$p$-subgroups and Sylow $p$-subgroups

Question

In several proofs and exercises I have read is used the fact that if one is given that $H$ is a $p$-subgroup of $G$, then $H$ is a subgroup of a Sylow $p$-subgroup. Whenever I ask, I am told that this is as a consequence of Sylow Theorems (I think that the second one). Why is this a consequence if the theorem just claims that all Sylow $p$-subgroups are conjugate? There cannot be a $p$-subgroup that isn't contained in any Sylow $p$-subgroup?

Answer 1

Let $P$ be an arbitrary $p$-subgroup of $G$ and let $S$ be a Sylow $p$-subgroup. Then $P$ acts on $G/S$ by left-multiplication. This is a $p$-group acting on a set of order prime to $P$, so there is a fixed point, say $gS$.

Since the sets $PgS$ and $gS$ are the same, we have $P \subset gSg^{-1}$. But $gSg^{-1}$ is a Sylow $p$-subgroup.

Answer 2

In addition a fun fact (stated without proof but with hints): if $P$ is a $p$-subgroup of $G$, then $\#\{S \in Syl_p(G): P \subseteq S\} \equiv 1$ mod $p$.

Hints: $P$ acts by conjugation on the set of Sylow $p$-subgroups; use $P \subseteq S$ if and only if $P \subseteq N_G(S)$ (here you need the statement that you were after in the first place (applied in $N_G(S)$)); use the number of Sylow $p$-subgroups is $\equiv 1$ mod $p$; apply the Orbit-Stabilizer Theorem.