Classification of minimal sets of properties proving a group is Abelian 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.
 A: Here is an answer to the second question. Call $S$ an abelian forcing set if $S[G]$ implies that $G$ is abelian. 
Proposition. If $S$ is an abelian forcing set then some finite subset of $S$ is an abelian forcing set. So there is no infinite minimal abelian forcing set. 
Proof. Given $n>0$, let $\phi_n$ be the statement
$$
\forall x\forall y\, (xy)^n=x^ny^n.
$$
This is a first order statement in the language of groups. 
Now suppose $S$ is an abelian forcing set. Let $\Sigma$ be the set consisting of $\phi_n$ for all $n\in S$, together with the axioms for groups. Let $\psi$ be the statement
$$
\forall x\forall y\, xy=yx.
$$
Then our assumption is that $\psi$ is a logical consequence of $\Sigma$. By the Compactness Theorem for first order logic, there is some finite subset $\Sigma_0$ of $\Sigma$ such that $\psi$ is a logical consequence of $\Sigma_0$. So if $S_0$ is the set of $n$ such that $\phi_n$ is in $\Sigma_0$, then $S_0$ is an abelian forcing set and a finite subset of $S$. 
A: See the Monthly article: Joseph A. Gallian and Michael Reid, Abelian Forcing Sets, Amer. Math. Monthly, Vol. 100, No. 6 (Jun. - Jul., 1993), pp. 580-582.
Gallian and Reid show that a set $S$ of integers is abelian forcing if, and only if, the greatest common divisor of the numbers $n(n−1)$, for $n\in S$ is equal to $2$.
They show how this makes it particularly easy to see that standard exercise examples, such as $S = \{2\}$, $S = \{-1\}$ and $S = \{ k, k+1, k+2 \},$ (for $k$ an integer) are all abelian forcing sets, and note that those first two are the only singletons $S$ that are abelian forcing.
