Let $S$ be a non-empty, possibly infinite, set of integers, all of which are greater than $1$. For a given group $G$, let $S[G]$ denote the collection of statements $$ \forall (n \in S, a \in G, b\in g) \,\,(ab)^n=a^nb^n $$ For some sets $S$, $S[G]$ is sufficient to prove $G$ is Abelian. For example, the set $\{2\}$ is sufficient, since $(ab)^2 = a^2b^2\implies ba = ab$.

Define a minimal abelian forcing (maf) set as any $S$ such that $S[G]$ proves $G$ is Abelian, but for no proper subset $T \subset S$ does $T[G]$ prove $G$ is Abelian.

Has anybody studied the question of classifying all minimal abelian forcing sets? And has anybody addressed the question of whether there exists an infinite maf?

For example, some mafs are:

- $S = \{2\}$
- $S = \{3,5\}$
- $S = \{n+3,n+4,n+5\}, n\in \Bbb Z^+$

The line of attack I have tried is to exploit the fact that $(ab)^n = a^nb^n$ implies that $n$-th powers commute with $(n-1)$-th powers and that $(ab)^n = a^nb^n \implies (ab)^{n-1} = b^{n-1} a^{n-1} $. This easily confirms the examples listed above, but it is hard to get a systematic test of whether a given set $S$ is maf.

I realize that using the word "forcing" has jargon implications. But I felt it was better than minimal abelian *proving* since the abbreviation of that would be map and that would be more confusing.