[proofplan]
We first prove that pointwise addition is closed on $\operatorname{Hom}_R(M,N)$: the sum of two $R$-module homomorphisms is again an $R$-[module homomorphism](/page/Module%20Homomorphism). We then define the zero element as the zero map and the additive inverse of a homomorphism by pointwise negation, verifying in each case that the resulting map is $R$-linear. Finally, all abelian group laws are checked by evaluating two functions at an arbitrary element of $M$ and using the abelian group laws in the underlying additive group of $N$.
[/proofplan]
[step:Prove that pointwise sums of module homomorphisms are module homomorphisms]
Let $f,g\in \operatorname{Hom}_R(M,N)$. Define the map $h:M\to N$ by
\begin{align*}
h(m)=(f+g)(m)=f(m)+g(m)
\end{align*}
for every $m\in M$.
We verify that $h$ is an $R$-module homomorphism. Let $m_1,m_2\in M$. Since $f$ and $g$ are additive homomorphisms and addition in $N$ is associative and commutative,
\begin{align*}
h(m_1+m_2)=f(m_1+m_2)+g(m_1+m_2)
\end{align*}
\begin{align*}
=f(m_1)+f(m_2)+g(m_1)+g(m_2)
\end{align*}
\begin{align*}
=f(m_1)+g(m_1)+f(m_2)+g(m_2)
\end{align*}
\begin{align*}
=h(m_1)+h(m_2).
\end{align*}
Let $r\in R$ and $m\in M$. Since $f$ and $g$ respect scalar multiplication and scalar multiplication in $N$ distributes over addition,
\begin{align*}
h(rm)=f(rm)+g(rm)=r f(m)+r g(m)=r(f(m)+g(m))=r h(m).
\end{align*}
Thus $h\in\operatorname{Hom}_R(M,N)$, so pointwise addition is closed on $\operatorname{Hom}_R(M,N)$.
[guided]
We need to prove that the operation described in the statement actually lands back in $\operatorname{Hom}_R(M,N)$. This is the closure axiom for the proposed group operation. Let $f,g\in \operatorname{Hom}_R(M,N)$, and define $h:M\to N$ by
\begin{align*}
h(m)=(f+g)(m)=f(m)+g(m)
\end{align*}
for every $m\in M$.
To prove $h\in\operatorname{Hom}_R(M,N)$, we must verify additivity and compatibility with the left $R$-action. Let $m_1,m_2\in M$. Since $f$ and $g$ are $R$-module homomorphisms, they are additive, so
\begin{align*}
f(m_1+m_2)=f(m_1)+f(m_2)
\end{align*}
and
\begin{align*}
g(m_1+m_2)=g(m_1)+g(m_2).
\end{align*}
Substituting these identities into the definition of $h$ gives
\begin{align*}
h(m_1+m_2)=f(m_1+m_2)+g(m_1+m_2)
\end{align*}
\begin{align*}
=f(m_1)+f(m_2)+g(m_1)+g(m_2).
\end{align*}
The target module $N$ is an abelian group under addition, so we may reassociate and commute the middle terms:
\begin{align*}
f(m_1)+f(m_2)+g(m_1)+g(m_2)=f(m_1)+g(m_1)+f(m_2)+g(m_2).
\end{align*}
Using the definition of $h$ again, this becomes
\begin{align*}
h(m_1+m_2)=h(m_1)+h(m_2).
\end{align*}
Now let $r\in R$ and $m\in M$. Since $f$ and $g$ are $R$-module homomorphisms, they respect scalar multiplication:
\begin{align*}
f(rm)=r f(m)
\end{align*}
and
\begin{align*}
g(rm)=r g(m).
\end{align*}
Therefore
\begin{align*}
h(rm)=f(rm)+g(rm)=r f(m)+r g(m).
\end{align*}
The scalar multiplication on $N$ distributes over addition, so
\begin{align*}
r f(m)+r g(m)=r(f(m)+g(m)).
\end{align*}
Since $h(m)=f(m)+g(m)$, we obtain
\begin{align*}
h(rm)=r h(m).
\end{align*}
Thus $h$ is additive and $R$-linear, hence $h\in\operatorname{Hom}_R(M,N)$. This proves that pointwise addition is a well-defined binary operation on $\operatorname{Hom}_R(M,N)$.
[/guided]
[/step]
[step:Construct the zero homomorphism and pointwise additive inverses]
Let $0_N$ denote the zero element of the additive group of $N$. Define the zero map $0:M\to N$ by
\begin{align*}
0(m)=0_N
\end{align*}
for every $m\in M$.
For $m_1,m_2\in M$, the identity $0_N+0_N=0_N$ gives
\begin{align*}
0(m_1+m_2)=0_N=0_N+0_N=0(m_1)+0(m_2).
\end{align*}
For $r\in R$ and $m\in M$, the module identity $r0_N=0_N$ gives
\begin{align*}
0(rm)=0_N=r0_N=r0(m).
\end{align*}
Thus $0\in\operatorname{Hom}_R(M,N)$.
Now let $f\in\operatorname{Hom}_R(M,N)$. Define the map $-f:M\to N$ by
\begin{align*}
(-f)(m)=-f(m)
\end{align*}
for every $m\in M$. For $m_1,m_2\in M$,
\begin{align*}
(-f)(m_1+m_2)=-f(m_1+m_2)=-(f(m_1)+f(m_2)).
\end{align*}
Since $N$ is an abelian group under addition,
\begin{align*}
-(f(m_1)+f(m_2))=(-f(m_1))+(-f(m_2))=(-f)(m_1)+(-f)(m_2).
\end{align*}
For $r\in R$ and $m\in M$, using $f(rm)=r f(m)$ and the module identity $r(-n)=-(rn)$ for $n\in N$,
\begin{align*}
(-f)(rm)=-f(rm)=-(r f(m))=r(-f(m))=r(-f)(m).
\end{align*}
Thus $-f\in\operatorname{Hom}_R(M,N)$.
[/step]
[step:Verify the abelian group laws pointwise]
We now verify the group axioms for pointwise addition on $\operatorname{Hom}_R(M,N)$.
Let $f,g,k\in\operatorname{Hom}_R(M,N)$ and let $m\in M$. Associativity follows from associativity of addition in $N$:
\begin{align*}
((f+g)+k)(m)=(f(m)+g(m))+k(m)=f(m)+(g(m)+k(m))=(f+(g+k))(m).
\end{align*}
Since this holds for every $m\in M$, we have $(f+g)+k=f+(g+k)$.
The zero homomorphism is an identity element. For every $m\in M$,
\begin{align*}
(f+0)(m)=f(m)+0_N=f(m)
\end{align*}
and
\begin{align*}
(0+f)(m)=0_N+f(m)=f(m).
\end{align*}
Hence $f+0=f$ and $0+f=f$.
The pointwise negative map is an additive inverse. For every $m\in M$,
\begin{align*}
(f+(-f))(m)=f(m)+(-f(m))=0_N=0(m)
\end{align*}
and
\begin{align*}
((-f)+f)(m)=(-f(m))+f(m)=0_N=0(m).
\end{align*}
Hence $f+(-f)=0$ and $(-f)+f=0$.
Finally, commutativity follows from commutativity of addition in $N$. For every $m\in M$,
\begin{align*}
(f+g)(m)=f(m)+g(m)=g(m)+f(m)=(g+f)(m).
\end{align*}
Therefore $f+g=g+f$.
We have shown that pointwise addition is closed on $\operatorname{Hom}_R(M,N)$, has the zero homomorphism as identity, gives every homomorphism a pointwise additive inverse, and is associative and commutative. Hence $\operatorname{Hom}_R(M,N)$ is an abelian group under pointwise addition.
[/step]
