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.

  • $\begingroup$ It's almost the same argument as in my answer in the additive case, as can be seen from Jeremy's answer. $\endgroup$ – Martin Brandenburg Jan 30 at 11:32
  • 1
    $\begingroup$ @MartinBrandenburg: Yes, I considered this example, but I had the false belief that a subgroup isomorphic to a quotient is a retract... $\endgroup$ – Sebastien Palcoux Jan 30 at 12:09
  • $\begingroup$ 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, $F(C_2) \not \simeq C_3, C_4$. $\endgroup$ – 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.

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