Characterization of P-groups By Sum of Element Orders

TL;DR

This paper characterizes certain finite p-groups using the sum of element orders function, providing new insights into their structure based on this numerical invariant.

Contribution

It introduces a novel characterization of finite p-groups through the sum of element orders, linking group structure to this specific numerical property.

Findings

Identifies conditions under which p-groups are characterized by their sum of element orders

Provides classifications of p-groups based on the value of ψ(G)

Establishes relationships between group order and sum of element orders

Abstract

Let $G$ be a finite group. Then we denote $\psi(G) = \sum_{x\in G}o(x)$ where $o(x)$ is the order of the element $x$ in $G$. In this paper we characterize some finite $p$-groups ($p$ a prime) by $\psi$ and their orders.

Equations

$$\psi (P)$$

$$\psi (P)=\sum _{x\in X}\psi (xN)=\psi (N)+|N|\sum _{1\ne x\in X}o(x)$$

$$\psi (P)$$

$$1<{\Omega }_1(P)<{\Omega }_2(P)<...<{\Omega }_m(P)=P.$$

$$\psi (P)=\psi ({\Omega }_{m-t}(P))+\sum _{i=1}^t|{\Omega }_{m-i}(P)|p^{m-i+1}(\frac{|{\Omega }_{m-i+1}(P)|}{|{\Omega }_{m-i}(P)|}-1)$$

$$\psi (Q)=\psi ({\Omega }_{m-t}(Q))+\sum _{i=1}^t|{\Omega }_{m-i}(Q)|p^{m-i+1}(\frac{|{\Omega }_{m-i+1}(Q)|}{|{\Omega }_{m-i}(Q)|}-1).$$

$$\sum _{i=1}^t|{\Omega }_{m-i}(P)|p^{m-i+1}(\frac{|{\Omega }_{m-i+1}(P)|}{|{\Omega }_{m-i}(P)|}-1)$$

$$=\sum _{i=1}^t|{\Omega }_{m-i}(Q)|p^{m-i+1}(\frac{|{\Omega }_{m-i+1}(P)|}{|{\Omega }_{m-i}(Q)|}-1),$$

$$\psi ({\Omega }_{m-t}(P))-\psi ({\Omega }_{m-t}(Q))$$

$$\psi (P)$$

$$\psi ({\Omega }_i(P))$$

$$\psi (P)=\psi (M)+|M|p^m(\frac{|P|}{|M|}-1)=\psi (N)+|N|p^m(\frac{|Q|}{|N|}-1)=\psi (Q).$$

$$h(x)=\{\begin{array}{ll}f(x)& \text{ x∈Ωm−1(P)},\\ g(x)& \text{otherwise}.\end{array}$$

Full text

1. Introduction and Main results

In what follows all groups are finite and $p$ is a prime.

Given a finite group $G$, let $\psi(H)=\sum_{x\in H}o(x)$ for $H\subseteq G$, where as usual, $o(x)$ is the order of the element $x$. In this note, we ask what information about some classes of $p-$groups $G$ can be recovered if we know both $\psi(G)$ and $|G|$. The starting point for the function $\psi$ is given by the paper [1] which investigates the maximum of $\psi$ among all groups of the same order. In [2] the authors determined the structure of the groups which have the minimum sum of the element orders on all groups of the same order.

Let $CP_{2}$ be the class of finite groups $G$ such that $o(xy)\leq max\{o(x),o(y)\}$ for all $x,y\in G$. We denote $\Omega_{i}(G)=<\{x\in G|\ x^{p^{i}}=1\}>$ for all $i\in\mathbb{N}$. Now we state the first main result as follows.

Theorem 1.1**.**

Suppose that $P$ and $Q$ are contained in $CP_{2}$ of the same order $p^{n}$. Then the following statements are equivalent:

1. (1)

$\psi(P)=\psi(Q)$. 2. (2)

$|\Omega_{i}(P)|=|\Omega_{i}(Q)|$* for all $i\in\mathbb{N}$.* 3. (3)

$\psi(\Omega_{i}(P))=\psi(\Omega_{i}(Q))$* for all $i\in\mathbb{N}$.*

Note that the class $CP_{2}$ of $p$-groups is more large than the class of abelian $p$-groups, regular $p-$groups ( See Theorem 3.14 of [6],II, page 47) and $p-$groups whose subgroup lattices are modular ( see Lemma 2.3.5 of [5]). Moreover by the main theorem in [7], we infer that powerful $p-$groups for $p$ odd also belong to $CP_{2}$.

The following is the second main result.

Theorem 1.2**.**

Let $P$ and $Q$ be two finite $p-$groups of the same order and $\Omega_{m-1}(P)\neq P$, where $exp(P)=p^{m}$. If $exp(P)>exp(Q)$, then $\psi(P)>\psi(Q)$.

In general, it is not true that if $P$ and $Q$ are $p$-groups of the same order such that $exp(P)>exp(Q)$ , then $\psi(P)>\psi(Q)$. For example consider $Q=(C_{4})^{4}$ and $P=D_{16}\times(C_{2})^{4}$. The authors would like to thank Prof. E. Khukhru for giving this example.

But if $exp(P)=exp(Q)$, then we have the following.

Theorem 1.3**.**

Let $P$ and $Q$ belong to $CP_{2}$ of the same order and the same exponent $p^{m}$. Also suppose that $|\Omega_{m-i}(P)|=|\Omega_{m-i}(Q)|$ for $i=1,2,..,t$. If $|\Omega_{m-t-1}(P)|<|\Omega_{m-t-1}(Q)|$, then $\psi(P)>\psi(Q)$.

As an application of Theorems 1.1 and 1.2 we have the following.

Theorem 1.4**.**

Let $P$ and $Q$ belong to $CP_{2}$ of the same order $p^{n}$. Then $\psi(P)=\psi(Q)$ if and only if there is a bijection $f:P\rightarrow Q$ such that $o(f(x))=o(x)$ for all $x\in P$.

2. Proof of the main results

Lemma 2.1**.**

Let $P$ be a finite $p-$group, $exp(P)=p^{m}$ and $M=\Omega_{m-1}(P)\neq P$. Then $\psi(P)=\psi(M)+|M|p^{m}(\frac{|P|}{|M|}-1)$.

Proof.

Suppose that $X$ is a left transversal to $M$ in $P$ containing identity element. For all $m\in M$ and $1\neq x\in X$, we have $o(xm)=p^{m}$. Therefore $\psi(xM)=|M|p^{n}$ for all $1\neq x\in X$. This completes the proof. ∎

Theorem 2.2**.**

Let $P$ and $Q$ be two finite $p-$groups of order $p^{n}$ and $exp(P)=p^{m}$. If $\Omega_{m-1}(P)\neq P$ and $exp(P)>exp(Q)$, then $\psi(P)>\psi(Q)$.

Proof.

Let $M=\Omega_{m-1}(P)$. Then

[TABLE]

Since $exp(Q)\leq p^{m-1}$, we have $\psi(Q)<p^{n}p^{m-1}<\psi(P)$. ∎

We observe that if a finite group $G$ belongs to $CP_{2}$, then for every $x,y\in G$ satisfying $o(x)\neq o(y)$ we have $o(xy)=max\{o(x),o(y)\}$.

We shall need the following theorem about the groups belonging to $CP_{2}$.

Theorem 2.3**.**

( See Theorem D in [4]) A finite group $G$ is contained in $CP_{2}$ if and only if one of the following statements holds:

1. (1)

$G$* is a *$p-$group and $\Omega_{n}(G)=\{x\in G\ |\ x^{p^{n}}=1\}$. 2. (2)

$G$* is a Frobenius group of order $p^{\alpha}q^{\beta}$, $p<q$, with kernel $F(G)$ of order $p^{\alpha}$ and cyclic complement.*

In the sequel assume that $P$ and $Q$ are $p$-groups belonging to $CP_{2}$.

Lemma 2.4**.**

If $|\Omega_{1}(P)|=p^{r}$, then $\psi(P)=1-p+p^{r+1}\psi(\frac{P}{\Omega_{1}(P)})$.

Proof.

Suppose that $\Omega_{1}(P)=N$. Then we have $<x>\cap N\neq 1$ for all $1\neq x\in P$, Since $<x^{\frac{o(x)}{p}}>$ is a subgroup of $<x>\cap N$ . Let $X$ be a left transversal to $N$ in $P$ such that $1\in X$. Suppose that $1\neq x\in X$. Then $o(x)\geq p^{2}$ since $N$ does not contain $x$. If $y\in N$, then by Theorem 2.3 part one $exp(N)=p$ and so we have $o(xy)=o(x)$. This implies that

[TABLE]

If $1\neq x\in X$, then $<x>\cap N\neq 1$ which follows that $o(x)=po(xN)$. Hence

[TABLE]

Since $|N|=p^{r}$, we have $\psi(N)=p^{r+1}-p+1$, which completes the proof. ∎

Lemma 2.5**.**

If $\psi(P)=\psi(Q)$, then $|\Omega_{1}(P)|=|\Omega_{1}(Q)|$.

Proof.

Suppose that $\Omega_{1}(P)=N$ and $\Omega_{1}(Q)=M$. If $|N|=p^{r}$ and $|M|=p^{t}$, then it follows from previous lemma that $p^{r+1}\psi(\frac{P}{N})=p^{t+1}\psi(\frac{Q}{M})$. If $r+1<t+1$, then $p^{r+1}\psi(\frac{P}{N})\equiv 0\ (\ mod\ p^{t+1})$. Since $\frac{P}{N}$ is a $p-$group, we have $\psi(\frac{P}{N})=1+kp$ and so $\psi(\frac{P}{N})=1+kp\equiv 1\equiv 0\ (\ mod\ p)$, a contradiction. Thus $r=t$. ∎

Lemma 2.6**.**

We have $\Omega_{i}(\frac{P}{\Omega_{1}(P)})=\frac{\Omega_{i+1}(P)}{\Omega_{1}(P)}$ for all $i\in\mathbb{N}$.

Proof.

Since $\frac{\Omega_{i+1}(P)}{\Omega_{1}(P)}\leq\Omega_{i}(\frac{P}{\Omega_{1}(P)})$, it is enough to show that $\Omega_{i}(\frac{P}{\Omega_{1}(P)})\leq\frac{\Omega_{i+1}(P)}{\Omega_{1}(P)}$. Suppose that $t\Omega_{1}(P)\in\Omega_{i}(\frac{P}{\Omega_{1}(P)})$. Then $t^{p^{i}}\in\Omega_{1}(P)$ and since $exp(\Omega_{1}(P))=p$, we have $t^{p^{i+1}}=1$. Therefore $t\in\Omega_{i+1}(P)$ and so $t\Omega_{1}(P)\in\frac{\Omega_{i+1}(P)}{\Omega_{1}(P)}$.

∎

Corollary 2.7**.**

$\frac{P}{\Omega_{1}(P)}$* belongs to $CP_{2}$.*

Proof.

It follows from previous lemma that $\Omega_{i}(\frac{P}{\Omega_{1}(P)})=\frac{\Omega_{i+1}(P)}{\Omega_{1}(P)}$, for all $i\in\mathbb{N}$. Since $P$ belongs to $CP_{2}$, we have $\Omega_{i+1}(P)=\{x\in G\ |\ x^{p^{i+1}}=1\}$ for all $i\in\mathbb{N}$. Since $x^{p^{i}}\in\Omega_{1}(P)$, we see $\frac{\Omega_{i+1}(P)}{\Omega_{1}(P)}=\{x\Omega_{1}(P)\in\frac{P}{\Omega_{1}(P)}\ |\ x^{p^{i}}\Omega_{1}(P)=\Omega_{1}(P)\}$ for all $i\in\mathbb{N}$ by Theorem 2.3 and so $\frac{P}{\Omega_{1}(P)}$ is contained in $CP_{2}$. ∎

Theorem 2.8**.**

Let $P$ and $Q$ have the same order $p^{n}$ and the same exponent $p^{m}$ and suppose that $|\Omega_{m-i}(P)|=|\Omega_{m-i}(Q)|$ for $i=0,1,2,..,t$. If $|\Omega_{m-t-1}(P)|<|\Omega_{m-t-1}(Q)|$, then $\psi(P)>\psi(Q)$.

Proof.

If $P\in CP_{2}$ and $exp(P)=p^{m}$, then for all $i<m$, $\Omega_{i}(P)\neq P$ and

[TABLE]

Note that for all $i\leq j$, we have $\Omega_{i}(\Omega_{j}(P))=\Omega_{i}(P)$. Using Lemma 2.1 we can get

[TABLE]

and

[TABLE]

Since

[TABLE]

[TABLE]

it is enough to prove that $\psi(\Omega_{m-t}(P))>\psi(\Omega_{m-t}(Q))$. Suppose that $|\Omega_{m-t-1}(Q)|=p^{a}|\Omega_{m-t-1}(P)|$, where $a\geq 1$. By Lemma 2.1, we have

[TABLE]

This completes the proof. ∎

Using Lemmas 2.4 and 2.5 we can propose another proof for Corollary 6 in [3].

Corollary 2.9**.**

Let $P$ and $Q$ be abelian $p-$groups of the same order. Then $\psi(P)=\psi(Q)$ if and only if $P\cong Q$.

Proof.

It is sufficient to show that if $\psi(P)=\psi(Q)$, then $P\cong Q$. We prove this by induction on $|P|$. Base step of induction is trivial. Let $|\Omega_{1}(P)|=p^{t}$ and $|\Omega_{1}(Q)|=p^{r}$. It follows from Lemma 2.4 that

$\psi(P)=1-p+p^{r+1}\psi(\frac{P}{\Omega_{1}(P)})$ and $\psi(Q)=1-p+p^{t+1}\psi(\frac{Q}{\Omega_{1}(Q)})$. We have $r=t$ by Lemma 2.5 . Therefore $\psi(\frac{P}{\Omega_{1}(P)})=\psi(\frac{Q}{\Omega_{1}(Q)})$. So we have $\frac{P}{\Omega_{1}(P)}\cong\frac{Q}{\Omega_{1}(Q)}$ by induction hypothesis which implies that $P\cong Q$.

∎

The above result is not true for regular $p-$groups or $p-$ groups of nilpotent class 2. For example there exists regular $3-$group $P$ such that $|P|=27$ and $exp(P)=3$, but $P$ is not abelian, so $\psi(P)=79=\psi((C_{3})^{3})$ but $P$ is not isomorphic to $(C_{3})^{3}$.

Now we are ready to prove Theorem 1.1.

Theorem 2.10**.**

Suppose that $P$ and $Q$ have the same order. Then the following statements are equivalent:

1. (1)

$\psi(P)=\psi(Q)$. 2. (2)

$|\Omega_{i}(P)|=|\Omega_{i}(Q)|$* for all $i\in\mathbb{N}$.* 3. (3)

$\psi(\Omega_{i}(P))=\psi(\Omega_{i}(Q))$* for all $i\in\mathbb{N}$.*

Proof.

$(1)\Rightarrow(2)$. We prove by induction on $|P|$. Suppose that $P$ and $Q$ are contained in $CP_{2}$ and $\psi(P)=\psi(Q)$. It follows from Lemma 2.4 that $\psi(P)=1-p+p^{r+1}\psi(\frac{P}{\Omega_{1}(P)})$ and $\psi(Q)=1-p+p^{t+1}\psi(\frac{Q}{\Omega_{1}(Q)})$ where $|\Omega_{1}(P)|=p^{r}$ and $|\Omega_{1}(Q)|=p^{t}$. Since $\psi(P)=\psi(Q)$, we obtain $\ r=t$ by Lemma 2.5 and so $\psi(\frac{P}{\Omega_{1}(P)})=\psi(\frac{Q}{\Omega_{1}(Q)})$. By corollary 2.7 we have $\frac{P}{\Omega_{1}(P)}$ and $\frac{Q}{\Omega_{1}(Q)}$ are in $CP_{2}$. Since $|\frac{P}{\Omega_{1}(P)}|=|\frac{Q}{\Omega_{1}(Q)}|$, the induction assumption yields that $|\Omega_{i}(\frac{P}{\Omega_{1}(P)})|=|\Omega_{i}(\frac{Q}{\Omega_{1}(Q)})|$ for all $i\in\mathbb{N}$. Therefore $|\Omega_{i}(P)|=|\Omega_{i}(Q)|$ by Lemmas 2.3 and 2.4.

$(2)\Rightarrow(1)$. Let $exp(P)=p^{m}$. By Theorem 2.2, we have $exp(Q)=p^{m}$. Since $P$ and $Q$ are contained in $CP_{2}$, we have $exp(\Omega_{j}(P))=exp(\Omega_{j}(Q))=p^{j}$ for all $j\in\mathbb{N}$. But

[TABLE]

where the second equality holds by the hypothesis (2).

$(2)\Rightarrow(3)$. Since $|\Omega_{i}(P)|=|\Omega_{i}(Q)|$ for all $i\in\mathbb{N}$, we have $exp(P)=exp(Q)$. Let $exp(P)=p^{m}$. Since $P$ and $Q$ are contained in $CP_{2}$, we have $exp(\Omega_{j}(P))=exp(\Omega_{j}(Q))=p^{j}$ for all $j\in\mathbb{N}$. So

[TABLE]

$(3)\Rightarrow(2)$. Since $\psi(\Omega_{i}(P))=\psi(\Omega_{i}(Q))$ for all $i\in\mathbb{N}$, we have $exp(P)=exp(Q)=p^{m}$. Let $M=\Omega_{m-1}(P)$ and $N=\Omega_{m-1}(Q)$. By Lemma 2.1 we have

[TABLE]

Since $\psi(M)=\psi(N)$, we obtain that $|N|=|M|$. By repeated use of this technique we shall reach the claimed. This completes the proof. ∎

Finally we prove the last main result.

Theorem 2.11**.**

Let $P$ and $Q$ have the same order $p^{n}$. Then $\psi(P)=\psi(Q)$ if and only if there is a bijection $f:P\rightarrow Q$ such that $o(f(x))=o(x)$ for all $x\in P$.

Proof.

It is clear that if there is a bijection $f:P\rightarrow Q$ such that $o(f(x))=o(x)$ for all $x\in P$, then $\psi(P)=\psi(Q)$. Conversely suppose that $\psi(P)=\psi(Q)$. We proceed by induction on $n$. Base step is trivial. By Theorem 2.2 we have $exp(P)=exp(Q)=p^{m}$. It follows from Theorem 2.10 that $\psi(\Omega_{m-1}(P))=\psi(\Omega_{m-1}(Q))$ and so by inductive hypothesis there is a bijection $f:\Omega_{m-1}(P)\rightarrow\Omega_{m-1}(Q)$ such that $o(f(x))=o(x)$ for all $x\in\Omega_{m-1}(P)$. Theorem 2.10 follows that $|\Omega_{m}(P)|-|\Omega_{m-1}(P)|=|\Omega_{m}(Q)|-|\Omega_{m-1}(Q)|$ and hence there is a bijection $g$ from $\Omega_{m}(P)-\Omega_{m-1}(P)$ to $\Omega_{m}(Q)-\Omega_{m-1}(Q)$. Define $h$ from $P$ to $Q$ by

[TABLE]

It is easily seen that $h$ is a bijection from $P$ to $Q$ such that $o(h(x))=o(x)$ for all $x\in P$, as wanted. ∎