Show $\text{Hom}_R(M,N)$ is an abelian group Define a commutative ring $R$ where $M$ and $N$ are left $R$-modules, and denote by $\text{Hom}_R(M,N)=\{f:M\to N|f(rm)=r\cdot f(m)\text{ }\forall r\in R,m\in M\}$. Then, we want to prove that $\text{Hom}_R(M,N)$ is an $R$-module.
To do this, we need to show first that $\text{Hom}_R(M,N)$ is an abelian group under addition, and then that an action of $R$ on $\text{Hom}_R(M,N)$ denoted by $rm$ for all $r\in R$ and all $m\in M$ satisfies three properties of distributivity and associativity.
I am stuck on the first part: I know that we define addition on functions, but I am unclear on how thorough this proof needs to be (i.e. do we need to show all four group axioms, or just the abelian feature?), and in particular how to show $\text{Hom}_R(M,N)$ is commutative.
I want start with $f,g\in\text{Hom}_R(M,N)$ and see that $(f+g)(x)=(g+f)(x)$, but I don't see how this follows immediately from the way $\text{Hom}_R(M,N)$ has been defined, and I'm not sure if I need to use $rm$ instead of $x$ and invoke the linear homogeneity of $f$ and $g$.
Thanks!
Edit: $R$ also has an identity element.

Hint:
$\DeclareMathOperator\Hom{Hom}\Hom_R(M,N)$ is a subset of the abelian group $\;N^M$ (set of all maps from $M$ to $N$), so all you  to prove is it's a subgroup, i.e. it's not empty,  the sum of two linear maps is linear, and the opposite of a linear map is linear.Proving in a similar way that the product of a linear map by a scalar is linear requires $R$ to be commutative.
