Let $\mathsf{C}$ be a category. We call $\mathsf{C}$ binormal if it has a null object, has all equilizers and coequilizers, all monomorphisms are kernels and all epimorphisms are cokernels (whereby a kernel means an equalizer of a morphism and a zero morphism, and cokernel dually). Then my question is:



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> Can we make any binormal category a preadditive category?
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Here I'm using [wikipedia's definition](https://en.wikipedia.org/wiki/Preadditive_category) of preadditive category.


I know that a binormal category which has all binary products must be abelian. I'm just wondering if we can drop out the requirement of the existence of products.



Does a binormal category always admit an additive structure?
No. Let $C$ be a category constructed as follows. Let $G$ be a group and let $BG$ be the one-object category with automorphisms $G$. Then adjoin to $BG$ a zero object. This means that $C$ has two objects, $0$ (the zero object) and another object we'll call $c$. There are zero morphisms $0 \to 0, 0 \to c, c \to 0, c \to c$, and the only other morphisms are the automorphisms $g \in G : c \to c$. 


By construction, $C$ has a zero object. Any parallel pair of maps has equalizer either $0$ or $c$, and similarly for the coequalizer. The monomorphisms and epimorphisms are precisely the elements of $G$, which are both kernels and cokernels of $0 : c \to c$. So $C$ is binormal.


$C$ can be made into an $\text{Ab}$-enriched category if and only if there is a division ring $D$ whose group of units is $G$, and there are many obstructions to having this property. For example, if $G$ is finite, then so is $D$. By Wedderburn's little theorem, $D$ is a finite field, and so $G$ must be a finite cyclic group of order $q - 1$ for some prime power $q$. More generally, every finite abelian subgroup of $G$ must be cyclic. 


