# 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:
$\displaystyle N_G(N_G(P)) = N_G(P)$. Show that equation isn't true for some subgroup H (not sylow) of $\displaystyle S_4$ (symmetric group), meaning $\displaystyle N_G(N_G(H)) \neq N_G(H)$.
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 $\displaystyle S_4$, created by the cycle (1234).
(a) Show that $\displaystyle C_G(H)=H$ for $\displaystyle C_G(H)= \{g \in G \| gh=hg \forall h \in H \}$
(b) Show that $\displaystyle N_G(H)$ 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:
$\displaystyle N_G(N_G(P)) = N_G(P)$. Show that equation isn't true for some subgroup H (not sylow) of $\displaystyle S_4$ (symmetric group), meaning $\displaystyle N_G(N_G(H)) \neq N_G(H)$.
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 $\displaystyle H=S_4 = \{ \sigma\in S_4 | \sigma (4) = 4\}$ i.e. all permutations that fix 4. Then it is seem that $\displaystyle |H| = 6$ (in fact $\displaystyle H$ is naturally isomorphic to $\displaystyle S_3$). Now since $\displaystyle H$ is not normal it means $\displaystyle N(H) \not = S_4$. If $\displaystyle N(H) \not = H$ i.e. if there is $\displaystyle \sigma \in S_4 - H$ so that $\displaystyle \sigma H \sigma^{-1} = H$ then it means $\displaystyle H$ is propely contained in $\displaystyle N(H)$. By Lagrange's theorem it would mean $\displaystyle N(H) = A_4$ since $\displaystyle |N(H) | = 12$. But $\displaystyle N(A_4) = S_4$. Therefore, $\displaystyle N(N(H)) \not = N(H)$. 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.