This is a continuation of Are these exactly the abelian groups? I would like to consider another condition on a group and see if it implies commutativity. The condition is

$$(\forall A,B\subseteq G)(\forall g\in G)(\exists x,y\in G)\ AgB=xABy.$$

Again, abelian groups satisfy this condition -- it's enough to take $x=g$ and $y=1.$

Using an argument similar to Hurkyl's in the other question, I've been able to prove that if two elements $g,h$ of such a group $G$ do not commute then they must generate the quaternion group. I've also checked that the quaternion group does not satisfy the condition, but this is not enough: the condition could still hold in $G$ even if it doesn't hold in $\langle g,h\rangle.$

I haven't been able to do anything more. Considering other equations just left me with loads of unmanageable cases in which I simply got lost.


If, in the quaternion group, you can find an example where $$\left\vert AgB\right\vert\neq\left\vert AB\right\vert,$$ then $A,B,g$ will contradict the condition in any group containing the quaternion group.

This is probably not the smallest example, but I found $A=B=\{i,j,k\}$, $g=i$, where $$AB=\{-1,i,-i,j,-j,k,-k\}$$ has seven elements, but $$AgB=\{1,-1,i,-i,-j,-k\}$$ has six elements.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.