Let $G$ be a finite group and let $V = \{f : G → C\}$ be the vector space
of all complex valued functions on G. Define a representation of G on V by
$(ρ_gf)(h) = f(g^{-1}h), h ∈ G.$
Prove that (V, ρ) is isomorphic to the left regular representation $\mathbb{C}$G.
My attempt (not working):
If two representations are isomorphic then we need to construct an intertwining operator between them. Let $T: \mathbb{C}G → V$ be $T(e_x)=f(x), x \in G$, where $\{e_x\}_{x\in G}$ denotes the basis of $\mathbb{C}$G. However this is just one of the many examples of T that I have tried, all failing to work give the required identity:
$T \circ \rho_g^L(e_x)=\rho_g\circ T(e_x) $.
Any help would be appreciated in particular understanding the construction of these operators.
Best Answer
The left regular representation is a map $\rho^L:G\to GL(\mathbb{C}G)$, defined on the basis $\{e_x\mid x\in G\}\subset\mathbb{C}G$ by $$\rho^L(g)e_x=e_{gx}.$$
On the other hand, there is a basis of coordinate functions $\{\delta_x:G\to\mathbb{C}\mid x\in G\}\subset V$, where $$ \delta_x(h)=\begin{cases}1&\mbox{if }x=h\\0&\mbox{otherwise.}\end{cases} $$ Now, since $$ (g.\delta_x)(h)=\delta_x(g^{-1}h)=\begin{cases}1&\mbox{if }x=g^{-1}h\\0&\mbox{otherwise.}\end{cases} $$ we deduce that $g.\delta_x=\delta_{gx}$ (since $x=g^{-1}h\iff gx=h$). Therfore, we obtain a representation $\rho:G\to GL(V)$ given by $\rho(g)\delta_x=\delta_{gx}$.
Let $T:\mathbb{C}G\to V$ be the linear map $T(e_x)=\delta_x$. Then, $T$ is the intertwiner you are looking for.