# A variation of Sylow's counting theorem

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?

$$H$$ 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 $$|X|\equiv|\mathrm{Syl}_p(G)|\equiv 1\pmod{p}$$.