Normal Representation

Consider some group $G$ or order $n$. To build the normal representation,

1. Associate each permutation to an $n$ dimension vector. Each transformation/object is represented by a vector with a single non-null component $$ G = \{ e, g_{1}, g_{2}, \dots, g_{n-1} \} $$ $$ e \leftrightarrow v_{0} = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \qquad g_{1} \leftrightarrow v_{1} = \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix} \qquad g_{n-1} \leftrightarrow v_{n-1} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} $$
2. Determine the matrices for each element of $G$ which reproduces the transformation using the associated vectors. $$ g_{1}g_{2} = e \to D(g_{1}) v_{2} = v_{0} $$ The resulting representation is not necessarily an irrep, but it is a representation nonetheless.

$$ \begin{array}{c|ccc} \mathbb{Z}_{2} & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \end{array} $$$$ e = 0 \leftrightarrow v_{0} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \qquad a = 1 \leftrightarrow v_{1} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \qquad b = 2 \leftrightarrow v_{2} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} $$$$ D(e) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$$$ ae = 1 \to D(a)v_{0} = v_{1} \qquad aa = b \to D(a)v_{1} = v_{2} \qquad ab = e \to D(a)v_{2} = v_{0} $$$$ D(a) = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} $$$$ be = b \to D(b)v_{0} = v_{2} \qquad ba = e \to D(b)v_{1} = v_{0} \qquad bb = a \to D(b)v_{2} = v_{1} $$$$ D(b) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{pmatrix} $$

And so, the normal representation of $\mathbb{Z}_{4}$ is constructed.