A question on the sum of element orders of a finite group - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T12:15:02Z http://mathoverflow.net/feeds/question/74425 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74425/a-question-on-the-sum-of-element-orders-of-a-finite-group A question on the sum of element orders of a finite group Marius Tarnauceanu 2011-09-03T08:27:11Z 2011-09-05T09:19:40Z <p>Let <em>G</em> be a nontrivial finite group. Is it true that the sum of the orders of all elements of <em>G</em> is not divisible by the order of <em>G</em>? </p> http://mathoverflow.net/questions/74425/a-question-on-the-sum-of-element-orders-of-a-finite-group/74426#74426 Answer by Giorgio Mossa for A question on the sum of element orders of a finite group Giorgio Mossa 2011-09-03T08:43:43Z 2011-09-03T20:02:12Z <p>Edit: I misunderstood the question, I'll try to fix here. I don't have the complete answer but I'll try to give a partial answer: let $G$ be a group of order $|G|$ and for each $d \mid |G|$ let $n_d$ indicate the number of elements of order $d$ in $G$; then if $|G|$ is even $|G| \nmid \sum_{d \mid |G|}n_d d$. Indeed we have that if $d$ is a odd divisor of $|G|$ (not equal to $1$) either $n_d=0$ or exists a odd prime numeber $p$ such that $p-1 \mid n_d$ and so $n_d$ is even, on the other hand if $d$ is even clearly $n_d d$ is also even and so $\sum_{1 \ne d \mid |G|} n_d d$ must be even. Thus $\sum_{d \mid |G|}n_d d$ is odd and so $|G| \nmid \sum_{d \mid |G|} n_d d$, because by hypothesis $|G|$ is even.</p> http://mathoverflow.net/questions/74425/a-question-on-the-sum-of-element-orders-of-a-finite-group/74441#74441 Answer by Denis Chaperon de Lauzières for A question on the sum of element orders of a finite group Denis Chaperon de Lauzières 2011-09-03T16:15:57Z 2011-09-03T16:15:57Z <p>It is false in general, for instance there's a group of order $3\cdot 5\cdot 7=105$ with sum of orders equal to $1785=3\cdot 5\cdot 7\cdot 17$. (In Magma, it is the first of the two groups of order 105 in the "small groups" database).</p> <p>However it is true for all groups of even order, because the sum of orders of elements is always odd (this is shown by partitioning $G$ according to the equivalence relation $x\sim y$ if $x$ and $y$ generate the same cyclic subgroup, and using the fact that, for a positive integer $n\geq 1$, $n\varphi(n)$ is odd only if $n=1$.)</p>