Fixing Subgroup

Let $S\subseteq X$ be any subset. Then we can define the fixing subgroup of $S$:

$$\text{Fix}(S)=\{g\in G|g\cdot x=x\mathrm{\forall }x\in S\}$$

Which is equivalent to:

$$=\bigcap _{x\in S}\{g\in G|g\cdot x=x\}=\bigcap _{x\in S}{G}_x$$