Verify regular representation? Let $G$ be a finite group and let $V$ be the vector space of functions from $G$
to $\mathbb{C}$. For $g \in G$ and $f \in V$, let $R(g)(f)$ be the function
$$(R(g)f)(x) = f(xg^{-1}).$$
How can I show that this defines a representation of $G$ on $V$?
 A: It was already mentioned in the comments above that you "just need to check the axioms". Also, as mentioned in the comments above, your definition seems to be a bit off. I will give you a hint on how to get started by going with your definition. That way you can see what the problem is. You should be able to fix this.
Recall that a representation of a group is a homomorphism (of groups) $\phi: G \to GL(V)$ where $V$ is a vector space. You are considering the finite group $G$ and $V = \{f: G \to \mathbb{C}\}$ is the vector space. If you have a vector space, then you should already know that $GL(V)$ (the set of linear invertible transformations from $V$ to $V$) is a group under function composition.
You have a map 
$$
R: G \to GL(V)
$$
given by
$$
R(g)(f)(x) = f(xg^{-1}).
$$
You need to check a couple of things now. First, you need to check that for each $g$, $R(g)$ actually is a map in $GL(v)$. So, is it linear and invertible? This is not hard, and I leave you with the details.
Now, you have to check that $R$ is a homomorphism. That is, you need to show that $R(gh) = R(g)\circ R(h)$.
Ok, so 
$$
R(gh)(f)(x) = f(x(gh)^{-1}) = f(x\color{red}{h^{-1}g^{-1}})
$$
and
$$
[R(g)\circ R(h)](f)(x) = R(g)[R(h)(f)](x) = R(h)(f)(xg^{-1}) = f(x\color{red}{g^{-1}h^{-1}}).
$$
Now you see what the problem is. So you could try to correct this by changing the definition above. 
You could also try to see if $R(g)(f)(x) = f(g^{-1}x)$ might define a representation.
