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 $n$.

  • $\begingroup$ I think that there are too many functors to classify them. $\endgroup$ Jan 31, 2020 at 10:26
  • $\begingroup$ 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. $\endgroup$ Jan 31, 2020 at 10:32
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    $\begingroup$ 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. $\endgroup$ Jan 31, 2020 at 10:38
  • $\begingroup$ @MartinBrandenburg Is there a non-constant functor $F$ with $F(C_1) \neq C_1$? $\endgroup$ Jan 31, 2020 at 11:43
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    $\begingroup$ 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 $n,m \geq 0$. $\endgroup$ Jan 31, 2020 at 11:57

1 Answer 1


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.

  • $\begingroup$ From the other questions I assume that $C_0 = \mathbb{Z}$ belongs to $\mathsf{Cyc}$. $\endgroup$ Jan 31, 2020 at 10:38
  • $\begingroup$ Is there a functor $F$ with $F^2(C_1) \neq F(C_1)$? $\endgroup$ Jan 31, 2020 at 15:25
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    $\begingroup$ If $G$ is the functor in my update, and $FA=\mathbb{Z}/7\times GA$ then $F0=\mathbb{Z}/7$ and $F^20=\mathbb{Z}/21$. $\endgroup$ Jan 31, 2020 at 15:34

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