Classification of minimal sets of properties proving a group is Abelian

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.

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 $S$.