Q:

Homomorphism in the group of permutations of a set $X$

Let $G$ be a group acting on a set $X$. Let $g\in G$. A map $\theta_{g}:X \rightarrow X$ is defined by setting $\theta_{g}(x)=g\cdot x $ for all $x\in X$.
Let $S_x{}$ be the group of permutations of $X$
Show that the map $f:G \rightarrow S_{x}$ given by $f(g)=\theta_{g}$ is a homomorphism.
I know if $f(g_{1}g_{2})=f(g_{1})f(g_{2})$ so $\theta_{g_{1}g_{2}}=\theta_{g_1}\theta_{g_2}$ then can I say $(g_1g_2) \cdot x=(g_1)(g_2)\cdot x$? I feel like I'm missing something here.

A:

I'm not sure if you are missing something or perhaps it was too trivial for you to write it down explicitely. :D
Anyway what is the group operation on $S_X$? It is the function composition. So in order to show that $f$ is a group homomorphism you have to show that $\theta_{g_1g_2}=\theta_{g_1}\circ\theta_{g_2}$ where $\circ$ is the function composition. You show equality of functions by showing that they are equal on every element. So pick $x\in X$. Then
$$\theta_{g_1g_2}(x)=(g_1g_2)\cdot x=g_1\cdot(g_2\cdot x)=\theta_{g_1}(g_2\cdot x)=\theta_{g_1}(\theta_{g_2}(x))=(\theta_{g_1}\circ\theta_{g_2})(x)$$
So first equality by definition. Second since the action of $G$ on $X$ is associative. The rest of equalities are by definition.
Since $x$ was chosen arbitrarly, then $\theta_{g_1g_2}=\theta_{g_1}\circ\theta_{g_2}$. $\Box$
BTW "Permutations" is not necessarly a good word if you are dealing with possibly infinite set $X$ (most commonly a permutation is a bijective function that "moves" only finitely many elements). Just "bijections" will do.

