Why is $\mathbf{Groups}$ not a pre-additive category

Question

Definition. A category $\mathscr{C}$ is pre-additive if every $Hom(A,B)$ is equipped with a binary operation making it an (additive) abelian group for which the distributive law holds: for all $f,g \in Hom(A,b)$,

1. if $p: B \rightarrow B'$, then $$p(f+g) = pf+pg \in Hom(A,B')$$
2. if $q:A' \rightarrow A$, then $$(f+g)q = fq+gq \in Hom(A', B).$$

I'm reading Avanced Modern Algebra, Part I by Joseph J. Rotman and on exercise B-4.3 it asks to prove that Groups is a not a pre-additive category with the following hint:

Hint. If $G$ is not abelian and $f,g \colon G \rightarrow G$ are homomorphisms, show that the function $x \mapsto f(x)g(x)$ may not be a homomorphism.

My attempt: If Groups is pre-additive, then $Hom(G,G)$ has a ring structure. Thus for all $f \in Hom(G,G)$, $f0 = 0 = 0f$.

Now let us consider the group of premutations $S_5$. We have that $Z(S_5) = 1$. Furthermore, since $G/Z(G) \cong Inn(G)$ then $S_5 \cong Inn(S_5)$.

Since $0 \in Hom(G,G)$ commutes with all homomorphisms in $Hom(G,G)$, in particular it commutes with all the inner automorphisms of $G$ (with all $f \in Inn(G)$). Thus $0$ must be the identity since $Z(Inn(S_5)) = 1$. But this cannot be since for any homomorphism different than the identity $f = f1 = f0 = 0 = 1$, a contradiction. $\square$

I'm pretty sure I overcomplicated it since I didn't use the Hint. And I don't see how this proof helps in the next exercise (that is a a follow up to this one) that asks to prove that Rings and ComRings are not pre-additive categories. This exercise is in the first section of the introduction to category theory so no sophisticated answers should be expected.

Question: How do I solve this exercise with the hint? I'm absolutely clueless as to where such a hint could help.

Answer 1

The hint is only about disproving the validity of the "evident" pre-additive structure with pointwise operations. Your (EDIT: attempted) solution is much better, as it is about disproving the existence of any pre-additive structure.

Alternatively, notice that in every pre-additive category the canonical map $X \sqcup Y \to X \times Y$ from the binary coproduct to the binary product (in case these exist) is an isomorphism. Which clearly fails for groups.

The category of rings is not pre-additive since there are no zero morphisms. For example, there is no ring homomorphism $\mathbb{Q} \to \mathbb{F}_2$.

The category of non-unital rings has zero morphisms but is also not pre-additive for the same argument as for groups above. See here for a description of the coproduct of rings, the non-unital case is even easier to describe.

Answer 2

There is a mistake in your argument.

You say that whatever the $0$ element of $\mathrm{Hom}(G,G)$ is, it must commute with every element of $\mathrm{Hom}(G,G)$. This is correct.

You then say that this means that the $0$ element must commute with every element of $\mathrm{Inn}(G)$. This is also correct.

Then you claim that it must be the identity morphism, because $Z(\mathrm{Inn}(G))=\{\mathrm{id}_G\}$. This does not follow.

You know that $0$ centralizes every element of $\mathrm{Inn}(G)$. If it lies in $\mathrm{Inn}(G)$ (or in fact, if it lies in $\mathrm{Aut}(G)$, since your $G$ has no outer automorphism), then it is true that it must lie in the center of $\mathrm{Inn}(G)$. But you have no warrant for claiming that this is the case. You have not shown that the $0$ element of $\mathrm{Hom}(G,G)$ must in fact be an automorphism, let alone an inner automorphism.

For example, the trivial endomorphism $z\colon G\to G$ with $z(g)=e$ for all $g\in G$ satisfies that $zf=fz=z$ for all $f\in\mathrm{Hom}(G,G)$. So it commutes with all elements of $\mathrm{Hom}(G,G)$ without being in $\mathrm{Inn}(G)$. It commutes/centralizes those elements, but it does not lie in its center (because it's not an automorphism). So how do you know that the zero element of the additive structure is not this particular endomorphism? (One can prove it isn't, but you have not done so) How do you know that the zero element of the additive structure is necessarily an automorphism?

Answer 3

The hint is only about a specific pre-additive structure. In this case there is another elementary argument for this specific structure (and perhaps this was intended by the hint): Let $f\colon G\rightarrow G$ and $g\colon G\rightarrow G$ be any elements of $Hom(G,G)$. Suppose that the category of groups is pre-additive. Then $Hom(G,G)$ is an abelian group with the operation $(f+g)(x)=f(x)\circ g(x)$. Since we are not assuming that $G$ is abelian, we may rewrite the RHS just by $(f+g)(x)=f(x)g(x)$. Since $Hom(G,G)$ is a group with the above operation, it is closed with respect to this operation, i.e., the map $x\mapsto f(x)g(x)$ is a group homomorphism from $G$ to $G$. Now let $G=\langle a,b\rangle$ be the free group on two generators. By the universal property of a free group, the following definitions yield group homomorphisms from $G$ to $G$: \begin{align*} f \colon \langle a,b\rangle & \rightarrow \langle a,b\rangle, a\mapsto a^2, \; b\mapsto b, \\ g \colon \langle a,b\rangle & \rightarrow \langle a,b\rangle, a\mapsto a, \; b\mapsto b^2. \end{align*} Let $x=ab$. Since $x\mapsto f(x)g(x)$ is a group homomorphism, we have the equivalences \begin{align*} f(ab)g(ab) & = (f(a)g(a))(f(b)g(b)) \\ (a^2b)(ab^2) & = (a^2a)(bb^2) =a^3b^3. \end{align*} This is a contradiction, since both words are reduced but not equal.