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), 29782980). Is it true a similar inequality for the product of element orders of $G$?

1$\begingroup$ Have you checked it for small examples? I personally have no intuition otherwise for why this should/should not be true... $\endgroup$ – Igor Rivin Dec 3 '11 at 10:40

$\begingroup$ Yes, it seems to be true. $\endgroup$ – Marius Tarnauceanu Dec 3 '11 at 11:09

$\begingroup$ 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, 970972. $\endgroup$ – 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 $f(x)=\log x$.

$\begingroup$ It's an interesting idea. I will try to give a complete proof. Thanks! $\endgroup$ – Marius Tarnauceanu Dec 7 '11 at 8:03

5$\begingroup$ The proof given in that Monthly solution seems to me to be at best incomplete. I agree (in the notation of that problem) that the sets S_d for divisors d of k do not consume too much of the set G_k, but what about sets S_d where d < k but d is not a divisor. Perhaps they involve some elements of G_k. Does anyone see a way to repair this problem? $\endgroup$ – Marty Isaacs Jul 19 '12 at 19:52

$\begingroup$ The proof seems to only want to define $S_d$ for $d \le n$ with $d$ a divisor of $n$. Am I missing something? $\endgroup$ – Aaron Meyerowitz Aug 9 '12 at 6:24

$\begingroup$ @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 $S_k$. $\endgroup$ – Tom De Medts Aug 9 '12 at 13:28