Prove that σ ⋅ ai = aσ(i) for all σ $\in S_n$ and for all $a_i \in X$ is an action of $S_n$ on the set $X$. Let $X = \{a_1,a_2,...,a_n\}$ be a set with $n$ elements. Prove that $\sigma \cdot a_i = a_{\sigma(i)}$ for all $σ \in S_n$ and for all $a_i \in X$ is an action of $S_n$ on the set $X$.
Having some trouble setting it up. Thanks in advance.
 A: We have to prove that we have a homomorphism from $S_n$ to $\DeclareMathOperator{\Sym}{Sym}\Sym X$, the symmetric group on $X$.  
Define $\phi:S_n\to\Sym X$ by $\phi(\sigma)(a_i)=\sigma\cdot a_i=a_{\sigma(i)}$.  
Take $\sigma_1,\sigma_2\in S_n, a_i\in X$.  Then $\phi(\sigma_1\sigma_2)(a_i)=\sigma_1\sigma_2\cdot a_i=a_{\sigma_1\sigma_2(i)}=\sigma_1\cdot a_{\sigma_2(i)}=\phi(\sigma_1)(a_{\sigma_2(i)})=\phi(\sigma_1)(\sigma_2\cdot a_i)=\phi(\sigma_1)\phi(\sigma_2)(a_i)$.
Thus $\phi(\sigma_1\sigma_2)=\phi(\sigma_1)\phi(\sigma_2)$.
A: Firstly, $S_n$ is a group. Secondly, $a_{\sigma(i)}\in X$, because $\sigma(i)\in \{1,\dots,n\}$. Therefore, $(\sigma,a_i)\mapsto \sigma\cdot a_i:=a_{\sigma(i)}$ defines indeed a map $\mathcal{A}\colon S_n\times X \to X$, and we are in the right setting to ask ourselves whether $\mathcal{A}$ is an action, namely if it fulfils action's axioms. Let's see:


*

*$\iota\cdot a_i=a_{\iota(i)}=a_i, \space\forall a_i\in X$

*$(\sigma\tau)\cdot a_i=a_{(\sigma\tau)(i)}=a_{\sigma(\tau(i))}=\sigma\cdot a_{\tau(i)}=\sigma\cdot(\tau\cdot a_i), \space\forall \sigma,\tau\in S_n, \forall a_i\in X$
So, yes, that's an action.
