We have defined the regular expression $R$ as an $n$-dim expression of a group $G$ with order $n$ with:
\begin{align}R: G &\rightarrow GL_n(\mathcal{C}) \\ g &\rightarrow R(g) \phantom{aaaaaa}\text{with}\\ R_{kj}(g_i) &= \left\{ \begin{matrix} 1 \phantom{a}\text{if } g_kg_j^{-1} = g_i \\ 0 \phantom{aa111aaq}\text{else} \end{matrix}\right. \end{align}
I am struggling with proofing that this is indeed a representation, i.e that $(R(g_i)R(g_j) = R(g_ig_j) \phantom{a}\forall g_i, g_j\in G$ . Calculating $R(g_i)$ for example for $S_3$ all the relations hold true, as expected, but I really don't see why this holds true in general. I see, that all $R(g_i)$ (at least for $S_3$) are permutation matrices, which makes perfectly sense, but again, I am struggling proofing this from the above formula.
Edit: there are a lot of indices in need, so I'm denoting the order of the group $G$ by $N$, sorry!
Working out the product at any entry, we see for $g_m,g_n\in G$ and $1\leq i,j\leq N$ that $$ [R(g_m)R(g_n)]_{ij} = \sum_{\lambda=0}^nR_{i\lambda}(g_m)R_{\lambda j}(g_n) $$ By definition, $R_{i\lambda}(g_m)=1$ iff $g_ig_\lambda^{-1}=g_m$ and likewise $R_{\lambda j}(g_n)=1$ iff $g_\lambda g_j^{-1}=g_n$. These both have to be satisfied for the summand to be nonzero. These equations imply $g_m^{-1}g_i=g_\lambda=g_ng_j$ (which also shows that at most one summand can be nonzero).
Therefore, $[R(g_m)R(g_n)]_{ij}=1$ iff $g_m^{-1}g_i=g_ng_j$. Rewriting this equation, this is iff $g_ig_j^{-1}=g_mg_n$. This is exactly the condition for which $R_{ij}(g_mg_n)=1$. Therefore, $R(g_m)R(g_n)=R(g_mg_n)$, as desired.