# Classification of the functors on the category of cyclic groups

Let $$\mathsf{Grp}$$ be the category of groups and let $$\mathsf{Cyc}$$ be the subcategory of cyclic groups.
As seen in the posts here and there (and their answers), a functor $$F: \mathsf{Cyc} \to \mathsf{Cyc}$$ is a very structured/restrictive notion, we are then lead to wonder whether there exists such a functor which is non-equivalent to the identity or the trivial functor, or if there is a such functor with $$F(C_1) \not \simeq C_1$$. As pointed out by Martin Brandenburg and Jeremy Rickard, $$C_1$$ is a retract of $$F(C_1)$$, so that $$F(C_1)$$ must be a retract of $$F^2(C_1)$$, and more generally, $$F^n(C_1)$$ is a retract of $$F^{n+1}(C_1)$$, which means that $$F^{n+1}(C_1)$$ is isomorphic to a semidirect product $$F^n(C_1) \ltimes N_n$$; now $$F^{n+1}(C_1)$$ is a cyclic group, so the semidirect product is in fact a direct product and moreover $$gcd(|F^n(C_1)|,|N_n|) = 1$$.

Question: What are the functors on the categroy of cyclic groups?

Remark: $$Aut(-)$$ is not such a functor because $$Aut(C_8) \simeq C_2 \times C_2$$ (and $$Aut^2(C_8) \simeq S_3$$).

In his answer, Neil Strickland provides examples of functors $$F$$ with $$F(C_1) \not \simeq C_1$$ and with $$F^2(C_1) \not \simeq F(C_1)$$, but with $$F^3(C_1) \simeq F^2(C_1)$$.

Bonus question: Is there a functor $$F: \mathsf{Cyc} \to \mathsf{Cyc}$$ such that $$F^{n+1}(C_1) \not \simeq F^n(C_1)$$ for all $$n$$?

Remark: If so, the sequence $$(F^n(C_1))_n$$ cannot be periodic (for $$n$$ large enough), because then (as shown above) $$F^{n+1}(C_1) \simeq F^{n}(C_1) \times N_n$$ with $$|N_n|>1$$ for all $$n$$.

• I think that there are too many functors to classify them. – Martin Brandenburg Jan 31 at 10:26
• The first step is to describe the morphisms in $\mathsf{Cyc}$ by generators and relations and thereby to give a description of functors into any category. But it will be quite complicated and not easy to simplify even for simple target categories. – Martin Brandenburg Jan 31 at 10:32
• Another remark: $\mathsf{Cyc}$ has at least two additional structures, for example it is $\mathbb{Z}$-linear (aka preadditive) and symmetric monoidal. It is much easier to classify the functors which preserves one of both of these structures. – Martin Brandenburg Jan 31 at 10:38
• @MartinBrandenburg Is there a non-constant functor $F$ with $F(C_1) \neq C_1$? – Sebastien Palcoux Jan 31 at 11:43
• The first question is answered by Neil below, the second: the tensor product of abelian groups. We have $C_n \otimes C_m = C_{\mathrm{gcd}(n,m)}$ for $n,m \geq 0$. – Martin Brandenburg Jan 31 at 11:57

I'll use additive notation, and I'll assume that you are only considering finite cyclic groups. Let $$\mathbf{Cyc}_p$$ be the category of cyclic $$p$$-groups. Given $$i,j\geq 0$$ we can define $$Q(p;i,j)\colon\mathbf{Cyc}\to\mathbf{Cyc}_p$$ by $$Q(p;i,j)(A)=\{a\in p^iA\colon p^ja=0\}$$. We can also define a constant functor $$C(p;i)\colon\mathbf{Cyc}\to\mathbf{Cyc}_p$$ by $$C(p;i)(A)=\mathbb{Z}/p^i$$. Now suppose we have a collection of functors $$F_p$$, one for each prime $$p$$, each of the form $$Q(p;i,j)$$ or $$C(p;i)$$, and that only finitely many of the functors $$F_p$$ are constant. Then the group $$F(A)=\prod_pF_p(A)$$ is cyclic for all $$A$$, so we get a functor $$F\colon\mathbf{Cyc}\to\mathbf{Cyc}$$. I don't know if that gives all possible functors, but it certainly gives a reasonably rich supply of them.
As a very specific example, the functor $$F(A)=(A/2A)\times(\mathbb{Z}/3)$$ is non-constant with $$F(0)\neq 0$$.
UPDATE: Here's a more exotic example. If $$X$$ is a based set of size $$1$$ or $$3$$, there is a unique group structure for which the basepoint is the identity. If $$A$$ is cyclic of order $$1$$ or $$7$$ then we can impose an equivalence relation with $$a\sim a^2\sim a^4$$ for all $$a$$, and then $$A/\sim$$ has size $$1$$ or $$3$$ with basepoint $$0$$ and so has a group structure. This construction gives a functor on the category of groups of order $$1$$ or $$7$$, and we can compose with $$A\mapsto A/7$$ to get a functor $$\mathbf{Cyc}\to\mathbf{Cyc}_3$$. I am sure that there are many variations on this theme.
• From the other questions I assume that $C_0 = \mathbb{Z}$ belongs to $\mathsf{Cyc}$. – Martin Brandenburg Jan 31 at 10:38
• Is there a functor $F$ with $F^2(C_1) \neq F(C_1)$? – Sebastien Palcoux Jan 31 at 15:25
• If $G$ is the functor in my update, and $FA=\mathbb{Z}/7\times GA$ then $F0=\mathbb{Z}/7$ and $F^20=\mathbb{Z}/21$. – Neil Strickland Jan 31 at 15:34