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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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3  
Let $P$ act by conjugation on $E,$ and consider the fixed point set under this action. –  Geoff Robinson Jan 1 '12 at 0:32
    
Actually, it is also the case that $E \cap C$ has order coprime to $p,$ which requires a little more group theory. –  Geoff Robinson Jan 1 '12 at 0:51
    
@Geoff: The fixed point set is $E\cap C$. But the penny has not dropped yet. The only statement about fixed points I know is the Cauchy-Frobenius Lemma, which does not seem to be relevant here. –  Stefan Walter Jan 1 '12 at 1:09
4  
@Stefan: If $P$ is a $p$-group acting on a set $X$, and $X_0$ is the set of all points that are fixed by every element of $P$, then $|X_0|\equiv |X|\pmod{p}$. To see this, decompose $X$ into orbits; by the Orbit-Stabilizer Theorem, each nontrivial orbit has order divisible by $p$, so modulo $p$ the only contribution to $|X|$ are the 1-element orbits, i.e., $X_0$. Now use Geoff Robinson's hint to get the congruence. –  Arturo Magidin Jan 1 '12 at 1:45
    
@Arturo: Ah, yes. Thank you both. –  Stefan Walter Jan 1 '12 at 9:22
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1 Answer

up vote 4 down vote accepted

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 $p$.

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