Given $\text{Hom}(M,M')$ is abelian, does $M$ have to be abelian?

Question

I am doing a course on algebra (self-study) and it was left as an excercise to prove that if $M$ and $M'$ are abelian groups, then $(\text{Hom}(M,M'), +)$ is an abelian group, where the operation is defined as $(f+g)(m):=f(m)+g(m)$. I used the fact that $M'$ is abelian when proving that $\text{Hom}(M,M')$ is closed under $+$ and also when proving that it is commutative. However, I haven't used the fact that $M$ is abelian. Is this an extra hypothesis that I actually do not require?

Answer 1

Unless I'm misunderstanding the title and question body are converse statements.

In your question you state that you were asked to prove that if $M,M'$ are abelian groups then $\text{Hom}(M,M')$ is an abelian group with pointwise operations. As you've correctly seen from your proof, this is true as long as $M'$ is abelian. So to answer the question in the body, you are correct that you don't need the extra assumption that $M$ be abelian.

What you've titled the question is if the hom-set $\text{Hom}(M,M')$ is abelian is it necessary that $M$ is abelian, which is a different question.

Answer 2

Let $X$ be a set and $[X,M']$ denote the set of all functions $f : X \to M'$. Use the addition in $M'$ to define the sum $f + g$ of functions $f, g \in [X,M']$ pointwise, i.e. by $(f + g)(x) = f(x) + g(x)$. You can eeasily check that $([X,M],+)$ is an abelian group.

You see that the group structure of $M'$ is the only ingredient needed here.

If $M$ is a group (not necessary abelian), we can consider the subset $$ \text{Hom}(M,M') = \{ f \in [M,M'] \mid f \text{ is a group homomorphism}\} .$$

You can easily check that $\text{Hom}(M,M')$ is a subgroup of $[M,M']$. That is, $(\text{Hom}(M,M'),+)$ is an abelian group.

The group structure of $M$ does not play any role for the addition $+$ on $\text{Hom}(M,M')$, it is only used to define the subset $\text{Hom}(M,M')$ of $[M,M']$.

You can generalize the construction of $[X,M']$ to every group $(M',\circ)$ by defining $(f \circ g)(x) = f(x) \circ g(x)$. This produces a group $([X,M'],\circ)$. However, for a group $M$ in general $\text{Hom}(M,M')$ will not be closed under $\circ$.