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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1$\begingroup$ 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. $\endgroup$– user9072Commented Sep 3, 2011 at 12:35
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2 Answers
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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 $n=1$.)
answered Sep 3, 2011 at 16:15
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2$\begingroup$ Magma finds other counter-examples of order $357$ and $1785$, and these are the only three of order at most $2000$. $\endgroup$ Commented Sep 3, 2011 at 17:44
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$\begingroup$ 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? $\endgroup$ Commented Sep 5, 2011 at 9:07
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7$\begingroup$ @Marius, it's actually a complete answer - to the question you asked. $\endgroup$ Commented Sep 5, 2011 at 9:48
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$\begingroup$ 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). $\endgroup$ Commented Sep 6, 2011 at 5:07
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$\begingroup$ @Marius, then maybe you should accept the answer here and ask another question (including a link back to this question, of course). $\endgroup$ Commented Sep 6, 2011 at 13:38
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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.
answered Sep 3, 2011 at 8:43
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3$\begingroup$ 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 $\endgroup$ Commented Sep 3, 2011 at 8:48
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3$\begingroup$ Could Gerhard have meant the class equation? $\endgroup$– KConradCommented Sep 3, 2011 at 12:25
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$\begingroup$ Igor, I assume Gerhard means the class number mention here en.wikipedia.org/wiki/Conjugacy_class not the one there en.wikipedia.org/wiki/Class_number_formula $\endgroup$– user9072Commented Sep 3, 2011 at 12:29
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$\begingroup$ 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 $\endgroup$ Commented Sep 3, 2011 at 15:03