# Thread: Group Theory - Sylow and Conjugation

1. ## Group Theory - Sylow and Conjugation

A couple of questions I had hard time solving (and not succeeded).

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.

2. H is a subgroup of $S_4$, created by the cycle (1234).
(a) Show that $C_G(H)=H$ for $C_G(H)= \{g \in G \| gh=hg \forall h \in H \}$
(b) Show that $N_G(H)$ is a 2-sylow subgroup of G.

I have some difficulty by showing that too.

I will very appreciate any help.

2. 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.