# Prove that the isotropy group of G is a subgroup of G

This is one of those left-to-the-reader proofs. Since $es=s$, $e \in G_s$ so $G_s$ is not empty. Let $x,y \in G_s$, so $x$ fixes $s$ and $y$ fixes $s$, therefore $(xs)(ys)^{-1}=(xs)(s^{-1}y^{-1})=(xy^{-1})=ss^{-1}$ so $xy^{-1}s=s$ which implies $xy^{-1} \in G_s$. Now when I tried doing it without using the subgroup criterion I got stuck. Let $x,y \in G_s$, wts that $xy \in G_s$, then $(xy)s=s$. When I multiplied I got $xsy=s$, is there anything about group actions that will allow me to go from $xsy=s$ to $(xy)s=s$?

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Your proof is incorrect, because there is no such thing as $(ys)^{-1}$. The object $ys$ is in the set being acted on (not the group $G$), so it does not have an inverse. –  Ted Feb 11 '12 at 19:57

To do it without using Subgroup Criterion:

Let $G$ be a group. $S$ be a set. It can be empty if you wish. But interesting things happen when $S$ is non-empty!

Note that, if $G$ acts on a set, $S$, then the map $\cdot :G \times S \to S$ satisfies the following conditions:

• $e_G \cdot s=s$ for all $s \in S$ where $e_G$ is the identity in $G$
• $(xy) \cdot t=x \cdot (y \cdot s)$

Do you now see the proof?

1. As is easy to observe, the isotropy subgroup or the stabilizer of any element $s \in S$, denoted by $G_s$ is non-empty.

2. Now, let $x,y \in G_s$, you need to prove that $xy \in G_s$. Now, $(xy) \cdot s=x \cdot (y \cdot s)=x \cdot s=s$. This proves what we need to prove!

3. To prove the existence of inverse, here's how you'll have to go about:

\begin{align*}\text{Let x \in G_s. Now for its inverse, note:} x \cdot s&=s\\ x^{-1} \cdot (x.s)&=x^{-1} \cdot s \\ (xx^{-1})\cdot s&=x^{-1} \cdot s\\ \implies x^{-1} \cdot s &= s\end{align*}

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Even with the subgroup criterion I think that there is something amiss. A left operation of a group $G$ on a set $S$ is a map $G \times S \to S$ satisfying certain conditions. There is no assumption that $S$ has a multiplication or an inversion, and while it is true that one can define a corresponding right action by setting $sx = x^{-1}s$, that does not seem to be helpful here.

Let's try to prove that $x, y \in G_s$ implies $xy^{-1} \in G_s$, i.e. that $(xy^{-1})s = s$. From the axioms we know that $(xy^{-1})s = x(y^{-1}s)$. To find $y^{-1}s$, note that $ys = s$ and so $y^{-1}s = y^{-1}(ys)$. Do you see where to go from here?

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Yes, I do. Now I can see where I went wrong. Thank you. –  user2467 Feb 11 '12 at 19:26