Understanding correspondence between $RG$-modules and representations

Question

I'm reading part of a book that is talking about how there is a correspondence between representations of a group $G$ and $RG$-modules.

It says if $V$ is an $RG$-module , then the map $\rho :G \rightarrow \text{GL}(V)$ where $\rho(g) : v \mapsto gv $ is a representation since the map $\rho(g)$ is a module homomorphism for all $g \in G$. But this means that for any $r \in R$ we have $\rho(g)(rv)=g(rv)$ and $r \rho(g)(v)=r(gv)$ but why do we have that $g(rv)=r(gv)$? Since $V$ is an $RG$-module; in definitions I am accustomed to of what a module is there is no requirement for 'scalar multiplication' to be respected like this - in fact I thought that $V$ didn't have to be a module at all and just an abelian group. In these contexts (representation theory) do we just assume this property?

Answer 1

The group ring $RG$ consists of formal sums $\sum_{g \in G} r_g g$ where the coefficients $r_g \in R$ are elements of $R$ and all but finitely many of them are zero. Multiplication is defined by $$\left(\sum_{h \in G} r_h h \right) \left(\sum_{k \in G} s_k k \right)=\sum_{g \in G} t_g g$$ where $$t_g=\sum_{\substack{h, k \in G \\ hk=g}} r_h t_k.$$ Addition is defined by adding coefficients. You should check that this satisfies the axioms for a ring.

There is an injective ring homomorphism of $R$ into $RG$ defined by sending $r \in R$ to $r 1_G$, where $1_G \in G$ is the identity of $G$. It is frequently helpful (and more attractive) to abuse notation by writing simply $r$ for the image $r1_G$ of $r$ by this injection. Likewise, there is an injection of $G$ into $RG$ sending an element $g \in G$ to $1_R g$ where $1_R$ is the multiplicative identity of $R$, and for the same reasons of aesthetics we may sometimes abuse notation by writing simply $g$ for $1_R g$. One might note that the images of $1_R$ and $1_G$ by these injections are both equal to the multiplicative identity $1=1_R 1_G$ of $RG$, for which reason I, at least, always abuse notation by writing simply $1$ for all three objects.

The definition of multiplication in $RG$ implies $$(r 1_G)(1_R g)=r 1_R g=1_R r g=(1_R g) (r 1_G) \quad \hbox{for all $r \in R$ and $g \in G$,}$$ which I am sure you will agree looks much nicer when written simply as $$rg=gr.$$

Now, if $V$ is an $RG$-module, then via the injection $R \hookrightarrow RG$ it is an $R$-module: this follows from the general fact that given a homomorphism of rings $\phi:A \to B$ and a $B$-module $M$, one obtains an $A$-module structure on $M$ by defining $$am=\phi(a)m.$$ For $g \in G$ I will write, as you do, $\rho(g)$ for the map $v \mapsto gv$ from $V$ to $V$.

Finally, working in the ring $RG$, as noted above the products $rg$ and $gr$ are equal for all $g \in G$ and $r \in R$ (this follows from the definition of the ring structure). Therefore using the axioms for modules,

$$r\rho(g)(v)=r(gv)=(rg)v=(gr)v=g(rv)=\rho(g)(rv).$$

Answer 2

$G$ becomes represented by the $R$ module $V$ when you find a group homomorphism $G\to GL(V)$, which consists just of $R$ linear automorphisms of $V$. Quite often this is done with $R$ being a field like $\mathbb C$.

Now, $R[G]$ has the special property (a universal property, in fact) that for any ring $S$ with units $U(S)$, any any group homomorphism $G\to U(S)$, there is a unique ring homomorphism $R[G]\to S$ extending what the group homomorphism does.

Now what this says when $S=\mathrm{End}(V_R)$ is that a representation of $G$ by $V_R$ automatically gives a unique ring homomorphism $R[G]\to \mathrm{End}(V)$, but that is just another name for an $R[G]$ module structure on $V$.

Conversely if you take some ring homomorphism $R[G]\to \mathrm{End}(V_R)$ you know it is going to send units (like those in $G$) to units in $\mathrm{End}(V_R)$ (which are $GL(V_R)$) so by just restricting that ring homomorphism to $G$, you get a group representation of $G$ on $V_R$.

These two directions explain the connection between thinking about a representation as a group homomorphism versus as an $R[G]$ module action.