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Regarding to the question "Is there any non-abelian group with the property $AB=BA$?", now it is important for us to know that:

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

(b) If the answer of (a) is positive, then

  • is there any class of groups (e.g., solvable groups, free groups, CLT-groups, etc.) with the property?

  • is it true for all groups with oreder $\leq 16$?

($AB=\{ab: a\in A, b\in B\}$, and $|.|$ denotes the cardinal number)

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  • $\begingroup$ I would bet that the answer is no, and that it is not too hard to prove it! $\endgroup$ – Derek Holt Oct 8 '16 at 11:33
  • $\begingroup$ I checked that there are no examples of order at most $16$. $\endgroup$ – Derek Holt Oct 8 '16 at 12:59
  • $\begingroup$ Consider a group $G$ with elements $a, b \in G$ such that $ab = 1$ but $ba \neq 1$ (I think $G$ needs to be infinite for such a group to exist). Then with $A = \lbrace 1, a \rbrace$ and $B = \lbrace 1, b \rbrace$, $AB$ contains three elements while $BA$ contains four. $\endgroup$ – Bib-lost Oct 8 '16 at 14:51
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    $\begingroup$ @Bib-lost, not in a group: if $a b = 1$, then $b a = a^ {-1} (a b) a = a^{-1} a = 1$. $\endgroup$ – Andreas Caranti Oct 8 '16 at 14:54
  • $\begingroup$ @AndreasCaranti Of course, I was thinking of rings, my apologies. $\endgroup$ – Bib-lost Oct 8 '16 at 14:54
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This answer is complete once one appeals to Derek's comment below. (Edited after that comment.)

Suppose there are two elements $b, x \in G$ such that $b^{-1} x b \ne x, x^{-1}$.

Consider $A = \{ 1, x^{-1} \}$, $B = \{ b, b x \}$.

Then $A B = \{ b, b x, x^{-1} b, x^{-1} b x \}$ has four elements, while $B A = \{ b, b x^{-1}, b x, b \}$ has three.

Assume thus that for all $b, x \in G$ we have $b^{-1} x b \in \{ x, x^{-1} \}$. Then $G$ is Hamiltonian.

But the quaternion group $Q$ does not satisfy the assumption, as shown by Derek Holt in a comment below.

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  • $\begingroup$ "although it is complete if one appeals to Derek's comment."? What does that refer to? $\endgroup$ – quid Oct 8 '16 at 15:01
  • $\begingroup$ To be specific, for $G=Q$ with generators $a,b$, we can take $A=\{a,b,a^2 \}$, $B=\{1,ab,a^{-1}\}$. Then $|AB|=7$ and $|BA|=6$. $\endgroup$ – Derek Holt Oct 8 '16 at 15:03
  • $\begingroup$ @DerekHolt, great, with this comment it is complete. $\endgroup$ – Andreas Caranti Oct 8 '16 at 15:04
  • $\begingroup$ I guess this only finishes it for finite groups. It should not be hard to settle it for infintie groups. It would be sufficient to prove it for nonabelian quotient groups of the infinite metacyclic group $\langle a,b \mid a^{-1}ba=b^{-1} \rangle$. $\endgroup$ – Derek Holt Oct 8 '16 at 15:20
  • $\begingroup$ @DerekHolt, I believe that once one has proved that the group is Hamiltonian, there's only the quaternion group of order $8$ to be checked, as you did. $\endgroup$ – Andreas Caranti Oct 8 '16 at 15:56

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