Q:

$G$-graded vector space and module

Let $G$ be a finite group and let $k(G)$ be the set of functions on $G$ with values in a field in $k$.
I am reading a proof of the following fact: a $k(G)$ module is a $G$-graded vector space $V$.
The proof starts with writing the action as $\phi\cdot v=\sum_{u\in G} \phi(u) \beta_u(v)$, where $\phi \in k(G)$ and $\beta_u(v)$ are vectors in $V$.
But I don't know why we can write the action like this. How do I find such $\beta$?
I appreciate any help. 

A:

For $g \in G$ let $e_g \in k(G)$ be the function defined by $e_g(h) = 1$ if $h = g$ and $e_g(h) = 0$ if $h \neq g$. Then we can write $\phi = \sum_{u \in G} \phi(u) e_u$ for any $\phi \in k(G)$ (just evaluate at $h \in G$ to see that both sides give $\phi(h)$). So, if $V$ is a $k(G)$-module and $\phi \in k(G)$, then 
$$ \phi \cdot v = \sum_{u \in G} \phi(u) (e_u \cdot v)$$
for all $v \in V$ and we can set $\beta_u(v) = e_u \cdot v$.

