How Does the Normalizer of a Stabilizer Influence Transitivity on Fixed Points?

[math_grl]

Homework Statement

G acts transitively on S and let H be the stabilizer of s. Show that the normalizer of H, call it N, acts transitively on the fixpoints of H, call it F, where s is some element in S.

Homework Equations

Two different ways of showing this:
Either we show the orbit for any element in the fixpoints is the entire set of fixpoints
or we show that given any two fixpoints that there is an element in the normalizer such that it we can apply it to one of those elements and get the other one.

The Attempt at a Solution

There must be some trick or something small I'm missing.
Since G is tran. on S and F is a subset of S, then $$\forall s, t \in F, \exists g \in G$$ so that $$g s = t \Rightarrow s = g^{-1}t$$

Now $$ghs = ghg^{-1}t = h't = t$$ if and only if we can show that $$g \in N$$ since $$ghh'(gh)^{-1} = g(hh'h^{-1})g^{-1} \in N$$ whenever $$g \in N$$ and we have found our element in N, namely gh. I'm stuck, don't even know if I'm going in the right direction.

 

[kurt.physics]

The second approach: Let s, t \in F. Then \exists h \in H so that hs = s and ht = t. Then \exists n \in N so that nhn^{-1} = k \in H \Rightarrow n(hs) = nk(n^{-1}t) \Rightarrow nt = (nhn^{-1})t = kt = t. So we have found our n \in N so that when applied to one fixpoint we get the other one. I am really stuck and I don't know if I'm even going in the right direction. Any help would be appreciated. A:Let $F$ be the set of fixed points of $H$, and let $s,t \in F$. Since $G$ acts transitively on $S$, there exists $g \in G$ such that $g(s) = t$. Since $H$ is the stabilizer of $s$, it follows that $gHg^{-1}$ is the stabilizer of $t$. We also know that $gHg^{-1} \subseteq N$ since $N$ is the normalizer of $H$. Hence, $gHg^{-1} \in N$.Now, let $h \in H$ be arbitrary. Then $ghg^{-1} \in gHg^{-1} \subseteq N$. It follows that$$(ghg^{-1})(s) = g(h(g^{-1}(s))) = g(h(s)) = g(s) = t.$$Since $h \in H$ was arbitrary, we have shown that every element of $N$ maps $s$ to $t$. Hence, $N$ acts transitively on $F$.