Which subsets of the integers arise as the set of exponents of power endomorphisms for some group?

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A group $G$ is said to be an $n$-abelian group if the $n$th power map is an endomorphism of $G$.

Given a subset $S$ of the integers, three necessary conditions for $S$ to be the set of all integers $n$ for which $G$ is an $n$-abelian group for some fixed group $G$ are the following:

  1. $1 \in S$
  2. If $m,n \in S$, then also $mn \in S$.
  3. If $n \in S$, then also $1-n \in S$ (together with the first condition, this implies that $0 \in S$).

Question: Conversely, is the conjunction of the above three conditions also sufficient for $S$ to be the set of all integers $n$ for which $G$ is an $n$-abelian group for some fixed group $G$?

The idea, of course, is to let $G$ be the quotient of the free group $\langle x,y\rangle$ on two generators by the normal subgroup generated by all elements of the form $(xy)^n y^{-n} x^{-n}$ where $n \in S$.

The problem here is that conceivably, $G$ could be an $n$-abelian group for some $n \notin S$. The question then asks whether this problem will ever happen.

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A group $G$ is $n$-abelian if $(ab)^n = a^nb^n$ for all $n$.

The set you are looking at is

$$\mathcal{E}(G) = \{ n\in\mathbb{Z}\mid (ab)^n = a^nb^n\}.$$

That set forms a Levi system. This was studied by L.C. Kappe:

Your conditions are not quite enough. A set of integers is a Levi system if and only if it satisfies the following 5 conditions:

  1. $n,m\in W$ implies $nm\in W$.
  2. $n\in W$ implies $1-n\in W$.
  3. $0\in W$.
  4. There exists $w\in W$, $w\gt 0$, such that for all $n\in W$, $n^2\equiv n\pmod{w}$ and every integer congruent to $n$ modulo $w$ is in $W$.
  5. If the congruence classes of both $n$ and $n+1$ modulo $w$ lie in $W$, then $n\equiv 0\pmod{w}$.

Kappe proves that a set of integers is equal to $\mathcal{E}(G)$ for some $G$ if and only if it is a Levi system or it equals $\{0,1\}$.

See this mathoverflow answer for this and related concepts and references.