Show that $N_G(H) = H$

Question

Question :

Let $P \in Syl_p(G)$, and suppose that $N_G(P) \subseteq H$, where $H$ is a subgroup. Prove that $H = N_G(H)$.

We know that $H \subseteq N_G(H)$, please give me hint to show that $N_G(H) \subseteq H$.

Thank you.

Answer 1

Since $P \in Syl_p(G), P \in Syl_p(H)$, so if $x \in N_G(H)$, then $xHx^{-1} = H$, so $$ xPx^{-1} \in Syl_p(H) $$ Hence, $\exists y\in H$ such that $$ xPx^{-1} = yPy^{-1} $$ So $$ y^{-1}xP(y^{-1}x)^{-1} = P\Rightarrow y^{-1}x \in N_G(P) \subset H $$ Since $y \in H$, it follows that $x\in H$, and so $N_G(H) \subset H$.

Answer 2

Have you done the Frattini argument in your course? If $N \lhd G$ and $Q \in {\rm Syl}_{p}(N),$ then $G = NN_{G}(Q).$ This is because for any $g \in G, gQg^{-1}$ is also a Sylow $p$-subgroup of $N,$ so $gQg^{-1} = nQn^{-1}$ for some $n \in N.$ Consequently, $n^{-1}g \in N_{G}(Q)$ and $g \in NN_{G}(Q).$ The answer to the question is a special case of this, but you have to apply it with $N_{G}(H)$ in place of $G$,$H$ in place of $N,$ and $P$ in place of $Q.$ This is OK, since $H \lhd N_{G}(H)$ and $P$ is a Sylow $p$-subgroup of $H.$ We obtain $N_{G}(H) = HN_{N_{G}(H)}(P).$ But $N_{G}(P) \leq H,$ so this simplifies to $N_{G}(H) = H.$ (This is not significantly different from Prahlad's answer, but just wanted to point out the more general result).