Let $G$ be a finite group. Which are the values of $k$ for which if every two $k$subsets of $G$ commute then $G$ is commutative? Clearly, this holds for $k=1$ and $k=2$.

$\begingroup$ It's not true for $k=4$, because the product of any two subsets of size $4$ in the dihedral group of order $6$ is equal to the whole group. I think a similar argument with dihedral groups will rule out any larger $k$. So that leaves $k=3$. The property does not appear to hold for $k=3$ in small nonabelian groups, so I would guess that it is true for $k=3$. $\endgroup$ – Derek Holt Dec 12 '15 at 17:23

1$\begingroup$ Thus in the dihedral group of order 6, and all $k\ge 4$, the product of any two $k$element subsets is the whole group (for $k\ge 7$, this is just a $\forall x\in\emptyset$ sentence, thus tautological). $\endgroup$ – YCor Dec 12 '15 at 22:39
Here is a proof that the property holds for $k=3$. Let $G$ be a nonabelian group. Then $G$ contains an element $g$ with $g \ne g^{1}$ and $g \not\in Z(G)$. Choose $h$ such that $gh \ne hg$.
Not let $A=\{1,g,h\}$, $B=\{1,g^{1},h\}$. Then $AB= \{1,g,h,g^{1},h^2,gh,hg^{1}\}$ and $BA=\{1,g,h,g^{1},h^2,hg,g^{1}h\}$ (but there may be some repetitions in these lists of elements).
By the choice of $g,h$, we easily check that $gh \not\in BA$, so $AB \ne BA$.
Hence the property is true for $k=1,2,3$ and false for all larger $k$. There is no need to assume that $G$ is finite.