# A Little Fact From Group Actions

/Today we've got a little post on a little fact relating to group actions. I wanted to write about this not so much to emphasize its importance (it's certainly not a major result) but simply to uncover the intuition behind it.

**stabilizer**of an element $a\in A$ to be the set of all elements in $G$ which fix (

*stabilize*) $a$ and denote it by $G_a$: $$G_a=\{g\in G:g\cdot a=a\}.$$ It's a fact that $$G_{g\cdot a}=g G_a g^{-1}$$ which is to say, "the stabilizer of $g\cdot a\in A$ equals the conjugate by $g\in G$ of the stabilizer of $a\in A$." Now, if you're like me, you've got two looming questions:

*What?*and

*Why?*

## What?

*new*guy $g\cdot a$.

*How would you do it?*Here's how:

__Step 1:__ Send $g\cdot a$ back to $a$ by letting $g^{-1}$ act:

__step 2:__ Fix $a$ by letting any $b\in G_a$ act:

__step 3:__ Send $a$ back to $g\cdot a$ by letting $g$ act:

*Why?*).

## Why?

As usual, we prove the equality $G_{g\cdot a}=g G_a g^{-1}$ by showing containment both directions. To begin let $b\in G_a$ so that $b\cdot a=a$. We wish to show $gbg^{-1}\in G_{g\cdot a}$. But this is immediate: $$gbg^{-1}\cdot(g\cdot a)= (gbg^{-1}g)\cdot a = (gb)\cdot a=g\cdot (b\cdot a)=g\cdot a.$$ Since $gbg^{-1}$ fixes $g\cdot a$ it follows that $gbg^{-1}\in G_{g\cdot a}$ and so $gG_ag^{-1}\subseteq G_{g\cdot a}$.

For the other direction, we'll use a little trick. Let $h=g\cdot a$ so that $a=g^{-1}\cdot h$. Then by the above paragraph we know $$g^{-1}G_hg\subseteq G_{g^{-1}\cdot h}.$$ But the left hand side is equal to $g^{-1}G_{g\cdot a}g$, and the right hand side is equal to $G_a$. Hence $g^{-1}G_{g\cdot a}g\subseteq G_a$ and therefore $G_{g\cdot a}\subseteq g G_a g^{-1}$.

QED!