## In a finite group, Sylow subgroups induce Sylow subgroups in normal subgroups and their corresponding quotients

Let $G$ be a finite group, $P \in \mathsf{Syl}_p(G)$, and $N \leq G$ a normal subgroup. Use the conjugacy part of Sylow’s Theorem to prove that $P \cap N$ is a Sylow $p$-subgroup of $N$. Deduce that $PN/N$ is a Sylow $p$-subgroup of $G/N$.

Let $Q$ be a Sylow $p$-subgroup of $N$. $Q$ is a $p$-subgroup of $G$, so that $Q \leq xPx^{-1}$ for some $x \in G$. Then $Q \leq xPx^{-1} \cap N = xPx^{-1} \cap xNx^{-1}$ since $N$ is normal, hence $Q \leq x(P \cap N)x^{-1}$, thus $x^{-1}Qx \leq P \cap N$. Now $x^{-1}Qx$ is a subgroup of $N$, and has maximal $p$-power order in $N$, since $|x^{-1}Qx| = |Q|$. Since $P \cap N$ is a $p$-subgroup of $N$, $P \cap N$ is a maximal $p$-power subgroup of $N$, thus a Sylow $p$-subgroup.

Note that $PN/N \cong P/(P \cap N)$ by the Second Isomorphism Theorem, so that $PN/N$ is a $p$-subgroup of $G/N$. By the Third Isomorphism Theorem, $[G/N : PN/N] = [G : PN]$; since $P$ is a maximal $p$-power order subgroup of $G$, $p$ does not divide $[G : PN]$, so $p$ does not divide $[G/N : PN/N]$. Hence $PN/N$ is a Sylow $p$-subgroup of $G/N$.