Uniqueness of the operation for a preadditive category? 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?
 A: Preadditive structure is not unique in general. However, in the presence of a zero object and biproducts (i.e. binary products that are naturally isomorphic to binary coproducts), there is a unique enrichment over commutative monoids, and thus, at most one (pre)additive structure. See this post for more details.
A: Regarding uniqueness, an $\text{Ab}$-enriched category with one object is a ring. The question of whether the addition is unique is then the question of whether the additive structure on a ring is unique given its multiplicative structure, and we don't have this uniqueness in general; it suffices to consider any ring $R$ and any monoid automorphism of $R$ that doesn't respect the addition, then transfer the addition along the automorphism. 
For example, let $F$ be any field and consider the map $\alpha : F \to F$ which sends $0$ to $0$ and inverts everything else. This is a monoid automorphism, and for most $F$ this map doesn't respect addition (I think the smallest example is $\mathbb{F}_5$). For such a field $F$ the operation $a \oplus b = \alpha^{-1}(\alpha(a) + \alpha(b))$ then gives an alternative addition. 
A: 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.
A: 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!
