# Motivation in considering abelian groups in modules

I am leaning about modules and I did not understand what is the motivation behind an $$R$$ module $$M$$ as an abelian group in the definition of the module.

What difference does it make if we consider $$M$$ as non abelian group? Isn't considering only abelian groups much more restrictive than considering an arbitrary group?

• I can see a close vote indicating that the question is missing context. It would be nice if close voters can explain in comment here so that I can learn not to make same thing in the future. Aug 10 at 12:21

Let $$G$$ be a group, $$R$$ a ring and assume there is an "$$R$$-module structure on $$G$$" (something satisfying the usual axioms except we don't require $$G$$ to be abelian). Then, let $$x,y\in G$$ be arbitrary and consider that $$2.(xy)=(2.x)(2.y)=(1.x)(1.x)(1.y)(1.y)=x^2y^2$$ on one hand and $$2.(xy)=(1.xy)(1.xy)=(1.x)(1.y)(1.x)(1.y)=xyxy$$ on the other hand (we use the distributive law for the product $$xy$$ and the sum $$2=1+1$$ in both orders). This implies $$xy=yx$$. Since $$x,y$$ were arbitrary, it follows that $$G$$ is abelian. Thus, there was no loss of generality in requiring $$G$$ to be abelian to begin with.
Here's a more conceptual explanation of what just happened. The remarkable property of an abelian group $$A$$ is that the pointwise sum of two homomorphisms is again a homomorphism, which implies that the endomorphisms of $$A$$ form a ring $$\mathrm{End}(A)$$. It's an exercise to check that an $$R$$-module structure on $$A$$ is the same thing as a ring homomorphism $$R\rightarrow\mathrm{End}(A)$$. From this perspective, it is only natural to require an abelian group in the definition of an $$R$$-module. Now, if $$G$$ is just a group, we still have a multiplicative monoid $$\mathrm{End}(G)$$ and an $$R$$-module structure on $$G$$ is the same thing as a multiplicative monoid homomorphism $$R\rightarrow\mathrm{End}(G)$$ such that the addition in $$R$$ gets transformed into the pointwise multiplication in $$\mathrm{End}(G)$$ (in particular, the pointwise multiplication of two endomorphisms in the image has to be an endomorphism again). The image of this map is then a submonoid of $$\mathrm{End}(G)$$ that is closed under pointwise multiplication and forms a ring with pointwise multiplication as addition. In particular, the pointwise multiplication of $$\mathrm{id}_G$$ with itself, i.e. the squaring map, is an endomorphism. The above argument is just the good old exercise that if $$G\rightarrow G,\,x\mapsto x^2$$ is a homomorphism, then $$G$$ is abelian.
• Sure. In characteristic $2$, we have $2=0$, but all I used is that $2=1+1$, which is always true (by definition). In fact, if $R$ has characteristic $2$, then $x^2=2.x=0.x=1$ implies that every element of $G$ has order $2$ and is thus self-inverse. It follows directly for all $x,y\in G$ that $xy=x^{-1}y^{-1}=(yx)^{-1}=yx$. Aug 9 at 19:20