For a group-algebra $k[G]$ ($G$ finite), why is a $k[G]$-module the same as a $k$-representation of $G$? I'm reading the Atiyah-MacDonald book on Commutative Algebra. 
At the beginning of the module chapter on page 17, they make an example which I don't understand. Example 5) is:

$G$ = finite group, $A = k[G] =$ group-algebra of $G$ over the field $k$ (thus $A$ is not commutative, unless $G$ is). Then $A$-module $= k$-representation of $G$.

I'm far away from understanding how $k[G]$-module = $k$-representation of $G$. I try to unwind the definitions as follows:
After reading the Wikipedia articles on monoid rings and group rings, I come to the conclusion that a group algebra is a triple $(k[G], +, \star)$, where
\begin{align}
k[G] = \{ \phi: G \rightarrow R \mid \{g \in G \mid \phi(g) \neq 0 \}\ \text{finite} \}, \tag{1}
\end{align}
and where + is pointwise adition and $\star$ is the convolution
\begin{align}
(\phi \star \psi)(g) = \sum\limits_{kl = g} \phi(k) \psi(l).
\end{align}
According to the definition of a module on the same page (17) in Atiyah-MacDonald, a $k[G]$-module is a pair $(M, \mu)$, where $M$ is an (additive) Abelian group and $\mu$ is a map
\begin{align}
\begin{array}{rccl}
\mu: &k[G] \times M &\rightarrow &M \\
&(\phi, x) &\mapsto &\phi x := \mu(\phi, x)
\end{array}
\end{align}
such that 
\begin{align}
\phi(x + y) &= \phi x + \phi y, \\
(\phi + \psi) x &= \phi x + \psi x, \\
(\phi \psi) x &= \phi(\psi x), \\
1 x &= x.
\end{align}
Question 1: How is specifying a $k[G]$-module equivalent to specifying a $k$-representation of $G$?
Question 2: Why is finiteness of $G$ needed? Does it have something to do with the finiteness in definition (1) of $k[G]$?
 A: This equivalence is a very formal matter, and the finiteness of $G$ (nor any properties of the field $k$, nor that fact that $G$ is a group rather than just a monoid) does not play a role in it. The basic information is given in both cases by a a $k$-vector space $V$ equipped with a "multiplication" map $m:G\times V\to V$ that must satisfy


*

*associativity with respect to the first argument: $m(g_1,m(g_2,v))=m(g_1g_2,v)$ for all $g_1,g_2\in G$ and $v\in V$

*$k$-linearity with respect to the second argument: $m(g,\lambda v_1+\mu v_2)=\lambda m(g,v_1)+\mu m(g,v_2)$ for all $g\in G$, $v_1,v_2\in V$ and $\lambda,\mu\in k$.
If in this description you fix individual group elements $g$, then the second point says that $v\mapsto m(g,v)$ is a $k$-linear map $\rho_g:V\to V$, and then the first condition says that $\def\End{\operatorname{End}}\rho:G\to\End_k(V)$ is a monoid homomorphism, and by restriction of codomain a group homomorphism $G\to GL_k(V)$ if $G$ is a group, in other words this precisely describes a $k$-representation of $G$.
If one the other hand you start with $V$ as Abelian group, then you've got two additional structures: the $k$-vector space structure that defines scalar multiplications by $\lambda\in k$, and the multiplications $\rho_g$, which are compatible with the vector space structure (they are $k$-linear) and define a monoid action (the first point above). These are precisely the conditions needed to be able to define multiplication by formal $k$-linear combinations of elements of $G$ by the rule
$$
  \mu(\sum_i \lambda_i g_i), v)=\sum_i \lambda_i m(g,v)
$$
whenever the summation is finite (infinite linear combinations make no sense, unless all but finitely many coefficients are zero), and so that $\mu$ satisfies $\mu(\phi\psi,v)=\mu(\phi,\mu(\psi,v))$. Here multiplication of formal linear combinations of group elements is defined by extending the group multiplication by bilinearity:
$$
  \Bigl(\sum_i \lambda_i g_i\Bigr)\Bigl(\sum_j \kappa_j g_j\Bigr)=\sum_{i,j}\lambda_i\kappa_j g_ig_j
$$
But the collection of formal linear combinations with that multiplication (and the obvious $k$-vector space structure) is just the algebra $k[G]$, and we've just required that $\mu$ define a $k[G]$ module structure (the $k$-vector space axioms for $V$ are just the set of these requirements limited to scalars in the subalgebra $k=k[e]$ of $k[G]$.)
As you can see, requiring only finitely many elements of $G$ to have nonzero coefficient in an element of $k[G]$ is just something needed to make the formal expressions used have sense, and in particular to have a well defined multiplication on $k[G]$. They do not restrict $G$ itself in any way; indeed they are precisely there to be able to do this construction equally well when $G$ is infinite as when it is finite.
