Normalizers of Sylow subgroups are self-normalizing

Let H and K be a finite groups, P a Sylow p-subgroup of H, and H \leq K. If P is normal in H and H is normal in K, prove that P is normal in K. Deduce that if P \in \mathsf{Syl}_p(G) then N_G(N_G(P)) = N_G(P). (That is, normalizers of Sylow subgroups are self-normalizing.)

Let k \in K. Now kHk^{-1} = H since H is normal, so that conjugation by k is an automorphism of H. Since P is the unique subgroup of its order in H, P is characteristic in H, and we have kPk^{-1} = P. Thus P is normal in K.

Now let P be a Sylow subgroup of G. Note that N_G(P) is the \leq-largest subgroup of G in which P is normal. Now if N_G(P) \leq K is normal, we have P \leq K normal by the above argument, so that K \leq N_G(P). In particular, N_G(N_G(P)) = N_G(P).

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