# Existence of an addition bifunctor for the category of groups

Let $$\mathsf{Grp}$$ be the category of groups. A bifunctor $$A: \mathsf{Grp} \times \mathsf{Grp}\to \mathsf{Grp}$$ is an addition bifunctor if:

• $$A(C_n,C_m) \simeq C_{n+m}$$,
• $$A(C_0,G) \simeq A(G,C_0) \simeq G$$,

for every group $$G$$ and every $$n,m$$, with $$C_n$$ the cyclic group of $$n$$ elements if $$n>0$$, and $$C_0 \simeq \mathbb{Z}$$.

Question: Is there an addition bifunctor for the category of groups?
(or for the subcategory of countable groups)

Stronger question: Is there an addition bifunctor providing a monoidal structure?

This post is inspired by that one (without knowing whether there is an explicit link).

Multiplicative analogous: Existence of a multiplication bifunctor for the category of groups.

• How are you defining your category of cyclic groups (what are its morphisms), and how are you defining your addition and multiplication functors (what does each do to those morphisms)? – user44191 Jan 29 at 5:53
• @user44191 Your comment is relevant! – Sebastien Palcoux Jan 29 at 8:15
• I'm...strongly doubtful that there's a natural "addition" functor, and weakly doubtful that there's a natural "multiplication" functor. But if you do go with "set" and "map", then from what I can see, there's no interesting "structure" to $\tilde{A}$ - the extension can be defined completely arbitrarily. – user44191 Jan 29 at 9:24
• @MartinBrandenburg: We could ask whether the category $\mathcal{Grp}$ (or the subcategory of countable groups) admits a monoidal structure with unit $I \simeq C_0$ and $C_n \otimes C_m \simeq C_{n+m}$ (or with unit $I \simeq C_1$ and $C_n \otimes C_m \simeq C_{nm}$). – Sebastien Palcoux Jan 29 at 12:06
• Ok, I understand. A monoidal structure is even stronger than your initial requirement, but also more interesting. Now you got me ... ^^ – Martin Brandenburg Jan 29 at 16:13

The answer is no. Notice that $$A(-,C_1)$$ is a functor $$F : \mathsf{Grp} \to \mathsf{Grp}$$ with $$F(C_n) \cong C_{n+1}$$. But there is no such $$F$$. There is a split monomorphism $$C_1 \to C_2$$, hence $$F$$ would induce a split monomorphism $$C_2 \to C_3$$, contradiction.