Enumeration Of Finite Soluble Groups With

Abelian Sylow Subgroups

Geetha VenkataramanSt. Stephen’s College

Delhi-110007India

This is a pre-copyedited, author-produced PDF of an article acceptedfor publication in Quart. J. Math. Oxford following peer review. Theversion of record [Quart J. Math. Oxford (2), 48 (1997), 107-125] is

available online at:http://qjmath.oxfordjournals.org/content/48/1/107.full.pdf+html.

1 Introduction

A group is said to be an A-group if all its nilpotent subgroups are abelian.This paper deals with the enumeration of soluble A-groups of a given order.Throughout, we shall take logarithms to the base 2 unless stated otherwise.We follow the convention that 0 ∈ N. For any positive integer n we de-fine µ := µ(n) to be the largest integer such that pµ divides n for someprime p and λ := λ(n) to be the number of prime divisors of n includingmultiplicities.

A bound for the number of soluble A-groups was first given by G. A.Dickenson [2]. Let fA,sol(n) denote the number of isomorphism classes ofsoluble A-groups of order n. She showed that fA,sol(n) ≤ nc logn for someconstant c.

The bound proved by McIver and Neumann [8] for the number of A-groups of order n superseded the above result. Let fA(n) denote the numberof non-isomorphic A-groups of order n. They showed that fA(n) ≤ nλ+1.

In the same paper they also had the following conjecture regarding theenumeration of finite groups in general.

Conjecture 1.1 Let f(n) denote the number of (isomorphism classes of)

1

groups of order n. Then

f(n) ≤ n(227

+ε)λ2

where ε −→ 0 as λ −→∞.

The above conjecture was motivated by a result of Higman [6] and Sims[12] on p-group enumerations and a ‘folk-lore’ observation that when thereare many groups of a given order it is because of the rich variety of p-groups available to be Sylow subgroups rather than because there might bemany ways of fitting p-groups for various relevant primes together. In 1993L. Pyber [11] not only validated the sentiment behind the conjecture butalso proved a stronger version of Conjecture 1.1. He showed that the numberof groups of order n with specified Sylow subgroups is at most n75µ+16.

Pyber’s Theorem together with the Higman–Sims result on p-group enu-

merations shows that f(n) ≤ n227µ2+O(µ5/3).

Since there are less than m choices for an abelian group of order m,Pyber’s Theorem also has the corollary that fA(n) ≤ n75µ+17. From theproof we can in fact see that fA,sol(n) ≤ n40µ+2.

While, from the corollary to Pyber’s Theorem we have that for someconstants c > 0 and d > 0

fA(n) ≤ ncµ+d ,

the values c = 75 and d = 17, or even c = 40 and d = 2, are far fromrealistic.

We prove two main results in this paper. Both deal with bounds forfA,sol(n) but in very different ways.

We prove:

Theorem 1.1 Let n =∏ki=1 p

αii be the prime factorisation for n. Then

fA,sol(n) ≤ n2µ+6k∏i=1

pi5α2i .

Theorem 1.1 has the following obvious corollary.

Corollary 1.1fA,sol(n) ≤ n7µ+6 .

The second theorem we prove was motivated by the following conjecture.

Conjecture 1.2 There exists a constant c such that the number of groups

of order∏ki=1 p

αii with specified Sylow subgroups is at most

∏ki=1 p

cαi2

i .

2

Gaschutz tried to prove a stronger version of Conjecture 1.2 for solublegroups in 1969. The present form was suggested by Peter M. Neumann.However while the idea behind this conjecture was confirmed by Pyber’sTheorem, Conjecture 1.2 is false and the disproof is also due to Pyber[11]. The groups constructed by Pyber also turn out to be metabelian andhence are soluble. Thus Pyber’s disproof shows that one cannot improvethe bounds for fA,sol(n) to bounds with the same form as in Conjecture 1.2.However Pyber requires k (the number of distinct prime divisors of n) totend to infinity as n tends to infinity for his disproof.

This brings us to the second result we shall prove.

Theorem 1.2 For every positive integer k there exists a constant ck (de-pending only on k) such that for any k distinct primes p1, . . . , pk we have

fA,sol(n) ≤k∏i=1

pckα

2i

i

where n =∏ki=1 p

αii .

This gives us a result whose form is in some ways similar to that in Con-jecture 1.2 and also suggests that a modified form of Conjecture 1.2 may betrue, that is, one in which the constant c is no longer universal but depen-dent on the number of distinct primes dividing n. It is clear from Pyber’sdisproof that the sequence ck will have to be unbounded and the value wearrive at for ck is k + 14.

We also need to see how the bound in Theorem 1.2 compares with theone in Theorem 1.1. We show:

Proposition 1.1 Let ε > 0, c > 0 and M > 0 be given. Then for anypositive integer k ≥ 2, there exist infinitely many n with exactly k distinctprime divisors such that

nεµ∏ki=1 p

cα2i

i

> M .

It can be seen easily that the reverse is also true, namely that there areinfinitely many n with exactly k distinct prime divisors such that the boundin Theorem 1.1 is significantly better than that in Theorem 1.2.

The rest of the paper is organised as follows. Sections 2 and 3 dealwith results about soluble A-subgroups of the symmetric group on n lettersand the general linear group respectively. Section 4 has some basic resultsrequired for the main proofs and we prove Theorem 1.1, Theorem 1.2 andProposition 1.1 in Section 5. The last section contains a commentary on thebounds achieved.

3

2 Soluble subgroups of symmetric groups

In this section we prove propositions which give a bound on the order ofsoluble A-subgroups of Sym(n). We also deal with bounding the numberof transitive soluble A-subgroups of Sym(n). The proof of Proposition 2.1follows a path similar to that of John D. Dixon’s proof [3] that a maximalsoluble subgroup of Sym(n) has order at most 24(n−1)/3.

Proposition 2.1 Let G be a soluble A-subgroup of Sym(n). Then

(i) |G| ≤ 612(n−1);

(ii) if G has odd order then |G| ≤ 312(n−1).

Proof We shall prove the first part by induction on n. Let G be a solubleA-subgroup of Sym(Ω) where |Ω| = n. We assume that n ≥ 2 and considerthree cases.

Suppose that G is not transitive. Then G ≤ G1 × G2 where Gi is a solubleA-subgroup of degree ni and n = n1 + n2. So by the induction hypothesis|G| ≤ |G1| × |G2| ≤ cn1−1cn2−1 < cn−1 where c := 6

12 .

Suppose that G is transitive but not primitive. Then G is a subgroup of awreath product of some soluble A-groups G1 and G2 of degrees n1 and n2respectively. Further n = n1n2. Thus

|G| ≤ |G1 wr G2| ≤(cn1−1

)n2cn2−1 = cn−1 .

Suppose that G is primitive. Let M be a minimal normal subgroup of G.Since G is soluble, M is an elementary abelian p-group for some primep. So |M | = pd. As G is primitive, M is transitive. Further, becauseM is abelian, it acts regularly. Hence n = |Ω| = pd. Since F (G), theFitting subgroup of G, is also an abelian normal subgroup of G, by thesame reasoning we get that |F (G)| = n = pd. Note that F (G) contains Mand therefore we must have M = F (G). But since G is soluble we haveCG(F (G)) ≤ F (G). Further G is an A-group and hence for any Sylow p-subgroup P of G we have P ≤ CG(M) ≤ M ≤ P . Consequently M is theunique Sylow p-subgroup of G. Let H := Gα, the stabiliser of α ∈ Ω. SinceM is transitive we have G = HM and as M is regular, H ∩M = 1. SoH is a soluble p′-A-subgroup of G. Further H acts faithfully by conjugationon M and so we can consider H as a subgroup of GL(d, p), the group of allinvertible linear transformations on a vector space of dimension d over field

4

of p elements. Since |GL(d, p)| = (pd − 1)(pd − p) · · · (pd − pd−1) and |H| iscoprime to p, we have that |H| divides (pd − 1)(pd−1 − 1) · · · (p − 1). Thus

|H| ≤ pd(d+1)/2 ≤ n12(log(n)+1). Therefore

|G| = |M | |H| ≤ nn12(log(n)+1) = n

12(log(n)+3) .

It is not difficult to see that x12(log(x)+3) ≤ cx−1 for x ≥ 8. So for n ≥ 8 we

have|G| ≤ n

12(log(n)+3) ≤ cn−1 .

When n = pd for some prime p and n is less than 8 we can check easily byhand that the order of a primitive soluble A-subgroup in Sym(n) is less thanor equal to cn−1.

For the second part, we have three cases as above. We only need to lookat the third case in some detail. Let G be a primitive (soluble) A-group ofodd order in Sym(Ω), where |Ω| = n. Then as in the proof above G = HMwhere M is the unique Sylow p-subgroup of G, H is the stabiliser of a pointand n = pd = |M |. Further H is a p′-A-subgroup of odd order of GL(d, p).Now

|GL(d, p)| = pd(d−1)/2(pd − 1)(pd−1 − 1) · · · (p− 1) .

Since p is an odd prime we get that |H| divides ((pd − 1) · · · (p − 1))/2d.Thus

|H| ≤ pd(d+1)/2

2d≤ n

12((log(n)/ log(3)) + 1)

2.

Consequently |G| = |M | |H| ≤ 12 n

12((log(n)/ log(3)) + 3).

For x ≥ 12 we have 12 x

12((log(x)/ log(3)) + 3) ≤ 3

12(x−1). When n = pd for

some odd prime p and n is less than 12 we can easily verify that the orderof a primitive A-group of odd order in Sym(n) is at most 3

12(n−1).

Proposition 2.2 The number of transitive soluble A-subgroups of Sym(n)

is at most 6(n2−1)/4 2n log(n)+(log(n))3+(log(n))2+2 log(n).

Proof Let G be a primitive soluble A-subgroup of Sym(n). Then as in theproof of Proposition 2.1 we have that G = HM where M is the unique Sylowp-subgroup of G, H is the stabiliser of a point, n = pd = |M |. Further Hmay be regarded as a soluble p′-A-subgroup of GL(d, p) and the conjugacyclass of H in GL(d, p) determines the conjugacy class of G in Sym(n).

Let S be a Sylow π-subgroup of H for some prime divisor π of |H|. LetFp denote a finite field with p elements and let V := (Fp)

d. Using the fact

5

that S is a completely reducible abelian subgroup of GL(V) we can showthat S is generated by d elements. Thus by a theorem of Kovacs [7], H canbe generated by d+ 1 elements.

Let mpsa(n) denote the number of conjugacy classes of primitive solubleA-subgroups of Sym(n). Then

mpsa(n) ≤ the number of d+ 1 generator subgroups of GL(d, p)

≤ pd2(d+1)

≤ 2(log(n))2(log(n)+1) .

A transitive soluble (A-)subgroup G of Sym(n) is a subgroup of a wreathproduct (denoted by G) of some primitive soluble groups G1, . . . , Gk of de-grees n1, . . . , nk respectively. Further n = n1 · · ·nk and since G is an A-group, the groups Gi are also A-groups. Note that the conjugacy classes ofthe groups Gi determine the conjugacy class of G.

We know by Proposition 2.1 that any soluble A-subgroup of Sym(n) hasorder at most 6(n−1)/2. So |Gi| ≤ 6(ni−1)/2 for i = 1, . . . , k and hence byinduction on k we get that |G| ≤ 6(n−1)/2. A theorem of Peter M. Neumann,published in a paper by Peter J. Cameron, Ron Solomon and AlexandreTurull [1] states that any subgroup of Sym(n) can be generated by (n+1)/2elements. Thus once G has been chosen we have at most 6(n

2−1)/4 choicesfor G.

But by the first paragraph of this proof there are at most

k∏i=1

2(log(ni))3+(log(ni))

2

≤ 2(log(n))3+(log(n))2

choices for the groups Gi and hence for G (up to conjugacy). Now the size ofany conjugacy class is at most n! and so the number of choices for G cannotexceed

2n log(n)+(log(n))3+(log(n))2 .

Since there are at most n2 multiplicative partitions of n (see for example[5]), the result follows.

Note 2.1 Dixon’s bound [3] on the order of maximal soluble subgroups ofSym(n) is achieved asymptotically by Sym(4) wr Sym(4) wr Sym(4) · · ·.

With slight modifications to the proofs of Propositions 2.1 and 2.2 wecan show that the bound is achieved only for n = 1, 3 and that the order ofthe groups in question is strictly less than the given bounds otherwise. We

6

can construct soluble A-groups whose orders are about (2.04)n as n tendsto infinity by considering the following group.

Gm := A4 wrC5 wrC7 wr · · · wrCpm

where m ≥ 3, A4 is the alternating group on 4 symbols, Ck denotes the cyclicgroup of order k and pm is the mth prime. The degree of Gm is 4

∏mi=3 pi.

Similarly if we take C3 instead of A4 in the above example, we can get anA-group of odd order with degree n :=

∏mi=2 pi and size about (1.63)n.

3 Soluble subgroups of the general linear group

P. P. Palfy [10] has shown that an irreducible soluble group of maximalorder in GL(n, p) (p a prime) has order less than p3n. Thomas R. Wolf [14]has also shown independently that when G ≤ GL(n, q) (q a power of someprime p) and G is both soluble and completely reducible then |G| ≤ q3n. Weprove a result in this section in the same genre. We also find a bound onthe number of conjugacy classes of subgroups of GL(n, pd) that are maximalamongst soluble p′-A-subgroups.

Proposition 3.1 Let G be a soluble p′-A-subgroup of GL(n, q) where q :=pd for some prime p.

(i) If q is odd then |G| ≤ 612(n−1)qn;

(ii) if q is a power of 2 then |G| ≤ 312(n−1)qn.

Proof We shall prove the result by induction on n. Assume that the resultholds whenever the degree is less than n.

Let V := (Fq)n. Since G is a p′-group, by Maschke’s Theorem it is

completely reducible. So

G ≤ G1 × · · · ×Gk

where for each i, Gi is an irreducible soluble p′-A-subgroup of GL(ni, q) andn = n1 + · · · + nk. Thus if k ≥ 2 we get the result by using the inductionhypothesis on the Gi. So assume that k = 1, that is, G is irreducible.

We now consider the case when G is irreducible and primitive. LetF := F (G), the Fitting subgroup of G. Since G is an A-group, F is abelian.Let X be an irreducible FqF -submodule of V . By Clifford’s Theorem andthe fact that G is primitive we get

VF = X1 ⊕ · · · ⊕Xl (1)

7

where the Xi are conjugates of X and are isomorphic to X as FqF -modules.We note that F acts faithfully on X.

Let E denote the subalgebra generated by F in End(V ). Since the Xi

are isomorphic to X as FqF -modules, we can see that E acts faithfully(and irreducibly) on X. Thus if we restrict the action of E to X then bySchur’s Lemma, E is a (commutative) division ring and so E is a field andE ∼= Fqs , where s = dim(X). By (1) we have n = ls and since F ≤ E∗ (themultiplicative group of the field E), we have that |F | divides qs − 1.

Let g be an element in G and let t belong to E. Then t =∑ri=1 λihi for

some λi ∈ Fq and hi ∈ F . Let v belong to V . Then

tg(v) := gtg−1(v) = g(r∑i=1

λi hi(g−1(v)) = (

r∑i=1

λi ghig−1)(v) .

Since F is a normal subgroup of G, for each i, ghig−1 belongs to F and so

gtg−1 is in E. It is also easy to see that if t ∈ Fq ⊆ E then any element ofG commutes with t. Thus we have a homomorphism from G into GalFq(E)where GalFq(E) denotes the group of automorphisms of E that are identityon Fq. The kernel of this map is CG(E). But F ≤ E∗, hence

CG(E) = CG(E∗) ≤ CG(F ) ≤ F (since G is soluble) .

Therefore F = CG(E) and G/F ≤ GalFq(E). Now GalFq(E) ∼= Cs, thecyclic group of order s. So |G/F | divides s and hence |G| divides s (qs − 1).

Assume that q is odd. It is easy to see that s ≤ 2s−1 ≤ 612(s−1). Thus

we have |G| ≤ s (qs − 1) < 612(n−1) qn.

Now suppose that q is a power of 2. Since |G| is coprime to p = 2, wehave that G is then an irreducible primitive subgroup of odd order in GL(V ).

For s = 1 and values of s greater than or equal to 3, we have s ≤ 312(s−1).

So for these values of s we have |G| < 312(n−1) qn. Now let s = 2. Since |G|

is odd and divides s(qs − 1) we must have |G| ≤ qs − 1 < 312(n−1) qn. So

our result holds when G is an irreducible primitive soluble p′-A-subgroup ofGL(n, q).

We look at the remaining case where G is irreducible but imprimitive.Then G preserves a direct sum decomposition V = V1 ⊕ · · · ⊕ Vk and G ≤G1 wrG2 where dim(V1) = m, G1 is an irreducible primitive soluble p′-A-subgroup of GL(m, q) and G2 is a transitive soluble A-subgroup of Sym(k).We shall first do the case when q is a power of 2. In this case G has odd orderand so does G2. Therefore by Proposition 2.1 (ii) we have |G2| ≤ 3

12(k−1)

and by the above part |G1| ≤ 312(m−1) qm and the result follows.

8

When q is odd, |G1| ≤ 612(m−1) qm and by Proposition 2.1 (i), |G2| is at

most 612(k−1). Consequently we have our result by induction.

Corollary 3.1 Let G be a soluble p′-A-subgroup of GL(n, q) where q := pd

for some prime p. Then |G| ≤ q2n−1.

Proof Follows from Proposition 3.1 as 612 < 3 and 3

12 < 2.

Proposition 3.2 Let T := GL(n, q), where q is a power of p for some primep. Then the number of conjugacy classes of subgroups that are maximalamongst irreducible soluble p′-A-subgroups of T is at most q3n

225n−1.

Proof Let G be a subgroup of T such that it is maximal among irreduciblesoluble p′-A-subgroups of T . Let V := (Fq)

n and let F := F (G). Then byClifford’s Theorem

VF = Y1 ⊕ Y2 ⊕ · · · ⊕ Yrwhere Yi = lXi for all i and X1, . . . , Xr are irreducible FqF -submodules ofV . For all i, j there exists g in G such that Xig = Xj and the Xi form amaximal set of pairwise non-isomorphic conjugates for i = 1, . . . , r. FurtherG acts transitively on the Yi.

Now let Ki := CF (Yi) = CF (Xi). Then F/Ki acts faithfully on Yiand when its action is restricted to Xi, it acts faithfully and irreduciblyon Xi. Let Ei denote the subalgebra generated by F/Ki in End(Yi). SinceYi = lXi, it is not difficult to see that Ei acts faithfully on Xi when its actionis restricted to Xi. Now F/Ki is abelian and acts irreducibly on Xi andF/Ki ⊆ Ei. Thus Ei must act irreducibly on Xi. Hence by Schur’s Lemma,Ei is a (commutative) division ring and so Ei is a field and Ei ∼= Fqs wheres = dim(Xi). Note that n = lrs. Clearly F/Ki ≤ E∗i ≤ GL(Yi). Therefore

F ≤ F/K1 × F/K2 × · · · × F/Kr

≤ E∗1 × E∗2 × · · · × E∗r≤ GL(Y1)×GL(Y2)× · · · ×GL(Yr)

≤ GL(V ) .

It is clear that we need to find F only up to conjugacy. Let E := E∗1×· · ·×E∗r .It is not difficult to show that there is only one conjugacy class of subgroupsof this type in GL(V ). So up to conjugacy there is only one choice for E.Since E is a direct product of r isomorphic cyclic groups, any subgroup ofE can be generated by r elements. In particular, F can be generated by

9

r elements and so the number of choices for F up to conjugacy is at most

(qs − 1)r2

.Let Ti denote the subgroup 1 × · · · × F/Ki × · · · × 1 of GL(V ).

We shall abuse notation and assume that an element of Ti is of the formhKi. It is easy to see that for any g in G, (hKi)

g = hgKgi . Since Ei

is the subalgebra generated by F/Ki in End(Yi), it is also easy to seethat whenever g(Yi) = Yj we have Egi = Ej . Consequently G acts tran-sitively by conjugation on E1, . . . , Er. So we have a homomorphism fromG into Sym(r). Let N be the kernel of this map. Then G/N is a transi-tive soluble A-subgroup of Sym(r). So by Proposition 2.2 we have at most

6(r2−1)/4 2r log(r)+(log(r))3+(log(r))2+2 log(r) choices for G/N . We also note that

G normalises E and so G∩E is a normal abelian subgroup of G containingF . Therefore G ∩ E = F .

Now N acts by conjugation on each Ei. Thus for each i we have ahomomorphism from N into GalFq(Ei) with kernel Ni := CN (Ei). We claimthat ∩ri=1Ni = F . Since F = G ∩ E we have F ≤ ∩ri=1Ni. If g is in ∩ri=1Ni

and t is in F , then for all i, tKi = (tKi)g = tgKg

i . So for i = 1, . . . , r wehave Ki = Kg

i and t−1gtg−1 ∈ Ki. But ∩ri=1Ki = 1. So for all t in F wehave tg = gt. Thus g ∈ CG(F ) ≤ F . Hence we have proved our claim.

Therefore we have

N/F ≤ N/N1 × · · · ×N/Nr

≤ GalFq(E1)× · · · × GalFq(Er) .

Since GalFq(Ei) ∼= Cs we have that N/F can be generated by r elements

and so we have at most sr2

choices for N/F . We note that F ≤ N ≤ G ≤NT (F ). Once we have made our choice for F as a subgroup of T , we havedetermined NT (F ). We have a natural homomorphism φ from NT (F ) ontoNT (F )/F . So if we have made our choice for N/F then N is determined asa subgroup of NT (F ) by F and the inverse image of N/F under φ. FurtherN ≤ G ≤ NT (N). So we can determine G as a subgroup of NT (N) onceG/N is known.

Putting together the estimates for F , N/F , and G/N we get the fol-lowing. The number of conjugacy classes of subgroups that are maximalamongst irreducible soluble p′-A-subgroups of T is at most∑

(l,r,s)

(qs − 1)r2sr

26(r

2−1)/4 2r log(r)+(log(r))3+(log(r))2+2 log(r)

where (l, r, s) ranges over ordered triples of positive integers satisfying n = lrs.This part is devoted to simplifying the above expression. We note that

10

• (qs − 1)r2sr

2 ≤ q2sr2−r2 (since s ≤ qs−1);

• 6(r2−1)/4 ≤ 2(r

2−1).

Let β := (qs − 1)r2sr

26(r

2−1)/4 2r log(r)+(log(r))3+(log(r))2+2 log(r). Then wehave the following.

β ≤ q2sr2

2r log(r)+(log(r))3+(log(r))2+2 log(r)−1

≤ q2n22n log(n)+(log(n))3+(log(n))2+2 log(n)−1 .

It is easy to see that we have at most 256log(n)+log(6) choices for (l, r, s).

Therefore the number of conjugacy classes of subgroups that are maximalamongst irreducible soluble p′-A-subgroups of T is at most

q2n2

2n log(n)+(log(n))3+(log(n))2+ 176

log(n)+log(6)−1 .

For x ≥ 11 we have (log(x))3+(log(x))2+(17/6) log(x)+log(6) ≤ 6x and for1 ≤ n ≤ 10 we can verify that (log(n))3+(log(n))2+(17/6) log(n)+log(6) isat most 6n. Also for all n ≥ 1 we have log(n) ≤ n− 1. Thus the number ofconjugacy classes of subgroups that are maximal amongst irreducible solublep′-A-subgroups of T is at most q3n

225n−1.

Corollary 3.2 Let q := pd and let T := GL(n, q). Then the number of con-jugacy classes of subgroups that are maximal amongst soluble p′-A-subgroupsof T is at most q3n

226n−2.

Proof LetG be maximal amongst soluble p′-A-subgroups of T . By Maschke’sTheorem, G is completely reducible. ThusG = G1×· · ·×Gk where for each i,Gi is maximal amongst irreducible soluble p′-A-subgroups of Ti := GL(ni, q)and n = n1 + · · · + nk. Further the conjugacy classes of G1, . . . , Gk inT1, . . . , Tk respectively determine the conjugacy class of G in T .

Therefore the number of conjugacy classes of subgroups of T that aremaximal among soluble p′-A-subgroups of T is at most

∑(n1, . . . , nk)n1+···+nk=n

k∏i=1

q3n2i 25ni−1 ≤

∑(n1, . . . , nk)n1+···+nk=n

q3n2

25n−1 .

Since the number of ordered partitions of n is 2(n−1), the result follows.

11

4 Maximal soluble p′-A-subgroups

We need three more small results for the proofs of Theorems 1.1 and 1.2and these are presented in this section.

For a group G and a prime p, let Ms,p′,A(G) be the set consisting ofsubgroups of G that are maximal amongst soluble p′-A-subgroups of G andlet [Ms,p′,A(G)] denote the set of conjugacy classes of subgroups of G thatare maximal amongst soluble p′-A-subgroups of G.

Lemma 4.1 Let G be a finite group and let H be a subgroup of G. Then|Ms,p′,A(H)| ≤ |Ms,p′,A(G)|.

Proof Let M belong to Ms,p′,A(H). Then there exists M ∈Ms,p′,A(G) such

that M ≤ M . If M1 ∈Ms,p′,A(H) and M1 ≤ M then

T := 〈M,M1〉 ≤ M ∩H .

Consequently T is a soluble p′-A-subgroup of H and hence T = M = M1.Therefore |Ms,p′,A(H)| ≤ |Ms,p′,A(G)|.

Lemma 4.2 Let G be a finite group and N a normal p-subgroup of G. Then|[Ms,p′,A(G)]| ≤ |[Ms,p′,A(G/N)]|.

Proof Let M belong to Ms,p′,A(G). Since MN/N ∼= M/(M ∩N) ∼= M wehave that MN/N is a soluble p′-A-subgroup of G/N .

Now let T/N be any soluble p′-A-subgroup of G/N such that MN/N ≤T/N . Clearly T is soluble and N is the Sylow p-subgroup of T . So bythe Schur–Zassenhaus Theorem there exists a p′-subgroup H of T such thatT = HN . Clearly H is a Hall p′-subgroup of T . So by a theorem of PhilipHall (see [4, p. 231, 4.1]) we have that M ≤ Ht (where Ht := tHt−1) forsome t in T . But H is a soluble p′-A-subgroup of G and M ∈ Ms,p′,A(G),so M = Ht. Therefore T/N = HN/N = HtN/N = MN/N and thusMN/N ∈Ms,p′,A(G/N).

We can now define a map from [Ms,p′,A(G)] to [Ms,p′,A(G/N)] whichtakes [M ] in [Ms,p′,A(G)] to [MN/N ] in [Ms,p′,A(G/N)]. We shall show thatthis map is injective.

Suppose [M1N/N ] = [M2N/N ] for some M1 and M2 in Ms,p′,A(G). Thenthere exists g in G such that Mg

1N = M2N . It is easy to see that M2N is asoluble subgroup of G and by the Schur–Zassenhaus Theorem there existsb in M2N such that Mg

1 = M b2 . Consequently [M1] = [M2] and our result

follows.

12

For a group G and a prime p, let Op′(G) denote the largest normalp′-subgroup of G and let F (G) denote the Fitting subgroup of G.

Lemma 4.3 Let G be a soluble A-group of order∏ki=1 p

αii where the pi are

distinct primes and let Gi := G/Op′i(G). Then Gi is a soluble A-group and

(i) G ≤ G1 × · · · ×Gk as a subdirect product;

(ii) Op′i(Gi) = 1 and Gi is a semidirect product of its Sylow pi-subgroupby a p′i-subgroup and any Sylow pi-subgroup of Gi is isomorphic to aSylow pi-subgroup of G.

Proof Part (i) follows from the fact that⋂ki=1Op′i(G) = 1. It is obvious

that Op′i(Gi) = 1 and this implies that F (Gi) = Opi(Gi). Let Pi be aSylow pi-subgroup of Gi. Since Gi is a soluble A-group we have

Pi ≤ CGi(Opi(Gi)) = CGi(F (Gi)) = F (Gi) .

Thus Pi = Opi(Gi) and so Gi is an extension of a pi-subgroup by a p′i-subgroup. The result follows by the Schur–Zassenhaus Theorem.

Note 4.1 LetGi and Pi be as in Lemma 4.3 and letHi be a Hall p′i-subgroupof Gi. Then Gi = PiHi and Hi acts on Pi by conjugation. By Lemma 4.3(ii) this action is faithful. Thus Hi ≤ Aut(Pi).

Note 4.2 Let Φ(Pi) denote the Frattini subgroup of Pi and let Ai := AutPi.Then there exists a natural homomorphism from Ai into Aut(Pi/Φ(Pi)) withkernel a pi-group say Bi. If di := d(Pi) = d(Pi/Φ(Pi)) then Aut(Pi/Φ(Pi)) =GL(di, pi) and so Ai/Bi ≤ GL(di, pi).

5 Enumeration Of Soluble A-groups

We shall prove Theorems 1.1 and 1.2 in this section. Pyber’s paper [11] hasbeen the main source of inspiration for the proof of Theorem 1.1. Some ofthe methods used for enumerations in his paper have either been adaptedor used directly.

At the end of this section we present a proof of Proposition 1.1.

Proof of Theorem 1.1 Let G be a soluble A-group of order n and letGi := G/Op′i(G). Then by Lemma 4.3, G ≤ G1×· · ·×Gk. Further by Note4.1, for each i, we have Gi = PiHi where Pi is the Sylow pi-subgroup of Giand Hi is a p′i-A-subgroup of Ai := Aut(Pi). So there exists Mi ≤ Ai such

13

that Mi is maximal among soluble p′i-A-subgroups of Ai and Hi ≤Mi. LetGi := PiMi and let G := G1 × · · · × Gk. Then G ≤ G. By our constructionG is a soluble A-group.

We shall first enumerate the possibilities for G up to isomorphism andthen estimate the number of subgroups of G of order n up to isomorphism.For the former we need to count the number of Gi up to isomorphism. Butthis is less than or equal to∑

|Pi| = pαii

Pi abelian

|[Ms,p′i,A(Ai)]| . (2)

Fix Pi. Let di and Bi be as in Note 4.2 and let Ti := GL(di, pi). ByLemma 4.2 we have |[Ms,p′i,A

(Ai)]| ≤ |[Ms,p′i,A(Ai/Bi)]|.

But

|[Ms,p′i,A(Ai/Bi)]| ≤ |Ms,p′i,A

(Ai/Bi)|≤ |Ms,p′i,A

(Ti)| (by Lemma 4.1)

≤ pd2ii |[Ms,p′i,A

(Ti)]|

≤ p4d2ii 26di−2 (by Corollary 3.2)

≤ p4α2i

i 26αi−2 .

Thus

|[Ms,p′i,A(Ai)]| ≤ p

4α2i

i 26αi−2 . (3)

The number of choices for Pi is equal to the number of unordered par-titions of αi and this is at most 2αi−1. Thus by (2) and (3) the number ofchoices for Gi up to isomorphism is less than or equal to

p4α2i

i 27αi−3

and the number of choices for G is at most

k∏i=1

pi4α2i 27αi−3 .

Now let G be fixed. Let S1, . . . , Sk be a Sylow system for G. So Siis a Sylow pi-subgroup of G and for all i, j we have SiSj = SjSi. ThusG = S1S2 · · ·Sk. Further by a theorem of Philip Hall (see [4, p. 232, 4.3])

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there exist Q1, . . . , Qk, part of a Sylow system for G, such that Si ≤ Qi.Any two Sylow systems for G are conjugate. Consequently the number ofchoices for G as a subgroup of G (and up to conjugacy) is at most

|S1, . . . , Sk | Si ≤ Qi and |Si| = pαii | ≤k∏i=1

|Qi|αi

where Qi is a Sylow pi-subgroup of G and the Qi form part of a Sylowsystem for G.

For each i, we have Qi = Ri1 × · · · × Rik for some Sylow pi-subgroupsRij of Gj . Note that |Rii| = pαii . Let

Xj :=k∏

i = 1i 6=j

Rij .

Then Xj is a soluble p′j-A-subgroup of Gj = PjMj . So Xj is isomorphic toa subgroup of Mj . Further

Mj∼= MjBj/Bj ≤ Aj/Bj ≤ GL(dj , pj) .

Thus by Corollary 3.1

|Xj | ≤ |Mj | ≤ pj2dj−1 ≤ pj2αj−1 .

Let µ := µ(n). Then

k∏i=1

|Qi|αi =k∏i=1

k∏j=1

|Rij |αi

≤k∏t=1

|Rtt|αt) (∏i 6=j|Rij |)

µ

=k∏i=1

pα2ii

k∏j=1

|Xj |µ (since |Xj | =∏ki = 1i 6=j|Rij |)

≤k∏i=1

pα2i+2αiµ−µi

= n2µk∏i=1

pα2i−µi

≤ n2µ−1k∏i=1

pα2ii .

15

Putting together all the estimates we get

fA,sol(n) ≤ (# of choices for G) n2µ−1k∏i=1

piα2i

≤ n2µ−1k∏i=1

pi5α2i 27αi−3

≤ n2µ−1k∏i=1

pi5α2i+7αi .

Therefore

fA,sol(n) ≤ n2µ+6k∏i=1

pi5α2i .

The proof of Theorem 1.2 is similar to that of Theorem 1.1.

Proof of Theorem 1.2 Let p1, . . . , pk be any k distinct primes. Let G bea soluble A-group with |G| = n =

∏ki=1 p

αii and let Gi := G/Op′i(G).

Then by Lemma 4.3,

G ≤ (G1 × · · · ×Gk) := G .

Note that G is a soluble A-group. We shall first enumerate the possibilitiesfor G up to isomorphism and then estimate the number of subgroups of Gof order n up to isomorphism. For the former we need to count the numberof Gi up to isomorphism.

By Note 4.1, Gi = PiHi where Pi is the Sylow pi-subgroup of Gi and Hi

is a p′i-A-subgroup of Ai := Aut(Pi). Thus the number of choices for Gi upto isomorphism is at most∑

|Pi| = pαii

Pi abelian

# of choices for Hi as a subgroup of Ai .

Fix Pi and let di and Bi be as in Note 4.2. Now Hi is isomorphic toHiBi/Bi. It can be shown quite easily that any Sylow subgroup of HiBi/Bican be generated by di elements. Since HiBi/Bi is a soluble group, by atheorem of Kovacs [7] it can be generated by di + 1 elements and henced(Hi) ≤ di + 1. Further Hi ≤ M for some M ∈ Ms,p′i,A

(Ai). ClearlyM ∼= MBi/Bi and by Corollary 3.1,

|M | = |MBi/Bi| ≤ p2di−1i .

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Therefore the number of choices for Hi as a subgroup of Ai is at most∑M

(# of di + 1 generator subgroups of M) ≤∑M

|M |di+1

≤∑M

p2d2i+di−1i

≤∑M

p2α2i+αi−1

i

where M ranges over the elements of Ms,p′i,A(Ai).

We note that

|Ms,p′i,A(Ai)| ≤ p

α2ii |[Ms,p′i,A

(Ai)]|

≤ p5α2i

i 26αi−2 (by (3)) .

So the number of choices for Hi as a subgroup of Ai is at most

p5α2i+2α2

i+αi−1i 26αi−2 = p

7α2i+αi−1

i 26αi−2 .

The number of choices for Pi is equal to the number of unordered parti-tions of αi and this is at most 2αi−1. Thus the number of choices for Gi upto isomorphism is less than or equal to

p7α2i+αi−1

i 27αi−3

and the number of choices for G is at most

k∏i=1

p7α2i+αi−1

i 27αi−3 ≤k∏i=1

p7α2i+8αi−1

i .

We count the number of choices for G as a subgroup of G as in the proofof Theorem 1.1. Let G be fixed. Let S1, . . . , Sk be a Sylow system forG. So Si is a Sylow pi-subgroup of G and for all i, j we have SiSj = SjSi.Thus G = S1S2 · · ·Sk. Further there exists a Sylow system Q1, . . . , Qk,for G, such that Si ≤ Qi. Any two Sylow systems for G are conjugate.Consequently the number of choices for G as a subgroup of G (and up toconjugacy) is at most

|S1, . . . , Sk | Si ≤ Qi and |Si| = pαii | ≤k∏i=1

|Qi|αi

17

where Qi is a Sylow pi-subgroup of G and the Qi form a Sylow system forG.

Note that |Qi| divides pkαii . So∏ki=1 |Qi|

αi ≤∏ki=1 p

kα2i

i . Thus takinginto account all the estimates we get

fA,sol(n) ≤ (# of choices for G)k∏i=1

pikα2

i

≤k∏i=1

pi(k+7)α2

i +8αi−1 .

Consequently if we choose ck := k + 14 then we have

fA,sol(n) ≤k∏i=1

pckα

2i

i .

We conclude this section with the proof of Proposition 1.1.

Proof of Propositon 1.1 We may assume without loss of generality thatε < c. Let Sk be a set of natural numbers such that n ∈ Sk if and only ifthere exist distinct prime numbers p1, . . . , pk and natural numbers α1, . . . , αksuch that n =

∏ki=1 p

αii and

(i) α1 > max(cαi + 1)/ε , αi | i = 2, . . . , k.

(ii)∏ki=2 p

αii > M p

(c−ε)α21

1 .

It is not difficult to see that Sk is an infinite set. For we may chooseα2, . . . , αk arbitrarily. Once a choice of α2, . . . , αk is made we have infinitelymany choices for α1 such that Condition (i) is satisfied. We can then choosep1 arbitrarily. Then we have infinitely many choices for p2, . . . , pk such thatCondition (ii) is satisfied.

If n ∈ Sk and n =∏ki=1 p

αii then by Condition (i) we have µ(n) = α1.

Therefore

nεµ∏ki=1 p

cα2i

i

=

∏ki=1 p

εα1αii∏k

i=1 pcα2i

i

=pα2(εα1−cα2)2 · · · pαk(εα1−cαk)

k

p(c−ε)α2

11

>pα22 · · · p

αkk

p(c−ε)α2

11

(using Condition (i))

> M (using Condition (ii)) .

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Note 5.1 When k = 1 and n = pα for some prime p we have

fA,sol(n) = fA(pα) = p(α)

where p(α) is defined to be the number of unordered partitions of α. It iseasy to see that p(α) ≤ 2α−1 ≤ n.

6 Commentary

We have shown in Theorem 1.1 that fA,sol(n) ≤ n6µ+7. We however need toreflect on whether the bound we have is good.

As shown in Note 5.1, for any prime p and any α ∈ N, fA,sol(pα) ≤ pα. So

there exist infinitely many n with µ(n) unbounded such that fA,sol(n) ≤ n.Thus it is impossible to find a constant c > 0 such that ncµ ≤ fA,sol(n) forall n.

Let S be any class of A-groups and let fS(n) denote the number ofisomorphism classes of groups of order n in S. Further suppose that fS(n) ≤n for infinitely many n with µ(n) unbounded. We shall say that a constanta > 0 satisfies (*) with respect to S if there exists a constant b > 0 suchthat

(i) fS(n) ≤ naµ+b for all n;

(ii) Given ε > 0 there exist infinitely many n with µ(n) unbounded andsuch that n(a−ε)µ < fS(n).

It is not difficult to see that for any c > 0 (c 6= a), the constant c cannotsatisfy (*) with respect to S. Thus we can in this sense call a the rightleading term.

McIver and Neumann [8] and L. Pyber [11] have some lower bounds forfA,sol(n) of the form ncµ for certain specific types of n. But in both thecases we have 0 < c < 0.08.

On the other hand if An denotes the variety (see [9]) of all abelian groupsof exponent n and S denotes the join of the product varietiesApAq andAqAp(p, q distinct primes) then in this case we have a constant a > 0 satisfying(*) with respect to S (see [13]). This constant is however dependant on bothp and q and does not allow for generalisations. A survey of prime pairs upto 31 gave a = 0.141 for (p, q) = (11, 23) which is much below a = 6.

A correct leading term in the case of soluble A-groups should lead us tothe correct leading term for A-groups in general. As suggested in Pyber’s

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paper [11], this in turn should lead to better estimates for the error term inthe bound for f(n).

Acknowledgement: The results in this paper are part of the thesis pre-sented for a DPhil degree at the University of Oxford, under the supervisionof Dr P. M. Neumann, Queen’s College, Oxford. I owe much to him for hisencouragement, support and help.

References

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[2] Gabrielle A. Dickenson. On the enumeration of certain classes of solublegroups. Quart. J. Math. Oxford (2), 20:383–394, 1969.

[3] John D. Dixon. The Fitting subgroup of a linear solvable group. J.Austral. Math. Soc, 7:417–424, 1967.

[4] Daniel Gorenstein. Finite Groups. 1968.

[5] V. C. Harris and M. V. Subbarao. On product partitions of integers.Can. Math. Bull., 34:474–479, 1991.

[6] Graham Higman. Enumerating p-groups. I: Inequalities. Proc. LondonMath. Soc. (3), 10:24–30, 1960.

[7] L. G. Kovacs. On finite soluble groups. Math. Z, 103:37–39, 1968.

[8] Annabel McIver and Peter M. Neumann. Enumerating finite groups.Quart. J. Math. Oxford (2), 38:473–488, 1987.

[9] Hanna Neumann. Varieties of Groups. Springer-Verlag, Berlin Heidel-berg, 1967.

[10] P. P. Palfy. A polynomial bound for the orders of primitive solvablegroups. J. of Algebra, 77:127–137, 1982.

[11] L. Pyber. Enumerating finite groups of a given order. Annals Of Math-ematics, 137:203–220, 1993.

[12] Charles C. Sims. Enumerating p-groups. Proc. London Math. Soc. (3),15:151–166, 1965.

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[13] Geetha Venkataraman. Enumeration Of Types Of Finite Groups. PhDthesis, Department of Mathematics, University of Oxford, Oxford, 1993.

[14] Thomas R. Wolf. Solvable and nilpotent subgroups of gl(n, qm). Can.J. Math. (5), 34:1097–1111, 1982.

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