This is problem 1.45 of Rotman's Introduction to theory of groups.
Let $k$ be a field and denote the columns of the $n \times n$ identity matrix $E$ by $\epsilon_1,...,\epsilon_n$. A permutation matrix $P$ is obtained from E by permuting its columns, that is, the columns of $P$ are $\epsilon_{\alpha 1},..., \epsilon_{\alpha n}$ for some $\alpha \in S_n$. Prove that the set of all permutation matrices over $k$ is a group isomorphic to $S_n$.
So I called this set $\Bbb{P}$ and verified it is a where the inverse of every element $P$ is its transpose matrix $P^t$. However when trying to verify that $\Bbb{P} \cong S_n$ I got stuck. I tried by defining $f:S_n \to \Bbb{P} $ as
1)$f(\alpha)=P_{\alpha}$ where $P_{\alpha}$ is the matrix who´s columns are exactly $\epsilon_{\alpha 1},..., \epsilon_{\alpha n}$.
But $f$ defined in this manner is not a homomorphism.
2)$f(\alpha)=P^{t}_{\alpha}$ the same problem
3)$f(\alpha)=P_{\alpha^{-1}}$ the same problem.
So this guesses were in such a way randomly selected. Is there any way I can criteriously pursue the desired morphism?
Let $P$ denote the given set of permutation matrices. We have a map $f:P \to S_n$ given by $\pi e_i = e_{f(\pi)i}$, where $e_1, \dots, e_n$ is the standard basis of $k^n$. Clearly $f$ is a bijection, and it's immediate that $f(\pi \pi') = f(\pi)f(\pi')$. It follows that $P$ is a group, and the map $f$ is an isomorphism. (If it isn't clear, note that the group structure on $P$ follows from the fact that $\pi\pi' = f^{-1}(f(\pi)f(\pi'))$. In particular, inverses $\pi^{-1} = f(\pi^{-1})$ exist for all $\pi$.)