PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 101(115) (2017), 121–133 https://doi.org/10.2298/PIM1715121T

ON THE NUMBER OF SUBGROUPS OF A GIVEN EXPONENT IN A FINITE ABELIAN GROUP

Marius Tărnăuceanu and László Tóth

Abstract. This paper deals with the number of subgroups of a given exponent in a finite abelian group. Explicit formulas are obtained in the case of rank two and rank three abelian groups. An asymptotic formula is also presented.

1. Introduction

One of the most important problems of combinatorial abelian group theory is to determine the number of subgroups of a finite abelian group. This topic has enjoyed a constant evolution starting with the first half of the 20^{th} century. Since a finite abelian group is a direct product of abelian p-groups, the above counting problem can be reduced to p-groups. Formulas which give the number of subgroups of type μ of a finite p-group of type λ were established by Delsarte [4], Djubjuk [5] and Yeh [15]. An excellent survey on this subject together with connections to symmetric functions was written by Butler [2] in 1994.

Another way to find the total number of subgroups of finite abelian p-groups was described by Bhowmik [1] by using divisor functions of matrices. By invoking different arguments, formulas in the case of rank two p-groups were obtained by Călugăreanu [3], Tărnăuceanu [11, 12], Hampejs, Holighaus, Tóth, Wiesmeyr [6], Tóth [14] and for rank three p-groups by Hampejs, Tóth [7], Oh [8]. Note that the papers [6, 7, 14] include also direct formulas for the groups ℤ_{m} × ℤ_{n} and ℤ_{m} × ℤ_{n} × ℤ_{r}, respectively, where m, n, r ∈ ℕ^{∗} := {1, 2, …} are arbitrary.

The purpose of the current paper is to count the number of subgroups of a given exponent in a finite abelian p-group. Explicit formulas are obtained for rank two and rank three p-groups. The numbers of subgroups of exponent p, respectively p^{2} in an arbitrary p-group are also considered. We prove that if two finite abelian groups have the same number of subgroups of any exponent, then they are isomorphic. We also deduce compact formulas for the number of subgroups of a given exponent of the group ℤ_{m} ×ℤ_{n}, where m, n ∈ ℕ^{∗} are arbitrary. Furthermore, we obtain an exact formula for the sum of exponents of the subgroups of ℤ_{m} × ℤ_{n}, and an asymptotic formula for the arithmetic means of exponents of the subgroups of ℤ_{n} × ℤ_{n}.

2010 Mathematics Subject Classification: Primary 20K01; Secondary 20K27, 11N37.

Key words and phrases: finite Abelian group, subgroup, exponent.

Communicated by Zoran Petrović.

For the proofs we use two different approaches. The first one is based on the known formula for the number of subgroups of a given type in an abelian pgroup, given in terms of gaussian coefficients (Theorem 2.1). The second method, applicable only for rank two groups, uses the representation of the subgroups of ℤ_{m} × ℤ_{n} obtained by the second author in [14] (Theorem 3.1).

Most of our notation is standard and usually will not be repeated here. For basic notions and results on group theory we refer the reader to [10].

2. First approach

Let G be a finite abelian group of order n and G = G_{1} × G_{2} × ⋯× G_{m} be the primary decomposition of G, where G_{i} is a p_{i}-group (i = 1, 2, …, m). For every divisor d = p_{1}^{α_{1}} p_{2}^{α_{2}} …p_{m}^{α_{m}} of n we denote E_{d}(G) = {H ⩽ G : exp(H) = d}.

Since the subgroups H of G are of type H = H_{1} × H_{2} × ⋯ × H_{m} with H_{i} ⩽ G_{i} (i = 1, 2, …, m), we infer that

(2.1)  |E_{d}(G)| = ∏_{i=1}^{m} |E_{p_{i}^{α_{i}}} (G_{i})|.

Equality (2.1) shows that the problem of counting the number of subgroups of exponent d in G is reduced to p-groups. So, in this section we will assume that G is a finite abelian p-group, that is a group of type ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} × ⋯× ℤ_{p^{λ_{k}}} with λ_{1} ⩾ λ_{2} ⩾ ⋯⩾ λ_{k} ⩾ 1. In this case we will say that G is of type λ, where λ is the partition (λ_{1}, λ_{2}, …, λ_{k}, 0, …), and we will denote it by G_{λ}.

We recall the following well–known result, which gives the number of subgroups of type μ of G_{λ} (see [4, 5, 15]).

Theorem 2.1. For every partition μ ⪯ λ (i.e., μ_{i} ⩽ λ_{i} for every i ∈ ℕ^{∗}) the number of subgroups of type μ in G_{λ} is

α_{λ}(μ; p) = ∏_{i⩾1} p^{(a_{i} −b_{i} )b_{i+1}} \binom{a_{i} − b_{i+1}}{b_{i} − b_{i+1}} _{p},

where λ′ = (a_{1}, a_{2}, …, a_{λ_{1}} , 0, …), μ′ = (b_{1}, b_{2}, …, b_{μ_{1}} , 0, …) are the partitions conjugate to λ and μ, respectively, and

\binom{n}{k}_{p} = \frac{∏ _{i=1}^{n} (p^{i} − 1)}{∏ _{i=1}^{k} (p^{i} − 1) ∏ _{i=1}^{n−k} (p^{i} − 1)}

is the gaussian binomial coefficient (it is understood that ∏ _{i=1}^{m} (p^{i} − 1) = 1 for m = 0).

By using Theorem 2.1 a way to compute the number of subgroups of exponent p^{i} in G_{λ} can be inferred, namely

|E_{p^{i}} (G_{λ})| = ∑_{μ⪯λ, μ_{1} =i} α_{λ}(μ; p) (i = 0, 1, …, λ_{1}),

which is a polynomial in p with integer coefficients. If i ⩾ 1 and k ⩾ 2 are arbitrary, then the polynomial |E_{p^{i}} (G_{λ})| can not be given explicitly, but we will do this in some particular cases. Namely, we will consider the following cases: k ∈ {2, 3} and i ⩾ 1 arbitrary, i ∈ {1, 2} and k ⩾ 2 arbitrary.

We remark first that since ∑ _{H∈E_{p^{i}} (G_{λ} )} H ⩽ G_{λ} and ∑ _{H∈E_{p^{i}} (G_{λ} )} H is of type ℤ_{p^{i}} × ⋯× ℤ_{p^{i}} × ℤ_{p^{λr}} × ⋯× ℤ_{p^{λ_{k}}} , where r = min{j : λ_{j} < i}, we have

|E_{p^{i}} (G_{λ})| = |E_{p^{i}} (ℤ_{p^{i}} × ⋯× ℤ_{p^{i}} × ℤ_{p^{λr}} × ⋯× ℤ_{p^{λ_{k}}} )|

with the convention that the direct product ℤ_{p^{λr}} × ⋯× ℤ_{p^{λ_{k}}} is trivial for i ⩽ λ_{k}. Our first result gives a precise expression for |E_{p^{i}} (G_{λ})| in the case k = 2.

Proposition 2.1. Let G_{λ} = ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} with λ_{1} ⩾ λ_{2} ⩾ 1. Then

|E_{p^{i}} (G_{λ})| = \{ \begin{matrix}\frac{p^{i+1} +p^{i} −2}{p−1} , & 1 ⩽ i ⩽ λ_{2}, \\ \frac{p^{λ_{2} +1} −1}{p−1} , & λ_{2} < i ⩽ λ_{1}. \end{matrix}

Proof. We have λ = (λ_{1}, λ_{2}, 0, …) and consequently

λ′ = (2, 2, …, 2, 1, 1, …, 1, 0, …),

where the number of 2’s is λ_{2} and the number of 1’s is λ_{1} − λ_{2}. Then

|E_{p^{i}} (G_{λ})| = ∑_{j=0}^{i} N_{i,j} ,

where N_{i,j} denotes the number of subgroups of type (i, j) in G_{λ}, or equivalently in ℤ_{p^{i}} × ℤ_{p^{i}} if i ⩽ λ_{2} or in ℤ_{p^{i}} × ℤ_{p^{λ_{2}}} if λ_{2} < i ⩽ λ_{1}.

In the first case we obtain N_{i,j} = (p + 1)p^{i−j−1} for j = 0, 1, …, i − 1, and N_{i,i} = 1. Therefore

|E_{p^{i}} (ℤ_{p^{i}} × ℤ_{p^{i}} )| = 1 + ∑_{j=0}^{i−1} (p + 1)p^{i−j−1} = \frac{p^{i+1} + p^{i} − 2}{p − 1} .

In the second case we obtain N_{i,j} = p^{λ_{2} −j} for j = 0, 1, …, λ_{2}. Therefore

|E_{p^{i}} (ℤ_{p^{i}} × ℤ_{p^{λ_{2}}} )| = ∑_{j=0}^{λ_{2}} p^{λ_{2} −j} = \frac{p^{λ_{2} +1} − 1}{p − 1} .

This completes the proof. □

Example 2.1. We have

|E_{p^{i}} (ℤ_{p^{4}} × ℤ_{p^{2}} )| = \{ \begin{matrix}1, & i = 0, \\ p + 2, & i = 1, \\ p^{2} + 2p + 2, & i = 2, \\ p^{2} + p + 1, & i = 3 or i = 4. \end{matrix}

In particular, by summing all quantities |E_{p^{i}} (G_{λ})| (i = 0, 1, …, λ_{1}), we obtain the total number of subgroups of G_{λ} (see also [11, Prop. 2.9], [12, Th. 3.3]).

Corollary 2.1. The total number of subgroups of G_{λ} = ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} where λ_{1} ⩾ λ_{2} ⩾ 1, is

(2.2) \frac{1}{(p− 1)^{2}}[(λ_{1}− λ_{2}+ 1)p^{λ_{2} + 2} − (λ_{1}− λ_{2}− 1)p^{λ_{2} + 1} − (λ_{1}+ λ_{2}+ 3)p+ (λ_{1}+ λ_{2} + 1)].

Example 2.2. The total number of subgroups of ℤ_{p^{4}} × ℤ_{p^{2}} is 3p^{2} + 5p + 7.

Now consider the case of rank three p-groups. We need the following lemma.

Lemma 2.1. Let N_{i,j,ℓ} denote the number of subgroups of type (i, j, ℓ) (i ⩾ j ⩾ ℓ ⩾ 0) in G_{λ} = ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} × ℤ_{p^{λ_{3}}} with λ_{1} ⩾ λ_{2} ⩾ λ_{3} ⩾ 1. Then

N_{i,j,ℓ} = \{ \begin{matrix}p^{2i−2ℓ−3} (p + 1)(p^{2} + p + 1), & ℓ < j < i ⩽ λ_{3}, \\ p^{2(i−ℓ−1)} (p^{2} + p + 1), & ℓ < j = i ⩽ λ_{3} or ℓ = j < i ⩽ λ_{3}, \\ 1, & ℓ = j = i ⩽ λ_{3}, \\ p^{λ_{3} +i−2ℓ−2} (p + 1)^{2}, & ℓ < j ⩽ λ_{3} < i ⩽ λ_{2}, \\ p^{λ_{3} +i−2j−1} (p + 1), & ℓ = j ⩽ λ_{3} < i ⩽ λ_{2}, \\ p^{2λ_{3} +i−j−2ℓ−1} (p + 1), & ℓ ⩽ λ_{3} < j < i ⩽ λ_{2}, \\ p^{2(λ_{3} −ℓ)} , & ℓ ⩽ λ_{3} < j = i ⩽ λ_{2}, \\ p^{λ_{2} +2λ_{3} −j−2ℓ} , & ℓ ⩽ λ_{3} < j ⩽ λ_{2} < i ⩽ λ_{1}, \\ p^{λ_{2} +λ_{3} −2ℓ−1} (p + 1), & ℓ < j ⩽ λ_{3} ⩽ λ_{2} < i ⩽ λ_{1}, \\ p^{λ_{2} +λ_{3} −2ℓ} , & ℓ = j ⩽ λ_{3} ⩽ λ_{2} < i ⩽ λ_{1}. \end{matrix}

Proof. We use Theorem 2.1. We distinguish the following three cases: I. If i ⩽ λ_{3}, then N_{i,j,ℓ} is the number of subgroups of type (i, j, ℓ) in ℤ_{p^{i}} ×

ℤ_{p^{i}} × ℤ_{p^{i}} . Here we need to consider λ = (i, i, i, 0, …) with λ′ = (3, 3, …, 3, 0 …), where the number of 3’s is i, and μ = (i, j, ℓ, 0, …) with

μ′ = (3, 3, …, 3, 2, 2, …, 2, 1, 1, …, 1, 0, …),

where the number of 3’s is ℓ, the number of 2’s is j − ℓ, the number of 1’s is i − j. In the subcase ℓ < j < i ⩽ λ_{3} we deduce

N_{i,j,ℓ} = (p^{2})^{j−ℓ−1} p(p + 1)(p^{2})^{i−j−1} (p^{2} + p + 1) = p^{2i−2ℓ−3} (p + 1)(p^{2} + p + 1).

The subcases ℓ < j = i ⩽ λ_{3}, ℓ = j < i ⩽ λ_{3} and ℓ = j = i ⩽ λ_{3} are treated similar.

II. If λ_{3} < i ⩽ λ_{2}, then N_{i,j,ℓ} is the number of subgroups of type (i, j, ℓ) in ℤ_{p^{i}} × ℤ_{p^{i}} × ℤ_{p^{λ_{3}}} . We consider λ = (i, i, λ_{3}, 0, …) with λ′ = (3, 3, …, 3, 2, 2, …, 2, 0 …) where the number of 3’s is λ_{3}, the number of 2’s is i−λ_{3}, and μ, μ′ like in the case I.

For example, in the subcase ℓ = j ⩽ λ_{3} < i ⩽ λ_{2} we obtain

N_{i,j,ℓ} = (p^{2})^{λ_{3} −j} p^{i−λ_{3} −1} (p + 1) = p^{λ_{3} +i−2j−1} (p + 1).

III. If λ_{2} < i ⩽ λ_{1}, then N_{i,j,ℓ} is the number of subgroups of type (i, j, ℓ) in ℤ_{p^{i}} × ℤ_{p^{λ_{2}}} × ℤ_{p^{λ_{3}}} . We consider λ = (i, λ_{2}, λ_{3}, 0, …) with

λ′ = (3, 3, …, 3, 2, 2, …, 2, 1, 1, …, 1, 0 …),

NUMBER OF SUBGROUPS OF A GIVEN EXPONENT… 125

where the number of 3’s is λ_{3}, the number of 2’s is λ_{2} − λ_{3}, the number of 1’s is i − λ_{2}, and μ, μ′ like in the case I. □

Proposition 2.2. Let G_{λ} = ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} × ℤ_{p^{λ_{3}}} with λ_{1} ⩾ λ_{2} ⩾ λ_{3} ⩾ 1. Then

|E_{p^{i}} (G_{λ})| = \{ \begin{matrix}\frac{p^{2i−1} ((i+1)p^{4} +(i−1)p^{3} −p^{2} −(i+2)p−i)+3}{(p^{2} −1)(p−1)}  , & 1 ⩽ i ⩽ λ_{3}, \\ \frac{(λ_{3} +1)p^{λ_{3} +i} (p^{2} −1)(p+1)−2p^{2λ_{3} +2} +2}{(p^{2} −1)(p−1)}  , & λ_{3} < i ⩽ λ_{2}, \\ \frac{(λ_{3} +1)p^{λ_{2} +λ_{3} +1} (p^{2} −1)−p^{2λ_{3} +2} +1}{(p^{2} −1)(p−1)} , & λ_{2} < i ⩽ λ_{1}. \end{matrix}

Proof. We have |E_{p^{i}} (G_{λ})| = ∑ _{0⩽ℓ⩽j⩽i} N_{i,j,ℓ} and use Lemma 2.1. Case I. If i ⩽ λ_{3}, then

|E_{p^{i}} (G_{λ})|= N_{i,i,i} + ∑_{ℓ=0}^{i−1} ∑_{j=ℓ+1}^{i−1} N_{i,j,ℓ} + ∑_{ℓ=0}^{i−1} N_{i,i,ℓ} + ∑_{ℓ=0}^{i−1} N_{i,ℓ,ℓ}

= 1 + ∑_{ℓ=0}^{i−1} ∑_{j=ℓ+1}^{i−1} p^{2i−2ℓ−3} (p + 1)(p^{2} + p + 1) + 2 ∑_{ℓ=0}^{i−1} p^{2(i−ℓ−1)} (p^{2} + p + 1)

= 1 + \frac{p^{2} + p + 1}{(p^{2} − 1)(p − 1)} ((i − 1)p^{2i+1} − ip^{2i−1} + p) + 2(p^{2} + p + 1)\frac{p^{2i} − 1}{p^{2} − 1}

= \frac{p^{2i−1} ((i + 1)p^{4} + (i − 1)p^{3} − p^{2} − (i + 2)p − i) + 3}{(p^{2} − 1)(p − 1)} ,

by direct computations.

II. If λ_{3} < i ⩽ λ_{2}, then we have

|E_{p^{i}} (G_{λ})|= ∑_{ℓ=0}^{λ_{3}} ∑_{j=ℓ+1}^{λ_{3}} N_{i,j,ℓ} + ∑_{ℓ=0}^{λ_{3}} ∑_{j=λ_{3} +1}^{i−1} N_{i,j,ℓ} + ∑_{ℓ=0}^{λ_{3}} N_{i,i,ℓ} + ∑_{ℓ=0}^{λ_{3}} N_{i,ℓ,ℓ}

= ∑_{ℓ=0}^{λ_{3}} ∑_{j=ℓ+1}^{λ_{3}} p^{λ_{3} +i−2ℓ−2} (p + 1)^{2} + ∑_{ℓ=0}^{λ_{3}} ∑_{j=λ_{3} +1}^{i−1} p^{2λ_{3} +i−j−2ℓ−1} (p + 1)

+ ∑_{ℓ=0}^{λ_{3}} p^{2(λ_{3} −ℓ)} + ∑_{ℓ=0}^{λ_{3}} p^{λ_{3} +i−2ℓ−1} (p + 1)

= \frac{λ_{3}p^{i−2} (p + 1)^{2} + p^{λ_{3}} + p^{i} + p^{i−1}}{p^{λ_{3}} (p^{2} − 1)} (p^{2λ_{3} +2} − 1)

− \frac{p^{i−2} (p^{2λ_{3} +2} − (λ_{3} + 1)p^{2} + λ_{3})}{p^{λ_{3}} (p − 1)^{2}} + \frac{(p^{2λ_{3} +2} − 1)(p^{i−λ_{3} −1} − 1)}{(p − 1)^{2}} ,

which gives the above formula.

III. Finally, if λ_{2} < i ⩽ λ_{1}, then

|E_{p^{i}} (G_{λ})| = ∑_{ℓ=0}^{λ_{3}} ∑_{j=λ_{3} +1}^{λ_{2}} N_{i,j,ℓ} + ∑_{ℓ=0}^{λ_{3} −1} ∑_{j=ℓ+1}^{λ_{3}} N_{i,j,ℓ} + ∑_{ℓ=0}^{λ_{3}} N_{i,ℓ,ℓ}

= ∑_{ℓ=0}^{λ_{3}} ∑_{j=λ_{3} +1}^{λ_{2}} p^{λ_{2} +2λ_{3} −j−2ℓ} + ∑_{ℓ=0}^{λ_{3} −1} ∑_{j=ℓ+1}^{λ_{3}} p^{λ_{2} +λ_{3} −2ℓ−1} (p + 1) + ∑_{ℓ=0}^{λ_{3}} p^{λ_{2} +λ_{3} −2ℓ}

= \frac{(p^{2λ_{3} +2} − 1)(p^{λ_{2} −λ_{3}} − 1)}{(p^{2} − 1)(p − 1)} + \frac{p^{λ_{2} −λ_{3} +1}}{(p^{2} − 1)(p − 1)} λ_{3}p^{2λ_{3} +2} − (λ_{3} + 1)p^{2λ_{3}} + 1)

+ p^{λ_{2} −λ_{3}} \frac{p^{2λ_{3} +2} − 1}{p^{2} − 1} ,

leading to the given result. □

Note that Proposition 2.2 is valid also in the case λ_{3} = 0, when it reduces to Proposition 2.1.

Example 2.3. We have

|E_{p^{i}} (ℤ_{p^{4}} × ℤ_{p^{2}} × ℤ_{p^{2}} )| = \{ \begin{matrix}1, & i = 0, \\ 2p^{2} + 2p + 3, & i = 1, \\ 3p^{4} + 4p^{3} + 6p^{2} + 3p + 3, & i = 2, \\ 3p^{4} + 2p^{3} + 2p^{2} + p + 1, & i = 3 or i = 4. \end{matrix}

Corollary 2.2. The total number of subgroups of G_{λ} = ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} × ℤ_{p^{λ_{3}}} , where λ_{1} ⩾ λ_{2} ⩾ λ_{3} ⩾ 1, is

\frac{A}{(p^{2} − 1)^{2}(p − 1)} ,

where

A = (λ_{3} + 1)(λ_{1} − λ_{2} + 1)p^{λ_{2} +λ_{3} +5} + 2(λ_{3} + 1)p^{λ_{2} +λ_{3} +4}
− 2(λ_{3} + 1)(λ_{1} − λ_{2})p^{λ_{2} +λ_{3} +3} − 2(λ_{3} + 1)p^{λ_{2} +λ_{3} +2}
+ (λ_{3} + 1)(λ_{1} − λ_{2} − 1)p^{λ_{2} +λ_{3} +1} − (λ_{1} + λ_{2} − λ_{3} + 3)p^{2λ_{3} +4}
− 2p^{2λ_{3} +3} + (λ_{1} + λ_{2} − λ_{3} − 1)p^{2λ_{3} +2}
+ (λ_{1} + λ_{2} + λ_{3} + 5)p^{2} + 2p − (λ_{1} + λ_{2} + λ_{3} + 1).

Proof. The total number of subgroups of G_{λ} is

∑_{i=0}^{λ_{1}} |E_{p^{i}} (G_{λ})| = ∑_{0⩽i⩽λ_{3}} |E_{p^{i}} (G_{λ})| + ∑_{λ_{3} <i⩽λ_{2}} |E_{p^{i}} (G_{λ})| + ∑_{λ_{2} <i⩽λ_{1}} |E_{p^{i}} (G_{λ})|

and summing the quantities given in Proposition 2.2 we deduce the result. □

We remark that an equivalent formula to that given in Corollary 2.2 was obtained in [8, Cor. 2.2] by using different arguments. Corollary 2.2 is valid also in the case λ_{3} = 0, when it reduces to Corollary 2.1.

Example 2.4. The total number of subgroups of ℤ_{p^{4}} ×ℤ_{p^{2}} ×ℤ_{p^{2}} is 9p^{4} +8p^{3} + 12p^{2} + 7p + 9.

Next consider the number of subgroups of exponent p (that is, the number of elementary abelian subgroups) in G_{λ}, which equals the total number of nontrivial subgroups of (ℤ_{p})^{k}. Since (ℤ_{p})^{k} is a k–dimensional linear space over ℤ_{p}, the number in question is exactly the total number of nonzero subspaces, which is, as wellknown, ∑ _{i=1}^{k} \binom{k}{i}_{p} (Galois number). So we have the following result.

Proposition 2.3. Let G_{λ} = ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} × ⋯× ℤ_{p^{λ_{k}}} with λ_{1} ⩾ λ_{2} ⩾ ⋯⩾

λ_{k} ⩾ 1. Then |E_{p}(G_{λ})| = ∑ _{r=1}^{k} \binom{k}{r}_{p}.

We give a direct proof of this formula based on Theorem 2.1.

Proof. We use Theorem 2.1 in the case λ = (1, 1, …, 1, 0, …), where the number of 1’s is k, and μ = (1, 1, …, 1, 0, …), where the number of 1’s is r with 1 ⩽ r ⩽ k. Here λ′ = (k, 0, 0, …), μ′ = (r, 0, 0, …) and obtain that the number of subgroups of type μ is \binom{k}{r}_{p}, while the number of subgroups of exponent 1 is exactly

∑ _{r=1}^{k} \binom{k}{r}_{p}. □

Example 2.5. We have for any λ_{1} ⩾ λ_{2} ⩾ λ_{3} ⩾ λ_{4} ⩾ 1,

|E_{p}(ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} × ℤ_{p^{λ_{3}}} × ℤ_{p^{λ_{4}}} )| = p^{4} + 3p^{3} + 4p^{2} + 3p + 4.

Concerning the number of subgroups of exponent p^{2} in G_{λ}, we have the following formula.

Proposition 2.4. Let G_{λ} = ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} × ⋯× ℤ_{p^{λ_{k}}} with λ_{1} ⩾ λ_{2} ⩾ ⋯⩾

λ_{t} > λ_{t+1} = ⋯= λ_{k} = 1, where 0 ⩽ t ⩽ k is fixed (t = 0 if each λ_{j} is 1 and t = k if each λ_{j} is ⩾ 2). Then

|E_{p^{2}} (G_{λ})| = ∑_{\begin{matrix}1⩽r⩽t \\ 0⩽s⩽k−r \end{matrix}} p^{r(k−r−s)} \binom{k − r}{s} _{p}\binom{t}{r} _{p},

which is zero (empty sum) for t = 0.

Proof. Let t ⩾ 1. Use Theorem 2.1 for λ = (2, 2, …, 2, 1, 1, …, 1, 0, …), where the number of 2’s is t and the number of 1’s is k − t and

μ = (2, 2, …, 2, 1, 1, …, 1, 0, …),

where the number of 2’s is r and the number of 1’s is s with 1 ⩽ r ⩽ t, 0 ⩽ s ⩽ k−r. Now λ′ = (k, t, 0, …), μ′ = (r+s, r, 0, …) and obtain that the number of subgroups of type μ is p^{r(k−r−s)} \binom{k−r}{s}_{p}\binom{t}{r}_{p} and the number of subgroups of exponent p^{2} is deduced by summing over r and s. □

Example 2.6. (k = 4, t = 2) We have

|E_{p^{2}} (ℤ_{p^{4}} × ℤ_{p^{2}} × ℤ_{p} × ℤ_{p})| = p^{5} + 5p^{4} + 6p^{3} + 4p^{2} + 2p + 2.

In what follows, let G_{λ} = ℤ_{p^{λ_{1}}} × ℤ_{p^{λ_{2}}} × ⋯× ℤ_{p^{λ_{k}}} and G_{κ} = ℤ_{p^{κ_{1}}} × ℤ_{p^{κ_{2}}} × ⋯× ℤ_{p^{κ_{ℓ}}} be two finite abelian p-groups, where λ_{1} ⩾ λ_{2} ⩾ ⋯⩾ λ_{k} ⩾ 1 and κ_{1} ⩾ κ_{2} ⩾ ⋯⩾ κ_{ℓ} ⩾ 1. Assume that G_{λ} and G_{κ} have the same number of subgroups of exponent p^{i} for every i, i.e., |E_{p^{i}} (G_{λ})| = |E_{p^{i}} (G_{κ})| (i ⩾ 0). Then λ_{1} = κ_{1}. On the other hand, since the function f : ℕ^{∗} → ℕ^{∗}, f(k) = ∑ _{i=1}^{k} \binom{k}{i}_{p} is one–to–one, by Proposition 2.3 we infer that k = ℓ. Clearly, if k = 1 one obtains G_{λ} ≅ G_{κ}. The same thing can be also said for k = 2 and k = 3 by Propositions 2.1 and 2.2. Inspired by these remarks, we state and prove the following result.

Proposition 2.5. Two finite abelian p-groups G_{λ} and G_{κ} are isomorphic if and only if they have the same number of subgroups of exponent p^{i}, for every i ⩾ 0.

Proof. Let λ = (λ_{1}, λ_{2}, …, λ_{k}, 0, …) and κ = (κ_{1}, κ_{2}, …, κ_{ℓ}, 0, …), where λ_{k}, κ_{ℓ} > 0, be partitions such that for p-groups G_{λ} and G_{κ} one has |E_{p^{i}} (G_{λ})| = |E_{p^{i}} (G_{κ})|, for every i ⩾ 0. As noted before, λ_{1} = κ_{1} and k = ℓ hold true.

Let us define λ_{k+1} = κ_{k+1} = 0. We prove, by reverse induction on t ⩽ k + 1 that λ_{t} = κ_{t}. So, let us assume that λ_{i} = κ_{i} for all k + 1 ⩾ i ⩾ t + 1, and prove that λ_{t} = κ_{t}.

Suppose to the contrary that w.l.o.g. λ_{t} > κ_{t}. Since λ_{1} = κ_{1}, one has t > 1. Consider the partitions λ^{λ_{t}} and κ^{λ_{t}} defined with

λ^{λ_{t}} = (\underbrace{λ_{t}, …, λ_{t}}_{t}, γ_{t+1} , …, γ_{k}, 0, …),
κ^{λ_{t}} = (\underbrace{λ_{t}, …, λ_{t}}_{s}, κ_{s+1} , …, κ_{t}, γ_{t+1} , …, γ_{k}, 0, …),

where γ_{i} = λ_{i} = κ_{i}, for i ⩾ t + 1, and s = max{j : κ_{j} ⩾ λ_{t}}. Note that by our assumption s < t, and since κ_{1} = λ_{1} ⩾ λ_{t} also s ⩾ 1.

From the definition of numbers α_{ω} (μ; p), it is clear that for any three partitions μ, σ, τ, which satisfy μ ⪯ σ ⪯ τ, we have α_{σ}(μ; p) ⩽ α_{τ}(μ; p).

Using this remark, the fact that κ^{λ_{t}} ⪵ λ^{λ_{t}} and α_{λ^{λ_{t}}} (λ^{λ_{t}} ; p) = 1, one has

1 + |E_{p^{λ_{t}}} (G_{κ})| = 1 + |E_{p^{λ_{t}}} (G_{κ^{λ_{t}}} )| = 1 + ∑_{μ⪯κ^{λ_{t}} , μ_{1} =λ_{t}} α_{κ^{λ_{t}}} (μ; p)
⩽ α_{λ^{λ_{t}}} (λ^{λ_{t}} ; p) + ∑_{μ⪯κ^{λ_{t}} , μ_{1} =λ_{t}} α_{λ^{λ_{t}}} (μ; p) ⩽ ∑_{μ⪯λ^{λ_{t}} , μ_{1} =λ_{t}} α_{λ^{λ_{t}}} (μ; p)
= |E_{p^{λ_{t}}} (G_{λ^{λ_{t}}} )| = |E_{p^{λ_{t}}} (G_{λ})|,

a contradiction. □

Corollary 2.3. Two arbitrary finite abelian groups are isomorphic if and only if they have the same number of subgroups of any exponent.

Finally, we note that another interesting problem is to find the polynomial |E_{p^{i}} (G_{λ})| in the case k = 4.

3. Second approach

We need the following result giving the representation of subgroups of the group ℤ_{m} × ℤ_{n}. For every m, n ∈ ℕ^{∗} let

J_{m,n} := \{(a, b, c, d, ℓ) ∈ (ℕ^{∗})^{5} : a | m, b | a, c | n, d | c, \frac{a}{b} = \frac{c}{d},

ℓ ⩽ \frac{a}{b}, gcd( ℓ, \frac{a}{b}) = 1\}.

Note that here gcd(b, d) · lcm(a, c) = ad and gcd(b, d) | lcm(a, c).

For (a, b, c, d, ℓ) ∈ J_{m,n} define

K_{a,b,c,d,ℓ} := \{( i\frac{m}{a}, iℓ\frac{n}{c} + j \frac{n}{d}) : 0 ⩽ i ⩽ a − 1, 0 ⩽ j ⩽ d − 1\}.

Theorem 3.1. [14, Th. 3.1] Let m, n ∈ ℕ^{∗}.

i) The map (a, b, c, d, ℓ) ↦ K_{a,b,c,d,ℓ} is a bijection between the set J_{m,n} and the set of subgroups of (ℤ_{m} × ℤ_{n}, +).

ii) The invariant factor decomposition of the subgroup K_{a,b,c,d,ℓ} is

K_{a,b,c,d,ℓ} ≃ ℤ_{gcd (b,d)} × ℤ_{lcm (a,c)} .

iii) The order of the subgroup K_{a,b,c,d,ℓ} is ad and its exponent is lcm(a, c).

Let s_{E}(m, n) stand for the number of subgroups of exponent E of the group ℤ_{m} × ℤ_{n}.

Proposition 3.1. For every m, n ∈ ℕ^{∗}, E | lcm(m, n) we have

(3.1)  s_{E}(m, n) = ∑_{\begin{matrix}i|m,j|n \\ lcm (i,j)=E \end{matrix}} gcd(i, j)

= \frac{1}{E} ∑_{\begin{matrix}i|m,j|n \\ lcm (i,j)=E \end{matrix}} ij. (3.2)

Proof. According to Theorem 3.1,

s_{E}(m, n) = ∑_{\begin{matrix}a|m \\ b|a \end{matrix}} ∑_{\begin{matrix}c|n \\ d|c \end{matrix}} ∑_{\begin{matrix}a/b=c/d=e \\ lcm (a,c)=E \end{matrix}} ϕ(e),

where ϕ is Euler’s totient function. This can be written (with m = ax, a = by, n = cz, c = dt) as

s_{E}(m, n) = ∑_{\begin{matrix}bxe=m \\ dze=n \\ e lcm (b,d)=E \end{matrix}} ϕ(e) = ∑_{\begin{matrix}ix=m \\ jz=n \end{matrix}} ∑_{\begin{matrix}be=i \\ de=j \\ e lcm (b,d)=E \end{matrix}} ϕ(e)

= ∑_{\begin{matrix}i|m \\ j|n \\ lcm (i,j)=E \end{matrix}} ∑_{e|gcd (i,j)} ϕ(e) = ∑_{\begin{matrix}i|m \\ j|n \\ lcm (i,j)=E \end{matrix}} gcd(i, j),

which is (3.1). Formula (3.2) is its immediate consequence. □

Remark 3.1. Proposition 2.1 is a direct consequence of the above result. The total number s(m, n) of subgroups of the group ℤ_{m} × ℤ_{n} is (see [6, Th. 3], [14, Th. 4.1])

(3.3)  s(m, n) = ∑_{i|m,j|n} gcd(i, j)

and (3.1) shows the distribution of the number of subgroups according to their exponents. Formula (2.2) can be obtained also by using (3.3).

Example 3.1. The total number of subgroups of ℤ_{12}×ℤ_{18} is s(12, 18) = 80 and we have s_{1}(12, 18) = 1, s_{2}(12, 18) = 4, s_{3}(12, 18) = 5, s_{4}(12, 18) = 3, s_{6}(12, 18) = 20, s_{9}(12, 18) = 4, s_{12}(12, 18) = 15, s_{18}(12, 18) = 16, s_{36}(12, 18) = 12.

Corollary 3.1 (m = n). For every n ∈ ℕ^{∗} and E | n,
(3.4)  s_{E}(n, n) = ∑_{\begin{matrix}i|E,j|E \\ gcd (E/i,E/j)=1 \end{matrix}} gcd(i, j),

which equals the number of cyclic subgroups of the group ℤ_{E} × ℤ_{E}.

The fact that s_{E}(n, n) equals the number of cyclic subgroups of the group ℤ_{E} × ℤ_{E}, but without deriving formula (3.4) is [6, Th. 8], proved by using different arguments.

Proof. In the case m = n, for every E | n we have by (3.1),

s_{E}(n, n) = ∑_{\begin{matrix}i|n,j|n \\ lcm (i,j)=E \end{matrix}} gcd(i, j) = ∑_{\begin{matrix}ia=E,jb=E \\ lcm (E/a,E/b)=E \end{matrix}} gcd(i, j) = ∑_{\begin{matrix}ia=E,jb=E \\ gcd (a,b)=1 \end{matrix}} gcd(i, j),

giving (3.4), which equals the number of cyclic subgroups of the group ℤ_{E} × ℤ_{E} by [6, eq. (16)]. □

Proposition 3.2. For every m, n ∈ ℕ^{∗} the sum of exponents of the subgroups of ℤ_{m} × ℤ_{n} is σ(m)σ(n), where σ(k) = ∑ _{d|k} d.

Proof. By (3.2) the sum of exponents of the subgroups of ℤ_{m} × ℤ_{n} is

∑_{E|lcm (m,n)} Es_{E}(m, n) = ∑_{E|lcm (m,n)} E · \frac{1}{E} ∑_{\begin{matrix}i|m,j|n \\ lcm (i,j)=E \end{matrix}} ij = ∑_{\begin{matrix}i|m,j|n \\ lcm (i,j)|lcm (m,n) \end{matrix}} ij

= ∑_{i|m} i ∑_{j|n} j = σ(m)σ(n).  □

By Proposition 3.2, the arithmetic mean of exponents of the subgroups of ℤ_{m} × ℤ_{n} is σ(m)σ(n)/s(m, n), where s(m, n) is given by (3.3). Now consider the case m = n. Let A(n) stand for the arithmetic mean of exponents of the subgroups of ℤ_{n} × ℤ_{n}. We have

(3.5)  A(n) = \frac{σ(n)^{2}}{s(n)} , where s(n) := s(n, n).

Recall that a function f : ℕ^{∗} → ℂ is said to be multiplicative if f(nn′) = f(n)f(n′) whenever gcd(n, n′) = 1. It is well known that the sum-of-divisors function σ is multiplicative. The function s(n) = ∑ _{i,j|n} gcd(i, j) is also multiplicative, as shown by the following direct proof: Let gcd(n, n′) = 1. Then

s(nn′) = ∑_{i,j|nn′} gcd(i, j) = ∑_{\begin{matrix}a,b|n \\ a′,b′|n′ \end{matrix}} gcd(aa′, bb′)

= ∑_{a,b|n} gcd(a, b) ∑_{a′,b′|n′} gcd(a′, b′) = s(n)s(n′).

We conclude that the function A given by (3.5) is multiplicative and and for every prime power p^{ν},

A(p^{ν}) = \frac{(p^{ν+1} − 1)^{2}}{p^{ν+2} + p^{ν+1} − (2ν + 3)p + 2ν + 1} ,

cf. Corollary 2.1.

Since the exponent of every subgroup of ℤ_{n} × ℤ_{n} is a divisor of n, whence ⩽ n, we deduce that A(n) ⩽ n (n ∈ ℕ^{∗}). For the function n ↦ f(n) := A(n)/n ∈ (0, 1] the series taken over the primes

∑_{p} \frac{1 − f(p)}{p} = ∑_{p} \frac{p − 1}{p^{2}(p + 3)}

is convergent, and it follows from a theorem of H. Delange (see, e.g., [9]) that the function f has a nonzero mean value M given by

M := lim_{x→∞} \frac{1}{x} ∑_{n⩽x} f(n) = ∏_{p} ( 1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{f(p^{ν})}{p^{ν}}

= ∏_{p} ( 1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{(p^{ν+1} − 1)^{2}}{p^{2ν}(p^{ν+2} + p^{ν+1} − (2ν + 3)p + 2ν + 1)} .

the products being over the primes.

We prove the following more exact result.

Proposition 3.3. We have

(3.6)  ∑_{n⩽x} A(n) = \frac{M}{2} x^{2} + O(x log^{3} x).

Proof. Let f(n) = ∑ _{d|n} g(d) (n ∈ ℕ^{∗}), that is g = μ ∗ f in terms of the Dirichlet convolution, where μ is the Möbius function. Here g(p^{ν}) = f(p^{ν})−f(p^{ν−1} ) for every prime power p^{ν} (ν ∈ ℕ^{∗}). Note that

A(p^{k}) = \frac{(∑ _{i=0}^{k} p^{i})^{2}}{∑ _{i=0}^{k} (2i + 1)p^{k−i}} (k ⩾ 0),

so if we denote S = ∑ _{i=0}^{ν−1} p^{i} and T = ∑ _{i=0}^{ν−1} (2i + 1)p^{ν−1−i} , we have

g(p^{ν}) = \frac{(Sp + 1)^{2}}{p^{ν}(T p + 2ν + 1)} − \frac{S^{2}}{p^{ν−1} T} = \frac{2ST p + T − pS^{2}(2ν + 1)}{p^{ν}T (T p + 2ν + 1)} .

Since S ⩽ T and T ⩽ (2ν − 1)S, we have
|g(p^{ν})| < \frac{max\{2ST p + T − pS^{2}(2ν − 1), pS^{2}(2ν + 1) − 2ST p\}}{p^{ν}T (T p + 2ν + 1)}

⩽ \frac{1}{p^{ν}} max\{ \frac{ST p + T}{T (T p + 1)}, \frac{pS^{2}(2ν − 1)}{T^{2}p} \}

⩽ \frac{1}{p^{ν}} max\{1, 2ν − 1\} = \frac{2ν − 1}{p^{ν}} ,

valid for every prime power p^{ν} (ν ∈ ℕ^{∗}). Hence, |g(n)| ⩽ τ(n^{2})/n for every n ⩾ 1, where τ(k) stands for the number of positive divisors of k.

We deduce that

∑_{n⩽x} f(n) = ∑_{de⩽x} g(d) = ∑_{d⩽x} g(d) ∑_{e⩽x/d} 1 = ∑_{d⩽x} g(d)(x/d + O(1))

= x ∑_{d=1}^{∞} \frac{g(d)}{d} + O( x ∑_{d>x} \frac{|g(d)|}{d}) + O( ∑_{d⩽x} |g(d)|)

= Mx + O( x ∑_{d>x} \frac{τ(d^{2})}{d^{2}}) + O( ∑_{d⩽x} \frac{τ(d^{2})}{d}) ,

where in the main term the coefficient of x is M by Euler’s product formula. It is known that ∑ _{n⩽x} τ(n^{2}) = cx log^{2} x + O(x logx) with a certain constant c and partial summation shows that ∑ _{n>x} τ(n^{2})/n^{2} = O((log^{2} x)/x), ∑ _{n⩽x} τ(n^{2})/n = O(log^{3} x). Therefore,

(3.7)  ∑_{n⩽x} f(n) = Mx + O(log^{3} x).

Now (3.6) follows from (3.7) by partial summation. □

Formula (3.6) and its proof are similar to those of [13, Th. 3.1.3].

Acknowledgements. The authors thank the referee for very careful reading of the manuscript, many useful comments and for the proof of Proposition 2.5.

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Faculty of Mathematics (Received 02 07 2015) “Al. I. Cuza” University (Revised 14 03 2016) Iaşi Romania

tarnauc@uaic.ro

Department of Mathematics University of Pécs Pécs Hungary

ltoth@gamma.ttk.pte.hu