Second Sylow Theorem

Theorem
Let $P$ be a Sylow $p$-subgroup of the finite group $G$.

Let $Q$ be any $p$-subgroup of $G$.

Then $Q$ is contained in a conjugate of $P$.

Some sources call this the third Sylow theorem.

Proof
Let $P$ be a Sylow $p$-subgroup of $G$.

Let $\mathbb S$ be the set of all distinct $G$-conjugates of $P$: $\mathbb S = \left\{{g P g^{-1}: g \in G}\right\}$.

Let $h * S$ be defined as $\forall h \in P, S \in \mathbb S: h * S = h S h^{-1}$

From Conjugate of a Set, this is a group action for $S \le G$ - it is a simple matter to show it is closed for $S \in \mathbb S$:

So, consider the orbits of $\mathbb S$ under this group action.

$h * P = h P h^{-1} = P$, so $\operatorname{Orb} \left({P}\right) = \left\{{P}\right\}$.


 * We now show that $P$ is the only element of $\mathbb S$ such that $\left|{\operatorname{Orb} \left({S}\right)}\right| = 1$.

If $g P g^{-1}$ has one element in its orbit, then $\forall x \in P: x \left({g P g^{-1}}\right) x^{-1} = g P g^{-1}$.

Thus $\forall x \in P: g^{-1} x g \in N_G \left({P}\right)$ such that $\left|{g^{-1} x g}\right| = \left|{x}\right|$.

Thus $P_1 = g^{-1} P g$ is a $p$-subgroup of $N_G \left({P}\right)$.

As $P$ and $P_1$ have the same number of elements, $P_1$ is a Sylow $p$-subgroup of $N_G \left({P}\right)$.

Hence $P_1 = P$ by Normalizer of Sylow P-Subgroup, so $g P g^{-1} = P$.

Thus $P$ is the only element of $\mathbb S$ whose orbit has length $1$.

Thus, for any $g \notin P$, $\left|{\operatorname{Orb} \left({g P g^{-1}}\right)}\right|$ under conjugation by elements of $P$ has orbit greater than $1$.

By the Orbit-Stabilizer Theorem, these orbit lengths are all congruent to $0$ modulo $p$.

Thus $\left|{\mathbb S}\right| \equiv 1 \left({\bmod\, p}\right)$.


 * Next we consider orbits of $\mathbb S$ under conjugation by elements of $Q$.

Since every orbit has length a power of $p$, the above conclusion shows there is at least one orbit of length $1$.

So there is an element $g$ such that $\forall x \in Q: x \left({g P g^{-1}}\right) x^{-1} = g P g^{-1}$.

As previously, $g^{-1} Q g \in N_G \left({P}\right)$ and so by Normalizer of Sylow P-Subgroup, $g^{-1} Q g \subseteq P$ so $Q \subseteq g P g^{-1}$ as required.