Show that $G_{s}$ is a normal subgroup of $G$ Definition: $G_{s}:=\{g \in G: g.s=s\}$
My attempt is the following:
We take $g \in G$, and we consider this two sets:
$$gG_{s}:=\{gh:h\in G_{s}  \}$$
$$G_{s}g:=\{hg :h\in G_{s}\}$$
and we will prove that $gG_{s}=G_{s}g$, then let $x\in gG_{s}$, so $x=gh$ for some $h \in G_{s}$ ; given that $g \in G$ then $g^{-1}x=h$.
But then the things are not good because this equality does not help very much to show that $x=hg$ there has to be a better way to do this can you help me to prove this please? thank you :)
 A: If $g \in G_s$, we want to show that $hgh^{-1} \in G_s$ for any $h \in G$.
$$hgh^{-1}s=hg(h^{-1}s)=h(h^{-1}s)=s.$$

Edit for clarification: the statement is true if $G_s$ is defined to be the subgroup of group elements $g$ such that $g.s=s$ for any $s \in S$, by the above argument.
However, the notation $G_s$ is suggestive of another subgroup known as the stabilizer of $s$; it is the set of group elements $g$ such that $g.s=s$ for this particular element $s$. In general, this subgroup is not normal (see Hagen von Eitzen's answer). It would be best if you made sure which definition of $G_s$ you really want.
A: Let $g\in G_s$ and $h\in G$ then we want to show that $$hgh^{-1}\in G_s$$
in other words $hgh^{-1}(s)=s$ for any $s$, but $g\in G_s$ so $gh^{-1}=h^{-1}$ therefore
$$hgh^{-1}(s)=h(gh^{-1})(s)=(hh^{-1})(s)=1\cdot s=s$$
A: Your claim is simply not correct in general.
Let $G=S_5$ acting by definition on $\{1,2,3,4,5\}$ and let $s=5$. Then $G_s\approx S_4$ which is not normal in $S_5$. (Works with $S_3$ and $S_2$ as well).
More generaly: If $h\notin G_s$ and $hs=t$, say, with $t\ne s$. Then $hG_sh^{-1}=G_t$ and for most group actions this is $\ne G_s$.
