# A recursive formula for the product of element orders of finite abelian groups

Subhrajyoti Saha

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Subhrajyoti Saha. A recursive formula for the product of element orders of finite abelian groups. Communications in Mathematics, 2023, Volume 32 (2024), Issue 1 (1), 10.46298/cm.10996. hal-03997516v2

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Submitted on 28 Mar 2023

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# A recursive formula for the product of element orders of finite abelian groups

Subhrajyoti Saha

**Abstract.** Let $G$ be a finite group and let $\psi(G)$ denote the sum of element orders of $G$; later this concept has been used to define $R(G)$ which is the product of the element orders of $G$. Motivated by the recursive formula for $\psi(G)$, we consider a finite abelian group $G$ and obtain a similar formula for $R(G)$.

## 1 Introduction

Let $G$ be a finite group. For any non-empty subset $S$ of $G$, let $\psi(S)$ denote the sum of element orders of $S$. This has been introduced in [2] and later in [4], the notion $R(G)$ was introduced which stands for the product of element orders of $G$. In the same paper, a formula for computing $R(G)$ when $G$ is a finite abelian group was obtained. In [3], [5], an explicit recursive formula for computing $\psi(G)$ were obtained in case $G$ is abelian. Motivated by these results, in this paper, we obtain a similar recursive formula for computing $R(G)$ when $G$ is a finite abelian group.

Throughout this paper, we let $\varphi(n)$ denote the Euler totient function of the positive integer $n$ and let $p$ denote a prime number. A cyclic group of order $n$ will be denoted by $C_n$ whereas $C_p^{(r)}$ will denote the elementary abelian $p$-group of rank $r$. We always assume $G$ to be finite. For a group $G$ and element $x \in G$, the notation $o(x)$ denotes the order of $x$. For any group $G$, we take

$$R(G) = \prod_{x \in G} o(x).$$

For a group $G$, the notation $\exp(G)$ denotes the exponent of $G$ which is the smallest positive integer $z$ such that $g^z = 1_G$ for all $g \in G$ where $1_G$ is the identity element of $G$; without any ambiguity we will denote this identity element as 1.

MSC 2020: 20D60, 20K01

Keywords: Finite group; Abelian group; Element order

Affiliation:

Monash University, Australia and University of Liverpool, UK

E-mail: subhrajyotimath@gmail.com
## 2 Explicit formulas for finite abelian groups

In this section, we obtain explicit recursive formula for $R(G)$ where $G$ is a finite abelian group. We present this in different cases starting from a finite cyclic group. This format is inspired by [3, Section 2]. We will then consider the direct product of a finite cyclic $p$-group and a (not necessarily abelian) $p$-group. Finally, we will consider a the most general case of finite abelian groups. The proofs of our results in this section are motivated by the methods used in [3]. We begin with the following important preliminary results.

**Theorem 2.1 ([4, Proposition 1.1]).** Let $G_1, G_2, \dots, G_k$ be finite groups having co-prime orders and $G \cong G_1 \times G_2 \times \dots \times G_k$. Then

$$R(G) = \prod_{i=1}^{k} R(G_i)^{n_i}, \quad \text{where} \quad n_i = \prod_{j=1, j \neq i}^{k} |G_j|, i = 1, 2, \dots, k.$$

**Lemma 2.2 ([3, Lemma 2.6]).** Let $H \cong C_{p^{r_1}} \times C_{p^{r_2}} \times \dots \times C_{p^{r_n}}$ where $1 \le r_1 \le r_j$ for all $j$ with $2 \le j \le n$. Then for any $i \in \{1, \dots, r_1\}$, there are $(p^i)^n - (p^{i-1})^n$ elements of $H$ of order $p^i$

The following lemma, motivated by [3, Lemma 1.1], follows easily from the fact that $C_n$ has exactly $\varphi(d)$ elements of order $d$ for each divisor $d$ of $n$.

**Lemma 2.3.** Let $n$ be any positive integer. Then

$$\psi(C_n) = \sum_{d|n} d\varphi(d) \quad \text{and} \quad R(C_n) = \prod_{d|n} d^{\varphi(d)}$$

### 2.1 Finite Cyclic Groups

Let $G$ be a cyclic group of order $n$. Then we know that $G \cong C_{m_1} \times \dots \times C_{m_k}$, where $m_1, \dots, m_k$ are co-prime to each other and $n = m_1 \dots m_k$.

**Lemma 2.4.** Let $G$ be a cyclic group of order $p^n$ where $p$ is a prime number and $n$ is a positive integer, then $R(G) = p^{\binom{p+np^{n+2}-(n+1)p^{n+1}}{p(p-1)}}$.

*Proof.* Using Lemma 2.3, we get $R(G) = \prod_{r=1}^{n} p^{\varphi(p^r)r} = p^{\sum_{r=1}^{n} \varphi(p^r)r}$. So we have $R(G) = p^z$ where $z = \sum_{r=1}^{n} r(p^r - p^{r-1}) = (1 - \frac{1}{p}) \sum_{r=1}^{n} rp^r$. Now the sum $\sum_{r=1}^{n} rp^r$ is an arithmetic-geometric series with common difference 1 and common ratio $p$. Thus

$$\sum_{r=1}^{n} rp^{r} = \frac{p + np^{n+2} - (n+1)p^{n+1}}{(p-1)^2}.$$

This shows that $z = \frac{p+np^{n+2}-(n+1)p^{n+1}}{p(p-1)}$. So we get $R(G) = p^{\binom{p+np^{n+2}-(n+1)p^{n+1}}{p(p-1)}}$. $\square$
**Lemma 2.5.** Let $G$ be a cyclic group of order $s = p_1^{r_1} \dots p_k^{r_k}$ where the $p_i$ are distinct primes with $r_i \ge 1$ for $i = 1, \dots, k$. Then

$$R(G) = \prod_{i=1}^{k} p_i \left( \frac{p_i+r_i p_i^{r_i+2} - (r_i+1) p_i^{r_i+1}}{p_i (p_i-1)} \right) n_i, \quad n_i = p_i^{-r_i} \prod_{j=1}^{k} p_j^{r_j}.$$

*Proof.* We know that $G \cong C_{p_1^{r_1}} \times \dots \times C_{p_k^{r_k}}$ and $p_i$ are all distinct primes. Hence we can apply Theorem 2.1. Thus by Lemma 2.4, we arrive at the required result. $\square$

## 2.2 Direct product of a cyclic p-Group and a p-Group

In this section, we obtain a recursive formula for $R(G)$ where $G$ is a direct product of a finite cyclic $p$-group and any $p$-group. We begin with the following lemma from [3, Lemma 2.3].

**Lemma 2.6.** If $H$ and $K$ are $p$-groups, then $o((x_1, x_2)) = \max\{o(x_1), o(x_2)\}$ for all $H$ and $K$.

We now prove the following using techniques motivated by [3].

**Proposition 2.7.** Let $G = C_{p^r} \times H$ where $r \ge 1$, and $H$ is a $p$-group with $\exp(H) \ge p^r$. Let $N_j$ be the number of elements in $H$ that have order $p^j$. Then

$$R(G) = \begin{cases} p^{p-1} R(H)^{p^r} \prod_{i=2}^{r} \left( \prod_{j=1}^{i-1} (p^{i-j})^{N_j} \right)^{p^i - p^{i-1}}, & \text{if } r > 1 \\ p^{p-1} R(H)^p, & \text{if } r = 1 \end{cases}$$

*Proof.* Note that $G$ is a finite group whose elements are of the form $(x, y)$ where $x \in C_{p^r}$ and $y \in H$. We now partition $G$ based on the order of the elements in the first component. In particular we have, $G = \cup_{k=0}^{r} F_k$ where $F_k = \{(x_1, x_2) \in G \mid o(x_1) = p^k\}$. Since the $F_i \cap F_j = \emptyset$ for $i \ne j$, we have $R(G) = \prod_{k=0}^{r} R(F_k)$. Now let $x_1 \in C_{p^r}$ with $o(x_1) = p^i$ for some $i$ with $0 \le i \le r$. For each such $x_1$, define $F_{i,x_1} = \{(x_1, x_2) \mid x_2 \in H\}$. Then we have $F_i = \cup_{x_1 \in C_{p^r}, o(x_1)=p^i} F_{i,x_1}$ and $R(F_i) = \prod_{x_1 \in C_{p^r}, o(x_1)=p^i} R(F_{i,x_1})$ for $i = 0, 1, \dots, r$. There are $(p^i - p^{i-1})$ elements of order $p^i$ in $C_{p^r}$, see Lemma 2.2. As a result,

$$R(F_i) = R(F_{i,x_1})^{(p^i - p^{i-1})}.$$

Taking $i = 0$, we have $F_0 = F_{0,1}$ and thus, $R(F_0) = R(H)$. For $i = 1$, each element $(x_1, x_2)$ in $F_{1,x_1}$ has order same as $o(x_2)$ except for $(x_1, 1)$ which has order $p$. Thus $R(F_{1,x_1}) = pR(H)$. So $R(F_1) = (pR(H))^{p-1}$.

If $r = 1$ then $R(G) = R(F_0)R(F_1) = p^{p-1}R(H)^p$.

Now consider $r > 1$ and let $i \in \{2, \dots, r\}$. For $(x_1, x_2) \in F_{i,x_1}$ with $o(x_2) = p^j$, we have
$o((x_1, x_2)) = p^i$ if $j < i$ and $o((x_1, x_2)) = p^j$ if $j \ge i$. If $r' = \exp(H)$ then

$$
\begin{align*}
R(F_{i,x_1}) &= \prod_{j=0}^{i-1} (p^i)^{N_j} \prod_{j=i}^{r'} (p^j)^{N_j} \\
&= p^i \prod_{j=1}^{i-1} (p^i)^{N_j} \prod_{j=i}^{r'} (p^j)^{N_j} \\
&= p^i \prod_{j=1}^{r'} (p^j)^{N_j} \prod_{j=1}^{i-1} (p^{i-j})^{N_j} \\
&= p^i R(H) \prod_{j=1}^{i-1} (p^{i-j})^{N_j}
\end{align*}
$$

Thus $\prod_{i=2}^{r} R(F_i) = R(H)^{(p^r-p)} \prod_{i=2}^{r} \left( \prod_{j=1}^{i-1} (p^{i-j})^{N_j} \right)^{p^i-p^{i-1}}$ and finally the result follows from a direct calculation $R(G) = R(F_0) R_l(F_1) \prod_{i=2}^{r} R_l(F_i)$. $\square$

## 2.3 Finite abelian groups

We can now state how to compute $R(G)$ for any finite abelian group $G$. In view of Theorem 2.1, the following is a direct application of Proposition 2.7 and Lemma 2.2.

**Theorem 2.8.** Let $G$ be a finite abelian group with $G \cong H_1 \times \cdots \times H_k$ where each $H_i$ is an abelian $p_i$-group and $p_i$ are distinct primes for $i = 1, \dots, k$. Then

$$R(G) = R(H_1) \dots R(H_k)$$

where $R(H_i)$ for $i = 1, \dots, k$ are computed as follows:

a) If $H_i \cong C_{p_i^n}$ then

$$R(H_i) = p_i^{\left(\frac{p_i+n p_i^{n+2}-(n+1) p_i^{n+1}}{p_i(p_i-1)}\right)}.$$

b) If $H_i \cong C_{p_i^{r_1}} \times C_{p_i^{r_2}} \times \cdots \times C_{p_i^{r_n}}$, where $1 \le r_1 \le r_2 \le \cdots \le r_n$, and $r_1 + \cdots + r_n = r$ then $R(H_i)$ can be determined recursively as follows

i) If $r_1 > 1$ then

$$R(H_i) = p_i^{p_i-1} R(C_{p_i^{r_2}} \times \cdots \times C_{p_i^{r_n}})^{p_1} \prod_{z=2}^{r_1} \left( \prod_{j=1}^{z-1} (p_i^{z-j})^{N_j} \right)^{p_i^z - p_i^{z-1}},$$

$$\text{where } N_j = \left( (p_i^j)^{n-1} - (p_i^{j-1})^{n-1} \right)$$

ii) If $r_1 = 1$ then

$$R(H_i) = p_i^{p_i-1} R(C_{p_i^{r_2}} \times \cdots \times C_{p_i^{r_n}})^{p_i}.$$
## 3 Some Examples

In this final section we compute some examples using Theorem 2.8.

**Example 3.1.** We compute $R(C_p^{(r)} \times C_{p^n})$ where $n$ and $r$ are positive integers. By part a) of Theorem 2.8 we have

$$R(C_{p^n}) = p^{\left(\frac{p+np^{n+2}-(n+1)p^{n+1}}{p(p-1)}\right)}.$$

Then by part b) of Theorem 2.8 we have

$$R(C_p \times C_{p^n}) = p^{p-1} p^{\left(\frac{p+np^{n+2}-(n+1)p^{n+1}}{p(p-1)}\right)} = p^{\left(\frac{(p-1)^2+(p+np^{n+2}-(n+1)p^{n+1})}{(p-1)}\right)}.$$

Similarly

$$R(C_p \times C_p \times C_{p^n}) = p^{\left(\frac{(p-1)^2(1+p)+p(p+np^{n+2}-(n+1)p^{n+1})}{(p-1)}\right)}.$$

Thus inductively it is easy to show that

$$R(C_p^{(r)} \times C_{p^n}) = p^{\left(\frac{(p-1)^2(1+p+\dots+p^{r-1})+p^{r-1}(p+np^{n+2}-(n+1)p^{n+1})}{(p-1)}\right)}.$$

Thus we have

$$R(C_p^{(r)} \times C_{p^n}) = p^{\left(\frac{(p-1)(p^{r-1})+p^{r-1}(p+np^{n+2}-(n+1)p^{n+1})}{(p-1)}\right)}.$$

The next example is an application of Theorem 2.8 in the case where $r_1 > 1$.

**Example 3.2.** In this example we compute $R(C_{p^2} \times C_{p^n})$ where $n$ is a positive integers. By part a) of Theorem 2.8 we have

$$R(C_{p^n}) = p^{\left(\frac{p+np^{n+2}-(n+1)p^{n+1}}{p(p-1)}\right)}.$$

Then by part b) of Theorem 2.8 we have

$$R(C_{p^2} \times C_{p^n}) = p^{p-1} p^{\left(\frac{p+np^{n+2}-(n+1)p^{n+1}}{p(p-1)}\right)} p^{p(p-1)(p^{n-1}-1)}.$$

This gives

$$R(C_{p^2} \times C_{p^n}) = p^{\left(\frac{(p-1)^2(p^{n+1}-p)+p(p+np^{n+2}-(n+1)p^{n+1})}{p-1}\right)}.$$

In the following example we compute $R(C_p^{(r)} \times C_{p^2} \times C_{p^n})$ where $n$ and $r$ are positive integers with $n \ge 2$.

**Example 3.3.** In example 3.2 we have computed $R(C_{p^2} \times C_{p^n})$. Then by part b) of Theorem 2.8 we have

$$R(C_p \times C_{p^2} \times C_{p^n}) = p^{p-1} p^{\left(\frac{(p-1)^2(p^{n+1}-p)+p(p+np^{n+2}-(n+1)p^{n+1})}{p-1}\right)}.$$
This gives

$$R(C_p \times C_{p^2} \times C_{p^n}) = p^{\left( \frac{(p-1)^2(1+p-p^2+p^{n+1})+p^2(p+np^{n+2}-(n+1)p^{n+1})}{p-1} \right)}$$

Then inductively one can show that

$$R(C_p^{(r)} \times C_{p^2} \times C_{p^n}) = p^{\left( \frac{(p-1)^2(1+p+...+p^r-p^{r+1}+p^{n+r})+p^{r+1}(p+np^{n+2}-(n+1)p^{n+1})}{p-1} \right)}$$

where $n$ and $r$ are integers with $n \ge 2$.

Note that formulas obtained in Examples 3.1, 3.2 and 3.3 are straightforward to compute even when $n$ and $r$ are large. For any finite abelian $p$-group, an explicit formula for computing $R(G)$ was obtained in [4, Theorem 1.1]. By expanding and simplifying the formula obtained in [4, Theorem 1.1] one can compare the computations for $R(C_p^{(r)} \times C_{p^2} \times C_{p^n})$; the recursive formula used in Example 3.3 provides a more efficient method for obtaining $R(C_p^{(r)} \times C_{p^2} \times C_{p^n})$.

## References

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[2] Amiri H., Jafarian Amiri S. M. and Isaacs I. M.: Sums of element orders in finite groups. *Communications in Algebra* 37 (9) (2009) 2978–2980.

[3] Chew C. Y., Chin A. Y. M. and Lim C. S.: A recursive formula for the sum of element orders of finite abelian groups. *Results in Mathematics* 72 (4) (2017) 1897–1905.

[4] Tarnauceanu M.: A note on the product of element orders of finite abelian groups. *Bulletin of the Malaysian Mathematical Sciences Society. Second Series* 36 (4) (2013) 1123–1126.

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Received: April 12, 2021

Accepted for publication: June 25, 2022

Communicated by: Pasha Zusmanovich