A variation of Sylow's counting theorem

Question

Let $G$ be a finite group, $p$ a prime number dividing the order of $G$ and $H$ a finite $p$-subgroup of $G$. Let $n_p(G,H)$ be the number of Sylow $p$-subgroups of $G$ containing $H$. I want to prove that $n_p(G,H)\equiv 1 \pmod{p}$.

Here is my attempt: Denote by $\operatorname{Syl}_p(G)$ the set of Sylow $p$-subgroups of $G$ and let $$ X = \{P\in \operatorname{Syl}_p(G) \, | \, H\leq P\}. $$ As $H$ is a $p$-subgroup of $G$, it is contained in some Sylow $p$-subgroup of $G$, so $X$ is non empty. Fix $Q\in X$. Let $N_G(H)$ the normalizer of $H$ in $G$ and consider the group $K=Q\cap N_G(H)$. Then $K$ acts on $X$ by conjugation, and $Q$ is a fixed point of this action. Denote by $X_0$ se set of fixed points of this action. As $K$ is a $p$-group, we have that $$ n_p(G,H) = |X|\equiv |X_0| \pmod{p}. $$ Thus all I need to prove is that $X_0=\{Q\}$, but I'm stuck here. How can I show (if possible) that $Q$ is the unique fixed point of the action of $K$ on $X$? Is there another approach?

Answer 1

$H$ acts by conjugation on $\mathrm{Syl}_p(G)$. Let $P\in\mathrm{Syl}_p(G)$ be a fixed point of this action, i.e. $H\le N_G(P)$. Since $P$ is the only Sylow $p$-subgroup of $N_G(P)$, it follows that $H\le P$, i.e. $P\in X$. Now by orbit counting we get $|X|\equiv|\mathrm{Syl}_p(G)|\equiv 1\pmod{p}$.