Why $\Psi(H) \le \exp(G)$ for any subgroup $H$ of a finite $p$-group $G$ where $\Psi(H)=\frac{\Sigma_{h\in H}|h|}{|H|}$?

Question

I was reading this paper. I need help to understand a claim.

Let's take $G$ which is a finite $p$-group. For any subgroup $H\le G$ we define $\Psi(H)$ to be the average order of the elements of $H$ that is $$\Psi(H)=\frac{\Sigma_{h\in H}|h|}{|H|}$$

On page 4 of the paper, it is claimed that

$\Psi(H) \le \exp(G)$ for any subgroup $H$ of $G$.

Can anyone please help me to understand why the claim is true? And is this true for any finite group, not just finite $p$-group?

Thanks in advance.

Answer 1

Hmm, it seems to be true fairly trivially, so I hope I'm not misunderstanding. For a $p$-group the exponent of $G$ is just the highest order among elements in $G$ by definition. So $|H| \cdot \Psi(H) = \Sigma_H |h| \leq \Sigma_H \exp(G) = |H| \cdot \exp(G)$ and thus $|H| \cdot \Psi(H) \leq |H| \cdot \exp(G)$. Canceling gets what you want. Equality occurs in any elementary $p$-group and a non-trivial cyclic subgroup.

The argument seems to work for any finite group, since in that case $\exp(G)$ is at least as big as $|h|$ for any $h \in H$ (or any $g \in G$).