Maximal subgroup growth of a few polycyclic groups
We give here the exact maximal subgroup growth of two classes of polycyclic groups. Let Gk = ⟨x1, x2, . . . , xk | xixjxi⁻¹ for all i < j⟩, so Gk = ℤ⋊(ℤ⋊(ℤ⋊• • •⋊ℤ)). Then for all integers k ≥ 2, we calculate mn(Gk), the number of maximal subgroups of Gk of index n, exactly. Also, for infinitely...
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nasplib_isofts_kiev_ua-123456789-1887492025-02-09T20:50:58Z Maximal subgroup growth of a few polycyclic groups Kelley, A. Wolfe, E. We give here the exact maximal subgroup growth of two classes of polycyclic groups. Let Gk = ⟨x1, x2, . . . , xk | xixjxi⁻¹ for all i < j⟩, so Gk = ℤ⋊(ℤ⋊(ℤ⋊• • •⋊ℤ)). Then for all integers k ≥ 2, we calculate mn(Gk), the number of maximal subgroups of Gk of index n, exactly. Also, for infinitely many groups Hk of the form ℤ² ⋊ G₂, we calculate mn(Hk) exactly. This paper was done with the support of the Student Collaborative Research grant at Colorado College. We would like to thank the referee for a detailed referee report that helped us improve this paper. 2021 Article Maximal subgroup growth of a few polycyclic groups / A. Kelley, E. Wolfe // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 2. — С. 226-235. — Бібліогр.: 9 назв. — англ. 1726-3255 DOI:10.12958/adm1506 2020 MSC: 20E07. https://nasplib.isofts.kiev.ua/handle/123456789/188749 en Algebra and Discrete Mathematics application/pdf Інститут прикладної математики і механіки НАН України |
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We give here the exact maximal subgroup growth of two classes of polycyclic groups. Let Gk = ⟨x1, x2, . . . , xk | xixjxi⁻¹ for all i < j⟩, so Gk = ℤ⋊(ℤ⋊(ℤ⋊• • •⋊ℤ)). Then for all integers k ≥ 2, we calculate mn(Gk), the number of maximal subgroups of Gk of index n, exactly. Also, for infinitely many groups Hk of the form ℤ² ⋊ G₂, we calculate mn(Hk) exactly. |
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Kelley, A. Wolfe, E. |
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Kelley, A. Wolfe, E. Maximal subgroup growth of a few polycyclic groups Algebra and Discrete Mathematics |
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Kelley, A. Wolfe, E. |
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Kelley, A. |
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Maximal subgroup growth of a few polycyclic groups |
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Maximal subgroup growth of a few polycyclic groups |
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Maximal subgroup growth of a few polycyclic groups |
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Maximal subgroup growth of a few polycyclic groups |
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Maximal subgroup growth of a few polycyclic groups |
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maximal subgroup growth of a few polycyclic groups |
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Інститут прикладної математики і механіки НАН України |
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Maximal subgroup growth of a few polycyclic groups / A. Kelley, E. Wolfe // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 2. — С. 226-235. — Бібліогр.: 9 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
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AT kelleya maximalsubgroupgrowthofafewpolycyclicgroups AT wolfee maximalsubgroupgrowthofafewpolycyclicgroups |
| first_indexed |
2025-11-30T16:16:16Z |
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2025-11-30T16:16:16Z |
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1850232672117325824 |
| fulltext |
© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 32 (2021). Number 2, pp. 226ś235
DOI:10.12958/adm1506
Maximal subgroup growth
of a few polycyclic groups∗
A. Kelley and E. Wolfe
Communicated by R. A. Grigorchuk
Abstract. We give here the exact maximal subgroup growth
of two classes of polycyclic groups. Let Gk = ⟨x1, x2, . . . , xk |
xixjx
−1
i xj for all i < j⟩, so Gk = Z⋊ (Z⋊ (Z⋊ · · ·⋊Z)). Then for
all integers k ⩾ 2, we calculate mn(Gk), the number of maximal
subgroups of Gk of index n, exactly. Also, for inőnitely many groups
Hk of the form Z2 ⋊G2, we calculate mn(Hk) exactly.
Introduction
Let G be a őnitely generated group. We denote by an(G) the number
of subgroups of G of index n (which is necessarily őnite), and we denote
by mn(G) the number of maximal subgroups of G of index n. Subgroup
growth is the study of the growth of different subgroup counting functions
in groups, such as an(G), mn(G), and sn(G) :=
∑n
k=1
ak(G).
People have made great progress in understanding subgroup growth.
One highlight is the classiőcation of all őnitely generated groups for which
an(G) is bounded above by a polynomial in n (see chapter 5 in [7]). Also,
Jaikin-Zapirain and Pyber made a signiőcant advance in [3], where they
give a łsemi-structural characterizationž of groups G for which mn(G) is
bounded above by a polynomial in n.
∗This paper was done with the support of the Student Collaborative Research grant
at Colorado College.
2020 MSC: 20E07.
Key words and phrases: maximal subgroup growth, polycyclic groups, semidirect
products.
https://doi.org/10.12958/adm1506
A. Kelley and E. Wolfe 227
For calculating the word growth in a group with polynomial growth, the
degree is given by a nice, simple formula. However, for subgroup growth, it
is often very challenging, given a group G of polynomial subgroup growth,
to calculate its degree of polynomial growth deg(G):
deg(G) = inf{α | an(G) ⩽ nα for all large n} = lim sup
log an(G)
log n
.
Similarly, for groups G with polynomial maximal subgroup growth, it is
often difficult to determine mdeg(G), where
mdeg(G) = inf{α | mn(G) ⩽ nα for all large n} = lim sup
logmn(G)
log n
.
Progress has been made in both areas. In [9], Shalev calculated deg(G)
exactly for certain metabelian groups and for all virtually abelian groups.
In [6], the őrst author calculated mdeg(G) for some metabelian groups,
and in [4] he does so for all virtually abelian groups.
The groups G for which mdeg(G) is known are rare, and rarer still are
groups for which an exact formula for mn(G) is known. In [2], Gelman
gives a beautiful, exact formula for an(BS(a, b)), assuming gcd(a, b) = 1,
where BS(a, b) is the Baumslag-Solitar group having presentation ⟨x, y |
y−1xay = xb⟩. Gelman’s argument can be easily modiőed to give an
exact formula for mn(BS(a, b)), where again gcd(a, b) = 1. (Alternatively,
a different argument, that explains why gcd(a, b) = 1 is such a nice
assumption, is given by the őrst author in [5].)
Since there are so few groups G for which mn(G) has been calculated,
this paper gives exact formulas for two inőnite classes of polycyclic groups.
For k ⩾ 2, consider the group Gk with presentation
⟨x1, x2, . . . , xk|xixjx
−1
i xj for all i < j⟩.
Then Gk has the form Z⋊ (Z⋊ (Z⋊ · · ·⋊Z)), where the ith Z, reading
from right to left, is generated by xi. Note that the Hirsch length of Gk
is k, and so if i ≠ j, then Gi ̸∼= Gj . In Theorem 3, we calculate mn(Gk)
exactly for k ⩾ 2.
Let G2 be as above. Note that G2 is the Baumslag-Solitar group
BS(1,−1), also known as the fundamental group of the Klein bottle. We
will write G2 = Z ⋊ Z as ⟨b⟩⋊ ⟨a⟩ instead of ⟨x2⟩⋊ ⟨x1⟩. For k ∈ Z, we
will deőne the group Hk, which is of the form Z2 ⋊G2. The generator a
acts (by conjugation) on Z2 by multiplication by the matrix A = ( 0 1
1 0
),
and the generator b acts (by conjugation) on Z2 by multiplication by the
228 Maximal subgroup growth
matrix Bk =
(
0 1
−1 k
)
. Then in Theorem 9, we calculate mn(Hk) exactly
for all k ∈ Z. A consequence of this theorem is that among the groups
Hk, there are inőnitely many that are pairwise non-isomorphic. Also, it is
interesting that mdeg(H2) = 2, but mdeg(Hk) = 1 for all k ̸= 2.
One reason for studying the two families {Gk}k⩾2 and {Hk}k∈Z is that
the őrst author thinks that it might be possible to extend the methods
of [6] to apply to the class of polycyclic groups. In particular, it might
be feasible to give an exact formula for mdeg(G) when G is a group of
the form Ak ⋊ (Ak−1 ⋊ (Ak−2 ⋊ . . . ⋊ A1)), where each Ai is a őnitely
generated abelian group. Another reason why we chose the particular
inőnite families we did is that (besides G2) they appeared to be the easiest
such groups to work with that aren’t of the form A2 ⋊A1.
1. Groups of the form Z ⋊ (Z ⋊ (Z ⋊ · · · ⋊ Z))
For a group G = N ⋊ H with N abelian, to calculate mn(G), it is
useful to consider the H-module structure given by G on N . See Lemma
5 from [6].
Let Gk be as in the introduction, and let G1 = Z. For a group G and
N a G-module, recall that a function δ : G → N is called a derivation (or
a 1-cocycle) if δ(gh) = δ(g)+g ·δ(h) for all g, h ∈ G. The set of derivations
from G to N is denoted Der(G,N). In the following lemma, we will use
the fact that if δ ∈ Der(G,N), then for g ∈ G, we have δ(g−1) = −g−1δ(g)
which follows from the fact that δ(g−1g) = δ(1) = 0.
Lemma 1. Let S be a Gk-module. There is a one-to-one correspon-
dence between the set Der(Gk, S) and the set ∆ of all functions δ :
{x1, x2, . . . xk} → S satisfying
(1− x−1
j )δ(xi) = (−xi − x−1
j )δ(xj) for all i, j with i < j. (∗)
Proof. If δ ∈ ∆, then exercise 3(a) in [1] (pg. 90) (or Lemma 2.20 from [4]),
gives us a unique derivation δ : Fk −→ S, where Fk is the free group on k
generators and the action of Fk on S is the induced action. So by slight
abuse of notation, by taking δ in ∆, we mean the derivation of the free
group Fk that corresponds to the map δ ∈ ∆.
Let δ be an element either of Der(Gk, S) or of ∆. We will show that
δ(xixjx
−1
i xj) = 0 for all i < j if and only if (∗) holds. Fix i and j with
A. Kelley and E. Wolfe 229
i < j. Then
δ(xixjx
−1
i xj) = δ(xi) + xiδ(xj) + xixjδ(x
−1
i ) + xixjx
−1
i δ(xj)
= δ(xi) + xiδ(xj)− xixjx
−1
i δ(xi) + x−1
j δ(xj)
= δ(xi) + xiδ(xj)− x−1
j δ(xi) + x−1
j δ(xj)
= (δ(xi)− x−1
j δ(xi))− (−xiδ(xj)− x−1
j δ(xj))
= (1− x−1
j )δ(xi)− (−xi − x−1
j )δ(xj)
If δ ∈ Der(Gk, S), then δ(xixjx
−1
i xj) = 0 because δ(1) = 0, and so (∗)
holds. Conversely, if δ ∈ ∆, then that (∗) holds implies δ(xixjx
−1
i xj) = 0
for all i < j, in which case Lemma 2.19 from [4], which is basically exercise
4(a) in [1] (pg. 90), gives a derivation δ from Gk to S.
Lemma 2. Consider Z/pZ, a simple Gk-module, where each generator
xi of Gk acts on Z/pZ by multiplication by −1. Then
|Der(Gk,Z/pZ)| =
{
2k if p = 2
p if p ̸= 2.
Proof. If p = 2, then the action of Gk on Z/2Z is trivial, and so
|Der(Gk,Z/2Z)| = |Hom(Gk,Z/2Z)|, which is 2k.
Next, suppose p ≠ 2. The action of (1−x−1
i ) and (−xi−x−1
j ) on Z/pZ
is multiplication by 2, which is invertible since p ≠ 2. So (∗) from Lemma 1
becomes 2δ(xi) = 2δ(xj) for all i < j, which simpliőes to δ(xi) = δ(xj)
for all i < j.
So by Lemma 1, we may choose a derivation by picking δ(xk) to be
any element of Z/pZ, and then letting δ(xi) = δ(xk) for all i < k. Thus
|Der(Gk,Z/pZ)| = |Z/pZ| = p.
Theorem 3. Let Gk be as above. Then
mn(Gk) =
1 + (k − 1)n if n is a prime with n > 2
2k − 1 if n = 2
0 if n is not prime.
(∗∗)
Proof. Consider N , the subgroup of Gk generated by xk. Then N ∼= Z,
and N ⊴ Gk with Gk/N ∼= Gk−1. So, N is a Gk−1-module. Since Gk
∼=
N ⋊Gk−1, Lemma 5 from [6] gives us
mn(Gk) = mn(Gk−1) +
∑
N0
|Der(Gk−1, N/N0)|
230 Maximal subgroup growth
where the sum is over all maximal submodules N0 of N of index n. Of
course, the maximal submodules of N are precisely the subgroups of prime
index. Thus if n is not prime, then mn(Gk) = 0; this follows by induction
on k.
Fix a prime p. For both cases p > 2 and p = 2, we proceed by induction
on k.
First, let p > 2, and let k = 1. Then mp(G1) = 1 = 1 + (k − 1)p.
Assume (∗∗) holds for k = a. Then mp(Ga) = 1 + (a − 1)p. Consider
k = a+ 1. We have
mp(Ga+1) = mp(Ga) +
∑
N0
|Der(Ga, N/N0)|.
By Lemma 2,
∑
N0
|Der(Ga, N/N0)| = p. So mp(Ga+1) = 1+(a−1)p+
p = 1 + (a+ 1− 1)p, the desired result.
Finally, let p = 2, and let k = 1. Then m2(G1) = 1 = 21 − 1. Assume
(∗∗) holds for k = a. Then m2(Ga) = 2a − 1. Consider k = a+ 1. Then
m2(Ga+1) = m2(Ga)+ |Der(Ga,Z/2Z)|. By Lemma 2, |Der(Ga,Z/2Z)| =
2a. Thus m2(Ga+1) = 2a − 1 + 2a = 2a+1 − 1, the desired result.
2. Some groups of the form Z2 ⋊ (Z ⋊ Z)
Next, we will deőne the groups Hk, which are of the form Z2⋊ (Z⋊Z).
We will write G2 = Z ⋊ Z as ⟨b⟩⋊ ⟨a⟩ instead of ⟨x2⟩⋊ ⟨x1⟩. Recall that
G2 = ⟨a, b|aba−1b⟩. To form a group of the form Z2 ⋊ (Z ⋊ Z), what
we need is an action of G2 on Z2, and so, we just need to őnd matrices
A,B ∈ GL2(Z) such that ABA−1B = I2. With this, we can say that
the action (by conjugation) of the generator a on Z2 is multiplication
by the matrix A, and the generator b acts (by conjugation) on Z2 by
multiplication by the matrix B.
We will take A =
(
0 1
1 0
)
. Let B =
(
w x
y z
)
. Then ABA−1B =
(
y2 + wz yz + xz
wy + xw wz + x2
)
, and we will őnd solutions which make this equal
I2. We thus have y2 + wz = 1, wz + x2 = 1, wy + xw = 0 (equivalently,
w = 0 or x+ y = 0), and yz + xz = 0 (equivalently, z = 0 or x+ y = 0).
Also, we have wz − xy = ±1. One way to solve these equations is to let
w = 0. Then x, y = ±1. Take x = 1. If we take y = −1, then z can be any
integer.
A. Kelley and E. Wolfe 231
Let the group Hk be the group formed when we take B to be Bk =
(
0 1
−1 k
)
. Our choice of A and B is in part to make calculating mn(Hk)
be as simple as possible, but other choices could also be considered.
For a module M , we let N ⩽ M denote that N is a submodule of M .
Lemma 4. Consider Z2 to be a G2-module as above. Let M be a maximal
submodule of Z2. Then pZ2 ⩽ M for some prime p.
Proof. First, recall that every maximal subgroup of a polycyclic group
has prime power index; this follows, for example, from the proof of Result
5.4.3 (iii) in [8].
Let Hk be as above, so Hk = Z2 ⋊ G2. We claim that M yields
a maximal subgroup of Hk with index equal to [Z2 : M ]. Indeed, we
have that Hk/M ∼= (Z2/M) ⋊ G2. Thus Z2/M has a complement in
(Z2/M) ⋊ G2, which by Lemma 3 of [6] must be maximal and of index
[Z2 : M ]. Then just take its preimage in Hk.
Since M yields a maximal subgroup of Hk with index equal to [Z2 : M ]
and Hk is polycyclic, we must have [Z2 : M ] = pj for some prime p.
Therefore, pjZ2 ⩽ M . Consider the group Z2/pjZ2. Its Frattini subgroup
is pZ2/pjZ2, and therefore, pZ2/pjZ2 +M/pjZ2 = M/pjZ2. And hence
pZ2 +M = M , that is, pZ2 ⩽ M .
For a module N and for ni ∈ N for i = 1, . . . , t, the submodule they
generate is denoted ⟨n1, n2, . . . , nt⟩. For a prime p, consider the submodule
Mp,w of Z2, where Mp,w =
〈
( p
0
) ,
(
0
p
)
,w
〉
, with w ∈ Z2. We will assume
w /∈ pZ2. Then Mp,w is a proper (and hence maximal) submodule of Z2
if and only if the image of w in Z/pZ2 is an eigenvector of both matrices
A and Bk, considered as elements of GL2(Fp).
Let v = ( 1
1
) and u =
(
1
−1
)
, and let Mp = Mp,v and Mp,−1 = Mp,u.
Of course, v and u are eigenvectors of A ∈ GL2(Fp) for all p. (And v ≡ u
(mod 2). Hence, M2 = M2,−1.)
Lemma 5. With the above notation, we have that Mp is a maximal
submodule of Z2 if and only if p | k − 2. Also, Mp,−1 is a maximal
submodule of Z2 if and only if p | k + 2. Further, Mp,w is not a proper
submodule of Z2 unless Mp,w = Mp or Mp,−1. Thus, pZ2 is a maximal
submodule of Z2 if and only if p ∤ (k − 2)(k + 2). Finally there are no
maximal submodules of Z2 besides (the appropriate) Mp, Mp,−1, and pZ2.
Proof. Let v and u be as above, but consider them as elements of Z2/pZ2.
We have that Bkv =
(
1
k−1
)
, and so Bkv = λv for some λ ∈ Z if and only
if k − 1 ≡ 1 (mod p), i.e. if and only if p | k − 2.
232 Maximal subgroup growth
Also, Bku =
(
−1
−1−k
)
, and since multiples of u are characterized by
the sum of their coordinates being 0 (mod p), Bku = λu for some λ ∈ Z
if and only if −1− 1− k ≡ 0 (mod p), which is equivalent to p | k + 2.
That no other Mp,w is a proper submodule of Z2 follows from the fact
that any eigenvector of A is a multiple of v or of u.
Next, let p ∤ (k − 2)(k + 2). Since neither Mp nor Mp,−1 is a maximal
submodule of Z2 and neither is any other Mp,w, we have that pZ2 is
a maximal submodule of Z2. And if p | (k − 2)(k + 2), then p | k − 2 or
p | k + 2, in which case Mp or Mp,−1 is a proper submodule of Z2 that
properly contains pZ2.
The őnal statement of this lemma follows from the previous parts of
this lemma, together with Lemma 4; indeed it follows from Lemma 4 that
any maximal submodule is either pZ2 or of the form Mp,w.
For a module N , we let m̃n(N) denote the number of maximal sub-
modules of N of index n.
Corollary 6. We have
m̃n(Z
2) =
1 if n is a prime p, and p | (k − 2)(k + 2)
1 if n = p2 for some prime p, and p ∤ (k − 2)(k + 2)
0 otherwise.
Proof. Note that if p | k − 2 and p | k + 2, then k − 2 ≡ k + 2 (mod p),
whence p = 2. And using the previous notation, recall that M2 = M2,−1.
This corollary then follows from Lemma 5.
Lemma 7. Consider G2 with presentation ⟨a, b | aba−1b⟩, as described
above. Let S be a G2-module. Then there is a one-to-one correspondence
between the set Der(G2, S) and the set of functions δ : {a, b} −→ S satisfying
(1− b−1)δ(a) = (−a− b−1)δ(b). (∗ ∗ ∗)
Proof. This follows from Lemma 1.
Lemma 8. Fix k, and let Z2 have the G2-module structure given above.
For a given prime p, deőne S as either Z2/Mp, Z
2/Mp,−1 or Z2/pZ2 such
that S is simple (see Lemma 5). Any simple quotient of Z2 must be some
such S. Then
|Der(G2, S)| =
|S|2 = p2 if p | k − 2
|S| = p if p | k + 2 and p > 2
|S| = p2 if p ∤ (k − 2)(k + 2).
A. Kelley and E. Wolfe 233
Proof. That any simple quotient of Z2 is some such S follows from
Lemma 5.
Let δ ∈ Der(G2, S) (to be speciőed later). The element 1−b−1 in (∗∗∗)
from Lemma 7 acts on δ(a) by multiplication by the matrix I2 −B−1
k =
(
1−k 1
−1 1
)
which has determinant 2− k, and the element −a− b−1 acts by
multiplication by the matrix −A−B−1
k =
(
−k 0
−2 0
)
.
First, suppose p ∤ k − 2. Then by Lemma 5, either Mp,−1 or pZ2 is
a maximal submodule of Z2, depending on whether or not p | k+2. Notice
that if p = 2, then p | k + 2 implies p | k − 2, and hence if p | k + 2, then
p > 2 (because we are assuming here that p ∤ k − 2).
In this case, I2 −B−1
k is invertible, considered as a 2× 2 matrix over
Fp. Hence (∗ ∗ ∗) from Lemma 7 may be written as
δ(a) = (I2 −B−1
k )−1(−A−B−1
k )δ(b).
And so in this case, we are free to choose δ(b) to be any element of
S, and then this determines what δ(a) must be. Thus we would have
|Der(G2, S)| = |S|. If p | k + 2, then S = Z2/Mp,−1, and |S| = p. If
p ∤ k + 2, then S = Z2/pZ2, and |S| = p2.
Next, suppose that p | k−2. Then I2−B−1
k ≡
(
−1 1
−1 1
)
and −A−B−1
k ≡
(
−2 0
−2 0
)
(mod p). Also, p | k−2 implies (by Lemma 5) that Mp is a maximal
submodule of Z2. By Corollary 6, we have that m̃p(Z
2) ⩽ 1, and thus
S = Z2/Mp.
We have that {( i
0
) : 0 ⩽ i < p} is a complete set of representatives of
Z2/Mp. Then letting δ(a) = ( i
0
)+Mp and δ(b) =
(
j
0
)
+Mp, we have that
equation (∗∗∗) from Lemma 7 holds because (I2−B−1
k ) ( i
0
) =
(
−i
−i
)
∈ Mp
and (−A − B−1
k )
(
j
0
)
=
(
−2j
−2j
)
∈ Mp. And so both (1 − b−1)δ(a) and
(−a− b−1)δ(b) are the trivial element of Z2/Mp. Therefore, in this case,
we have |S|2 options for a derivation from G2 to S.
Theorem 9. We have
mn(Hk) =
n2 + n+ 1 if n is prime, and n | k − 2
2n+ 1 if n is prime, and n | k + 2 with p > 2
n+ 1 if n is prime, and n ∤ (k − 2)(k + 2)
n if n = p2 for some prime p, and
p ∤ (k − 2)(k + 2)
0 otherwise.
234 Maximal subgroup growth
Proof. Consider Z2 ⊴ Z2 ⋊G2. By Lemma 5 from [6], we have
mn(Hk) = mn(G2) +
∑
N0
|Der(G2,Z
2/N0)| (1)
where the sum is over all maximal submodules N0 of Z2 of index n. Also,
by Theorem 3, we have
mn(G2) =
{
n+ 1 if n is a prime
0 otherwise.
(2)
We have that Lemma 8 and Corollary 6 together imply that
∑
N0
|Der(G2,Z
2/N0)| =
n2 if n is prime, and n | k − 2
n if n is prime and n | k + 2 with n > 2
n if n = p2 for some prime p, and
p ∤ (k − 2)(k + 2)
0 otherwise.
(3)
The present theorem follows from (1) by adding (2) and (3).
For n ∈ Z, let π(n) denote the set of prime numbers dividing n. Then
a consequence of Theorem 9 is that for i, j ∈ Z, if π(i − 2) ̸= π(j − 2)
or π(i + 2) ̸= π(j + 2), then Hi ̸∼= Hj . Also, note that mdeg(H2) = 2
(because π(0) is the set of all primes), and for all k ̸= 2, mdeg(Hk) = 1.
Acknowledgment
We would like to thank the referee for a detailed referee report that
helped us improve this paper.
References
[1] K. Brown. Cohomology of groups. Springer-Verlag, New York, 1982.
[2] E. Gelman. (2005). Subgroup growth of Baumslag-Solitar groups. J. Group Theory,
8(no. 6), 801ś806.
[3] A. Jaikin-Zapirain and L. Pyber. (2011). Random generation of őnite and proőnite
groups and group enumeration. Ann. of Math. (2), 173(no. 2):769ś814.
[4] A. Kelley. Maximal Subgroup Growth of Some Groups. PhD thesis, State University
of New York at Binghamton, (2017).
[5] A. Kelley. (2020). Subgroup growth of all Baumslag-Solitar groups. New York J.
of Math., 218ś229; http://nyjm.albany.edu/j/2020/26-11.html.
http://nyjm.albany.edu/j/2020/26-11.html
A. Kelley and E. Wolfe 235
[6] A. Kelley. Maximal subgroup growth of some metabelian groups. To appear in
Comm. Algebra. Preprint (2018), https://arxiv.org/abs/1807.03423.
[7] A. Lubotzky and D. Segal. Subgroup growth. Birkhauser Verlag, Basel, 2003.
[8] D. Robinson. A course in the theory of groups. Springer-Verlag, New York, second
edition, 1996.
[9] A. Shalev. (1999). On the degree of groups of polynomial subgroup growth. Trans.
Amer. Math. Soc., 351(no. 9):3793ś3822.
Contact information
Andrew James
Kelley
Department of Mathematics and Computer
Science
14 E. Cache La Poudre St.
Colorado Springs, CO, 80903, USA
E-Mail(s): akelley2500@gmail.com
Elizabeth Ciorsdan
Dwyer Wolfe
Elizabeth Wolfe 1848
Colorado College
902 N Cascade Ave
Colorado Springs, CO 80946, USA
E-Mail(s): e_wolfe@coloradocollege.edu
Received by the editors: 03.12.2019
and in őnal form 04.01.2021.
https://arxiv.org/abs/1807.03423
mailto:akelley2500@gmail.com
mailto:e_wolfe@coloradocollege.edu
A. Kelley and E. Wolfe
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