$AA^{-1} \subseteq A^{-1}A$ for every infinite subset $A$ of $G$

Question

Let $G$ be an infinite group, and for every $A\subseteq G$ put $A^{-1}=\{ a^{-1}:a\in A\}$.

Is it true that if $AA^{-1} \subseteq A^{-1}A$ for every infinite subset $A$ (containing $1$) then $G$ is abelian?

As a special case, we can start with $AA^{-1} =A^{-1}A$ (to check that whether it is true or not). Note that the property holds for all symmetric subsets (i.e., $A= A^{-1}$) and all $A$ such that $A^{-1}\subseteq N_G(A)$.

This is true, but $AA^{-1} \subseteq A^{-1}A$ does not imply $AA^{-1} = A^{-1}A$ (also we have another similar question when $A^{-1}A \subseteq AA^{-1}$) — M.H.Hooshmand, Mar 26 '16 at 06:10

score 4 · Accepted Answer · answered Mar 26 '16 at 09:07

4

There are infinite non-abelian groups $G$ with the property that, for any two elements $x$ and $y$, either $xy=yx$ or $x^2=y^2$. For example, take $G$ to be the direct product of a quaternion group of order $8$ and infinitely many cyclic groups of order $2$.

If $G$ is such a group and $A\subseteq G$, then for any $x,y\in A$, either $xy^{-1}=y^{-1}x$ or $xy^{-1}=x^{-1}y$, so $AA^{-1}\subseteq A^{-1}A$.

answered Mar 26 '16 at 09:07

Jeremy Rickard

32,279

Therefore $AA^{-1}=A^{-1}A$, for all subsets $A$ in the above infinite non-abelian group, right? – M.H.Hooshmand Mar 26 '16 at 09:36

$AA^{-1} \subseteq A^{-1}A$ for every infinite subset $A$ of $G$

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