Show that if the prime $p$ divides $|G|$, then $|X|$ is divisible by $p$.

Question

Question :

Let $p$ be a prime number that divides the order of the finite group $G$. Let $X$ = $\bigcup_{P \in Syl_p(G)}P$. Show that $|X|$ is divisible by $p$.

Answer 1

Lemma Let $G$ be a finite group and $p$ be a prime divding $|G|$. Let $H$ be a $p$-subgroup of $G$ and $P \in Syl_p(G)$. Then $H \cap C_G(P)=H\cap Z(P)$.

Proof It is clear that $H \cap Z(P) \subseteq C_H(P)=H \cap C_G(P)$. Conversely, observe that $C_H(P)$ is a $p$-subgroup (it is a subgroup of $H$!) and it normalizes (even centralizes) $P$. So $C_H(P)P$ is a $p$-subgroup containing $P$, and since $P$ is Sylow, this can only be the case if $C_H(P)P=P$, that is $C_H(P) \subseteq P$. So $C_H(P) \subseteq C_G(P) \cap P=C_P(P)=Z(P)$ and of course $C_H(P) \subseteq H$.

Proposition Let $G$ be a finite group and $p$ be a prime divding $|G|$. Let $X=\bigcup_{P \in Syl_p(G)}P$. Then $|X| \equiv 0$ mod $p$.

Proof Let $S \in Syl_p(G)$ and let $S$ act on $X$ by conjugation. Let $Y=\{x \in X: s^{-1}xs=x$ for all $s \in S\}$, the set of fixed points under the action. By the Orbit-Stabilizer Theorem and the fact that $S$ is a $p$-group, it is evident that $|X| \equiv |Y|$ mod $p$. Let us analyze the set $Y$ by applying the Lemma. $Y$ is the set of elements of $X$ that centralize $S$: $$Y=C_X(S)= X \cap C_G(S) = \bigcup _{P \in Syl_p(G)}(P \cap C_G(S))= \bigcup _{P \in Syl_p(G)}(P \cap Z(S)) = X \cap Z(S) \subseteq Z(S).$$But obviously $Z(S) \subseteq Y$, and we conclude $Y=Z(S)$. Since $S$ is a non-trivial $p$-group, $Z(S)$ is non-trivial, in particular $|Y|=|Z(S)| \equiv 0$ mod $p$, so $p | |X|$.