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Let $G$ be a finite group such that has some elements of order $2p$, where $p$ is an odd prime. Let $H$ be the set of all elements of order $2p$ in $G$. We can show $G$ acts on $H$ by conjugation. So $H$ cut to partition by orbits of this action. Hence $H=\bigcup_{x\in H}x^G$, where these conjugacy classes are distinct. It follows that $|H|=\sum_{x\in H}|x^G|$. How we can conclude that $p\nmid(\sum_{x\in H}|x^G|)$?

Caleb Stanford
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user204248
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  • $p=1$ is a counterexample to this. – Matt Samuel Jan 02 '15 at 19:13
  • @MattSamuel maybe $p$ should be odd prime? – Jihad Jan 02 '15 at 19:15
  • @Jihad that would make more sense, but it wasn't specified. – Matt Samuel Jan 02 '15 at 19:16
  • yes $p$ is odd prime. – user204248 Jan 02 '15 at 19:17
  • There are other issues though. $|H|\neq \sum_{x\in H}{|x^G|}$ unless every element is the only element in its orbit. If we modify it so the equality does hold by summing over one element in each orbit, why not just just phrase the question as "prove that $p\nmid |H|$?" Or should it be $p\nmid \sum_{X\in H^G}{|X|^2}$, which is what the sum is really equal to? – Matt Samuel Jan 02 '15 at 19:23
  • This is very similar to a question (from user mita) I answered on Mathoverflow today. The answer to this question is almost exactly the same. The number of elements of order $2p$ in $G$ is coprime to $p$ if and only if the number of involutions in $C_{G}(P)$ is coprime to $p,$ where $P$ is a Sylow $p$-subgroup of $G$ ( at least when $p$ divides $|G|$ and is odd). – Geoff Robinson Jan 02 '15 at 19:25
  • And of course you have to sum the size of the distinct orbits. – Geoff Robinson Jan 02 '15 at 19:30
  • why you said that $|H|\neq\sum_{x\in H}|x^G|$ unless every element is the only element in its orbit? please explain more it. you want to say it is impossible that $H$ to be union of some orbits, which has more than one element? – user204248 Jan 02 '15 at 19:35
  • $\sum_{x\in H}{|x^G|}$ is a sum over all elements of $H$ of the size of the orbit the element is in. For example, if there were only one orbit and the orbit had size $n$, then you would be adding a string of $n$ $n$'s, obtaining $n^2$. $|H|$ is equal to the sum over all orbits of the size of the orbit, not the sum over all elements. – Matt Samuel Jan 02 '15 at 20:05
  • yes, I was written that conjugacy classes are distinct. another question if the conjugacy class sizes are same? – user204248 Jan 02 '15 at 20:12

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