Upper Bounds for Totally Symmetric Sets
aa r X i v : . [ m a t h . G R ] F e b UPPER BOUNDS FOR TOTALLY SYMMETRIC SETS
KEVIN KORDEK, QIAO LI, AND CALEB PARTIN
Abstract.
Totally symmetric sets are a recently introduced tool for studying homomorphisms betweengroups. In this paper, we give full classifications of totally symmetric sets in certain families of groups andbound their sizes in others. As a consequence, we derive restrictions on possible homomorphisms betweenthese groups. One sample application of our results is that any homomorphism of a braid group to a directproduct of solvable groups must have cyclic image. Introduction
Given any two groups G and H , understanding all the possible homomorphisms G −→ H is a naturallyimportant but technically difficult task. This motivates using a carefully-chosen subset of the information inthe given groups to constrain and eventually classify homomorphisms. As a simple example, cyclic groupsand subgroups map to cyclic images. Analogously, the motivation for using totally symmetric sets to studyhomomorphisms follows this pattern. Definition 1.1.
A totally symmetric set of a group G is a finite subset { x , x , ..., x n } of G that satisfiestwo key properties: • The elements of the subset commute pairwise. • Any permutation of the subset can be achieved through conjugation: for any element of the sym-metric group on n letters σ ∈ S n , there exists some g ∈ G such that x σ ( i ) = gx i g − for all 1 ≤ i ≤ n .Loosely speaking, a totally symmetric set is a subset of elements in a group characterized by commu-tativity and conjugation properties, and these properties are preserved under group homomorphisms. Thus,the image of a totally symmetric set under a homomorphism is also a totally symmetric set. An even strongerversion of this statement holds: the image of a totally symmetric set under a group homomorphism must bea totally symmetric set of the same size or a singleton. This fundamental lemma, shown in section 2, is amajor obstruction to the possible homomorphisms between two groups.Totally symmetric sets are first defined by Kordek and Margalit in the context of the braid group[KM19]. The fact that odd Artin generators of the braid group form a totally symmetric set is exploited incharacterizing homomorphisms from certain braid group to other certain braid groups. The same fact hasbeen used in the work of the authors along with Chudnovsky, where they derive a lower bound on the size ofnon-cyclic quotients of the braid group [CKLP20]. It has also made an appearance in the work of Caplingerand Kordek [CK20], the work of Scherich and Verberne [SV20], and the work of Chen and Mukherjea [CM20].All of these results vitally depend on the properties of totally symmetric sets. Overview.
In this paper, we will develop and present some theory about totally symmetric sets as well asa full classification of them in several families of groups. In Section 2, we give motivating examples of totallysymmetric sets and present a proof for the fundamental lemma. In Section 3, we give a full classificationof totally symmetric sets in a few special families of groups. In Section 4, we investigate upper boundson sizes of totally symmetric sets in various products of groups. In Section 5, we give constant boundson sizes of totally symmetric sets in groups of odd order and solvable groups through the action of thestabilizer. Finally, in Section 6, we present a table of sizes of totally symmetric sets and derive classificationsof homomorphisms.
Date : Monday 15 th February, 2021.
Acknowledgments.
The majority of this work was completed during the summer of 2019 during theGeorgia Institute of Technology Mathematics REU under the guidance of Dan Margalit and Kevin Kordek.The authors would like to thank Dan Margalit for his mentorship and guidance throughout this project,Alice Chudnovsky for useful discussions during the Georgia Tech Mathematics REU, and Santana Afton forhelpful conversations on free products.2.
Fundamental Lemma and Examples
Totally symmetric sets are a powerful aid to the study of homomorphisms due to the following funda-mental lemma due to Kordek and Margalit [KM19]:
Lemma 2.1 (The Fundamental Lemma of Totally Symmetric Sets) . Let ϕ : G −→ H be a homomorphismbetween two groups, and S ⊂ G a totally symmetric set of size n . Then ϕ ( S ) is a totally symmetric set in H with | ϕ ( S ) | = n or | ϕ ( S ) | = 1 .Proof. We will first show that ϕ ( S ) is a totally symmetric set of H given that S is a totally symmetric set of G . Let S = { x , x , . . . , x n } ⊂ G . It is clear that the elements of the set { ϕ ( x ) , ϕ ( x ) , . . . , ϕ ( x n ) } pairwisecommute, so we just need to show they also satisfy the conjugation requirement. Given σ ∈ S n , let g σ ∈ G be the element inducing the permutation given by σ on S , i.e. x σ ( i ) = g σ x i g − σ for all i . Applying ϕ gives ϕ ( x σ ( i ) ) = ϕ ( g σ ) ϕ ( x i ) ϕ ( g σ ) − . Thus, ϕ ( g σ ) achieves the same conjugation for ϕ ( S ) as g σ does for S .Now for the second part of the lemma, suppose that S = { x , x , . . . , x n } , and ϕ ( x ) = ϕ ( x ). We needto show that ϕ ( x ) = ϕ ( x i ) for all i . If ϕ ( x ) = ϕ ( x ), then we know that ϕ ( x x − ) = e , and x x − ∈ ker ϕ .Since S is a totally symmetric set we know that there exists an element h ∈ G such that hx h − = x and hx h − = x i . Therefore, x x − i = ( hx h − )( hx − h − ) = hx x − h − . Since ker ϕ is a normal subgroup, weknow that it is closed under conjugation and x x − i ∈ ker ϕ , so ϕ ( x ) = ϕ ( x i ). Thus if S does not mapinjectively to a totally symmetric set of size n , then it must map to a set of size 1. (cid:3) An immediate corollary of this lemma is any totally symmetric set of a subgroup H G is a totallysymmetric set in G by the inclusion map.Totally symmetric sets or their maximum sizes can provide insight about the possible homomorphismsbetween two groups. Going forward, we denote the maximum size of a totally symmetric set in a group G by S ( G ). We will discuss this in much greater detail and rigor in section 6. Before diving deep into the theory,we present two primary motivating examples to show what a totally symmetric set looks like in the wild. Example 1.
The motivating example for totally symmetric sets comes from the braid group, B n . The braidgroup is generated by a collection of half-twists, σ , σ , . . . , σ n − , the Artin generators. Take the set of oddArtin generators, X = { σ i − } ⌊ n ⌋ i =1 . One can deduce this is a totally symmetric set from the commutation relations of the braid group andthe change-of-coordinates principle from mapping class group theory [FM12, Section 1.3]. This example isespecially useful, since the commutator subgroup of the braid group, B ′ n , is normally generated by σ σ − ;see [Lin04]. From the fundamental lemma of totally symmetric sets above, we know that if B n maps to agroup G with S ( G ) < ⌊ n ⌋ , then the homomorphism collapses X to a singleton. Thus B ′ n is in the kernel ofthe homomorphism, and the map factors through the abelianization of B n , which is Z for all n .This fact that all homomorphisms from B n to a group G with S ( G ) < ⌊ n ⌋ are cyclic becomes a powerfultool in studying homomorphisms from the braid group. This is the driving force behind the recent resultson braid groups from the authors, Chudnovsky, and Kordek [CKLP20], Kordek and Margalit [KM19] andCaplinger and Kordek [CK20]. Example 2.
Another example of large totally symmetric sets comes from the symmetric group S n . Takethe standard homomorphism ϕ : B n → S n where σ i ( i i + 1). From the fundamental lemma, we knowthat the image of the totally symmetric set in Example 1 consisting of odd Artin generators will also be atotally symmetric set. Therefore, the set of disjoint transpositions { (2 i − i ) } ⌊ n ⌋ i =1 is a totally symmetricset. We will later see that these two examples are rare instances of large totally symmetric sets, since in themany classifications following this section, most families of groups will have a constant upper bound on thesize of their totally symmetric sets. PPER BOUNDS FOR TOTALLY SYMMETRIC SETS 3 Classifying Totally Symmetric Sets in Specific Groups
In this section, we will provide upper bounds for the sizes of totally symmetric sets of free groups,dihedral groups and a subset of the Baumslag–Solitar groups. These results will naturally lead to the boundsgiven on general products of groups in section 4.Recall that one of the defining conditions of a totally symmetric set is that all of its elements pairwisecommute. Thus, a natural place to start when searching for totally symmetric sets is one where all theelements commute, i.e. abelian groups.
Proposition 3.1.
Let A be an abelian group and S ⊂ A a totally symmetric set, then | S | = 1 . Hence, theonly totally symmetric sets of A are all the singleton subsets.Proof. From the definition of a totally symmetric set, we know that any permutation of the elements of S must be achieved through conjugation in A . Since A is abelian, conjugation is always a trivial action, thusthe only achievable permutation is the identity. Since all permutations must be possible, we can concludethat | S | = 1. (cid:3) It is perhaps disappointing to the reader to start out on such an uninteresting example, but this shouldmotivate us to search elsewhere for interesting totally symmetric sets. At the risk of over-correcting, we caninvestigate the totally symmetric sets of a highly non-abelian group, the free group.
Theorem 3.2.
Let F be the free group on two generators. For all G F , S ( G ) = 1 .Proof. To prove this theorem, we just need to show that S ( F ) = 1, that is, F only has trivial totallysymmetric sets.First, we will check when any two elements in F commute. If two elements a, b ∈ F commute, we knowthat they must generate an abelian subgroup. The Nielsen–Schreier theorem states that any subgroup of afree group is free, and the only abelian free group is F ∼ = Z . This implies that both of these elements arepowers of the generator of Z . If Z ∼ = h x i , then a = x n and b = x m .For a and b to be members of a totally symmetric set, there must exist an element h ∈ F such that hx n h − = x m and hx m h − = x n . Take the first of these equalities and raise it to the m th power, to obtainthe following: ( hx n h − ) m = ( x m ) m = x m . Using our second conjugation equality, we can manipulate the above expression to see that:( hx n h − ) m = hx n · m h − = ( hx m h − ) n = ( x n ) n = x n . Thus, x n = x m and n = ± m . This reduces our problem to two distinct cases.If n = m , then these are the same element and our totally symmetric set consists of a singleton. Onthe other hand, if n = − m , then hx n h − = x − n and hx − n h − = x n . Combining these equalities shows usthat h x n h − = x n , implying h commutes with x n and therefore h and x n can both be expressed as someelement y to a power.Suppose that h = y a and x n = y b . Moreover, we can assume that y is chosen to be minimal in thesense that it cannot be written as another element z c , where c >
1. This follows from the residual finitenessof the free groups. Hence, a must be even and h = y a/ . Therefore, h commutes with x n , and the equation hx n h − = x − n reduces to x n = x − n . This implies that n = 0, and thus F has trivial totally symmetricsets. (cid:3) Note since every free group on a finite or countably infinite number of generators is a subgroup of F , it follows that all free groups have trivial totally symmetric sets. Another consequence is that totallysymmetric sets under quotients will not be well-behaved, as every group is a quotient of some free group.From these first two examples, we can make an observation: The two properties that define a totallysymmetric set are inherently at odds with each other. On the one hand, we want to find subsets of elementsthat all pairwise commute, so groups that are “more abelian” seem to be the natural place to find large setsof this nature. Yet we also need conjugation to be efficacious in our group in order to achieve the full setof permutations, but conjugation is trivial when elements commute. So if we want to find examples of largetotally symmetric sets, we need to find groups that intuitively achieve this abelian, non-abelian balance thatlarge totally symmetric sets would require. KEVIN KORDEK, QIAO LI, AND CALEB PARTIN
After seeing two important classes of groups that both have trivial totally symmetric sets, we nowpresent our first example of a family of groups exhibiting non-trivial totally symmetric sets.
Theorem 3.3.
The maximal size of a totally symmetric subset of the dihedral group D n is 2 for all n ≥ . Furthermore, every totally symmetric set of size 2 must take the form { r i , r − i } or { sr i , sr i + n } for ≤ i ≤ n − , with the latter only occurring if n is divisible by 4.Proof. We will use the following presentation for the proof: D n = h r, s | r n = 1 , s = 1 , srs = r − i . Hence,any element of D n can be written as s ǫ r i where ǫ = 0 , ≤ i ≤ n −
1. We first check the necessaryconditions for any two elements of the group to commute with each other. Any r i and r j commute bydefinition, but for elements sr i and sr j , we use the following implications to show that if they commute, i = j or j = i + n : sr i sr j = sr j sr i = ⇒ s r − i r j = s r − j r i = ⇒ r j − i = r i − j . Thus, j − i ≡ i − j mod n, or 2( j − i ) ≡ i = j , or j − i = n = ⇒ j = i + n , which can onlyoccur when n is even. Therefore, the only nontrivial case of two elements of this form commuting is sr i and sr i + n . From the definition of conjugation and the fact that s = 1, two conjugate elements must have thesame exponents on s . Hence, we do not consider any further cases and the two cases discussed above aredisjoint from each other.We will now check the conditions necessary for conjugation. The following shows two elements of theform r i and r j are conjugate if and only if they are inverses: sr k r i = r j sr k = ⇒ sr k + i = sr k − j = ⇒ k + i = k − j = ⇒ i = − j. This gives the first totally symmetric set { r i , r − i } . For our other candidates, elements of the form sr i and sr i + n are conjugate if and only if i = j + n . sr j sr i = sr i + n sr j = ⇒ s r i − j = s r j − i − n = ⇒ i − j = j − i − n ⇒ i − j ) ≡ n . This implies that i − j = n or i = j + n , which can only occur if n is divisible by 4. If two elements of thisform are conjugate, there are no other elements in their conjugancy class, as sr ( i + n )+ n = sr i . We have thusobtained a full classification of the totally symmetric sets of D n , and S ( D n ) = 2.Note: This proof extends to the infinite dihedral group as well D ∞ = Z ⋉ Z , showing that its totallysymmetric sets also have a maximal size of 2. (cid:3) The classification in the above proof relies on the convenient presentation of the dihedral group. Whilethis direct proof approach will not work in general, there is another class of groups where we can explicitlycompute totally symmetric sets from the group presentation: Baumslag–Solitar groups. For nonzero integers m, n , the Baumslag–Solitar group BS ( m, n ) is defined by the following presentation: h a, b | ba m b − = a n i . Theorem 3.4.
The maximal size of a totally symmetric subset of the Baumslag–Solitar group BS (1 , n ) is1 when n = − and 2 when n = − .Proof. The group BS (1 , n ) has the one-relation presentation h a, b | bab − = a n i . This relation is equivalentto ba = a n b , which allows us to write any element of this group in the form a i b j . We can use this fact tocheck necessary conditions for different elements of the group to commute and be conjugates. If elements a i b j and a x b y commute, we have the following implication: a i b j a x b y = a x b y a i b j = ⇒ a i a xn j b j b y = a x a in y b j b y . This gives the condition that i + xn j = x + in y . We couple this with necessary conditions for conjugation.If elements a i b j and a x b y are conjugated to each other by a x b y , we have that a x b y a i b j = a x b y a x b y = ⇒ a x a in y b y b j = a x a x n y b y b y . Hence, conjugation give us two conditions: x + in y = x + x n y and y + j = y + y = ⇒ j = y . Usingthe latter condition, we can combine this with the conditions for commutativity to obtain the followingstatement: elements a x b y and a i b y (as j = y ) will commute and be conjugate when i + xn y = x + in y .Rearranging and solving for i , we obtain an expression for i in terms of x, y, n : i = x (1 − n y )1 − n y . PPER BOUNDS FOR TOTALLY SYMMETRIC SETS 5
This implies that i = x , which makes the two elements the same, except when the denominator is 0. For thedenominator to be 0, we have three cases to check. It must either be the case that y = 0, y = 0 and n = 1,or y = 2 m for some integer m and n = − y = 0, we only need to check when two elements a x and a y are conjugate. By definition of atotally symmetric set there exists an element h swapping a x and a y through conjugation: ha x h − = a y and ha y h − = a x . We derive conditions on the exponents by raising the former equality to the power of y :( ha x h − ) y = a y y = ⇒ ha xy h − = a y = ⇒ ( ha y h − ) x = a y = ⇒ ( a x ) x = a y . So a x = a y and x = ± y . If x = y , then the two elements are the same. Otherwise, if x = − y , we insteadhave ha x h − = a − x . As previously discussed, we can represent our general element h as the element b i a j .Plugging this in gives b i a x b − i = a − x , which becomes a xn i = a − x , so x (1 + n i ) = 0. Either x = 0 or n i = − n = − i is an odd integer. Thus, wecan examine this case by assuming n = − b m +1 a j . This gives b m +1 a x b − (2 m +1) = a − x . Using the conjugation relation, we can see that conjugating a x an odd number oftimes by b gives a − x , so the equation is always true. Hence, when n = −
1, two elements of the form a x , a − x form a totally symmetric set of size 2.The other case to consider is when y = 0. This breaks down into two subcases, the first of which is n = 1.When n = 1, the relation for BS (1 , n ) becomes the commuting relation for a and b , implying that BS (1 , n = − y = 2 m for m ∈ Z . Thus, we need to check when two elements of the form a x b m and a y b m commute and are conjugate.Using our general commuting condition from above, we substitute 2 m for j and y : i + x ( − m = x + i ( − m .This reduces to i + x = x + i , which is always true.We now must check that for any two elements a x b m and a y b m , there exists an element h that swapsthem through conjugation, i.e. ha x b m h − = a y b m and ha y b m h − = a x b m . As above, we raise the formerequation to y and simplify:( ha x b m h − ) y = ( a y b m ) y = ⇒ ha xy b my h − = a y b my = ⇒ ( ha y b m h − ) x = a y = ⇒ ( a x ) x = a y . Therefore, x = y and x = ± y . Again, if x = y , then the elements are equal, otherwise if x = − y , we canwrite h as b i a j to obtain the following: b i a j a x b m a − j b − i = a − x b m a ( x + j )( − i b m + i a − j b − i = a − x b m a ( x + j )( − i +( − j )( − m + i b m = a − x b m a x ( − i b m = a − x b m . Thus, the elements a x b m and a − x b m form a totally symmetric set. Any other element in such a totallysymmetric set would have to satisfy the above conditions with a x b m and a − x b m , which is impossible foran element not equal to either. (cid:3) For the groups above, we are able to directly compute the totally symmetric sets by exploiting theexistence of their normal forms. In general, we want to be able to compute the totally symmetric sets ofgroups without having to rely on assumptions such as the existence of normal forms or explicit descriptionsof centralizers and conjugacy classes. For many groups, such a direct computation will not be possible, sowe need to introduce more powerful techniques.4.
Totally Symmetric Sets in Products of Groups
A natural question that arises in studying totally symmetric sets is how they behave under variousproducts, such as direct products, free products, and semi-direct products. To this end, we will prove thefollowing theorem:
Theorem 4.1.
Let G and H be groups. S ( G × H ) = max { S ( G ) , S ( H ) } S ( G ∗ H ) = max { S ( G ) , S ( H ) } KEVIN KORDEK, QIAO LI, AND CALEB PARTIN
The next step after understanding how totally symmetric sets behave under direct and free productsis to explore semi-direct products or other non-trivial group extensions. While we will not classify all totallysymmetric sets in semi-direct products, we will generalize our results on dihedral groups from the previoussection.
Proposition 4.2.
For p prime and m any integer where p | m , S ( Z p ⋉ Z m ) = 2 . Direct and Free Products.
While we may expect that taking the direct product of two groupsprovides a way of creating larger totally symmetric sets, we begin by showing that direct products do notcreate larger totally symmetric sets.
Lemma 4.3.
Let G and H both be groups and S a subset of G × H . If there exists elements ( x , y ) , ( x , y ) , ( x , y ) ∈ S , with x = x , y = y , and y any element in H , then S cannot be a totally symmetric subset of G × H .Proof. Suppose that S is a totally symmetric subset of G , and it contains elements of the form above.Since S is a totally symmetric set, we know that there is a permutation φ that sends ( x , y ) to ( x , y )and ( x , y ) to ( x , y ), and there is a conjugating element ( h G , h H ) that achieves this permutation. By thefundamental lemma, the image of S under the projection map π G is a totally symmetric subset of G . Fromthe permutation φ , we have ( h G , h H )( x , y )( h G , h H ) − = ( x , y ) and ( h G , h H )( x , y )( h G , h H ) − = ( x , y ).Using these equations, we derive a contradiction that h G conjugates x to two different elements. To do this,we first apply π G to both sides of the first equation. This yields: π G (( h G , h H )( x , y )( h G , h H ) − ) = π G (( h G , h H )) π G (( x , y )) π G (( h G , h H ) − )= h G x h − G = π G (( x , y )) = x . Similarly, applying π G to both sides of the second equation, we find that: π G (( h G , h H )( x , y )( h G , h H ) − ) = π G (( h G , h H )) π G (( x , y )) π G (( h G , h H ) − )= h G x h − G = π G (( x , y )) = x . Since h G conjugates x to itself and x , we have a contradiction. Therefore, S cannot be a totally symmetricsubset of G × H . (cid:3) Corollary 4.4.
Let G and H be groups, and let S = { ( x , y ) , . . . , ( x k , y k ) } ⊆ G × H be a totally symmetricset. All the x i are equivalent, or they are all distinct. The same holds for y i . Moreover, S ( G × H ) =max { S ( G ) , S ( H ) } .Proof. By Lemma 4.3, if a set S is a totally symmetric subset of G × H , it cannot simultaneously containelements of the form ( x , y ) , ( x , y ) and ( x , y ). There are thus three possibilities for the elements of S :(1) S contains elements of the form ( x , y ) and ( x , y ) but none of the form ( x , y ). We also can’tcontain elements of the form ( x , y ′ ) for some y ′ = b , since the labels could easily be swapped betweenthe elements. This implies that a totally symmetric set in this case will take the form S = { ( x , y ) , . . . , ( x n , y n ) } . All the x i are distinct elements of G and each y i is any element of H .(2) We allow elements of the form ( x , y ) and ( x , y ) but none of the form ( x , y ). This is equivalentto the previous case by relabeling.(3) S contain elements of the form ( x , y ) and ( x , y ), but none of the form ( x , y ). This implies thata set in this case will take the form S = { ( x , y ) , . . . , ( x , y n ) } , where x ∈ G and all the y i ∈ H .The above implies that a totally symmetric set S = { ( x , y ) , . . . , ( x k , y k ) } must have all the x i equiv-alent or all distinct. The same can be said of the y i by the symmetry of the direct product.Now that we know the totally symmetric sets of G × H must take on these specific forms, we can firstshow that S ( G × H ) ≤ max { S ( G ) , S ( H ) } . Any non-trivial totally symmetric subset of G × H will have PPER BOUNDS FOR TOTALLY SYMMETRIC SETS 7 either the x i or y i elements all be distinct. Suppose that we have a totally symmetric set S ′ such that allthe x i elements are all distinct. By Lemma 2.1, the set π G ( S ′ ) is a totally symmetric subset of G , in thiscase with the same cardinality as S ′ due to all the x i being distinct. The same can be said if S ′ had all the y i distinct and we mapped it to H by π H . Together these imply that a totally symmetric subset of G × H has cardinality bounded above by max { S ( G ) , S ( H ) } .To see that S ( G × H ) = max { S ( G ) , S ( H ) } , we will construct a totally symmetric subset of sizemax { S ( G ) , S ( H ) } in G × H . Without loss of generality, suppose that S ( G ) ≥ S ( H ) and let T = { x , . . . , x n } be a maximal totally symmetric subset of G . Consider the set T ×{ e H } ⊂ G × H . This is a totally symmetricsubset of G × H . Any two elements ( x i , e H ) and ( x j , e H ) commute. Additionally, any permutation on theelements of T ×{ e H } corresponds to a permutation of T . If g σ is the element of G that induces a permutationon T , then ( g σ , e H ) will be the element that induces the same permutation on T × { e H } . (cid:3) The following corollary will not be used again in the rest of the paper, but it describes totally symmetricsets in direct products if we ask all the elements to be distinct in each coordinate.
Corollary 4.5.
Let G and H be groups, and let S = { ( x , y ) , . . . , ( x k , y k ) } ⊆ G × H be a totally symmetricset. If all the x i are distinct and all they y i are distinct, then | S | = min { S ( G ) , S ( H ) } . As shown, totally symmetric sets in direct products of groups behave in a straightforward way. Wemight now ask whether the same behavior persists in additional group constructions, such as free products,semi-direct products, group extensions. While the latter two are much more difficult to answer, it turns outthat we can say the following about totally symmetric sets of the free product of two groups.
Proposition 4.6.
Let G and H be groups, and let G ∗ H denote their free product. Then S ( G ∗ H ) =max { S ( G ) , S ( H ) } . Moreover, any totally symmetric subset of G ∗ H is of the form wSw − , where w ∈ G ∗ H ,and S is a totally symmetric subset of G or H .Proof. The direction S ( G ∗ H ) ≥ max { S ( G ) , S ( H ) } is almost immediate. To show this let S be a totallysymmetric subset of either G or H . Since G and H both inject into their free product, we have that S is alsoa totally symmetric subset of G ∗ H . Thus, we see S ( G ∗ H ) ≥ max { S ( G ) , S ( H ) } . For the other direction,we will show that every totally symmetric set in G ∗ H comes from a totally symmetric subset of either G or H in an injective manner, thus proving S ( G ∗ H ) ≤ max { S ( G ) , S ( H ) } .To begin, let w and w be commuting elements of G ∗ H . Then w and w are both in the sameconjugate of a factor of G ∗ H or both powers of some element w ∈ G ∗ H [MKS04, Corollary 4.1.6]. The formerof these means that w and w are both contained in xGx − or xHx − for some element x ∈ G ∗ H . Thiscorollary proves an even stronger statement: if we have a set of pairwise commuting elements { x , . . . , x n } where only one is in the conjugate of one of the free factors, then all the elements must be in that sameconjugate.Hence, given a totally symmetric set S in G ∗ H , there are three possible forms of the elements of the set:(1) Conjugates of G : S = { wg w − , . . . , wg n w − } for g i ∈ G and w ∈ G ∗ H .(2) Conjugates of H : S = { wh w − , . . . , wh n w − } for h i ∈ H and w ∈ G ∗ H .(3) Powers of the same element: S = { v i , . . . , v i n } for v ∈ G ∗ H and i ≤ i ≤ · · · ≤ i n .We will first examine the first two of these three cases. Without loss of generality, suppose that wehave a totally symmetric set S of the first type, with all its elements in a conjugate of G . Since this is atotally symmetric set, any permutation of its elements can be achieved through conjugation. Let σ be apermutation of the elements of S and let w σ ∈ G ∗ H be an element of G ∗ H that achieves this permutationby conjugation. If σ ( i ) = j , then w σ wg w − w − σ = wg j w − , which implies w σ = wg σ w − and g σ g i g − σ = g j for g σ ∈ G . This combined with commutativity of g i shows S ′ = { g , . . . , g n } is a totally symmetric subsetof G with the same size as S . Therefore, any totally symmetric subset of the first two forms above can havesize at most max { S ( G ) , S ( H ) } .In the other case, all elements of S are powers of the same element v ∈ G ∗ H . We can assume v is not an element of a conjugate of one of the factors, since otherwise, all powers of it would be as well,which would then reduce to case (1) and (2). Moreover, all elements with torsion in a free product areconjugates of finite-order elements in one of the factors [MKS04, Corollary 4.1.4], implying v must haveinfinite order. Since our set S = { v i , . . . , v i n } is totally symmetric, there exists an element w ∈ G ∗ H KEVIN KORDEK, QIAO LI, AND CALEB PARTIN such that wv i w − = v i , wv i w − = v i , and w fixes all other element of S . This reduces to a similarsituation as in the proof of Theorem 3.2 and it has a similar solution. We can show that i = ± i by taking wv i w − = v i , and exponentiating both sides to the i th power:( v i ) i = ( wv i w − ) i = wv i i w − = ( wv i w − ) i = ( v i ) i . This implies v i = v i , so i = i and i = ± i . In either case we have now reduced our totally symmetricset S to either a singleton or { v i , v − i } , as permutations such as the one above must be possible for all pairsof elements in S .We now check if { v i , v − i } is a totally symmetric set. Again, we must have an element w ∈ G ∗ H suchthat wv i w − = v − i , and wv − i w − = v i . As we saw in the proof of Theorem 3.2, this implies that w commutes with v i . From above, either w and v i are in the same conjugate of one of the factors of G ∗ H ,or they are powers of the same element. If v i was a conjugate of an element in either G or H , v is also aconjugate of an element in G or H , contradicting our initial assumption. Therefore, v i = z a and w = z b for some element z ∈ G ∗ H . We can assume z is primitive, that it can’t be written as a power of anotherelement in G ∗ H .We now show that b is even and w = z b . Since w is a freely reduced word in G and in H , it eitherhas even or odd length. If w has even length, then its starting and ending elements cannot both be from thesame group. Thus, w is exactly the concatenation of w with itself, with no cancelling or reducing involved.Since z is reduced and primitive, w is a concatenation of b copies of z . Alternatively, suppose that w hasodd length, implying it ends and begins with an element from the same factor. If cancellation occurs whenconcatenating, w and z are both conjugates of the same element x ∈ G ∗ H : z = xz ′ x − , w = xw ′ x − .This reduces to proving that w ′ = z ′ b , where w ′ , z ′ cannot be written as conjugates of some shorter lengthelement. Thus, assume w and z are not conjugates of a smaller length element. When taking powers ofboth elements, the only reduction will be happening at the letters where they are concatenated, so we canconclude that w = z b .Since w and v i are both powers of z , w commutes with v i . This implies v i = v − i and thus, S is asingleton. Hence, there are no non-trivial totally symmetric sets of form (3), and all totally symmetric setsof a free product come from conjugates of totally symmetric sets in the factors. (cid:3) We can now see that totally symmetric sets behave in a reasonably nice way with regards to these basicproduct operations and have a full proof of Theorem 4.1. However, before we rejoice over this predictablebehavior of totally symmetric sets under products, it’s worth pointing out a disappointment: at this point,we still don’t have a way to construct larger totally symmetric sets from small ones on the group level: Themaximum size of totally symmetric sets under direct and free products are limited by the totally symmetricsets within its factors. It has been a theme throughout this paper that large totally symmetric sets are rare.4.2.
Semi-direct Product of Cyclic Groups.Lemma 4.7.
Let G be a group and S ⊂ G a totally symmetric subset of G such that g, g − ∈ S for some g ∈ G . Then S = { g, g − } .Proof. Label the elements of the totally symmetric set S as S = { g, g − , x , . . . , x n } for some non-negativeinteger n . Since S is a totally symmetric set, there exists an element h ∈ G such that hgh − = g and hg − h − = x i for one of the x i ∈ S . Inverting both sides of the former of these equalities shows hg − h − = g − , thus we must have that x i = g − . This process can be continued for all elements of S , hence, S = { g, g − } . (cid:3) Proposition 4.2.
For p prime and m any integer where p | m , S ( Z p ⋉ Z m ) = 2. Proof.
We use the presentation Z p ⋉ Z m = { s, r | r p = e, s m = e, srs − = r k } for the proof. A typical elementin the group is of the form r a s b , where a ∈ Z /p Z and b ∈ Z /m Z . We first investigate the conditions for two PPER BOUNDS FOR TOTALLY SYMMETRIC SETS 9 elements, r a s b and r x s y , to commute.: r a s b r x s y = r x s y r a s b r a − x s b r s = s y r a s b − y r a − x s b r s = s y r a s − y s b r a − x s b r s = r ak y s b r a − x − ak y ( s b r x s − b ) = er a − x − ak y r xk b = er a − x − ak y + xk b = e. By the group presentation, we obtain the following equation:(1) a − x − ak y + xk b = 0 mod p. We put this aside and discuss the conditions for the same two elements, r a s b and r x s y , to also beconjugates of each other. Before we go through the calculations, we take a detour to discuss s − rs . By srs − = r k , it follows that s − rs = r l , where l is the multiplicative inverse of k modulo p , which existsbecause p is prime. Suppose there exists some element r e s f that conjugates r a s b to r x s y . We can use thegroup presentation to show that the equation b + y = 0 mod m must hold: r e s f r a s b s − f r − e = r x s y r e s b s f s − b r a s b s − f r − e = r x s y r e s b s f r al b s − f r − e = r x s y r e s b r al b k f − e = r x s y r e s b r al b k f − e s − b s − y s b = r x r e +( al b k f − e ) k b s b + y = r x . Deciphering the exponents, we now have the following relations: b + y = 0 mod m , and e + ( al b k f − e ) k b = x mod p . Recall b, y ∈ Z /m Z and thus, 0 ≤ b, y < m . The relation b + y = m mod m then impliesthat either b = y = 0, or y = m − b .The case that b = y = 0 amounts to considering elements of the form r a and r x as candidate elementsof the totally symmetric set. The commuting relation (1) shows elements of this form always commute, andthe second of the two conjugation relations reduces to ak f ≡ x mod p given b = 0 and k b ≡ p .Therefore, choosing a conjugating element r e s f determines x , and the relation shows the choice of e will notaffect the conjugation. It is important to note that because f is an integer between 0 and m −
1, multiplechoices of f could lead to the same x . Let 0 ≤ f = f ′ ≤ m − k f ≡ k f ′ mod p . This implies that k f − f ′ ≡ p and thus, f and f ′ must differ by a multiple of p − f as above to determine an x , and let f ′ = f + n ( p − r e s f ′ acts on r x by conjugation in the same way that r e s f does, i.e. different choices of f that determine the same x will act the same by conjugation: r e s f ′ r x s − f ′ r − e = r e r xk f ′ r − e = r xk f + n ( p − = r xk f . In order for r a and r x to form a totally symmetric set, we need an element that conjugates r a to r x to also conjugate r x back to r a . Hence, we need xk f ≡ a mod p . Combining this with ak f ≡ x mod p , wehave that k f ≡ p and thus, k f ≡ ± p . This tells us that x = ± a , and one cannot make alarger totally symmetric set containing the elements r a and r x .For the other case, assume that y = m − b , and use (1) to derive an equation to determine the followingequations expressing x in terms of a and b : a − x − ak m − b + xk b = 0 mod px ( k b −
1) = a ( k m − b −
1) mod p. (2) If k b − = 0 mod p , then given a, b , there is a unique solution for x . Otherwise, we consider the casewhere k b − ≡ p .From above we know that x = ± a and thus we have the following three elements to consider for atotally symmetric set, r a s b , r a s − b , and r − a s − b . We know that we can construct two totally symmetric setsfrom these elements, namely { r a s b , r a s − b } and { r a s b , r − a s − b } , but we also need to check if all three elementscan form a totally symmetric set. Under our assumption that k b ≡ p , we can show that r a s b and r − a s − b are inverses: r a s b r − a s − b = r a ( s b rs − b ) − a = r a ( r k b ) − a = r a r − a = e. By Lemma 4.7, this inhibits all three elements from being in a totally symmetric set together, so themaximal size of a totally symmetric set in Z p ⋉ Z m is 2. (cid:3) The Stabilizer
In this section, we introduce the stabilizer of the totally symmetric set and use it to bound the car-dinality of totally symmetric sets in groups. In particular, we will show that the totally symmetric sets inodd-ordered finite groups and solvable groups are small.Let G be a group, S a totally symmetric subset of size n in G , and Stab G ( S ) the stabilizer of S in G under conjugation. We claim that Stab G ( S ) surjects onto S n , the symmetric group on n letters. The stabilizeracts on S by conjugation, so we can explicitly construct a homomorphism ϕ : Stab G ( S ) → Sym ( S ) ∼ = S n described by sending γ ∈ Stab G ( S ) to the automorphism of S given by s → γsγ − . By definition, everypermutation of S is realized as a conjugation by some γ ∈ Stab G ( S ), so this homomorphism is surjective.Equivalently, we can write this as a short exact sequence, where k denotes the kernel of ϕ :1 −→ k −→ Stab G ( S ) −→ S n −→ G is of finite order, | G | ≥ | Stab G ( S ) | = | k || S n | = n ! | k | . The kernel consists of elementsthat commute with all elements of the totally symmetric set, or equivalently the intersection of centralizersof each element. In particular, the subgroup generated by the totally symmetric set will be a subgroupof the kernel k by commutativity. We can relate the cardinality of the group G to the size of its totallysymmetric sets through bounding the size of the kernel. Using a similar technique, Chudnovsky and theauthors showed that for braid groups and their commutator subgroups, the required cardinality of G growssuper-exponentially as n increases [CKLP20].We use the existence of the surjection from the stabilizer to the symmetric group to bound the sizes oftotally symmetric sets in the following groups. Proposition 5.1.
Let G be a finite group with odd order. Then S ( G ) = 1 .Proof. Let | G | = 2 n + 1 for some n ∈ N , and let S ⊂ G be a totally symmetric set in G . From the shortexact sequence above, the order of Sym ( S ) must divide the order of Stab G ( S ). Moreover, since the order ofStab G ( S ) divides the order of the whole group G , we have that | S | ! | | G | and hence | S | ! | n + 1. Since for | S | ≥ | S | ! is even, we must have that | S | = 1. (cid:3) Theorem 5.2.
Let G be a solvable group. Then S ( G ) ≤ .Proof. Let S ⊂ G be a totally symmetric subset of G . Since Stab G ( S ) is a subgroup of G , and G is solvable,Stab G ( S ) is also solvable. Moreover, from the short exact sequence above, we know that Sym ( S ) is a quotientof Stab G ( S ), so it is solvable. For n ≥ S n is not solvable. Thus, | S | ≤ (cid:3) We do not know if this bound is sharp. In other words, it’s an open question whether there are totallysymmetric sets in solvable groups of size 3 or 4.6.
Corollaries for Homomorphisms
In this last section, we summarize the results and display various corollaries on possible homomorphismsbetween the groups discussed in this paper. The corollaries on the homomorphisms we present stem fromthe following two corollaries of the fundamental lemma of totally symmetric sets:
PPER BOUNDS FOR TOTALLY SYMMETRIC SETS 11
Corollary 6.1.
Let G and H be groups with S ( G ) > S ( H ) . No homomorphism f : G −→ H can be injective.Proof. Let S ⊂ G be a totally symmetric set of size S ( G ) and f : G −→ H be a homomorphism. From thefundamental lemma, S maps to a totally symmetric set in H of size S ( G ) or a singleton. Since S ( G ) >S ( H ) ≥
1, there are no totally symmetric sets in H of size S ( G ), and S maps to a singleton. Since S is nota singleton, f is not injective. (cid:3) This corollary tells us that S ( G ) serves as an obstruction to G being a subgroup of another group. Afurther corollary that has been used frequently in the recent results on braid groups using totally symmetricsets is the following: Corollary 6.2.
Let B n be the braid group on n strands for n ≥ and let G be a group such that S ( G ) < ⌊ n ⌋ ,then any homomorphism f : B n −→ G is cyclic, i.e. it factors through Z . A refinement of this statement using totally symmetric sets by Chudnovsky and the authors [CKLP20]and improved by Caplinger and Kordek [CK20] says that this is true if | G | < ⌊ n ⌋ ! · ⌊ n ⌋− . This bound wasrecently improved upon by Scherich and Verberne [SV20].The table below summarizes the bounds on totally symmetric sets derived in this paper. Applying theabove two corollaries gives us a list of groups which can have no injective homomorphisms between them.For example, Z p ⋉ Z mp cannot be a subgroup of the Baumslag–Solitar group BS (1 , n ). Likewise, we obtainmany example of groups that the braid groups only map cyclically to. For example, by Corollary 6.2 anyhomomorphism from B n where n ≥
10 to solvable groups must have cyclic image. Using the last line, wecan take arbitrary products of the groups in this table, and B n with n ≥
10 will map cyclically to it. Forexample, any homomorphism from B n into ( BS (1 , ∗ ( Z ⋉ Z )) × D has cyclic image. S ( G ) Group1 AbelianFree Group ( F n )Odd Order Group BS (1 , n ), n = 12 Dihedral ( D n ) Z p ⋉ Z np BS (1 , − ≤ { S ( G ) , S ( H ) } Direct Product, G × H Free Product, G ∗ H References [CK20] Noah Caplinger and Kevin Kordek. Small quotients of braid groups, 2020.[CKLP20] Alice Chudnovsky, Kevin Kordek, Qiao Li, and Caleb Partin. Finite quotients of braid groups.
Geometriae Dedicata ,pages 1–8, 2020.[CM20] Lei Chen and Aru Mukherjea. From braid groups to mapping class groups, 2020.[FM12] B. Farb and D. Margalit.
A Primer on Mapping Class Groups . Princeton mathematical series. Princeton UniversityPress, 2012.[KM19] Kevin Kordek and Dan Margalit. Homomorphisms of commutator subgroups of braid groups. arXiv preprintarXiv:1910.06941 , 2019.[Lin04] Vladimir Lin. Braids and permutations. arXiv preprint math/0404528 , 2004.[MKS04] Wilhelm Magnus, Abraham Karrass, and Donald Solitar.