Local commensurability graphs of solvable groups
LLocal commensurability graphs of solvable groups 1
Local commensurability graphsof solvable groups
Khalid Bou-Rabee ∗ Chen ShiOctober 15, 2018
Abstract
The commensurability index between two subgroups A , B of a group G is [ A : A ∩ B ][ B : A ∩ B ] .This gives a notion of distance amongst finite-index subgroups of G , which is encoded in the p-local commensurability graphs of G . We show that for any metabelian group, any component ofthe p -local commensurabilty graph of G has diameter bounded above by 4. However, no universalupper bound on diameters of components exists for the class of finite solvable groups. In theappendix we give a complete classification of components for upper triangular matrix groups inGL ( , F q ) . Let G be a group and p a prime number. The p-local commensurability graph of G has all finite-index subgroups of G as vertices, and edges are drawn between A and B if [ A : A ∩ B ][ B : A ∩ B ] isa power of p . This graph was first introduced by K.B. and D. Studenmund in [BRSc] where theyproved results regarding the diameters of components. The connected diameter of a graph is thesupremum over all graph diameters of every component of the graph. In [BRSc], it was shown thatevery p -local commensurability graph of any nilpotent group has connected diameter less than orequal to one. In contrast, they also showed that every nonabelian free group has connected diameterequal to infinity for all of their p -local commensurability graphs. Here we show that solvable groups,in a sense, fill in the space between free groups and nilpotent groups.Our first result shows that metabelian groups, like nilpotent groups, have a uniform bound onthe connected diameters of their p -local commensurability graphs. This behavior is peculiar sincemany group invariants (e.g., subgroup growth [LS03], word growth [Alp00], residual girth growth[BRC, BRSb]) do not distinguish metabelian groups in the class of non-nilpotent solvable groups. Theorem 1.
The connected diameter of the p-local commensurability graph of any metabeliangroup is at most . Theorem 1 follows quickly from the following more technical result. Theorem 2, for instance,demonstrates that there exists many solvable groups that are not metabelian yet have uniform boundson the connected diameters of their p -local commensurability graphs (e.g., the group of upper trian-gular matrices in GL ( n , F q ) , for n > q a prime). The proofs of Theorems 1 and 2 appear in § Theorem 2.
Let G be a finite group. If a p-Sylow subgroup of [ G , G ] is normal in G, then theconnected diameter of the p-local commensurability graph of G is at most 4. Our next major result shows that the above theorems cannot be extended to arbitrary solvablegroups. We prove this in § Theorem 3.
For any D > , there exists a finite solvable group whose -local commensurabilitygraph has connected diameter greater than D. ∗ K.B. supported in part by NSF grant DMS-1405609 and Cycle 46 PSC-CUNY Research Award a r X i v : . [ m a t h . G R ] D ec ocal commensurability graphs of solvable groups 2 It would be interesting to determine whether this result extends to other primes. As a corollaryof Theorem 3, we arrive at a new proof of [BRSc, Theorem 2] for the case p =
3. We note that in[BRSc] alternating groups are used in a significant way, while here we use solvable groups (whichescapes the use of the classification of finite simple groups). The proof of the corollary also appearsin § Corollary 4.
Let F be a nonabelian free group. The -local commensurability graph of F hasinfinite connected diameter. In the appendix, we give an explicit description of the p -local commensurability graphs for thegroup of upper triangular matrices in GL ( F q ) , as p and q varies over all primes.We end the introduction with a brief history account. While the article, [BRSc], began thestudy of local commensurability graphs with the goal of drawing fundamental group properties fromresidual invariants (a direction of much activity: [KT], [BRK12], [BRM11], [GK], [BRHP], [BRSa],[KM11], [Riv12], [Pat13], [LS03]), the study of graphs given by relations between subgroups goesback to work of B. Cs´ak´any and G. Poll´ak in the 1960’s [CP69]. The p -local commensurabilitygraphs are weighted pieces of what are called the intersection graphs of a group (see, for instance,[AHM15]). This graph has vertices consisting of proper subgroups where an edge is drawn betweentwo groups if their intersection is non-trivial. The weights we add to these graphs significantlychange the theory, and gives connections to A. Lubotzky and D. Segal’s subgroup growth function(which, for every n , gives the number of subgroups of index n ). Indeed, one can think of the theoryof commensurability graphs as an interface between the theories of intersection graphs and subgroupgrowths. Acknowledgements
The authors thank Michael Larsen and Daniel Studenmund for useful con-versations. In particular, K. B. is grateful to Michael Larsen for suggesting the use of p -containmentgraphs that appear in the proof of Theorem 3. Also, we are grateful to Daniel Studenmund for helpfulcorrections on an earlier draft. Notation
Let G be a group and p be a prime number. We denote the p -local commensurability graph of G by Γ p ( G ) . If H and M are two vertices in the same component of Γ p ( G ) , then we say H is p-connected to M . If H and M are adjacent in Γ p ( G ) , then we say H and M are p-adjacent . A path connecting H and M is called a p-path and is denoted by H − V − · · · − V m − M , where the length of this path isdefined to be m + H , V , . . . , V m , M are the vertices along the path. We need some technical results for our proofs. In the following, we assume that G is a finite groupwith derived series G = G (cid:66) G (cid:66) . . . given by G i = [ G i − , G i − ] for i = , , . . . . Lemma 5.
Let Q be a normal p-subgroup of G. Let V be a vertex of Γ p ( G ) , then V is p-adjacent tothe group V Q.Proof. We need to show that [ V : V ∩ V Q ] and [ V Q : V ∩ V Q ] are both powers of p . Clearly, we have V ⊂ V Q , then [ V : V ∩ V Q ] = [ V : V ] = = p .Now, recalling a basic fact from algebra, | V Q | = | V || Q || V ∩ Q | , ocal commensurability graphs of solvable groups 3 gives [ V Q : V ∩ V Q ] = [
V Q : V ] = | V Q || V | = | V || Q || V ∩ Q | | V | = | Q || V ∩ Q | . Since Q is a p -group, [ V Q : V ∩ V Q ] is a power of p . It follows that V is p -adjacent to V Q , asdesired.
Lemma 6.
Let Q be a normal p-subgroup of G. Suppose A , B < G and A and B are p-adjacent, thenAQ and BQ are p-adjacent.Proof.
We need to show that | AQ || AQ ∩ BQ | and | BQ || AQ ∩ BQ | are both powers of p . To this end, we compute | AQ || A ∩ B | = | A || Q || A ∩ B || A ∩ Q | . Since Q is a p -group, then [ Q : A ∩ Q ] = | Q || A ∩ Q | is power of p . Hence, | AQ || A ∩ B | is a power of p . Also, A ∩ B < AQ ∩ BQ < AQ . Then [ AQ : A ∩ B ] = [ AQ : AQ ∩ BQ ][ AQ ∩ BQ : A ∩ B ] is power of p . Therefore, [ AQ : AQ ∩ BQ ] is a power of p . By symmetry, [ BQ : AQ ∩ BQ ] is also apower of p . Lemma 7.
Let V and W be subgroups of G. If V is p-adjacent to W and N (cid:47)
G. Then V ∩ N isp-adjacent to W ∩ N.Proof.
Let π : G → G / N , then by [BRSc, Lemma 5], with H = V , K = V ∩ W , and N = V ∩ N , weget [ V : V ∩ W ] = [ π ( V ) : π ( V ∩ W )][ V ∩ N : V ∩ W ∩ N ] Since V is p-adjacent to W , then [ V : V ∩ W ] is a power of p . Hence, [ V ∩ N : V ∩ W ∩ N ] is a powerof p . Similarly, [ W ∩ N : W ∩ V ∩ N ] is a power of p .The previous lemma gives us a rigidity result: Lemma 8.
Let i be a natural number. Let Q be a p-subgroup of G that is normal in G (then V Qand W Q are groups) where Q ∩ G i is the p-Sylow subgroup of G i . If V Q is p-connected to W Q, thenV Q ∩ G i = W Q ∩ G i .Proof. Since Q ∩ G i is the p -Sylow subgroup of G i , we have that p does not divide | G i / ( G i ∩ Q ) | .Thus, every connected component of Γ p ( G i / ( G i ∩ Q )) is a singleton (see Proposition 13). In partic-ular, if V Q ∩ G i is p -connected to W Q ∩ G i , then because both these groups contain Q ∩ G i , we have V Q ∩ G i = W Q ∩ G i . To finish, note that G i (cid:47) G and V Q is p -connected to W Q . Thus,
V Q ∩ G i is p -connected to W Q ∩ G i by Lemma 7, which completes the proof. Proof of Theorem 2.
Let G be a finite group and p a prime number. Let Q be the p -Sylow subgroupof G = [ G , G ] (it is unique by assumption since all p -Sylow subgroups are conjugate). Let V and W be finite-index subgroups of G that are p -connected by a path in Γ p ( G ) such that V = V − V −· · · − V m = W . Now, by Lemma 5 and Lemma 6 we obtain a new path in Γ p ( G ) , V − V Q − V Q − · · · − V m Q − V m . Define A : = (cid:104) V Q , V Q , . . . , V m Q (cid:105) and B : = A / ( V Q ∩ G ) . Let π : A → B be the projection from A to B . By Lemma 8, ker π = V i Q ∩ G for all i = , . . . , m .Thus, ker π contains [ V i Q , V i Q ] . ocal commensurability graphs of solvable groups 4 Next, we claim that π ( V i Q ) is abelian for all i = , . . . , m . Indeed, if we restrict π to V i Q , weobtain a surjective homomorphism f : V i Q → π ( V i Q ) , with f ([ V i Q , V i Q ]) = [ π ( V i Q ) , π ( V i Q )] . By the previous paragraph, we have [ V i Q , V i Q ] is in the kernel of π , and so π ( V i Q ) is abelian asclaimed.Let V (cid:48) i : = π ( V i Q ) , then, as shown above, V (cid:48) i are abelian for all i = . . . , m . Let Π be the set of allprimes in the prime factorization of | V (cid:48) | that are not p . Let H be a Hall Π subgroup of V (cid:48) . The group V (cid:48) is abelian, so H and V (cid:48) ∩ V (cid:48) are normal in V (cid:48) . Thus, by [BRSc, Lemma 6], [ V (cid:48) : V (cid:48) ∩ V (cid:48) ∩ H ] is apower of p . Since | H | is coprime with p and [ V (cid:48) : V (cid:48) ∩ V (cid:48) ∩ H ] = [ V (cid:48) : H ][ H : V (cid:48) ∩ V (cid:48) ∩ H ] , it follows that V (cid:48) ∩ V (cid:48) contains H , which means V (cid:48) contains H as a Hall Π subgroup. Repeating thisprocess, we get H is a Hall Π -subgroup for V (cid:48) i , for all i = , . . . , m . Thus, H is of index a power of p inside all V (cid:48) i . In particular, we conclude that π − ( H ) is p -adjacent to both V Q and V m Q . So we geta path V − V Q − π − ( H ) − V m Q − V m of length 4, as desired.We are now ready to prove Theorem 1. Proof of Theorem 1.
Let G be a metabelian group and p a prime number. For any two vertices V , W in a component of Γ p ( G ) , let N be the normal core of V ∩ W . Let π : G → G / N be the quotient map.Note that G / N is a finite group.The p -Sylow subgroup Q of the abelian group, [ G / N , G / N ] , is unique. Hence, since [ G / N , G / N ] is normal in G / N , it follows that Q is normal in G / N . Then Theorem 2 applies to show that π ( V ) is p -connected to π ( W ) via a path of length at most 4 in Γ p ( G / N ) .[BRSc, Lemma 8, Part 1] now applies to show that V is connected to W in Γ p ( G ) via a path oflength at most 4. We are done since V and W were arbitrary vertices in Γ p ( G ) , and so the connecteddiameter of Γ p ( G ) is at most 4. We introduce a slightly modified graph, which we call the p-containment graph of a group G . Thisgraph has vertices the finite-index subgroups of G , and an edge is drawn between two vertices ifone is a p -power index subgroup of the other. We set cd p ( G ) to be the connected diameter of the p -containment graph of G . Lemma 9. cd ( Sym ) = . Moreover, any path in the -containment graph connecting (cid:104) ( , ) (cid:105) to (cid:104) ( , ) (cid:105) must have two consecutive vertices ∆ and ∆ such that ( i , j ) ∈ ∆ ∩ ∆ where i ∈ { , } andj ∈ { , } .Proof. We discovered this through [GAP15] computations. Please see Figure 2, the componentcontaining (cid:104) ( , ) (cid:105) is in the upper left corner. The geodesics connecting (cid:104) ( , ) (cid:105) to (cid:104) ( , ) (cid:105) are of theform (cid:104) ( , ) (cid:105) , (cid:104) ( , ) , ( , , k ) (cid:105) , (cid:104) ( i , k ) (cid:105) , (cid:104) ( i , k ) , ( i , j , k ) (cid:105) , (cid:104) ( , ) (cid:105) , where i ∈ { , } and j , k ∈ { , } . Removing back-tracking from any path connecting (cid:104) ( , ) (cid:105) to (cid:104) ( , ) (cid:105) reduces to one of the above geodesics. The lemma follows immediately.We remark that the corresponding statement in Lemma 10, below, for connected diameters of p -local commensurability graphs is not true. This is why we use p -containment graphs. ocal commensurability graphs of solvable groups 5 Figure 1: The 3-containment graph for Sym . The component containing (cid:104) ( , ) (cid:105) is in the upperleft corner. The vertices (cid:104) ( , ) (cid:105) and (cid:104) ( , ) (cid:105) are the two vertices inside the square. This image wascreated with the use of [Res15]. Lemma 10.
Let H , H , H and H be isomorphic copies of a group H and set ∆ = H × H × H × H and G = ∆ (cid:111) Sym , where the action of Sym on ∆ is the permutation action on the coordinates. Thencd ( G ) ≥ cd ( H ) + . Proof.
Let V , V , . . . , V d be a geodesic path in the 3-containment graph of H of length d : = cd ( H ) .Set E = (cid:104) V × V × V d × V d , ( , ) (cid:105) and E = (cid:104) V d × V d × V × V , ( , ) (cid:105) . These two vertices are connected via a path given by the vertices: E = (cid:104) V × V × V d × V d , ( , ) (cid:105) , (cid:104) V × V × V d × V d , ( , ) (cid:105) , (cid:104) V × V × V d × V d , ( , ) (cid:105) , ... (cid:104) V d × V d × V d × V d , ( , ) (cid:105) , (cid:104) V d × V d × V d × V d , ( , , ) , ( , ) (cid:105) , (cid:104) V d × V d × V d × V d , ( , ) (cid:105) , (cid:104) V d × V d × V d × V d , ( , , ) , ( , )) (cid:105) , (cid:104) V d × V d × V d × V d , ( , ) (cid:105) , (cid:104) V d × V d × V d − × V d − , ( , ) (cid:105) , (cid:104) V d × V d × V d − × V d − , ( , ) (cid:105) , ... (cid:104) V d × V d × V × V , ( , ) (cid:105) . ocal commensurability graphs of solvable groups 6 Thus, they are in the same component. To finish, suppose, for the sake contradiction, that there isa geodesic with endpoints E and E with length equal to d : E = A , A , . . . , A d = E . Let π bethe projection G → G / ∆ ∼ = Sym and φ i be the maps G → H i given by A (cid:55)→ A ∩ ∆ (cid:55)→ π i ( A ) , where π i : ∆ → H i is the projection map.Using Lemma 7 and [BRSc, Lemma 5], it is straightforward to see that φ i ( A i ) and π ( A i ) are bothpaths in the 3-containment graphs of H and Sym , respectively. Moreover, if π ( A i ) contains ( j , k ) ,then φ j ( A i ) = φ k ( A i ) must be equal. By Lemma 9, there must be i where π ( A i ) and π ( A i + ) containssome ( j , k ) where j ∈ { , } and k ∈ { , } , and hence for some (cid:96) , φ (cid:96) ( A ) , . . . , φ (cid:96) ( A d ) has a repeatedvertex, which is impossible as this path must be a geodesic of length d .For a group H , we denote the group constructed in the above lemma by BS ( H ) . Lemma 11.
Let n be a natural number. If cd ( G ) > n, then there exists a component of the -localcommensurability graph of G with graph diameter at least n.Proof. Let L be a geodesic in the 3-containment graph of length 2 n with endpoints E and E . Itis straightforward to see that L is in a connected component of the 3-local commensurability graph.Let J be a geodesic path in the 3-local commensurability graph given by the vertices E = V , V , . . . , V k = E . Then the vertices E = V , V ∩ V , V , V ∩ V , V , . . . , V k = E . give a path of length 2 k in the 3-containment graph. We must have, then, that 2 k ≥ n , giving k ≥ n .So we have found a component of the 3-local graph of graph diameter at least n .We are now ready to complete the proof. Proof of Theorem 3.
Let n a natural number be given. Consider the group BS ( BS ( BS ( · · · ( BS ( Sym ) · · · ) , where the number of BS applications is greater than 2 n . By Lemmas 10, cd of the resulting groupis greater than 2 n . Since Sym is solvable, and any solvable-by-solvable group is solvable, it followsthat the above group is solvable. Hence, by Lemma 11, the proof is complete. Proof of Corollary 4.
Let N > Q of with acomponent Ω of the 3-local commensurability graph with graph diameter greater than N . Any freegroup of rank 2 contains, as a normal subgroup, a free group of rank equal to | Q | . Call this subgroup ∆ . Then ∆ maps onto Q , so we are done by [BRSc, Lemma 8]. A Some examples
Here will give a complete classification of the components of p -local commensurability graphs ofthe subgroup of upper triangular matrices, P ( , q ) of GL(2, F q ), where q is some prime. By Theorem2 in this paper, the connected diameter of this graph is at most 4. Through our analysis, we will seethat the connected diameter is at most 2 (and this is sharp).Before stating our classification of components, we need a definition. Let H be component ofgraph Γ . We say H is a star graph if every edge in H is incident with a fixed vertex v c in H , andthese are the only edges in H . We call v c the center of H , and it is unique. With this definition inhand, we can now state the main result of this appendix: Theorem 12.
Let p , q be prime numbers. The components of Γ p ( P ( , q )) are complete or star. ocal commensurability graphs of solvable groups 7 We devote the rest of this section to proving Theorem 12. We begin with primes p that do notdivide the order of the group. Proposition 13.
Let G be a finite group. Then for any prime p, Γ p ( G ) is totally disconnected if andonly if p (cid:45) | G | .Proof. Set N = | G | . If p | N then there exists a nontrivial p -Sylow subgroup which is p -adjacent tothe trivial group. Now, if p (cid:45) N . Suppose, for the sake of a contradiction, that there exist A < B ≤ G such that A is of index a power of p in B . Then p divides | G | , which is impossible. It follows that Γ p ( G ) is totally disconnected.Note that by Proposition 13, if p does not appear in the prime factorization of order of G , then the p -commensurability graphs are totally disconnected. Thus, for the remainder, it suffices to handlethe case when p divides the order of P ( , q ) .Table 1: The two graphs above are Γ ( P ( , )) and Γ ( P ( , )) , respectively. For p = p = P ( , q ) with the semidirect product Z / q (cid:111) ( Z / ( q − ) × Z / ( q − )) , where Z / q corresponds to the subgroup (cid:28)(cid:18) (cid:19)(cid:29) and Z / ( q − ) × Z / ( q − ) corresponds to diagonal matri-ces in GL ( n , F q ) . Proposition 14.
All the subgroups of P ( , q ) ∼ = Z / q (cid:111) ( Z / ( q − ) × Z / ( q − )) are the form of (cid:104) Z / q , s i (cid:105) and (cid:104) b n s i b − n (cid:105) , where { s j } enumerates all subgroups of Z / ( q − ) × Z / ( q − ) , and b is thegenerator of Z / q, n ∈ Z / q.Proof. Let G = Z / q (cid:111) ( Z / ( q − ) × Z / ( q − )) and U be a subgroup of G . Also, let Z / q = (cid:104) b (cid:105) , b be the generator of Z / q . Let P be the q -Sylow subgroup of U . Then Z / q is the unique q -Sylowsubgroup of Z / q (cid:111) ( Z / ( q − ) × Z / ( q − )) . So either P = Z / q or P = { e } . ocal commensurability graphs of solvable groups 8 Then by Hall’s theorem, we have a Sylow system for U , S p , S p , . . . , S p k , | U | = p l p l . . . p l k k , where S p i S p j = S p j S p i . Let p = p , so S p = P . Consider S p S p . . . S p k , then | S p S p . . . S p k | = | U || P | . That is | S p S p . . . S p k | is not divisible by p . Consider, ∆ = Z / ( q − ) × Z / ( q − ) , then ∆ is a Hall-subgroup. Then again by Hall’s Theorem, there exist γ ∈ G , such that γ S p S p . . . S p k γ − < ∆ . So we have U = (cid:10) P , γ S ( i ) γ − (cid:11) , where S ( i ) = S p S p . . . S p k and we index all the subgroups of Z / ( q − ) × Z / ( q − ) by i ∈ I some index set.Since P is either the identity or the entire Z / q , and Z / q is normal in G , then the result follows.Before we proceed, we need a quick technical result. Lemma 15.
Let ∆ , ∆ be subgroup of some finite group Γ such that ∆ and ∆ are both nilpotentand ∆ is p-adjacent to ∆ . Then ∆ = ∆ ∩ ∆ contains the normal complement of the p-Sylowsubgroup of ∆ Proof.
Let ∆ , ∆ by two subgroup of a finite group and they both are nilpotent. Since ∆ is nilpotent,the normal complement of the p -Sylow subgroup of ∆ exists, call it N . By Hall’s Theorem, thenormal complement in ∆ ∩ ∆ of the p -Sylow subgroup of ∆ ∩ ∆ is conjugate to a subgroup of N ,and hence is a subgroup of N . Since elements from different p -Sylow subgroups commute with oneanother, we have that N ∆ ∩ ∆ is a proper subgroup of ∆ containing ∆ ∩ ∆ as a subgroup of index not a power of p . Hence, it is impossible for [ ∆ : ∆ ∩ ∆ ] = [ ∆ : N ( ∆ ∩ ∆ )][ N ( ∆ ∩ ∆ ) : ∆ ∩ ∆ ] to be a power of p . Lemma 16.
Let p be some prime. If p | | Z / ( q − ) × Z / ( q − ) | , then each component of Γ p ( P ( , q )) is complete.Proof. Let { s i } i ∈ I be an enumeration of all the subgroups of Z / ( q − ) × Z / ( q − ) . Let b be thegenerator of Z / q , and n ∈ Z / q . It is straightforward to see that [ (cid:104) Z / p , s i (cid:105) : (cid:10) Z / p , s i ∩ s j (cid:11) ] = [ s i : s i ∩ s j ] , for all i , j . Moreover, we claim that (cid:104) Z / q , s i (cid:105) is never p -adjacent to (cid:10) b n s j b − n (cid:11) , for all i , j . Indeed, suppose that (cid:104) Z / q , s i (cid:105) is p -adjacent to (cid:10) b n s j b − n (cid:11) . It is clear the q does not divides | (cid:10) b n s j b n (cid:11) | , for all j ∈ I . Note, [ P ( , q ) : (cid:104) Z / q , s i (cid:105) ] = q ( q − ) q | s i | = ( q − ) | s i | . Since q does not divide both ( q − ) , then q does not divide ( q − ) | s i | . Therefore by [BRSc, Lemma9], we will have that q (cid:45) q ( q − ) | (cid:104) b n s i b n (cid:105) | Which is impossible because q does not divide | (cid:10) b n s j b − n (cid:11) | . We conclude that the two types ofsubgroups appearing in Proposition 14 never appear in the same component.By our earlier claim we know that [ (cid:104) Z / p : s i (cid:105) : (cid:10) Z / p : s i ∩ s j (cid:11) ] = [ s i : s i ∩ s j ] , for all i , j . Thisimplies that (cid:104) Z / q , s i (cid:105) is p -adjacent to (cid:10) Z / q , s j (cid:11) if and only if s i is p -adjacent to s j . We know that allthe s i are subgroup of Z / ( q − ) × Z / ( q − ) which is an abelian group, then by [BRSc, Theorem 1]we know that the p -local commensurability graph are complete, then the component for (cid:104) Z / q , s i (cid:105) isalso complete. ocal commensurability graphs of solvable groups 9 To finish, we will show that all components containing subgroups of the form (cid:104) b n s i b − n (cid:105) are alsocomplete. We break up the proof into two cases depending on whether (cid:104) b n s i b n (cid:105) has cardinality thatis some power of p . Case 1 : | (cid:104) b n s i b n (cid:105) | = | s i | is some power of p . It is clear that for all s i , . . . s i k with order somepower of p , we have (cid:104) b n s i α b − n (cid:105) is p -adjacent to (cid:68) b m s i β b − m (cid:69) , for i ≤ i α , i β ≤ i k and any m , n .Now, let | s i | = p k and | s j | = p m a , were gcd ( p m , a ) = a (cid:54) =
1. Then | (cid:104) b n s i b − n (cid:105) | = p k and | (cid:10) b v s j b v (cid:11) | = p m a . Thus, we have | (cid:10) b n s i b − n (cid:11) ∩ (cid:10) b v s j b v (cid:11) | = p l where l ≤ k , m . It follows from this that A is not p -adjacent to B for all i , j , n , v . Therefore, by aboveargument we know that all the subgroups (cid:104) b n s i b − n (cid:105) with order some power of p form a completecomponent. Case 2 : Let | s i | not be a power of p , that is | s i | = p n a , where gcd ( p n , a ) = a (cid:54) =
1. Then wemay write s i = H × A , where H is a p -Sylow subgroup and | A | = a . It is clear that (cid:104) b n s i b − n (cid:105) is p -adjacent to (cid:104) b n Ab − n (cid:105) .That is, for any n and (cid:104) b n s i b − n (cid:105) whose order is not a power of p , then there exist a subgroup A n , i of s i such that (cid:104) b n s i b − n (cid:105) is p -adjacent to b n A n , i b − n .If there exist another group A (cid:48) which is p -adjacent to b n A n , i b − n , then it is easy to see that A (cid:48) is p -adjacent to (cid:104) b n s i b − n (cid:105) . Similarly, if there is another group B = (cid:10) b m s j b − m (cid:11) which is p -adjacent toany A (cid:48) as above, then by Lemma 15 we have B is p -adjacent to b n A n , i b − n , and hence is p -adjacent to (cid:104) b n s i b − n (cid:105) . It follows that the component containing (cid:104) b n s i b − n (cid:105) is complete, as desired.We are left with p = q , so the proof of Theorem 12 is completed with the next result. Proposition 17.
Each component of Γ p ( P ( , p )) is1. a star graph, or2. a complete graph with two vertices.Proof. By Proposition 14, we need only consider subgroups of the form (cid:104) Z / p , s i (cid:105) and (cid:10) b − n s j b n (cid:11) .We give a brief outline. We first show that any two subgroups (cid:104) b − n s i b n (cid:105) and (cid:10) b − m s j b m (cid:11) are not p -adjacent if i (cid:54) = j or n (cid:54) = m . Clearly, for any i and n , (cid:104) Z / p , s i (cid:105) adjacent to (cid:104) b − n s i b n (cid:105) . So, next, wewill show that (cid:104) Z / p , s i (cid:105) is not p -adjacent to any (cid:10) b − n s j b n (cid:11) for i (cid:54) = j . Finally, we show that thereis only one group of the form (cid:10) Z / p , s j (cid:11) that is adjacent to (cid:104) Z / p , s i (cid:105) for any fixed i . Thus, all thecomponents under consideration are star graphs with centers (cid:104) Z / p , s i (cid:105) , or are complete graphs withtwo vertices.Let A n = (cid:104) b − n s i b n (cid:105) and A m = (cid:10) b − m s j b m (cid:11) , A n (cid:54) = A m . Let p be a prime such that p (cid:45) ( q − ) . Soif A n and A m are adjacent, then [ A n : A n ∩ A m ] = p c , so c =
0. By symmetry, A n is not adjacent to A m since A n (cid:54) = A m .Now, as we note in the proof of Lemma 16, (cid:104) Z / p , s i (cid:105) is p -adjacent to (cid:10) Z / p , s j (cid:11) if and only if s i is p -adjacent to s j . Since q does not divide ( q − ) and every s i , s j has order which divide ( q − ) ,then we have s i is not p -adjacent to s j for all i , j . Hence (cid:104) Z / p , s i (cid:105) is not p -adjacent to (cid:10) Z / p , s j (cid:11) .Finally, we will show that any group of the form (cid:104) Z / p , s i (cid:105) that is adjacent to A n is unique. Clearly, (cid:104) Z / p , s i (cid:105) is p -adjacent to (cid:104) b n s i b − n (cid:105) . Now, if (cid:10) Z / p , s j (cid:11) is also p -adjacent to (cid:104) b n s i b − n (cid:105) , then we have [ (cid:10) b n s i b − n (cid:11) : (cid:10) Z / p , s j (cid:11) ∩ (cid:10) b n s i b − n (cid:11) ] = p k . Since p does not divides the order of (cid:104) b n s i b − n (cid:105) , then we must have k =
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