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.

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