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 $P$.

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 the conjugacy action:
 * $\forall h \in P, S \in \mathbb S: h * S = h S h^{-1}$

From Conjugacy Action on Subgroups is Group Action, this is a group action for $S \le G$.

To show it is closed for $S \in \mathbb S$:

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

Since $\forall S \in \mathbb S, \operatorname{Stab} \left({S}\right) \le P$, we have that:


 * $\left|{\operatorname{Stab} \left({S}\right)}\right| \mathrel \backslash \left|{P}\right|$

Therefore, by the Orbit-Stabilizer Theorem, these orbit lengths are all congruent to either $0$ or $1$ modulo $p$, since $P$ is a Sylow $p$-subgroup of $G$.

Note that this will imply, as we shall mark later on:
 * $\left|{\mathbb S}\right| \equiv 1 \pmod p$

Now, $h * P = h P h^{-1} = P$, 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 $\forall x \in P$ we have that:
 * $g^{-1} x g \in N_G \left({P}\right)$

From Order of Conjugate Element, we have 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 $1$.

From Stabilizer of Coset Action on Power Set, $P = N_G \left({P}\right)$.

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 $1$.

Hence:
 * $\left|{\mathbb S}\right| \equiv 1 \pmod p$

as promised.

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:
 * $g^{-1} Q g \subseteq N_G \left({P}\right)$

So by Normalizer of Sylow P-Subgroup:
 * $g^{-1} Q g \subseteq P$

Thus $Q \subseteq g P g^{-1}$ as required.

Also known as
Some sources call this the third Sylow theorem.