# A question on the product of element orders of a finite group

Let $G$ be a finite group of order $n$ and $\psi(G)$ be the sum of element orders of $G$. Then $\psi(G)\leq\psi(C_n)$, where $C_n$ is the cyclic group of order $n$ (see "Sums of element orders in finite groups", Comm. Algebra 37 (2009), 2978-2980). Is it true a similar inequality for the product of element orders of $G$?

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Have you checked it for small examples? I personally have no intuition otherwise for why this should/should not be true... – Igor Rivin Dec 3 '11 at 10:40
Yes, it seems to be true. – Marius Tarnauceanu Dec 3 '11 at 11:09
The sum of elements theorem was proved in 1991 - earlier than the 2009 paper cited here. Citation: Schmidt, F., Stong, R., & Lindsey, J. H. (1991). 6636. American Mathematical Monthly, 970-972. – Benjamin Dickman Jun 12 '14 at 7:33

Denoting the order of $g$ by $o(g)$, you can show that for any decreasing function $f$ the following inequality holds $$\sum_{g\in G}f(o(g))\geq \sum_{g\in \mathbb Z/n\mathbb Z}f(o(g)).$$ This is because one can actually construct a bijection $\sigma:G\to\mathbb Z/n\mathbb Z$ which satisfies $$o(\sigma(g))\geq o(g)$$ for all $g\in G$. The main ingredient is a classical theorem of Frobenius saying that when $k$ divides the order of a group, the number of elements of order dividing $k$ is divisible by $k$, then proceed by induction. An application of this exact idea is for example problem 10775 on the American Math Monthly. For your question we just need $f(x)=-\log x$.
The proof seems to only want to define $S_d$ for $d \le n$ with $d$ a divisor of $n$. Am I missing something? – Aaron Meyerowitz Aug 9 '12 at 6:24
@Aaron: The problem is with the $S_d$ with $d < k$, $d$ not a divisor of $k$, but $\gcd(d,k) \neq 1$. Those $S_d$'s might have taken away too many elements of $G_k$, leaving less than $\phi(k)$ elements to build $S_k$. – Tom De Medts Aug 9 '12 at 13:28