Isomorphism $\text {Rep}_{G,k}\cong \space \text {Mod}_{k[G]} $ Let $G$ be a  group and $k$ a field.
Proposition: There is an isomorphism of categories $F:\text {Rep}_{G,k}\rightarrow \space \text {Mod}_{k[G]} $.
I begun by proving that there is a functor $\text {Grp} \rightarrow \text {Alg}_{k}$ by extending the inclusion assignment $G \mapsto k[G] $ by the universal mapping property with $f : k[G] \rightarrow A $ where $A$ is a $k$-algebra. Now I hope construct a chain $\text{Rep}_{k,G}\rightarrow \text {Grp} \rightarrow \text {Alg}_{k} \rightarrow \text {Mod}_{k[G]}$ such that the chain is an isomorphism of functors. I am stuck at this point. I can't get a grip on how to show an isomorphism for objects and morphisms. Should I try to construct the reverse chain and show it it the "inverse" or is there a more efficient way to approach the problem?
 A: Actually more is true: Let $k$ be a commutative ring (think of a field if you want to) and $M$ be a monoid (e.g. a group). There is an isomorphism of categories $\mathsf{Rep}_{M,k} \cong \mathsf{Mod}(k[M])$ over $\mathsf{Mod}(k)$. This means: If $V$ is some $k$-module, then giving an action of the monoid $M$ on $V$ is the same as giving a $k$-linear action of the $k$-algebra $k[M]$ on $V$ (and similar with homomorphisms, see below). The proof is an immediate consequence of the universal property of the monoid algebra $k[M]$: An action of $M$ on $V$ is a monoid homomorphism $M \to \mathrm{End}_k(V)$. This corresponds to a $k$-algebra homomorphism $k[M] \to \mathrm{End}_k(V)$. But this is precisely a $k$-linear action of the $k$-algebra $k[M]$ on $V$.
Explicitly, if $M$ acts on $V$, then $k[M]$ acts on $V$ via $\bigl(\sum_m \lambda_m \cdot m\bigr) \cdot v := \sum_m \lambda_m \cdot (m \cdot v).$
If $V,W$ are $k[M]$-modules, then a $k$-linear map $f : V \to W$ is $k[M]$-linear iff it is $M$-linear (one usually says $M$-equivariant) - this follows easily from the explicit description in the last paragraph, or one useses again the universal property of $k[M]$.
A: This chain is not going to lead to the answer, I guess.
Let $V$ be a linear representation of $G$ over $k$, i.e. $G$ acts on the $k$-vector field $V$. Then let $F(V)$ be the naturally generated $k[G]$-module structure on $V$.
This mapping naturally extends to morphisms, and it admits an inverse: for a $k[G]$-module $M$ one can define both the $k$-vector space structure and the $G$-action in an obvious way, yielding the representation $F^{-1}(M)$.
