. 3) ofthe proof of Theorem 3.4 and using Lemma 2.2, we see that there are conjugates y i of x i for i = 1 , , y y y = ( x x x ) n − = 1, a contradiction to the fact that G wasa counterexample.The equivalence of (i) and (iii) is straightforward. (cid:3) Theorem 1.4 yields the following immediate consequence:
Corollary 5.8.
Let G be a finite group. Then the following statements are equivalent: (i) G admits no (2 , , p ) -triple for any prime p ≥ ; (ii) G is -solvable; (iii) Every composition factor of G is either cyclic or a Suzuki group. Examples with p = 2One of the key steps in proving Theorem 1.1 was Proposition 2.1. As we have observed,Proposition 2.1 fails for p = 2. Here we produce examples showing that in fact Theorem1.1 fails for p = 2 as well.Let E = 2 n − be the extraspecial group of type − of order 2 n for any n ≥
5. Itis well known that there is a non-split extension G of E such that G/E ∼ = H := O − n (2).Then G has a a complex irreducible character ϕ of degree 2 n which is irreducible andfaithful when restricted to E . For x ∈ G , let ¯ x := xE be the corresponding element of H .Let x i ∈ G of order 2 n i + 1 be acting on E/Z ( E ) with one nontrivial irreduciblesubmodule of dimension 2 n i (and trivial on a complement). It follows by [Gor, p. 372]that:(2) ϕ ( x i ) = − n − n i . Now we show that Theorem 1.1 fails for p = 2. Proposition 6.1.
In the above notation, choose n = 1 , n = n − , and n = n . (i) If y i ∈ G is such that h y i i is conjugate to h x i i for i = 1 , , , then y y y = 1 . IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 19 (ii)
There are conjugates z i ∈ G of x i such that h ¯ z , ¯ z , ¯ z i = H = O − n (2) , ¯ z ¯ z ¯ z = 1 , but the generating triple (¯ z , ¯ z , ¯ z ) of H does not lift to any triple ( t , t , t ) in G with ¯ t i = ¯ z i , | t i | = | z i | , and t t t = 1 .Proof. (i) Note that any generator of h x i i also fulfills the conditions imposed on x i . Hencewe may assume that y i ∈ x Gi . We use (1) to count the number N of triples in x G × x G × x G with product 1, and break the sum in (1) into three pieces. The first piece is the sum overthe irreducible characters whose kernel contains E . The second piece is the sum over thecharacters whose kernel is Z ( E ) and the third piece is the sum over all faithful characters.Let N denote the first sum and let N denote the third sum.First we note that any character β in the second sum is afforded by an induced modulefrom the stabilizer of a linear character of E/Z ( E ). Since x has no fixed points on E/Z ( E ), any such character β vanishes on x . Thus the second sum is 0.Next, by Gallagher’s theorem [I2, 6.17], the characters in the third sum are preciselythose of the form ϕλ where λ is an irreducible character of G/E (and they are all distinctfor distinct λ ). Applying (2), we now see that N = ϕ ( x ) ϕ ( x ) ϕ ( x ) ϕ (1) N = − N , whence N = 0.(ii) Note that ¯ x and ¯ x are regular semisimple elements of H . Hence, by Lemma 5.1,there exist z i ∈ G conjugate to x i for i = 1 , , z ¯ z ¯ z = 1. By [GS, 7.1],the elements ¯ z i generate H . Now consider any t i ∈ z i E = ¯ z i with | t i | = | z i | . By theSchur-Zassenhaus theorem, h t i i is conjugate to h z i i and hence to h x i i . It follows by (i)that t t t = 1. (cid:3) Note that one can construct similar examples for odd p with E extraspecial of exponent p of order p a and G/E ∼ = Sp a ( p ). However, in this case the extension is split and so G is not a Frattini cover.For the next example, let E and G as in the beginning of the section with n = 2 m ≥ m such that 2 m − p and p . This is possible for instance for m = 14, with p = 29 and p = 113. Alsofix a primitive prime divisor p of 2 m −
1, and choose n = n = m and n = n . Since E/Z ( E ) is a quadratic space of type − , one can check that, for i = 1 , ,
3, any nontrivial p i -element x i ∈ G acts irreducibly on a subspace of dimension 2 n i of E/Z ( E ) and triviallyon a complement. Arguing exactly as above, we see that there are no conjugates y i of x i in G with y y y = 1. Thus, there is no ( p , p , p )-triple in G , but G has a compositionfactor whose order is divisible by p p p .We know of no such counterexample with one of the primes being even. We conjecture: Conjecture 6.2.
Let q < p be odd primes and let G be a finite group. The followingstatements are equivalent: (i) G contains a composition factor whose order is divisible by pq ; and (ii) G contains a (2 , p, q ) -triple. The same arguments as in the proof of Theorem 1.4 show that (ii) implies (i), andthat a minimal counterexample to Conjecture 6.2 would be a quasisimple group G suchthat G has no simple sections of order divisible by pq . Moreover, we can assume that O ( G ) = O p ( G ) = O q ( G ) = 1. Note that the result holds for G = A n by Lemma 5.5 andso for the covering groups as well (checking 6 A directly).7. A Short Proof of the Feit-Tits Theorem
Let F be an algebraically closed field of characteristic p ≥
0. If S is a finite non-abeliansimple group, let m p ( S ) be the smallest positive integer n such that S is a section of somesubgroup of GL n ( F ). Also, let d p ( S ) be the smallest degree of a nontrivial representationof the covering group of S over F (i.e. the smallest nontrivial degree of a projectiverepresentation of S over F ). The following theorem was proved by Feit and Tits in [FT](and it was refined further by Kleidman and Liebeck [KL] using the classification of finitesimple groups). Here we give a short proof of the Feit-Tits theorem. Theorem 7.1.
Suppose m p ( S ) = d p ( S ) for a finite non-abelian simple group S . Then p = 2 and m p ( S ) = 2 n ( S ) , where n ( S ) is the smallest positive integer n such that S embedsin Sp n (2) .Proof.
1) Certainly, m p ( S ) ≤ d p ( S ). Also, suppose that p = 2 and S embeds in someSp n (2). Since GL n ( F ) contains a subgroup of the form ( C ◦ n + ) · Sp n (2), we see that m p ( S ) ≤ n . Thus m p ( S ) ≤ n ( S ) .Set m := m p ( S ) and let H ≤ GL m ( F ) = GL( V ) where H is a finite group with S a section of H . By passing to a subgroup, we may assume that H surjects onto S andno proper subgroup of H surjects onto S . Now, for any proper normal subgroup N of H , if N is not contained in a maximal subgroup M of G , then M N = H and so M/ ( M ∩ N ) ∼ = H/N . By the minimality of H , S is not a composition factor of N , henceit is a composition factor of H/N . It follows that a subgroup of M projects onto S , acontradiction. Thus every proper normal subgroup of H is contained in Φ( H ), and so H/ Φ( H ) ∼ = S . It also follows that H is perfect. Since m p ( S ) < d p ( S ) by the hypothesis,we see that Φ( H ) > Z ( H ).2) Clearly, we may assume that H is an irreducible subgroup of GL( V ). Now we choose N ≤ Φ( H ) to be a noncentral normal subgroup of H of smallest possible order. So N is an r -group for some prime r . If r = p , then N acts trivially on V by irreducibility, acontradiction (since H ≤ GL ( V )). Thus r = p .If N is abelian, then, since N is not central, m is at least the size of the smallest orbit of H on the set of nontrivial irreducible characters of N by Clifford’s theorem. This impliesthat some nontrivial homomorphic image H/K of H embeds in S m . By 1), S is a quotientof H/K , and so m = m p ( S ) ≤ m p ( S m ) ≤ m −
1, a contradiction. On the other hand, if r = 2 then N must be abelian by Proposition 2.1(ii). Hence p = r = 2. In particular, we IFTING IN FRATTINI COVERS AND A CHARACTERIZATION OF SOLVABLE GROUPS 21 see that O ′ (Φ( H )) ≤ Z ( H ) and the only noncentral normal subgroups of H contained inΦ( H ) are non-abelian 2-groups.Let M be any normal subgroup of H properly contained in N . Then M is central bythe minimality of N , and so M ≤ Z ( H ). In particular, Z ( N ) = N ∩ Z ( H ) and so itis cyclic by Schur’s lemma. Furthermore, Z ( N/Z ( N )) = N/Z ( N ), and N/Z ( N ) is anelementary abelian 2-group. Also, H acts irreducibly on N/Z ( N ), and the commutatormap defines an H -invariant nondegenerate alternating form on N/Z ( N ) (with values inΩ ( Z ( N )) ∼ = C ). Since O ′ (Φ( H )) ≤ Z ( H ) acts trivially on N/Z ( N ), the irreducibilityof H on N/Z ( N ) implies that Φ( H ) acts trivially on N/Z ( N ). Recall that S = H/ Φ( H )is simple. Now if S acts trivially on N/Z ( N ), then [ H, N ] ≤ Z ( N ) centralizes H andso [ H, N ] = [[
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