Let $R$ be a ring. Then a left $R$- module is an additive Abelian group M along with an operation

$R\times M \to M$ such that it satisfies -

• $r(x+y)=rx+ry$
• $(r+s)x=rx+sx$
• $(rs)x=r(sx)$
• $1_R x=x$

where $x,y \in M$ and $r \in R$

Now my question is why do we require a module to be an abelian group?. Can't we turn a non-abelian group into a module over some ring? Say if we have $G=S_3\ \text{or}\ GL_2(\Bbb{Z})$ and $R$ be some ring (commutative/noncommutative)

• Rather than being an arbitrary group action, modules are a generalization of vector spaces, except working over an arbitrary ring rather than a field. Working with a commutative object makes it easier and more logical to carry over the idea of subspaces, linear algebra, etc. Jun 10, 2015 at 21:25
• @anomaly I think we ask for $M$ to be commutative because we don't have a choice. Indeed, even if we did not require commutativity, it would follow from the OP's axioms anyway. Jun 10, 2015 at 21:28
• @Pierre-GuyPlamondon: With the given axioms, sure. I could imagine defining a different sort of "non-commutative module", though, which might be interesting but would have a very different theory than usual modules. Jun 10, 2015 at 21:37

Commutativity of the operation of $M$ is forced from the other axioms. Indeed, if $x,y\in M$, then $$0 = 0_R(x+y) = (1_R-1_R)(x+y) = 1_R(x+y) - 1_R(x+y) = x+y-x-y.$$
Thus, adding $y$, then $x$ on the right, we get $y+x=x+y$.
• why $0_Rx=0_M$? Jun 10, 2015 at 21:30
• Because $0_Rx = (0_R+0_R)x = 0_Rx + 0_Rx$. Substracting one copy of $0_Rx$ on both sides, you get the equality. Jun 10, 2015 at 21:31
Another perspective is the following. If $A$ is an abelian group, then the endomorphism monoid $\text{End}(A)$ naturally has the structure of a ring: multiplication is composition, and addition is pointwise addition. One way to describe what a (left) $R$-module structure on $A$ is is that it is a ring homomorphism $R \to \text{End}(A)$. (Compare: one way to describe what a group action on a set is is that it is a group homomorphism $G \to \text{Aut}(X)$.)
By contrast, if $A$ is not necessarily abelian, then $\text{End}(A)$ only has the structure of a monoid: the problem is that the pointwise sum of two endomorphisms need not be an endomorphism, so there's no candidate for addition available.