Let $G$ be a group generated by $x,y,z$ such that the product $xyz$ is in its centre; that is, $xyz$ commutes with any element $g\in G$.


  1. What is the "freest" example of such a group $G$? Is it isomorphic to any familiar group?
  2. Are there any other non-Abelian groups that satisfy this property?

I can prove that the inner homomorphism group of $G$ is isomorphic to the free group with two generators. Another fact that is not hard to deduce is that $xyz=yzx=zxy$. Not sure if this helps.

  • 4
    $\begingroup$ Since $x,y,xyz$ also generates your group, it's pretty easy to see it is just (free group on $x,y$)$\times$(free abelian group on $xyz$). $\endgroup$ – Steve D Mar 27 '18 at 20:59
  • $\begingroup$ Thank you @SteveD! I appreciate your comment! $\endgroup$ – Zuriel Mar 27 '18 at 21:00

So, your group is $\langle x, y, z \, | \, [x, xyz] = [y, xyz] = [z, xyz] = 1\rangle $. It has obvious homomorphism to $\langle x, y, z \, | \, xyz = 1\rangle$ with kernel $\Bbb Z$, so it's a central extension of free group on 2 generators. As you can construct a section (because free group, is, actually, free), central extension splits and your group is isomorphic to $F_2 \times \Bbb Z.$

edit: It's pretty obvious that any other group of this kind is a factor of $F_2 \times \Bbb Z$, and you can construct plenty of those.

| cite | improve this answer | |

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.