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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 $g\in G$.

Questions

  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.

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    $\begingroup$ 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
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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.

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