Group Theory - Sylow and Conjugation

Quote:

Originally Posted by pavelr

1. We showed that for every p-sylow group P in G, the following is true:
$N_G(N_G(P)) = N_G(P)$ . Show that equation isn't true for some subgroup H (not sylow) of $S_4$ (symmetric group), meaning $N_G(N_G(H)) \neq N_G(H)$ .
I don't fully understand how to do this without doing it brute-force, by applying all elements of G on the chosen H.

I did not actually work out the details for this problem, but I had an idea that you might want to try. Let $H=S_4 = \{ \sigma\in S_4 | \sigma (4) = 4\}$ i.e. all permutations that fix 4. Then it is seem that $|H| = 6$ (in fact $H$ is naturally isomorphic to $S_3$ ). Now since $H$ is not normal it means $N(H) \not = S_4$ . If $N(H) \not = H$ i.e. if there is $\sigma \in S_4 - H$ so that $\sigma H \sigma^{-1} = H$ then it means $H$ is propely contained in $N(H)$ . By Lagrange's theorem it would mean $N(H) = A_4$ since $|N(H) | = 12$ . But $N(A_4) = S_4$ . Therefore, $N(N(H)) \not = N(H)$ . I cannot gaurentte you this would work because I was kinda lazy to work out the details but this would be the first thing I would try to do.