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УДК 519.21
Examples of Groups with the Same Number of Subgroups of Every Index
Chris W. Parker*
School of Mathematics University of Birmingham Edgbaston
Birmingham B15 2TT, UK
Atapattu A. C. Kanchana^
Department of Mathematics & Computer Science Faculty of Natural Sciences Open University of Sri Lanka Nawala
Sri Lanka
Received 06.09.2013, received in revised form 06.10.2013, accepted 06.11.2013 In this note we show that certain non-isomorphic free amalgamated products with cyclic amalgamated subgroup and the same type have the same number of subgroups of every index.
Keywords: non-isomorphic free amalgamated products.
Introduction
For a group G and a natural number n, we denote the number of subgroups of G of index n by an(G). Suppose G and H are groups. Then, following Lubotzky and Segal [5], G and H are called isospectral if and only if an(G) = an(H) for all natural numbers n. Our main theorem is as follows.
Theorem 0.1. If A and A' are cyclic amalgams of the same type, then their universal completions are isospectral.
All the unexplained terminology in Theorem 0.1 is introduced in Section Two. For here it suffices to say that a cyclic amalgam is one in which the amalgamated subgroup is cyclic. It is very easy to manufacture examples of cyclic amalgams of the same type which have non-isomorphic universal completions. Indeed, if two amalgams of finite groups have the same type and are not isomorphic then their universal completions are not isomorphic. The smallest example is obtained by taking A1 = A2 isomorphic to the Frobenius group of order 20 and B cyclic of order 4. Then there are exactly two isomorphism classes of amalgams say A1 and A2 of type (F20, F2o, Z/4Z). Their universal completions have presentations
G(A1) = {x, y,z | x5 = y5 = z4 = 1, xz = x3, yz = y2)
and
G(A2) = {x, y, z | x5 = y5 = z4 = 1,xz = x2,yz = y2).
Theorem 0.1 asserts that these two (non-isomorphic) groups are isospectral. It is natural to ask about the subgroup lattice in each group. It turns out the they are not isomorphic. In fact,
* C.W.Parker@bham.ac.uk taackanchana@hotmail.com © Siberian Federal University. All rights reserved
using the computational algebra package Magma [1], it has been shown that the intersection of all subgroups of index 5 in the first group has index 1296000000 while in the second it has index 100. It follows easily that these groups have non-isomorphic subgroup lattices. It is also easy to see that G(Ai) has two normal subgroups of index 20 while G(A2) has six. I don't know whether G(Ai) and G(A2) have the same number of conjugacy classes of subgroups at a given index. For more information about subgroups of finite index in free amalgamated products see [8]. Theorem 0.1 has the following corollary.
Corollary 0.2. For each natural number k, there exist k pairwise non-isomorphic amalgams with isospectral universal completions.
Other instances of isospectral groups can be found in [3, Theorem 1.3] where du Sautoy, McDermott and Smith prove that the groups Z x Z and (x, y,t | [x, y],t2 = y, x4 = x-1} are
isospectral. Also, in [6], Mednykh shows that rg and r* are isospectral where
g
r = (Xl,...,Xg ,yi,...,yg | JJ [Xj,yj] = 1}
j=1
and
g
r* = (x1, . . . ,xg,yi, . . . ,yg 1 I}(xiyi)2}
i=i
are the fundamental groups of a closed orientable surface of genus g, respectively, a closed non-orientable surface of genus 2g.
This note originated from observations made about symmetric presentations and their accompanying progenitors in the monomial case (see [2]). It turns out that there are two progenitors of shape 7*7 :m F21 up to isomorphism (where F21 denotes the Frobenius group of order 21). One has presentation (x,y, z | x7 = y7 = z3 = 1,xz = x2,yz = y2} and the other has presentation (x, y, z | x7 = y7 = z3 = 1,xz = x2,yz = y4} (so they are cyclic amalgams of the same type (F21,F21,Z/3Z)). Experimenting with Magma [1] confirmed that the progenitors are not isomorphic (one has GL3(2) as an image the other does not) and, more strikingly, showed that they have an identical number of subgroups at each index up to 15 (with 39791 subgroups of index at most 15). The main result of this paper says that the groups are in fact isospectral.
Our notation is standard, but we mention that Sym(n) denotes the symmetric group of degree n and that, for groups A and B, Hom(A, B) is the set of homomorphisms from A to B. All our maps are written on the right.
1. Amalgams and completions
A group amalgam, or more simply an amalgam, is a quintuple A = (A1, A2, B, ^2) where A1 ,A2 and B are groups and, for i = 1,2, ^ : B ^ Aj are monomorphisms. Let A = (A1, A2, B, ) and A' = (A1, A2, B', ^1, ^2) be amalgams. Then A and A' have the same type provided there are group isomorphisms a1 : A1 ^ A1, a2 : A2 ^ A2 and 7 : B ^ B' satisfying Im(^1a1) = Im(7^1) and Im(^2a2) = Im(7^2). The amalgams A and A' are isomorphic if there are group isomorphisms a1 : A1 ^ A1, a2 : A2 ^ A2 and 7 : B ^ B' satisfying = 7^1 and ^2a2 = 7^2 (as homomorphisms with domain B). Obviously, isomorphic amalgams have the same type. Suppose that 7 G Aut(B). Then we define an amalgam AY = (A1,A2,B,^1,7^2). The triple of maps (IdAl, IdA2,IdB) demonstrates that A and AY have the same type. Often the type of an amalgam will be uniquely determined by specifying the groups A1, A2 and B. This is for example the case if, for i = 1, 2, any two subgroups of Aj isomorphic to B are conjugate in Aj. If this is the case we can denote the type of amalgam simply by (A1, A2,B) as we did in the introduction.
It is natural to ask how many isomorphism classes of amalgam there are of a given type. The answer is provided by the Goldschmidt Lemma. To state it we require a definition. Suppose that H < K. Then Aut(K, H) = NAut(K)(H)/CAut(K)(H) identified as a subgroup of Aut(H).
Lemma 1.1 (Goldschmidt Lemma). Suppose A = (A1 ,A2,B,^1,^2) is an amalgam and define
X1 = {^a^-1 I a € Aut(A1, (B)^1)> < Aut(B)
and
X2 = | P € Aut(A2, (B)^2)> < Aut(B).
Then the map
X17X2 ^ AY
defines a bijection between the set of (X1, X2)-double cosets in Aut(B) and isomorphism classes of amalgams of the same type as A.
Proof. See [4, (2.7)]. □
Notice that X1 and X2 both contain all the inner automorphisms of B and so the calculation of the double cosets in the Goldschmidt Lemma really takes place in the outer automorphism group of B. For the proof of Theorem 0.1 the important point in Lemma 1.1 is conveyed by the following corollary.
Corollary 1.2. If A = (A1, A2, B, ) is an amalgam and A' has the same type as A, then there exists 7 € Aut(B) such that A' is isomorphic to AY.
Let A = (A1, A2, B, ^2) be an amalgam. A representation of A into a group G is a pair of homomorphisms (^1,^2) where ^ € Hom(Aj, G), for i = 1, 2, such that = €
Hom(B, G). The triple
(((A1)V1 ^2)^2)^1,^2)
is called a completion of A (in G). A completion (G, of A is called universal provided
that given any completion (H, p1,p2) of A, there exists a unique n :€ Hom(G, H) such that Pi = n for i = 1,2. Universal completions of A exist, are unique up to isomorphism, and the group G in the universal completion can be identified with the free amalgamated product
G(A) = (A1 *A2)/{(6)^1(6-1 I b € B)
where A1 * A2 is the free product of A1 and A2 (see [7]). We note that because G(A) is the universal completion of A, every representation of A into a group G leads to a unique homomorphism from G(A) into G and vice versa.
Suppose that A = (A1, A2, B, ^2) is an amalgam. For a group G set
Hom(A, G) = {^ | ^ is a represention of A into G}.
Let 0 € Hom(B, G). Then we say (^1, ^2) € Hom(A, G) extends 0 if and only if 0 = = Put
Hom0 (A, G) = {^ € Hom(A, G) | ^ extends 0}.
Assume that 0 € Hom(B, G). Let no : B ^ B/ker 0 be the projection map, and 0 be the canonical isomorphism from B/ker 0 to (B)0 . Then 0 = no0. Define
6 € Hom(NG((B)0), Aut(B/ker 0))
by (x)6 = 0cx0 1 for all x € (B)0 where cx denotes the automorphism of (B)0 induced by conjugation by x. Finally, for 7 € Aut(B) such that (ker 0)7 = ker 0, define 7* so that the
no7* = Yno. So 7* € Aut( B/ker 0).
Lemma 1.3. Assume that 0 £ Hom(B, G) and 7 £ Aut(B) with (ker 0)7 = ker 0. If there exists x £ NG((B)0) such that (x)0 = 7-1 , then there exists a bijection between Hom0(A, G) and Hom0 (AY ,G).
Proof. We define two maps
and
a : Hom0 (A, G) ^ Hom0 (AY, G)
(^1,^2) ^ (^1 )
P : Hom0(AY, G) ^ Hom0 (A, G)
(^1,^2) ^ (^1,^2Cx-i).
Clearly a and p are inverse to each other, so if they are well-defined we are done. So we show that (^1,^2cx) £ Hom0(AY,G). We have
7^2 ^2 cx = Y0Cx =^7^0 0cx = 77T0_0cx0 10 _ _
= 7^0 (x)00 = 7^0 7-1 0 = 77-1ng 0 = 0 = 0.
Since = 0, a is well-defined. The proof that p is well-defined is similar. Thus the lemma holds. □
We say that A = (A1, A2, B, ^>2) is a cyclic amalgam if B is a cyclic group.
Theorem 1.4. Suppose that A = (A1, A2, B, ^2) is a cyclic amalgam. Then for all natural numbers n and for all 7 £ Aut(B), there is a bijection between Hom(A, Sym(n)) and Hom(AY, Sym(n)).
Proof. Let G = Sym(n). Suppose that 0 £ Hom(B, G). Since B is a cyclic group, (ker 0)7 = ker 0. Furthermore, the generators for (B)0 in G all have the same cycle type and so are conjugate in G. Thus d £ Hom(NG((B)0), Aut(B/ker 0)) is an isomorphism. In particular, for all 0 £ Hom(B,G), there exists x £ NG((B)0) such that (x)0 = 7-1 . It follows from Lemma 1.3 that for all 0 £ Hom(B, G), there is a bijection between Hom0(A, G) and Hom0(AY, G). Since Hom(A, G) = U0gHom(B,G) Hom0(A, G) and Hom(AY, G) = ]}0eHom(B,G) Hom0(AY, G) the theorem is true. □
Since there is a bijection between Hom(A, G) and Hom(G(A),G), we have the following corollary.
Corollary 1.5. For all natural numbers n and all 7 £ Aut(B), there is a bijection between Hom(G(A), Sym(n)) and Hom(G(AY), Sym(n)).
For a group G, let hn(G) = |Hom(G, Sym(n))|. The following result intertwines hn(G) and a„(G).
Lemma 1.6. Suppose that G is a group. Then
1 n-1 1 a«(G) = 7-7T7hn(G) — y] --— h„_fc(G)afc(G).
(n — 1)! z—' (n — k)!
Proof. See [5, Corollary 1.1.4]. □
Proof of the Main Theorem. Suppose that A and A! are cyclic amalgams of the same type. Then by Corollary 1.2 there exists 7 G Aut(B) such that A' is isomorphic to AY. It follows from Corollary 1.5 that h„(G(A)) = h„(G(A')) for all n. We have that ai(G(A)) = ai(G(A')) and so using Lemma 1.6 and induction gives us that a„(G(A)) = a„(G(A')) for all n. □
We now prove Corollary 0.2. Suppose that k is a natural number. Let p be a prime such that p — 1 > k2, then ф(р — 1) > k where ф is the Euler totient function. Now let A1 and A2 be Frobenius groups of order p(p — 1) with kernel of order p and let B be the cyclic group of order p — 1. Then, with X1 and X2 as in the Goldschmidt Lemma, X1 = X2 = 1 and so there are exactly |Aut(B)| = ф^ — 1) > k pairwise non-isomorphic amalgams of type (A1,A2,B). Now the corollary follows from Theorem 0.1.
References
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[2] R.T.Curtis, Monomial modular representations and construction of the Held group, J. Algebra, 184(1996), no. 3, 1205-1227.
[3] M.P.F. du Sautoy, J.J.McDermott, G.C.Smith, Zeta functions of crystallographic groups and analytic continuation, Proc. London Math. Soc., 79((1999), no. 3, 511-534.
[4] D.M.Goldschmidt, Automorphisms of trivalent graphs, Ann. of Math, 111(1980), no. 2, 377-406.
[5] A.Lubotzky, D.Segal, Subgroup growth, Progress in Mathematics, 212. Birkhauser Verlag, Basel, 2003.
[6] A.D.Mednykh, On the number of subgroups in the fundamental group of a closed surface, Comm. Algebra, 16(1988), no. 10, 2137-2148.
[7] J-P.Serre, Trees. Translated from the French by John Stillwell, Springer-Verlag, Berlin-New York, 1980.
[8] W.W.Stothers, Subgroups of finite index in a free product with amalgamated subgroup, Math. Comp., 36(1981), no. 154, 653-662.
Примеры групп с одинаковым числом подгрупп любого индекса
Крис В. Паркер Атапату А.С. Канчана
Мы показываем, что при определённых условиях неизоморфные свободные произведения с циклической амальгамой имеют и одинаковое число подгрупп каждого индекса.
Ключевые слова: неизоморфные свободные смешанные произведения.
