# Existence of a multiplication bifunctor for the category of groups

For $$\mathsf{Grp}$$ the category of groups, a bifunctor $$M: \mathsf{Grp} \times \mathsf{Grp}\to \mathsf{Grp}$$ is a multiplication bifunctor if:

• $$M(C_n,C_m) \simeq C_{nm}$$,
• $$M(C_1,G) \simeq M(G,C_1) \simeq G$$,

for every group $$G$$ and every $$n,m>0$$, with $$C_n$$ the cyclic group of $$n$$ elements.

Question: Is there a multiplication bifunctor for the category of groups?
(or for the subcategory of countable groups, or of finite groups)

Stronger question: Is there a multiplication bifunctor providing a monoidal structure?

This post is a multiplicative analogous of that additive one.

• It's almost the same argument as in my answer in the additive case, as can be seen from Jeremy's answer. – Martin Brandenburg Jan 30 at 11:32
• @MartinBrandenburg: Yes, I considered this example, but I had the false belief that a subgroup isomorphic to a quotient is a retract... – Sebastien Palcoux Jan 30 at 12:09
• We finally get the following funny result: let $F:\mathsf{Grp} \to \mathsf{Grp}$ be a functor, then $F^n(C_1)$ is a retract of $F^{n+1}(C_1)$. For example, if $F(C_1) \simeq C_2$, then $C_2$ is a retract of $F(C_2)$, so in particular, $F(C_2) \not \simeq C_3, C_4$. – Sebastien Palcoux Jan 30 at 15:30

No. $$C_1$$ is a retract of $$C_2$$, so $$M(C_2,C_1)\simeq C_2$$ would have to be a retract of $$M(C_2,C_2)\simeq C_4$$, which it isn't.