Frattini's argument

If ${\displaystyle G}$ is a finite group with normal subgroup ${\displaystyle H}$, and if ${\displaystyle P}$ is a Sylow p-subgroup of ${\displaystyle H}$, then

${\displaystyle G=N_{G}(P)H,}$

where ${\displaystyle N_{G}(P)}$ denotes the normalizer of ${\displaystyle P}$ in ${\displaystyle G}$, and ${\displaystyle N_{G}(P)H}$ means the product of group subsets.

The group ${\displaystyle P}$ is a Sylow ${\displaystyle p}$-subgroup of ${\displaystyle H}$, so every Sylow ${\displaystyle p}$-subgroup of ${\displaystyle H}$ is an ${\displaystyle H}$-conjugate of ${\displaystyle P}$, that is, it is of the form ${\displaystyle h^{-1}Ph}$ for some ${\displaystyle h\in H}$ (see Sylow theorems). Let ${\displaystyle g}$ be any element of ${\displaystyle G}$. Since ${\displaystyle H}$ is normal in ${\displaystyle G}$, the subgroup ${\displaystyle g^{-1}Pg}$ is contained in ${\displaystyle H}$. This means that ${\displaystyle g^{-1}Pg}$ is a Sylow ${\displaystyle p}$-subgroup of ${\displaystyle H}$. Then, by the above, it must be ${\displaystyle H}$-conjugate to ${\displaystyle P}$: that is, for some ${\displaystyle h\in H}$

${\displaystyle g^{-1}Pg=h^{-1}Ph,}$

and so

${\displaystyle hg^{-1}Pgh^{-1}=P.}$

Thus

${\displaystyle gh^{-1}\in N_{G}(P),}$

and therefore ${\displaystyle g\in N_{G}(P)H}$. But ${\displaystyle g\in G}$ was arbitrary, and so ${\displaystyle G=HN_{G}(P)=N_{G}(P)H.\ \square }$