Question: When working on problems in Rotman's Algebra, he asks us to show that **Groups** is not a preadditive category. If we could show that the binary operation on $\mathrm{Hom}(A,B)$ had to be $f + g \mapsto f(x)g(x)$ then I completely understand how to proceed. However, I cannot seem to find a way to show that this must be the binary operation on the hom set. Similarly, it makes sense that the "zero map" ought to be the identity of the operation, but I do not see how to show that either.

I thought we could use left and right distributivity to relate the hom sets, but couldn't make it work to get what I want.

My general question is whether the preadditive structure is unique, and if so, how do we show this?

Answer: 
In a linear (=preadditive) category products and coproducts coincide (if they exist; see the notion of a biproduct). In the category of groups, $C\_2 \times C\_2$ is finite, but $C\_2 \sqcup C\_2$ is infinite (for example since the dihedral group $D\_n$ is a quotient of $C\_2 \sqcup C\_2$ for every $n$). Thus, $\mathsf{Grp}$ is not linear.
The set $G:=\operatorname{Hom}(S\_3,S\_3)$ contains

* the zero map $x\mapsto 1$.
* three maps with $C\_3$ as kernel and one of the three 2-subgroups as image
* six isomorphisms coming from permutations of $\{1,2,3\}$

Assume we have an addition on $G$ that makes it an abelian group and that this addition distributes over composition.
Since $|G|=10$, we conclude that $G$ is cyclic, hence has exactly one element of order $2$.

The zero map must be the neutral element of $G$.
If $\phi$ is any of the maps with $\ker \phi=C\_3$, then $\phi\circ \psi=\phi$ unless $\psi=0$.
Therefore, if $\phi$ has order $k(\ge2)$, we find
$$ 0=\phi\circ0=\phi\circ(k\phi)=\phi\circ((k-1)\phi)+\phi\circ\phi=\phi+\phi$$
and hence conclude that $\phi$ has order $2$.
Thus $G$ has at least three elements of order $2$ - contradiction!