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I would like to prove the following:


Let $G$ be a finite group, $p$ a prime number and $P$ a Sylow $p$-subgroup of $G$. Let $E$ be the set of all $p$-regular elements of $G$ (i.e. elements whose order is not a multiple of $p$). Let $C$ be the centralizer of $P$ in $G$.

Then $|E|\equiv|E\cap C|$ (mod $p$).


Unfortunately, I don't know where to start. I have absolutely no intuition on why this statement should be true.

I'd be glad for anything to start me off.

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    FYI: You can also prove that $p \nmid \lvert E \rvert$. Proof: By Schur-Zassenhaus theorem, there is a $p'$-subgroup $H$ of $C$ such that $C = P \rtimes H$. Now $H = E \cap C$ and we conclude $\lvert E \rvert \equiv \lvert E \cap C \rvert = \lvert H \rvert \not\equiv 0 \pmod{p}$.2017-05-11

1 Answers 1

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Consider the action of $P$ on $E$ defined by $g*x=gxg^{-1}$ for $g\in P$ and $x\in E$. This is well defined, since the property of being $p$-regular is clearly preserved by group isomorphisms. The set of fixed points under this action (i.e. the elements of $E$ left fixed by all elements of $P$) is $C\cap E$ by the very definition of the centralizer.

The proof is finished by the following

Lemma: Let $P$ be a $p$-group acting on a finite set $E$. Let $E^P$ be the set of fixed points under this action. Then $|E|\equiv|E^P|\text{ mod }p.$

Proof: We have to show that the order of $E\backslash E^P$ is a multiple of $p$. By definition, $E^P$ is the union of the one-point orbits of the action. So the order of $E\backslash E^P$ is the sum of the orbit orders $>1$. By the orbit-stabilizer-theorem, the order of an orbit divides the order of $P$. Since $P$ is a $p$-group, it follows that orbit orders $>1$ have to be multiples of $p$. So the order of $E\backslash E^P$ is the sum of multiples of $p$ and therefore itself a multiple of $p$.