New York Journal of Mathematics
New York J. Math.24(2018) 429–442.
The topology of local commensurability graphs
Khalid Bou-Rabee and Daniel Studenmund
ABSTRACT. We initiate the study of the p-local commensurability graph of a group, wherepis a prime. This graph has vertices consisting of all finite-index subgroups of a group, where an edge is drawn betweenAandBif[A:A∩B]
and[B:A∩B]are both powers ofp. We show that any component of thep-local commensurability graph of a group with all nilpotent finite quotients is complete.
Further, this topological criterion characterizes such groups. In contrast to this result, we show that for any primepthep-local commensurability graph of any large group (e.g. a nonabelian free group or a surface group of genus two or more or, more generally, any virtually special group) has geodesics of arbitrarily long length.
CONTENTS
1. Preliminaries and basic facts 431
2. Nilpotent groups: The Proof of Theorem 1 434
3. Free groups: The Proof of Theorem 2 436
References 441
Let G be a group. The commensurability index of two commensurable sub- groupsA,B≤Gis[A:A∩B][B:A∩B].For a prime number p, the p-local com- mensurability graph ofG, denoted Γp(G), is the graph with vertices consisting of finite-index subgroups ofGwhere two subgroupsA,B≤Gare adjacent if and only if their commensurability index is a power of p. For a warm-up example, see Figure1.
The goal of this paper is to draw algebraic information ofGfrom the topology ofΓp(G).
Theorem 1. Let G be a finitely generated group. The following are equivalent:
(1) For any prime p, every component ofΓp(G)is complete.
(2) All of the finite quotients of G are nilpotent.
Received March 4, 2018.
2010Mathematics Subject Classification. Primary: 20E26 and 20E15; Secondary: 20B99 and 20F18.
Key words and phrases. commensurability, nilpotent groups, free groups, very large groups.
K.B. supported in part by NSF grant DMS-1405609.
D.S. supported in part by NSF grants DMS-1246989 and DMS-1547292.
ISSN 1076-9803/2018
429
The proof of Theorem 1 is in§2. The structure theory of solvable groups plays an important role in our proofs. Theorem1applies, for example, to Grigorchuk’s group [Gri83], which is a 2-group and therefore has only nilpotent finite quotients.
FIGURE1. Let Sym3be the symmetric group on 3 elements (note Sym3 is solvable and not nilpotent). The figure above displays Γ2(Sym3),Γ3(Sym3), andΓ5(Sym3)in that order. AllΓp(Sym3) for primesp>3 are discrete spaces.
In contrast to the above theorem, we show that components of the local com- mensurability graphs of free groups are far from complete:
Theorem 2. Let F be a rank two free group. For any prime p and N>0, there exist infinitely many geodesicsγ, each in a different component ofΓp(F), such that the length of eachγis greater than N.
We prove Theorem2in§3. A result of Robert Guralnick (which uses the classi- fication of finite simple groups) concerning subgroups of prime power index in a nonabelian finite simple group is used in an essential way in our proof [Gur83].
Moreover, in our proof we get a clean description of an entire component of the p-local commensurability graph of many finite alternating groups. See Figure 2, for example.
Our next result demonstrates that arbitrarily long geodesics in the p-local com- mensurability graph of a free group cannot possibly all come from a single com- ponent. We prove this at the end of§1.
Proposition 3. Let G be a finitely generated group. LetΩbe a connected compo- nent ofΓp(G). Then there exists C>0such that any two points inΩare connected by a path of length less than C. That is, the diameter ofΩis finite. Moreover if any vertex ofΩis a normal subgroup of G then the diameter ofΩis bounded above by 3.
As a consequence of Theorem 2 and Proposition 3, there exists components of the p-local commensurability graph of a nonabelian free group with no normal subgroups as vertices (see Corollary22at the end of§3).
Recall that a group is largeif it contains a normal finite-index subgroup that admits a surjective homomorphism onto a non-cyclic free group. Such groups enjoy the conclusion of Theorem2. See the end of§3for the proof.
Corollary 4. Let G be a large group. For any prime p and N >0, there exists infinitely many geodesicsγ, each in a different component ofΓp(G), such that the length of eachγis greater than N.
Experiments that led us to the above theorems were done using GAP [GAP15]
and Mathematica [W15].
The results of this paper are motivated by the authors’ interest in studying the metric properties of thecommensurability graphof a finitely generated groupG, denotedΓ(G)and defined to be the complete graph on the set of all finite-index sub- groups ofGwith edges weighted by the commensurability indices. The weighted path metric gives Γ(G) the structure of a metric space. This graph encodes, for instance, Alex Lubotzky and Dan Segal’ssubgroup growth[LS03] as the growth of balls in the graph-theoretic star of the vertex G in Γ(G). The study of local commensurability graphs Γp is analogous to studying the local subgroup growth functions, where one only considers subgroups of index a power of a fixed prime p. In a forthcoming paper we will investigate the full geometry of commensu- rability graphs and explore connections between local and global aspects. Note, however, that in this paper we consider a local commensurability graph as a metric space with the standard path metric as an abstract graph; in particular, a ‘geodesic’
is a path which minimizes the edge lengths of paths between its vertices.
This paper sits in the broader program of studying infinite groups through their residual properties, which is an area of much activity (see, for instance, [KT16], [BRK12], [BRM11], [GK17], [BRHP15], [BRS16], [KM11], [Riv12], [Pat13], [LS03]). Specifically, a similar object is studied in [AAH+15]. There a graph is constructed with vertices consisting of subgroups of finite index, and an edge is drawn between two vertices if one is a prime-index subgroup (the prime is not fixed) of the other. They show that for every groupG, their graph is bipartite with girth contained in the set{4,∞}and ifGis a finite solvable group, then their graph is connected.
Acknowledgements.We are grateful to Ben McReynolds and Sean Cleary for useful and stimulating conversations. An anonymous referee provided the simple proof of Lemma14that appears here. Another anonymous referee provided several clar- ifications and corrections to the exposition, including the proof of Lemma5.
1. Preliminaries and basic facts
In this section we record some basic facts that will be used throughout. We start with a couple of elementary results.
Lemma 5. Let N≤G be a normal subgroup andπ:G→G/N the quotient map.
For subgroups K≤H≤G we have
[H:K] = [π(H):π(K)][H∩N:K∩N].
Proof. Consider the action ofH∩N on the coset space H/K by left translation.
LetZbe the kernel of this action, and note thatZ≤K∩N. This action has number
of orbits equal to[π(H):π(K)], so Burnside’s lemma gives
|(H∩N)/Z| ·[π(H):π(K)] =
∑
g∈(H∩N)/Z
|(H/K)g|.
On the one hand, we clearly have
|(H∩N)/Z|= [H∩N:K∩N]· |(K∩N)/Z|.
On the other hand, a cosethK ∈H/K is fixed precisely by the conjugateh((K∩ N)/Z)h−1under the action of(H∩N)/Z, so we have
∑
g∈(H∩N)/Z
|(H/K)g|=
∑
hK∈H/K
|{g∈(H∩N)/Z|ghK=hK}|
=|H/K| |(K∩N)/Z|.
The desired result follows.
Lemma 6. Let N be a normal subgroup of G and p a prime. If A and N are both subgroups of index a power of p in G, then[G:A∩N]is also a power of p.
Proof. Letπ:G→G/Nbe the quotient map. Then[A:A∩N] =|π(A)|. Because G/Nis ap-group, it follows that[A:A∩N]is a power ofp. Therefore[G:A∩N] =
[G:A][A:A∩N]is a power ofp.
Our next couple of lemmas give control of local commensurability graphs under some maps.
Lemma 7. If G is a group,π:G→Q is a surjection, andγa path inΓp(G), then π(γ)is a path inΓq(Q)with length bounded above by the length ofγ.
Proof. IfK≤H≤Gthen[π(H):π(K)]divides[H:K]by Lemma5. Therefore adjacent vertices inγmap to adjacent vertices inπ(γ), or are possibly identified in
Γp(Q).
Lemma 8. Suppose G is a group and p is prime.
(1) If N is a normal subgroup of G, then the quotient mapπ:G→G/N in- duces an isometric graph embedding Γp(G/N)→Γp(G) as an induced subgraph.
(2) If H is a finite-index subgroup of G, then the inclusion i:H→G induces a graph embeddingΓp(H)→Γp(G)as an induced subgraph.
(3) If N is a finite-index normal subgroup of G, then the inclusion i:N → G induces an isometric graph embeddingΓp(N)→Γp(G)as an induced subgraph.
Proof. For1, ifπ:G→G/Nis a quotient map, then the assignmentK7→π−1(K) defines a graph embeddingΓp(G/N)→Γp(G) whose image is an induced sub- graph. This embedding is isometric by Lemma7.
For2, ifH≤Ghas finite index, then the assignmentK7→i(K)defines a graph embeddingΓp(H)→Γp(G)whose image is an induced subgraph.
For3, let NCGbe a finite-index subgroup, with assignmentφ :K7→i(K)de- fined over all subgroupsKinN. LetH1,H2∈φ(Γp(N))and letH1=J1, . . . ,Jn= H2be a path inΓp(G)fromH1toH2. Then for eachi=1, . . . ,n−1, we have that
[Ji:Ji∩Ji+1][Ji+1:Ji∩Ji+1]
is a power of p. By Lemma 7, π(J1), . . . ,π(Jn)is a path in Γp(G/N). Because J1≤N, this is a path of p-subgroups ofG/N. Therefore[Ji:Ji∩N]is a power of pfor alli=1, . . . ,n. Thus, by Lemma6applied toJi∩NandJi+1∩Ji, we have for i=1, . . . ,n−1,
[Ji:(Ji∩N)∩(Ji∩Ji+1)][Ji+1:(Ji+1∩N)∩(Ji∩Ji+1)], is a power ofp. Hence, fori=1, . . . ,n−1,
[Ji:N∩Ji][N∩Ji:N∩Ji∩Ji+1] = [Ji:N∩Ji∩Ji+1]
is a power of p giving that [N∩Ji :N∩Ji∩Ji+1] is a power of p, since above we showed that[Ji:N∩Ji]is a power of p. By a similar argument, we get that [N∩Ji+1:N∩Ji∩Ji+1]is a power of p, and thusN∩JiandN∩Ji+1are adjacent in Γp(G). It follows that the pathJ1, . . . ,Jn can be replaced by the path (which possibly has repeated vertices)J1=J1∩N,J2∩N, . . . ,Jn−1∩N,Jn∩N=Jn, which is entirely contained inΓp(N). It follows thatΓp(N)is a geodesic metric space in
the path metric induced fromΓp(G), as desired.
Note that the hypothesis of normality in3cannot be removed. For example, sup- pose S and T are disjoint sets with |S|=|T|=5 and consider the non-normal subgroup AltS×AltT ≤AltS∪T. It can be shown using Lemma18below that AltS
and AltT are in the same component ofΓ5(AltS∪T)but in different components of Γ5(AltS×AltT).
Our next lemma will lead to our first result concerning free groups.
Lemma 9. Let A,B be vertices inΓp(G). Suppose B shares an edge with A. If qk divides[G:A]for some prime q6=p then qkdivides[G:B].
Proof. In this case, we have
[G:A∩B] = [G:A][A:A∩B] = [G:B][B:A∩B].
Hence, if qk divides[G:A], then qk must divide [G:B]because[B:A∩B]is a
power of p.
Proposition 10. The p-commensurability graph of a free group has infinitely many components.
Proof. Any free group has subgroups{N1,N2,N3, . . .}with distinct prime indices {q1,q2,q3, . . .}. By the previous lemma, any vertex that is in the connected com- ponent ofNi has index divisible byqi. Thus, no path exists betweenNi andNj for
distincti,j.
We finish this section by proving a general result: for any groupG, any compo- nent ofΓp(G)has finite diameter.
Proof of Proposition3. LetGbe any group andΩa component ofΓp(G). Take any vertex A inΩand letN be the normal core ofA. Let π:G→G/N be the quotient map. LetD={BN:B∈Ω}. We claim that the diameter ofΓp(G)is less than|D|+2.
LetBbe a subgroup in Ω. LetV1, . . . ,Vm be a path inΓp(G) connectingAto B. Then by Lemma5,π(V1), . . . ,π(Vm)is a path inΓp(G/N)connectingπ(A)to π(B). Hence
V1N,· · ·,VmN
is a path connectingAtoBN, and soBNis an element ofΩ. Further, if[G:B] = nprwheregcd(n,pr) =1, then[G:BN] =npe by Lemma9. SinceB≤BN and [G:BN][BN:B] = [G:B], we get
npe[BN:B] =npk
and therefore[BN:B] = pk−e. HenceBN andBare adjacent inΓp(G). It follows that there is an edge from any element inΩto one inD, and so the diameter ofΩ is bounded above by the diameter of the subgraph induced byDplus 2. This gives the desired bound|D|+2.
IfΩcontains a normal subgroup as a vertex then we can pickA=Nin the above argument. ThereforeDis the set ofp-subgroups ofG/N. Any two such subgroups are connected by an edge, so the diameter ofΩis bounded above by 3.
2. Nilpotent groups: The Proof of Theorem1
We will prove Theorem1in two steps, as Propositions 12and15 below. For a finite nilpotent group G let Sp(G) denote the unique Sylow p-subgroup of G.
Recall thatGis the direct product of its Sylow subgroups.
Lemma 11. Suppose G=Sp1(G)× · · · ×Spk(G) for primes p1,· · ·,pk. Let πi: G→Spi(G) be the quotient map for each i. Then any subgroup H≤G has the form H=π1(H)× · · · ×πk(H).
Proof. Choose`1, . . . , `k so thatgp
`i
i =1 for allg∈Spi(G). ChooseNso that N p`11· · ·p`k−1k−1 ≡1 (mod p`kk).
Take anyh∈Hand writeh= (h1, . . . ,hk)forhi∈Spi(G)for alli. Then hN p
`1
1···p`k−1k−1 = (1, . . . ,1,hk).
Therefore(1, . . . ,1,hk)∈H, and so we may identifyπk(H)with a subgroup ofH.
Applying this argument to each other factor, the result follows.
Proposition 12. If G is a finitely generated group such that every finite quotient of G is nilpotent, then every component ofΓp(G)is complete for all p.
Proof. SupposeAandBare subgroups ofGin the same component ofΓp(G)for some prime p and take any path A=P0,P1, . . . ,Pn =B from A to B. Let N be a normal, finite-index subgroup ofGcontained inPi for everyi. ThenG/N is a
nilpotent group andπ(P0),π(P1), . . . ,π(Pn)is a path inΓp(G/N), whereπ:G→ G/Nis the quotient map.
LetPbe a finite set of primes so thatG/N=∏q∈PSq(G/N). By Lemma11we have decompositionsπ(Pi) =∏q∈PSq(π(Pi))for eachi. It is straightforward to see that
π(Pi)∩π(Pi+1) =
∏
q∈P
Sq(π(Pi))∩Sq(π(Pi+1))
for anyi, and so for j=ior j=i+1 we have [π(Pj):π(Pi)∩π(Pi+1)] =
∏
q∈P
[Sq(π(Pj)):Sq(π(Pi))∩Sq(π(Pi+1))].
Since π(Pi) and π(Pi+1) are adjacent in the p-local commensurability graph of G/N, it follows that Sq(π(Pi)) =Sq(π(Pi+1))for all iand all q6= p. Therefore Sq(π(A)) =Sq(π(B))for allq6=p, and so[π(A):π(A)∩π(B)][π(B):π(A)∩π(B)]
is a power of p. Because [K:L] = [π(K):π(L)]for any subgroupsL≤K≤G containingN, this shows thatAandBare adjacent inΓp(G).
Lemma 13. If Q is a finite solvable group that is not nilpotent then there is some prime p so that a connected component ofΓp(Q)is not complete.
Proof. LetΠbe the set of prime divisors of the order of the finite solvable group Q. For any primeq∈Πthere is a Hall subgroupHqso that[Q:Hq] =qkfor some kandqdoes not divide the order ofHq. BecauseQis not nilpotent, there is some prime pand a Hall subgroupHpso thatg−1Hpg6=Hpfor someg∈Q. Then both Hp andg−1Hpgare adjacent toQinΓp(Q), but there is no edge between Hpand
g−1HpginΓp(Q).
Lemma 14. If Q is a non-nilpotent finite group then Q contains a non-nilpotent solvable subgroup.
Proof. Suppose every solvable subgroup of Qwere nilpotent. Take any prime p and letSbe ap-Sylow subgroup ofQ. LetT ≤Sbe any nontrivial subgroup. Then NQ(T)/CQ(T) is a p-group. If it were not, then there would be an element x∈ NQ(T)−CQ(T)and a prime numberq6=pso thatxq∈CQ(T). Then the subgroup H≤Qgenerated byxandSwould have order with only two prime divisors, hence be solvable and therefore nilpotent. Sincex is in aq-Sylow subgroup of H, this would meanx∈CQ(T).
By Frobenius’ normal p-complement theorem, there is a normal subgroupN≤ Qof order prime topso thatQ=SN. Because this argument holds for anyp,Qis solvable by Hall’s theorem. This is a contradiction.
Proposition 15. Suppose G is a finitely generated group with a finite-index, nor- mal subgroup N such that G/N is not nilpotent. Then there is some p so that a component ofΓp(G)is not complete.
Proof. TakeGandNas above, letQ=G/Nand letπ:G→Qbe the quotient map.
IfQis solvable, then by Lemma13there is a prime pand subgroupsA,B≤Qin the same component ofΓp(Q)that are not adjacent. Thenπ−1(A)andπ−1(B)are
non-adjacent vertices in the same component ofΓp(G)by Lemma8, soΓp(G)is not complete.
Now consider the case that Qis not solvable. By Lemma 14 there is a non- nilpotent solvable subgroupS≤Q. By Lemma 13there is some prime pwith a componentΩofΓp(S)that is not complete. By Lemma8the componentΩfully embeds in a component ofΓp(G), which is therefore not complete.
3. Free groups: The Proof of Theorem2
LetF be the free group of rank two. Let p be a prime and N∈N be given.
By Lemma8, to prove Theorem2it suffices to find a finite quotient QofF with subgroupsA,B≤Qsuch that the length of any geodesic inΓp(Q)connectingAto Bis greater thanN. Our candidate forQis AltX, the alternating group on a setXof more thanpk>Nelements, and our candidates forAandBare conjugates of AltS
for a subsetS⊆Xwith pkelements.
We first need a couple technical group theoretic results. First, we give a descrip- tion of a connected component inΓp(AltX). This requires a simple lemma.
Lemma 16. If|T1∩T2| ≥2and|T1|,|T2| ≥4, thenhAltT1,AltT2i=AltT1∪T2. Proof. We prove this by induction on|T1∪T2|. The case thatT1=T2 is clear, so supposeT16=T2. The base case, when|T1|=|T2|=4 and|T1∩T2| ∈ {2,3}, follows by computation (we did this in [GAP15]). For the inductive step, suppose without loss of generality that x ∈T1\T2. By inductive hypothesis
AltT1\{x},AltT2
=
AltT1∪T2\{x}. Arguing similarly ifT2\T1is nonempty, we reduce to the case when T1∪T2\T1∩T2consists of at most two points. To finish, we claim that any 3-cycle on points inT1∪T2is inhAltT1,AltT2i. Letv1,v2,v3be distinct points inT1∪T2. If {v1,v2,v3} ⊆T1or{v1,v2,v3} ⊆T2, then we are done. Thus, by suitably relabeling, we may assumev1,v2∈T1andv3∈T2. Further, sinceT1∪T2\T1∩T2consists of at most two points, then by relabeling again, we may assumev2∈T2. Selectw1,w2∈ T1∩T2 that are distinct from v1, v2, and v3. Then, by the base case applied to Alt{v1,v2,v3,w1}≤AltT1 and Alt{v1,v2,v3,w2}≤AltT2, we obtain that Alt{v1,v2,v3,w1,w3}is contained inhAltT1,AltT2i, and hence the desired 3-cycle is found. This completes
the proof.
For any subsetS⊆X, we denote the symmetric group on S by SymS and the alternating group onSby AltS. For a subgroupP≤SymSwe define thesupportto be the complement of the fixed point set of the action ofPonS.
Lemma 17. Let p be a prime number and k an integer so that pk>4. Let X be a finite set, S⊆X , and P≤SymX a p-group with support disjoint from S. Let E be an index pj subgroup of AltS×P. If|S|=pk or|S|=pk−1, then we have the decomposition E=AltT×P0 for some P0≤P and some T ⊆S with|T|= pk or
|T|=pk−1.
Proof. Letπ: AltS×P→AltSbe the projection map. By Lemma5we have [AltS×P:E] = [AltS:π(E)][1×P:E∩(1×P)].
The left hand side of this equation is a power of p, so [AltS :π(E)] is a power of p. Because |S|=pk or|S|=pk−1 by assumption, Theorem 1(a) in [Gur83]
immediately implies that eitherπ(E) =AltS or|S|=pk andπ(E) =AltS\{v} for somev∈S. LetT denote the set such thatπ(E) =AltT. Letqbe 3 if p6=3 and qbe 2 if p=3. For the case p6=3, recall that AltT is generated by 3-cycles by elementary properties of alternating groups. In the case p=3, note that pk >6.
Because Alt6is generated by an element of order 2 and one of order 4, Lemma16 implies that AltT is generated by elements of order 2 or 4 in this case. Therefore in either case it follows that AltT is generated by elementsg1, . . . ,gk each with order dividing a power ofq. Sinceπmaps onto AltT, we have that for eachi=1, . . . ,k, there existsvi∈Psuch that(gi,vi)∈E. Sincevi∈P, we have that the order ofvi
is coprime withgi, hence asq6=p, there exists`such that (gi,vi)`= (gi,1).
It follows then that E contains all of AltT×1, and henceE =AltT×P0 where
P0≤P, as desired.
LetΩS,X be the component ofΓp(AltX)containing AltS, and letBS,X denote the set of subgroups inΩS,X isomorphic to AltT for some|T| ∈ {pk,pk−1}. For odd primesp, we get the following description:
Lemma 18. Let S⊆X be a set of cardinality pk for some odd prime p such that pk >4. Vertices of the componentΩS,X inΓp(AltX)consist of two classes of sub- groups:
Type 1.subgroups of the formhAltT,Pi, where|T|=pkand P≤AltX, and Type 2.subgroups of the formhAltT,Pi, where|T|=pk−1and P≤AltX. In either case, the subgroup is AltT×P, where P is a p-group with support in Tc. Moreover, for all primes p, if V is a vertex of Type 1 or Type 2, the set T is uniquely determined by V .
Proof. We first show uniqueness ofT. This implies that Type 1 and Type 2 are disjoint classes. LetV be a vertex with distinct decompositions AltTi×Pi with
|Ti|>3 and p-groupPiwith support inTic fori=1,2 such thatT16=T2. IfT1∩T2 is empty, then
[V : AltT1×AltT2×1][AltT1×AltT2×1 : AltT1×1] = [V : AltT1×1] =|P1|, and thus[AltT1×AltT2×1 : AltT1×1] =|AltT2|must be a power of p. But this is impossible as|AltT2|is either(pk)!/2 or(pk−1)!/2 for pk>4. Thus,T1 andT2 overlap. IfT16=T2then AltT1×1 cannot be normal inV because AltT2 acts transi- tively onT2. But AltT1×1 is clearly normal in AltT1×P1, so this is a contradiction.
ThereforeT1=T2.
Since elements inBS,X are of Type 1 or 2 and AltS itself lies inBS,X, it suffices to show that anyE that is adjacent to an element of Type 1 or 2 must itself be of Type 1 or 2.
LetEbe adjacent toV =AltT×PwherePis ap-group with support inTcand
|T|=pk or|T|=pk−1. ThenE∩V is a subgroup of AltT×Pof index a power
of p. By Lemma17, E∩V =AltT×P0 or E∩V =AltT\{v}×P0 whereP0 ≤P andv∈T. We will therefore assume without loss of generality that E contains AltT×1=AltT as a subgroup ofppower index.
Suppose thatE does not leaveT invariant. Let T1,T2,· · ·,Tk be the orbit of E acting onT and note thatE contains AltTi for eachi. SupposeTi∩Ti+1 has fewer than two elements for somei. The group AltTicontains AltTi\Ti∩Ti+1, which includes a permutation of order 2 since|Ti|>4. Hence E contains AltTi×Z/2Z≥AltTi. This is impossible, as AltTi is of index pk inE for an odd prime p. We therefore know that Ti∩Ti+1 has more than two elements for every i. Then by applying Lemma16we conclude thatEcontains AltT1∪T2∪···∪Tk. SinceE contains AltT as a subgroup of prime power index andT1∪· · ·∪Tk6=T1, it follows that|T1∪· · ·∪Tk|= pk and in factEcontains AltT1∪···∪Tk as a subgroup of indexp`for some`.
We may therefore assume, after replacingT withT1∪ · · · ∪Tk if necessary, that EleavesTinvariant. ThenE≤SymT×QwhereQis a group with support disjoint fromT. Letπ: SymT×Q→SymT be the projection onto the first coordinate. By Lemma5,[π(E): AltT]divides[E: AltT]and hence is a power ofp. It follows that π(E) =AltT, as AltT is a maximal subgroup of SymT of index two. Further, since AltT is normal, we apply Lemma5to the mapψ: AltT×Q→Qto see that|ψ(E)|
is a power ofp. Applying Lemma17we obtain the desired conclusion.
The primep=2 requires relaxing the conclusion of Lemma18, since any sym- metric group on three or more elements contains an alternating group of index 2.
Lemma 19. Let S⊆X be a set of cardinality 2k such that k>2. Vertices of the componentΩS,X inΓ2(AltX)consist of at least one of two types:
Type 1’.subgroups V such that AltT×1≤V ≤SymT×P, where|T|=2k and P≤AltX, and
Type 2’. subgroups V such that AltT×1≤V ≤SymT×P, where |T|= 2k−1and P≤AltX.
In either case, P is a2-group with support in Tc.
Proof. Since elements inBS,X are of Type 1’ or 2’, it suffices to show that anyE that is adjacent to an element of one of the types must itself be of one of the types.
LetE be adjacent to someV with AltT×1≤V ≤SymT×P wherePis a 2- group. BecauseV has index a power of 2 in SymT×P, we know thatE∩V also has index a power of 2 in SymT×P. Since AltT×P is a normal subgroup of SymT×P, we have by Lemma6that(AltT×P)∩E∩V has index a power of 2 in SymT×P, and hence in AltT×P. By Lemma17,E∩V∩(AltT×P) =AltT×P0 orE∩V∩(AltT×P) =AltT\{v}×P0 whereP0≤Pandv∈T. We conclude that AltT×1 or AltT\{v}×1 has index a power of 2 inE∩V∩(AltT×P), and hence has index a power of 2 inE. We will therefore assume without loss of generality that Econtains AltT×1=AltT as a subgroup with index a power of 2, where|T|=2k or|T|=2k−1.
Suppose thatE does not leaveT invariant. Let T1,T2,· · ·,Tk be the orbit of E acting onT and note thatE contains AltTi for eachi. SupposeTi∩Ti+1 has fewer
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FIGURE2. ΩS,X with|S|=5 and|X|=7. The coloring gives the types and the numbers give the valence of each vertex. This figure was generated using GAP [GAP15] and Mathematica [W15]
than two elements for somei. The group AltTicontains AltTi\Ti∩Ti+1, which includes a permutation of order 3 because|Ti|>3. HenceE contains AltTi×Z/3Z≥AltTi. This is impossible, as AltTi is of 2 power index in E. We therefore know that Ti∩Ti+1has more than two elements for everyi. Then by applying Lemma16we conclude that E contains AltT1∪T2∪···∪Tk. SinceE contains AltT as a subgroup of prime power index andT1∪ · · · ∪Tk6=T1, it follows that|T1∪ · · · ∪Tk|=2kand in factE contains AltT1∪···∪Tk as a subgroup of index 2`for some`.
We may therefore assume, after replacingT withT1∪ · · · ∪Tk if necessary, that E leavesT invariant. Then AltT×1≤E≤SymT×QwhereQis a 2-group with
support disjoint fromT, as desired.
Note that groups of Type 1 and Type 2 are of Type 1’ and Type 2’ respectively.
The next result allows us to restrict attention to geodesics inBS,X when computing distances there.
Lemma 20. Let S⊆X be a set of cardinality pk>4for some prime p and integer k. Then BS,X is a geodesic metric space in the path metric induced fromΩS,X. Proof. We first need a local fact. LetV1,V2be two adjacent vertices inΩS,X. If p is odd, then by Lemma18we haveVi=AltTi×Pi, where|Ti|=pkor|Ti|=pk−1 and the support ofPi is disjoint fromTi fori=1,2. If p=2, then by Lemma19, AltTi×1≤Vi≤SymTi×Pi, where|Ti|=pk or|Ti|=pk−1 and the support ofPi is disjoint from Ti for i=1,2. We claim that in either case, AltT1 and AltT2 are connected by an edge inΓp(AltX).
SinceV1andV2are adjacent, we have that
[V1:V1∩V2][V2:V1∩V2]
is a power of p. Thus, when p is odd, Lemma 17 applied twice along with the uniqueness in Lemma18gives thatV1∩V2is AltU×PwhereU⊆T1∩T2satisfies
|U|=pk or|U|= pk−1 and P≤P1∩P2. Thus it is straightforward to see that AltT1 is adjacent to AltT2.
When p=2, setHi=AltTi×1 andΛ=V1∩V2. ThenHi is normal inVi, thus Hi∩Λis normal inΛ. Since[SymTi×Pi:Hi]is a power of 2 and
[SymTi×Pi:Vi][Vi:Hi] = [SymTi×Pi:Hi],
we get [Vi:Hi]is a power of 2. Since Hi is normal inVi, Lemma6 implies that [Vi:Hi∩Λ]is a power of 2. Further, as[Vi:Λ]is a power of 2 and
[Vi:Λ][Λ:Hi∩Λ] = [Vi:Hi∩Λ]
we conclude that[Λ:Hi∩Λ]is a power of 2 fori=1,2. Thus, applying Lemma6 toH1∩ΛCΛandH2∩ΛCΛ, we have thatH1∩H2∩Λhas index a power of 2 in Λ. As
[Vi:Λ][Λ:H1∩H2∩Λ] = [Vi:H1∩H2∩Λ],
it follows that[Vi:H1∩H2∩Λ]is a power of 2. Because[Vi:Hi]is also a power of 2 (shown above) and
[Vi:Hi][Hi:H1∩H2∩Λ] = [Vi:H1∩H2∩Λ]
we have[Hi:H1∩H2∩Λ]is a power of 2 for eachi. By applying Theorem 1(a) in [Gur83] and the uniqueness in Lemma18, we haveH1∩H2∩Λis AltSfor some S⊆T1∩T2with|S|=pkor|S|=pk−1. Thus, AltT1is adjacent to AltT2, as claimed.
Now let γ be a path in ΩS,X that, except for its endpoints, is entirely in the complement ofBS,X. Enumerate the vertices ofγin the order they are traversed,
V1,V2, . . . ,Vm,where AltTi×1≤Vi≤SymTi×Pifor alli=1, . . . ,m Then by the previous claim, we may form a new path (after throwing out repeated vertices)
AltTi
1,AltTi
2, . . . ,AltTin.
that is entirely contained inBS,X and has the same endpoints asγ. It follows that
BS,X is geodesic inΩS,X, as desired.
Proposition 21. Let S⊆X be a set of cardinality pk >4for some prime p and integer k. There exists V,W ∈BS,X such that any path inΩS,X connecting V to W has length at least pk−max{0,2pk− |X|}.
Proof. By Proposition20, it suffices to show that there existsV,W∈BS,X such that any path inBS,X has length greater than|X| −pk. LetO1,O2⊆Xwith|O1∩O2| ≤ max{0,2pk−|X|}and either|Oi|=pkor|Oi|=pk−1 for eachi. LetE1,E2, . . . ,Em
be distinct vertices in a non-back-tracking path inBS,X connecting AltO1 to AltO2. LetT1,T2, . . . ,Tm be subsets of X such that Ei=AltTi for i=1, . . . ,m. For each i=1, . . . ,m, we have one of three cases:
(1) Eiis Type 1 andEi+1is Type 1: In this case,|Ti+1∩Ti|=|Ti| −1=pk−1.
(2) Eiis Type 1 andEi+1is Type 2: In this case,Ti+1⊂Tiand|Ti+1|=|Ti|−1= pk−1.
(3) Eiis Type 2 andEi+1is Type 1: In this case,Ti+1⊃Tiand|Ti+1|=|Ti|+1= pk.
(4) Eiis Type 2 andEi+1is Type 2: This case never occurs, as[AltT: AltU]is not a power ofpfor any proper subsetU⊂T with|T|=pk−1.
Thus, we see that for each i, we see that Ti and Ti+1 differ by moving, adding, or removing at most one element. It follows that m≥ pk− |O1∩O2| ≥ pk−
max{0,2pk− |X|}.
Proof of Theorem2. LetF be a rank two free group andpa prime. GivenN>0, choosekso thatpk>Nandpk>4. For any finite setX with|X|>2pk, letγX be a path of length pkinΓp(AltX)guaranteed by Proposition21. Then pulling backγX
over any surjectionπ:F→AltX produces a path of length pkinΓp(F)by Lemma 8. By Lemma 9, setsX1 and X2 with relatively prime cardinalities will produce
geodesics in different components ofΓp(F).
Proof of Corollary4. LetGbe a large group,pa prime, andN>0. Since a finite- index subgroup of a nonabelian free group is nonabelian, there exists a normal finite-index subgroup H≤G that surjects ontoF, the free group of rank 2. By Lemma7 and Theorem 2, there exists verticesV,W ∈Γp(H) such that any path connecting them inΓp(G)has length greater thanN. The result now follows from Lemma8, asΓp(H)isometrically embeds intoΓp(G).
Corollary 22. Let G be a large group and p be a prime. There exists a connected component ofΓp(G)that does not contain any normal subgroup.
Proof. By Proposition3, any component ofΓp(G)containing a normal subgroup as a vertex has diameter at most 3. By Corollary4, there are components ofGwith
arbitrarily long geodesics.
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This paper is available via http://nyjm.albany.edu/j/2018/24-24.html.
