I am trying to understand the equivalence between group representations, $(V, \rho)$, and left modules over the group ring $F[G]$.
Can you explain explicitly why it is the same?
My progress:
Consider a group $G$. If $(V, \rho)$ is a representation of $G$, we can take $V$ to be a left module over $F[G]$ by defining: $gv = \rho(g)v$. So given a representation we can get a left module over $F[G]$.
The other direction is more confusing. Given a left module $M$ over $F[G]$, what is the vector space? Is $M$ necessarily a vector space? (The concept of a module is rather new to me).
Best Answer
I always think this stuff is done incredibly formally, as the other answers suggest. Basically the idea is that we are lazy and want to ignore the rho!
Ok so we have a representation $\rho: G \longrightarrow GL(V)$. So each $\rho(g)$ is a linear map on $V$.
So instead of writing $\rho(g)v$ as the linear map applied to $v\in V$, why not just write $gv$, and declare this as a "linear action" of $G$ on $V$ (of course there are certain axioms etc that follow but the upshot is that we get something like a "module over a group", a $G$-module. You can add in scalars to get a ring $F[G]$ and in this way you get an actual module over a ring, a $F[G]$-module).
This is ALL that is going on here!
(So to go the opposite way we just define $\rho(g)$ to be the linear map $v\mapsto gv$, and then we get a representation of $G$.)