I've been reading some notes on Representation Theory of finite groups and now I'm trying to answer the following question:
Let $G$ be a finite group, $F$ any field, $\tau: G \to S_{n} $ be a group homomorphism. Define $\sigma: G \to GL(n,F)$ by defining $\sigma(g)$ to be the matrix having $(i,j)$ entry $1$ if $i=(\tau(g))(j)$ and $0$ otherwise. Prove that $\sigma$ is a representation.
So as far as I understand we need to prove that $\sigma$ is a (group) homomorphism, i.e. that $\sigma(gh)=\sigma(g)\sigma(h)$, for some $g,h \in G$, where on the right hand side, in this case, the operation is matrix multiplication. We know that the matrix $\sigma(gh)$ has $(i,j)$ entry $1$ if $i=(\tau(gh))(j)=(\tau(g)\circ\tau(h))(j)$ and $0$ otherwise, but I don't seem to able to see how the homomorphism follows from this. Can someone help, please?
Let $(e_1,\ldots,e_n)$ be the canonical basis of $\mathbb{R}^n$. The matrix $\sigma(g)$ is the matrix, with respect to that basis, of the linear map $\Sigma(g)\colon\mathbb{R}^n\longrightarrow\mathbb{R}^n$ such that $\Sigma(g)(e_i)=e_{\tau(g)(i)}$ (that is, $\Sigma(g)(e_i)=e_j$ with $j=\tau(g)(i)$). Now, if you have $g,h\in G$, then\begin{align}\Sigma(gh)(e_i)&=e_{\tau(gh)(e_i)}\\&=e_{\tau(g)(\tau(h)(e_i))}\text{ (because $\tau$ is a homomorphism)}\\&=\Sigma(g)(e_{\tau(h)(e_i)})\\&=\Sigma(g)\bigl(\Sigma(h)(e_i)\bigr)\\&=(\Sigma(g)\circ\Sigma(h))(e_i).\end{align}This take place for each $i\in\{1,2,\ldots,n\}$ and therefore $\Sigma(gh)=\Sigma(g)\circ\Sigma(h)$.