Is there any non-abelian group with the property $AB=BA$?

Question

Is there any finite (resp. infinite) non-abelian group of order $\geq 8$ such that $AB=BA$ for all subsets $A, B$ with $|A|\geq 3$ and $|B|\geq 3$?

($AB=\{ab: a\in A, b\in B\}$)

Answer 1

Let $G$ be any nonabelian group and choose $a,b\in G$ are such that $ab\neq ba$. Choose distinct elements $c,d\in G\setminus\{a,bab^{-1}\}$, and choose distinct elements $e,f\in G\setminus\{b,a^{-1}ba,c^{-1}ba,d^{-1}ba\}$. This is possible since any nonabelian group has at least $6$ elements. Let $A=\{a,c,d\}$ and $B=\{b,e,f\}$. Our choice of $c,d,e,$ and $f$ exactly guarantees that $ba\not\in AB$ (we chose $c$ and $d$ such that $cb,db\neq ba$, and then we chose $e$ and $f$ such that $ae,ce,de,af,cf,df\neq ba$). Since $ba\in BA$, this means $AB\neq BA$.

(More generally, we could similarly choose $A$ and $B$ to have $n$ elements as long as $|G|\geq 2n$.)