Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 Jan 1 '12 at 1:09
@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 Jan 1 '12 at 9:22
up vote 6 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 $p$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.