We know that $n!$ permutations on a set of $n$ elements is a group. A proof of this statement starts by considering a set $S=\lbrace a_1, a_2, ..., a_n\rbrace$ and $S'=\lbrace p:~p~\text{is a permutation on $n$ elements of $S$}\rbrace$. Now, our goal is to show that $S'$ is a group with respect to multiplication of permutations. It is easy to show that $(S', \cdot)$ is closed and associative. We are left to show how identity exists in $S'$? I encountered the following proof, but still confused: Let $I_a=\begin{pmatrix} & a_1 & a_2 &....a_n \\ & a_1 & a_2 &....a_n \end{pmatrix} \quad$, $I_b=\begin{pmatrix} & b_1 & b_2 &....b_n \\ & b_1 & b_2 &....b_n \end{pmatrix} \quad$ and $\alpha=\begin{pmatrix} & a_1 & a_2 &....a_n \\ & b_1 & b_2 &....b_n \end{pmatrix} \quad$; all these three elements belong to $S'$ and $I_a=I_b$ (since $(a_1~a_2...a_n)=(b_1~b_2...b_n)$). Now, $I_a\cdot \alpha=\begin{pmatrix} & a_1 & a_2 &....a_n \\ & a_1 & a_2 &....a_n \end{pmatrix}\begin{pmatrix} & a_1 & a_2 &....a_n \\ & b_1 & b_2 &....b_n \end{pmatrix} \quad=\alpha.$ Similarly, $\alpha\cdot I_b=\begin{pmatrix} & a_1 & a_2 &....a_n \\ & b_1 & b_2 &....b_n \end{pmatrix}\begin{pmatrix} & b_1 & b_2 &....b_n \\ & b_1 & b_2 &....b_n \end{pmatrix} \quad=\alpha.$
Therefore, $I\alpha=\alpha I=\alpha?$ My confusion is why do we take here only, $\alpha\cdot I_b$, but not $I_b\cdot \alpha$ (the product seems not compatible in this case) or, $I_a\cdot \alpha$, but not $\alpha\cdot I_a $ (the product seems not compatible in this case). Can some one help me to understand this better or, any simpler proof of this?