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If $|G|=p^{a}m$ with $p\nmid m$, a Sylow subgroup for $p$ is a subgroup of order $p^a$. The Sylow theorems assert: such subgroups exist; every $p$-subgroup is contained in a Sylow subgroup; all Sylow $p$-subgroups are conjugate; and their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$.

For groups of order p squared q are not simple, if $p>q$, then $n_p\mid q$ and $n_p\equiv 1(\mathrm{mod}p)$ force $n_p=1$. Its unique Sylow subgroup is a nontrivial proper normal subgroup. Suppose instead $p<q$. If either Sylow count is one, the conclusion already follows. Otherwise $n_p=q$, while $n_q$ is either $p$ or $p^2$. The value $p$ is impossible because $0<p-1<q$. Thus $n_q=p^2$, and $q\mid (p^2-1)=(p-1)(p+1)$. Primality and $q>p$ imply $q\mid p+1$, hence $q=p+1$. The only consecutive primes are $p=2,q=3$.

In that exceptional order-$12$ case, four distinct Sylow $3$-subgroups contribute eight distinct nonidentity elements: their intersections are trivial. Only three nonidentity elements remain. Every subgroup of order four must contain exactly those remaining three, so there can be only one Sylow $2$-subgroup, contradicting the assumption that both counts were nontrivial. Therefore no group of order $p^{2}q$ with distinct primes is simple.

For the final factorization, normality of $H$ ensures $g^{-1}Pg$ is a Sylow $p$-subgroup of $H$. By Sylow conjugacy within $H$, choose $h\in H$ such that $g^{-1}Pg=h^{-1}Ph$. Then $k=g{h}^{-1}$ satisfiesThus $k\in N_G(P)$ and $g=kh$. We have proved the Frattini argumentThe normaliser factorization includes the case $P=\{1\}$, when its normalizer is all of $G$.