# A functor $\mathsf{Grp} \to \mathsf{Grp}$ with $\mathbb{Z} \mapsto G$

Given a group $$G$$, does there exist a functor $$\square \tilde{\otimes} G: \mathsf{Grp} \to \mathsf{Grp}$$ with the following properties?

• $$\mathbb{Z} \tilde{\otimes} G \cong G$$

• $$\square \tilde{\otimes} G$$ preserves all direct sums, direct products and free products

The idea is that $$\square \tilde{\otimes} G$$ is a "base change from $$\mathbb{Z}$$ to $$G$$". I'm particularly interested in the cases $$G = \mathbb{Q}$$ and $$G = \mathbb{R}$$.

• What have you tried? Why has it not worked? What are the "obvious" guesses and why don't they work? Is there any reason you want to find a functor that satisfies these specific properties? It feels a bit random. Commented Apr 25 at 3:46
• @JonathanBeardsley It's definitely not random, and I think this question is already interesting enough to not have additional context. (+1) Commented Apr 25 at 3:48
• My point is that I can I think of several things that might work but it seems like a waste of effort if we have no clue what the OP originally asked. Commented Apr 25 at 3:53
• This feels unlikely. For example, one easily verifies that $\mathbb{F}_n \mathbin{\tilde{\otimes}} G = G^{\ast n}$ for all $1 \leq n \leq \infty$. But while $\mathbb{F}_\infty \hookrightarrow \mathbb{F}_2$, it is not true that $G^{\ast \infty} \hookrightarrow G \ast G$ for all $G$. (Say, $G = \mathbb{Z}/2\mathbb{Z}$.) While this doesn’t directly prove such functors cannot exist, it does make it unlikely, in my opinion. Commented Apr 25 at 8:24
• Here is an idea for how to define such a functor: math.stackexchange.com/questions/4907258/… Commented Apr 28 at 17:55