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 $$S \in \mathbb S$$:

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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.