I am trying to solve the following question:
Let $G$ be a group of finite order and let $P$ be a p-subgroup of $G$. Then
$$|\{Q \in Syl_p(G) : P \subset Q\} | \equiv 1 \pmod{p}$$
A couple of thoughts about this. We know that the set we are trying to find the size of is not empty (because of Sylow Theorems). Moreover, if $P$ itself is a Sylow p-subgroup, then the result follows trivially. Likewise, if $P$ is a normal subgroup of $G$ then every Sylow-p subgroup of $G$ contains it, which then forces LHS to be $n_p$ and we are done.
I have tried the following approach. Let $P$ act on $Syl_p(G)$ by conjugation. Certainly, if $P \subset Q$, then $Q$ is a fixed point under the action. If, we could show that $Q$ is a fixed point if and only if $P\subset Q$, then, by applying the orbit-stabilizer theorem, we would have
$$|\{Q \in Syl_p(G) : P \subset Q\} | = |Syl_p(G)^{P}| \equiv |Syl_p(G)| \pmod{p}$$
where $|Syl_p(G)^{P}|$ denotes the fixed points under the action of $P$ on $Syl_p(G)$, and we would be done.
I seem to be very close, however, I am having trouble proving that forward implication in the aforementioned if and only if statement. That is, proving that if $Q$ is a fixed point under the action, then $P$ sits inside it. Is this even true? If not, how can we prove this statement?
Also, in more generality, can we ever get a precise number for such Q, depending on P? Other than the trivial examples for when P is normal or P is itself Sylow-p?
Any ideas are welcome.
Best Answer
Your approach will work, here is a hint for the step you are having trouble with.
Hint: if $Q$ is a Sylow $p$-subgroup of $G$, note that $Q$ is the unique Sylow $p$-subgroup of $N_G(Q)$.