I try to solve the following problem:
Fix a prime $p$, and suppose that a subgroup $H \subset G$ (of a
finite group $G$) has the property that $C_G(x) \subset H$ for every
element $x \in H$ having order $p$ ($C_G(x)$ is the centralizer
of $x$ in $G$). Show that $p$ cannot divide both $|H|$ and $|G:H|$.
My attempt is the following.
Assume $p \mid |H|$.
Then, by Cauchy's theorem, there exists an element $x \in H$ of order $p$.
Let $O$ be the conjugacy class in $G$ containing $x$.
Then $|O| = |G:C_G(x)|$.
By assumption, $C_G(x) \subset H$, hence $|G:C_G(x)| =|G:H||H:C_G(x)|$.
Assume moreover that $p \mid |G:H|$.
Then we have that $p |O|$….
I wanted to deduce a contradiction.
How to solve the problem?
Best Answer
Here is a somewhat different solution (making use of actions of groups) than the one you are referring to of Arturo Magidin. Any reference below is to the book of M. Isaacs, Finite Group Theory.
Assume the prime $p$ divides $|H|$. Then a $P \in Syl_p(H)$ exists, with $P$ non-trivial. We will show that in fact $P \in Syl_p(G)$, implying $p \nmid |G:H|$.
Put the set $\Omega=\{\text {elements of order $p$ in $P$} \}$. By Problem $(1A.8)(b)$ $|\Omega| \equiv -1$ mod $p$. Let $N_G(P)$ act by conjugation on $\Omega$ and let $x_1, \ldots, x_k$ be the representatives of the different orbits under this action. Then $$|\Omega|=\sum_{i=1}^k|N_G(P):C_{N_G(P)}(x_i)|$$ Since for each $i$, $x_i \in H$, and using the premiss, we have $$C_{N_G(P)}(x_i)=C_G(x_i) \cap N_G(P) \subseteq H \cap N_G(P)=N_H(P) \subseteq N_G(P).$$ This means that $|N_G(P):N_H(P)|$ divides $|N_G(P):C_{N_G(P)}(x_i)|$ for each $i$, hence, using the sum above, $|N_G(P):N_H(P)| \mid |\Omega|$. In particular, since $|\Omega| \equiv -1$ mod $p$, $p \nmid |N_G(P):N_H(P)|$. In addition, because $P$ is Sylow in $H$, $\ p \nmid |N_H(P):P|$. So, $p$ does not divide $|N_G(P):P|=|N_G(P):N_H(P)| \cdot |N_H(P):P|$. Now we can apply Problem $(1A.10)$: $|G:P| \equiv |N_G(P):P|$ mod $p$, so $p$ does not divide $|G:P|$, whence $P$ is Sylow in $G$, as wanted. $\square$