MHB Proof of Gρ(x)=ρGxρ−1 in Symmetric Groups

[mathjam0990]

Let G be a subgroup of Sym(X) and ρ ∈ G. Prove that G_ρ(x) = ρG_xρ⁻¹, where ρG_xρ⁻¹ = {ρgρ⁻¹|g ∈ Gx}

What I Know: I need to somehow prove the left is contained in the right and the right is contained in the left.

What I Have Done: Well based on the definition of a stabilizer G_x I assumed that G_ρ(x) = {g'(p(x))=p(x) | g' w/in G} Sorry, I'm not sure if that is correct or not.

Then for (Left contained in right): For all g' w/in G, g'(p(x)=p(x) ---> (g'p)(x)=p(x) ---> pg'p^-1p^-1p(x)=p(p(x))p^-1p^-1 (multiplied both sides by pp^-1p^-1)
--->pg'p^-1(x)=x (by associativity for a group) and that is contained in pgp^-1 which equals ρGxp⁻¹

Haven't attempted the "right contained in the left" because I'm not even sure if I even did the initial part correctly, I feel like I messed up. Any help please??

Thank you!

 

[Euge]

Hi mathjam0990,

Based on what you've written for $G_{\rho(x)}$, I take it that you haven't yet grasped the definition of the stabilizer. If $G$ is a group acting (on the left) on a set $S$, then for each $s\in S$, the stabilizer $G_s$ of $s$ is defined as the set of all elements of $G$ which fix $s$, i.e., $\{g\in G : gs = s\}$. In your prompt, you should have stated that $x\in X$.

Now if $a\in G_{\rho(x)}$, then $a\in G$ and $a(\rho(x)) = \rho(x)$. Thus $\rho^{-1}a\rho(x) = \rho^{-1}\rho(x) = x$; since $G$ is a subgroup of $\operatorname{Sym}(X)$ and both $a,\rho\in G$, then $\rho^{-1}a\rho\in G$. Therefore $\rho^{-1}a\rho\in G_x$, say $\rho^{-1}a\rho = b$ for some $b\in G_x$. Then $a = \rho b\rho^{-1}\in \rho G_x\rho^{-1}$.

Conversely, if $a\in \rho G_x \rho^{-1}$, then there is a $g\in G_x$ such that $a = \rho g \rho^{-1}$. Since $G$ is a subgroup of $\operatorname{Sym}(X)$ and $\rho, g\in G$, then $\rho g \rho^{-1}\in G$, i.e., $a\in G$. As $g(x) = x$, $a(\rho(x)) = \rho(g(x)) = \rho(x)$. Consequently $a\in G_{\rho(x)}$.

 

[Deveno]

Some things to keep in mind:

for any group $G$, and any element $g \in G$, the map $x \mapsto g^{-1}xg$ is an automorphism of $G$ (an isomorphism $G \to G$), called the inner automorphism induced by $g$.

So, if $H$ is any subgroup of $G$, then $gHg^{-1}$ is *also* a subgroup of $G$ (isomorphic to $H$).

In particular, if $H$ is the stabilizer of some element in a set $S$ that $G$ acts on, say $H = G_a$, for some $a \in S$, it seems that $gG_ag^{-1}$ must also be a stabilizer of "something" (some perhaps different element of $S$).

But what? Let us see if we can discover this. Let $h \in G_a$, so $h(a) = a$. We will try to find out which $s \in S$ (if any) that $ghg^{-1}$ stabilizes.

From $ghg^{-1}(s) = s$ we have (letting $g^{-1}$ act on both sides):

$hg^{-1}(s) = g^{-1}(s)$, that is: $h(g^{-1}(s)) = g^{-1}(s)$. So $g^{-1}(s)$ must be an element of $S$ that $H = G_a$ stabilizes. We only know of one such element, $a$ (this is the only bit that requires a little cleverness).

So if we set $g^{-1}(s) = a$, we know this will work. Now we can solve for $s$, by having $g$ act on both sides:

$g(g^{-1}(s)) = g(a)$
$(gg^{-1})(s) = g(a)$
$e(s) = g(a)$
$s = g(a)$.

Clearly, then, $gG_ag^{-1}$ stablizes $g(a)$. This shows that $gG_ag^{-1} \subseteq G_{g(a)}$.

I'll repeat this finding, for emphasis:

CONJUAGTES $ghg^{-1}$ of stabilizers $h$ of $a\in S$, stabilize the IMAGE $g(a)$.

To show $G_{g(a)} \subseteq gG_ag^{-1}$, we just work backwards:

Suppose $x \in G$ is such that $x(g(a)) = g(a)$.

Then $xg(a) = g(a)$, so $g^{-1}xg(a) = g^{-1}g(a) = a$. That is, $g^{-1}xg$ stabilizes $a$.

From $g^{-1}xg \in G_a$, we have $x = exe = (gg^{-1})x(gg^{-1}) = g(g^{-1}xg)g^{-1} \in gG_ag^{-1}$, and that's the end of this short television episode.