Are the normalizers of Sylow p-subgroups isomorphic?

Question

Let $G$ be a finite group and $A,B \in \text{Syl}_p G$, for some prime $p$. Is it always true that the normalizers $N_G(A)\cong N_G(B)$? I just need a hint to get started, because I don't know where to begin...

If $A^g=B$, then $N_G(A)^g=N_G(B)$. – Levent May 14 '16 at 08:12 — Levent, May 14 '16 at 08:12

score 1 · Accepted Answer · answered May 14 '16 at 08:24

1

Note that if $gAg^{-1}=B$ and $h$ normalizes $B$, then $hgAg^{-1}h^{-1}=B$, hence $$g^{-1}hgAg^{-1}h^{-1}g=A$$ What does this tell you?

answered May 14 '16 at 08:24

Matt Samuel

58,164

$N_G(B)=N_G(gAg^{-1})= {k\in G : k^{-1}gAg^{-1}k=gAg^{-1}} = {k\in G : g^{-1}k^{-1}gAg^{-1}kg=A}$. By substituting $l=g^{-1}kg$, the last set is ${glg^{-1}: l\in G \land l^{-1}Al=A} = gN_G(A)g^{-1}$. Is it correct? I thought it was more difficult :) – Human May 14 '16 at 09:02
@Human yep. Easier than it looks. – Matt Samuel May 14 '16 at 13:56

score 0 · Answer 2 · answered Aug 18 '17 at 14:44

Additional fun fact: now that you know that $N_G(P)^g=N_G(P^g)$, you can ask yourself the question: how many different conjugates $N_G(P)^g$ of the Sylow $p$-group $P$ are there anyway? Well, in general, basic group theory tells you that a subgroup $H$ has $|G:N_G(H)|$ many different conjugates. So, in our case this is $|G:N_G(N_G(P))|$. However, it is a well-known fact that for a Sylow $p$-subgroup $N_G(N_G(P))=N_G(P)$, see for example here. So the answer to the question is $|G:N_G(P)|$, which is exactly the number of different Sylow $p$-subgroups!

Are the normalizers of Sylow p-subgroups isomorphic?

2 Answers2