most recent 30 from http://mathoverflow.net 2013-05-22T13:44:35Z http://mathoverflow.net/feeds/question/82547 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82547/a-question-on-the-product-of-element-orders-of-a-finite-group Marius Tarnauceanu 2011-12-03T10:29:57Z 2011-12-20T06:44:34Z

<p>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 $G$? </p>

http://mathoverflow.net/questions/82547/a-question-on-the-product-of-element-orders-of-a-finite-group/82554#82554 Gjergji Zaimi 2011-12-03T12:30:47Z 2011-12-20T06:44:34Z

<p>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 <a href="http://www.pitt.edu/~gmc/ch1/node7.html" rel="nofollow">theorem of Frobenius</a> 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 <a href="http://www.jstor.org/stable/2695368" rel="nofollow">problem 10775</a> on the American Math Monthly. For your question we just need $f(x)=-\log x$.</p>