Kernel of $\phi:G \rightarrow \operatorname{Sym}(S)$ Group actions

Question

$\operatorname{Sym}(S) == \text{All permutations of the set }S$.

Prove $\ker(\phi)=\bigcap_{x\in S}G_x$ where $G_x$ is the stabilizer of $x$. Let $$\phi(a) =\lambda_a(x)=ax \text{ where } x\in S $$

I'm having trouble with this problem, especially because I'm new to group actions. My attempt: $$\ker(\phi)=\{b\in G\mid \phi(b)=(1)\}=\{b\in G\mid bx=(1)\}$$

$$\bigcap_{x\in S}G_x=\{ a\in G\mid\forall x \in S ,ax=x\}$$

$\subseteq$

$\text{Let } k \in \ker(\phi)\rightarrow kx=(1) \rightarrow \cdots \text{ somehow }\cdots kx=x\rightarrow k\in \bigcap_{x\in S}G_x$

$\supseteq$

Let $b\in \bigcap_{x\in S}G_x \rightarrow bx=x \rightarrow \cdots \rightarrow bx=(1)\rightarrow b \in \ker(\phi) $

I am totally loss, can I get some hints?

Answer 1

Note that $\phi:G\to{\rm Sym}\, S$ sends elements of $G$ to functions, more specifically it sends elements of $G$ to permutations of $S$. For each $g\in G$, thus, $\phi(g)$ is a permutation of $S$, and $G$ acts on $S$ by $g\cdot s=\phi(g)s$.

Now $g\in\ker\phi$ iff $\phi(g)={\rm id}$. This happens iff $g\cdot s=\phi(g)s={\rm id}s=s$ for any $s$, and this happens iff $g\in {\rm stab}(s)$ for any choice of $s$, so $g\in\bigcap\limits_{s\in S}{\rm stab}(s)$ iff $g\in \ker \phi$.

Answer 2

From what you have asked, I take it that $a \in G$ and $x \in S$. Recall that the stabilizer of $x$ in $G$ is given by $G_x = \{a \in G: ax= x\}$. But then this means that $$\ker{\phi}= \{a \in G: ax= x \textit{ }\forall x \in S\}= \cap_{x \in S} G_x,$$ as needed.