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Let G be a nontrivial finite group. Is it true that the sum of the orders of all elements of G is not divisible by the order of G?

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Since you are new to the site and in case you are not aware: you can edit the question itself (link below the question) to include such a clarification directly in the main text. – user9072 Sep 3 '11 at 12:35
up vote 20 down vote accepted

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).

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 $n=1$.)

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Magma finds other counter-examples of order $357$ and $1785$, and these are the only three of order at most $2000$. – Denis Chaperon de Lauzières Sep 3 '11 at 17:44
It's a good (partial) answer. Thanks. Can be classified/characterized the finite groups whose orders divide the sum of the orders of their elements? – Marius Tarnauceanu Sep 5 '11 at 9:07
@Marius, it's actually a complete answer - to the question you asked. – Gerry Myerson Sep 5 '11 at 9:48
You are right, it's a complete answer to the initial question. Now, I am interested to say something about the finite groups whose orders divide the sum of the orders of their elements (for example, if this class contains some abelian groups). – Marius Tarnauceanu Sep 6 '11 at 5:07
@Marius, then maybe you should accept the answer here and ask another question (including a link back to this question, of course). – Gerry Myerson Sep 6 '11 at 13:38

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.

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He probably means 1 + 2 + 2 + 2 + 3 + 3, which is not 0 mod 6. However the trivial group is an example where the sum is divisible by the order of the group. You might be able to use the class number formula to establish his claim for other groups. Gerhard "Ask Me About System Design" Paseman, 2011.09.03 – Gerhard Paseman Sep 3 '11 at 8:48
Class number formula?? – Igor Rivin Sep 3 '11 at 9:58
Could Gerhard have meant the class equation? – KConrad Sep 3 '11 at 12:25
Igor, I assume Gerhard means the class number mention here not the one there – user9072 Sep 3 '11 at 12:29
Until my memory improves (or my copy of Herstein's Topics in Algebra ever comes back to me), let's go with quid's assessment. (I thought there was a 1 in the formula.) Apologies for any confusion I have caused. Gerhard "Misses His Copy Of Herstein" Paseman, 2011.09.03 – Gerhard Paseman Sep 3 '11 at 15:03

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