Frattini's Argument

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $K$ be a finite normal subgroup of $G$, and $p$ a prime which divides the order of $K$.

Let $P$ be a Sylow $p$-subgroup of $K$, and $N_G \left( P \right)$ the normalizer of $P$ in $G$.

Then:
 * $G = N_G \left({P}\right) \circ K = K \circ N_G \left({P}\right)$

Proof
Let $g \in G$.

Since $K$ is normal in $G$, and $P \subset K$, the conjugate $g \circ P \circ g^{-1}$ of $P$ is also a subset of $K$.

From Inner Automorphism is Automorphism, $g \circ P \circ g^{-1}$ is a subgroup of $K$ of the same order as $P$.

Thus $g \circ P \circ g^{-1}$ is also a Sylow $p$-subgroup of $K$.

By Second Sylow Theorem:
 * $\exists k \in K: g \circ P \circ g^{-1} = k \circ P \circ k^{-1}$

Then:
 * $g^{-1} \circ k \circ P \circ k^{-1} \circ g = P$

and so by definition of normalizer:
 * $k^{-1} \circ g \in N_G \left({P}\right)$

We can then write $g$ as
 * $g = k \circ \left({k^{-1} \circ g}\right) \in K \circ N_G \left({P}\right)$

Since $K$ is normal:
 * $N_G \left({P}\right) \circ K = K \circ N_G \left({P}\right)$

by Subset Product with Normal Subgroup as Generator.