Group homomorphisms form an abelian group 
Let $\mathrm{Hom}(G_1, G_2) = \{f:G_1\to G_2 \mid f $ is a morphism $\}$. Show that $(\mathrm{Hom}(G_1, G_2), +)$ is an abelian group. 

I had problems just with the closure property! How do you show that $f+g$ is a morphism? I should add that $(G_1,\cdot)$ and $(G_2,\cdot)$ are abelian groups?

Show that it has the required properties. In particular,
$$\begin{align}(f+g)(x+y)&=f(x+y)+g(x+y)\\&=f(x)+f(y)+g(x)+g(y)\\&=f(x)+g(x)+f(y)+g(y)\\&=(f+g)(x)+(f+g)(y)\end{align}$$
