4
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Aug 10 at 12:21

1 Answer 1

8
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Does your argument hold for characteristic 2 rings? $\endgroup$ Aug 9 at 18:38
  • 1
    $\begingroup$ 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 $xy=x^{-1}y^{-1}=(yx)^{-1}=yx$. $\endgroup$
    – Thorgott
    Aug 9 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.