Past exam of the algebra course of the University of Oxford / 2008 / B / 6

A Sylow subgroup for the prime $p$ is a maximal $p$-subgroup; in a finite group of order $p^{a}m$ with $p\nmid m$ it has order $p^a$. Sylow theorems assert existence, containment of every $p$-subgroup in a Sylow $p$-subgroup, conjugacy of all Sylow $p$-subgroups, and the counting conditions $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

Since $N$ is a normal subgroup, $NP$ is a subgroup andThe final index is prime to $p$. Since $N\cap P$ is a $p$-subgroup, its index in $N$ being prime to $p$ means it contains the entire $p$-part of $|N|$. Hence $N\cap P$ is Sylow in $N$.

The quotient $PN/N$ is a $p$-group, being the image of $P$, andIts index is prime to $p$, so $PN/N$ is Sylow in $G/N$.

If $P\le A$ and $Q\le B$ are Sylow $p$-subgroups, their product has the full $p$-part of $|A\times B|$, hence is Sylow. Conversely, for any Sylow $R\le A\times B$, its two projections are $p$-subgroups and lie in some Sylows $P,Q$. Then $R\subseteq P\times Q$, and equality follows by order. Distinct pairs give distinct subgroups because their projections recover $P$ and $Q$. Thus