Group acting on a set

Question

Let $A$ be a set, and let $G$ be any subgroup of $S_A$. $G$ is a group of permutations of $A$; we say it is a group acting on the set $A$. Assume here that $G$ is a finite group. If $u \in A$, the orbit of $u$ (with respect to $G$) is the set $$O(u)=\{g(u): g \in G\}.$$ Prove that $G_u = \{g \in G: g(u)=u\}$ is a subgroup of $G$.

Any advice? Thanks a lot.

Answer 1

Let $H$ be a group, whose operation is denoted by $\star$, and whose identity element is $e$.

By definition, a subset $K\subseteq H$ is a subgroup of $H$ when

• $e\in K$.
• For any $a,b\in K$, we also have $(a\star b)\in K$.
• For any $a\in K$, we also have $a^{-1}\in K$.

Prove that $G(u)$ satisfies these requirements. Remember, the group operation of your group $G$ is composition of permutations, so the identity element is the trivial permutation (i.e. the one that does nothing). Also, remember that (by the definition of a group action), for any $g_1,g_2\in G$ and any $u\in A$, we have that $g_1(g_2(a))=(g_1\circ g_2)(a)$.