Showing $gG_ag^{-1}=G_{ga}$

Question

Let a group $G$ act on a set $S$ with $s \in S$. Then the orbit of $s$ induced by $G$ is $$Gs = \{gs: s \in S \}$$

The stabilizer of $s$ induced by $G$ is $$G_s=\{g \in G: gs =s\}$$

The conjugation group of $H \leq G$ by $g$ is the group $$\{gkg^{-1}: k \in H\}$$

Now, let $G$ be a group, let $S$ be a set with $a \in S$, and suppose $G$ acts on $S$. We must prove that $gG_ag^{-1} = G_{ga}$. The first direction is easy. Let $gkg^{-1} \in gG_ag^{-1}$. Then

$$(gkg^{-1})(ga)=gk(g^{-1}g)a=gka = ga$$

Thus, $gG_ag^{-1} \subset G_{ga}$. The other direction is not as easy; I need help with it. First, we let $m \in G_{ga}$. Then $m$ fixes $ga$, i.e. $mga = ga$. But I'm not sure where to go from here. I'm thinking of something like

$$mga = 1_G mga 1_G^{-1} = 1_G ga 1_G^{-1} =ga = ... ?$$

Any help is greatly appreciated, thanks.

Answer 1

Since $m(ga) = ga$, $g^{-1}(mg)a = a$. Then $g^{-1}mg \in G_a$, hence $m \in g G_a g^{-1}$. Since $m$ was an arbitrary element of $G_{ga}$, the reverse inclusion holds.

Answer 2

A slightly different perspective. The map $h\mapsto ghg^{-1}$ sends $G_a$ to $G_{ga}$, because if $ha=a$, then $ghg^{-1}(ga)=gh(a)=ga$. On the other hand, by exactly the same reasoning, the map $h\mapsto g^{-1}hg$ maps $G_{ga}$ to $G_{g^{-1}(ga)}=G_a$. The two maps are inverses to each other, and therefore must induce isomorphisms between the two sets.

Answer 3

Straight from set-builder notation: \begin{alignat}{1} G_{ga} &= \{g'\in G\mid g'(ga)=ga\} \\ &= \{g'\in G\mid (g^{-1}g'g)a=a\}. \\ \tag 1 \end{alignat} Now, call $g'':=g^{-1}g'g$; then, $g'=gg''g^{-1}$, and $(1)$ yields: \begin{alignat}{1} G_{ga} &= \{gg''g^{-1}\in G\mid g''a=a\} \\ &= g\{g''\in G\mid g''a=a\}g^{-1} \\ &= gG_ag^{-1}. \\ \tag 2 \end{alignat}

Answer 4

You want to prove that $G_{ga}\subseteq gG_ag^{-1}$.

Pick $h\in G_{ga}$. You want to find $k\in G_a$ such that $h=gkg^{-1}$ (and such $k$ would be unique). Hence you need $k=g^{-1}hg$, so what's necessary to prove is that $g^{-1}hg\in G_a$. We're assuming that $$ h(ga)=ga $$ by definition of stabilizer. Hence $$ g^{-1}(hga)=g^{-1}ga $$ that's the same as saying that $$ g^{-1}hg\in G_a $$