4
$\begingroup$

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.

Improve this question
$\endgroup$
9
  • 5
    $\begingroup$ 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)? $\endgroup$
    – user44191
    Commented Jan 29, 2020 at 5:53
  • $\begingroup$ @user44191 Your comment is relevant! $\endgroup$
    – Sebastien Palcoux
    Commented Jan 29, 2020 at 8:15
  • 2
    $\begingroup$ 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. $\endgroup$
    – user44191
    Commented Jan 29, 2020 at 9:24
  • 1
    $\begingroup$ @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 $C_n \otimes C_m \simeq C_{nm}$). $\endgroup$
    – Sebastien Palcoux
    Commented Jan 29, 2020 at 12:06
  • 2
    $\begingroup$ Ok, I understand. A monoidal structure is even stronger than your initial requirement, but also more interesting. Now you got me ... ^^ $\endgroup$
    – Martin Brandenburg
    Commented Jan 29, 2020 at 16:13

1 Answer 1

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .