Group Actions: Verify a Bijective Correspondence This is an old exam problem: Given an action of $G$ on $X$, we can define $\varphi: G \to S_X$ by the rule $\varphi(g) = \sigma_g$, where $\sigma_g$ is left multiplication by $g \in G$. Prove that this establishes a bijective correspondence between the set of all actions of $G$ on $X$ and the set of all homomorphisms from $G$ to $S_X$.
Is this question simply asking to verify that $\varphi$ is a bijection? If so, I'm at a little of a loss as to how to verify injectivity. Any advice?
 A: Let $\varphi:G\rightarrow S_X$ be a homomorphism. Then we can define a group action of $G$ on $X$ by $g.x:=\varphi(g)(x)$. Conversely, given a group action of $G$ on $X$, we can define a homomorphism $\psi:G\rightarrow S_X$ by $\psi(g)(x):=g.x$ (technically, it should be checked that these define group actions and homomorphisms but this isn't a hard check and since our definition is fairly natural it's most likely a trivial check). We simply need to verify this is a bijective correspondence. Let's call the map from morphisms to actions $\mathscr{A}$ and the map from actions to morphisms $\mathscr{M}$.
One way to verify this is a bijective correspondence is by showing they are inverse to one another. So take $\varphi:G\rightarrow S_X$ a homomorphism. Then $\mathscr{M}(\mathscr{A}(\varphi(g)))=\mathscr{M}(g.x)=\psi(g)$. We'll check that $\varphi(g)=\psi(g)$ as a permutation on $X$ for every $g\in G$. But this is true by definition as $\psi(g)=g.x=\varphi(g)(x)$ so that they are equal as permutations on $X$ and therefore as elements of $S_X$. This shows $\phi=\psi$ as a homomorphism.
The other direction is similar.
A: I think the equations asks you to prove that given an action of $G$ on $X$ ,$g\star x $, then we can find a homomorphism from $G \to S_{X}$ (the homomorphism you gave works fine!). and given a homomorphism from $G$ to $S_{X}$, you can find an action of $G$ on $X$. 
