# Stabilizers (Group Theory)

1. Mar 4, 2009

### zcdfhn

Suppose that G acts on the set X. Prove that if g $$\in$$ G, x $$\in$$ X then StabG(g(x)) = g StabG(x) g-1.

Note: g StabG(x) g-1 by definition is {ghg-1 : h $$\in$$ StabG(x)}

My attempt at the problem is:
Let a $$\in$$ StabG(g(x)), then a(g(x)) = g(x) by definition.
Also Let b$$\in$$ StabG(x), then b(x) = x by definition.

and then I am completely stuck. Please guide me with this proof, I have tried for a couple hours.

2. Mar 4, 2009

### Hurkyl

Staff Emeritus
What is the most typical method used to prove that some set S is equal to some other set T?

3. Mar 4, 2009

### zcdfhn

So I have to prove that one set contains the other, and that the other contains the one set, but I still need a push in the right direction.

Oh ok i figured it out you check a(g(x)) = g(x) and you also check ghg^-1(g(x)) = g(x)

Thank you so much

Last edited: Mar 4, 2009
4. Mar 4, 2009

### Hurkyl

Staff Emeritus
Right. So now that you know the outline of the proof, just fill in the steps. e.g. one half of the proof is to show that $\mathrm{Stab}_G(g(x)) \subseteq g \cdot \mathrm{Stab}_G(x) \cdot g^{-1}$ -- I'll get you started:

Suppose $a \in \mathrm{Stab}_G(g(x))$
...
...
...
Therefore $a \in g \cdot \mathrm{Stab}_G(x) \cdot g^{-1}$

Note that you've already filled in the second step in your opening. The second to last step should be easy as well....

(I'm operating under the assumption you haven't done this yet, since the work you presented isn't along these lines... and I believe that doing this really should suggest something to try)

5. Mar 4, 2009

### zcdfhn

I actually still have no clue, I haven't even figured out the second the last step that you mentioned.

So from the assumption that a$$\in$$ StabG(g(x)), you can say a(g(x)) = g(x), but then I'm still trying to think out how to show that a is included in ghg^-1

Would you do a(ghg^-1), just a guess, but still no leads

6. Mar 4, 2009

### Hurkyl

Staff Emeritus
Well, you already said that $g \cdot \mathrm{Stab}_G(x) \cdot g^{-1} = \{ g h g^{-1} \mid h \in \mathrm{Stab}_G(x) \}$, didn't you? So the claim
$a \in g \cdot \mathrm{Stab}_G(x) \cdot g^{-1}$​
should be logically equivalent to
there exists an $h \in \mathrm{Stab}_G(x)$ for which $a = ghg^{-1}$​

(Now, if such an h really does exist, what would it have to be...?)

Note that nothing clever or insightful was involved here -- all I'm doing is carefully unfolding the definitions. You already told me how to write $g \cdot \mathrm{Stab}_G(x) \cdot g^{-1}$ in set-builder notation (i.e. $\{ \cdot \mid \cdot \}$ form) -- all I did after that was to write the definition of $\in$ for sets presented in such a way. It takes a while to get used to doing things with such precision, but it really is extremely helpful.

7. Mar 4, 2009

### zcdfhn

So is this at all correct for the first half of the proof:

Suppose a $$\in$$StabG(g(x)), then a(g(x)) = g(x)
Then suppose h$$\in$$StabG(x), then h(x) = x
if a=ghg-1, then h = g-1ag = g-1g = e = identity
so since a = ghg-1=geg-1 = e = identity

so a = e$$\in$$g StabGg-1

I'm confused why i end up with a = h = e, though, so I have a feeling i messed up