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$\langle a,b \,|\, ab = ba\rangle$ is the fundamental group of the torus. Consider $\langle x,y,z\,|\, xyz = zyx\rangle$. We have the homeomorphism $f(a) = zx, f(b) = zy^{-1}$. Is there an isomorphism? I think there should be because $\langle x,y,z\,|\, xyz = zyx\rangle$ is the fundamental group of the object from this question which is supposed to be homeomorphic to the torus.

Am I misunderstanding something? If not how can I find the explicit isomorphism, in particular I want to know what $x,y,z$ are mapped to.

Edit: @Desperado comment resolved my confusion, the hexagon I was thinking about had all vertices identified to each other while the one in the post I linked does not. Thanks!

They are definitely not isomorphic, as they have different abelianization. The group ⟨x,y,z|xyz=zyx⟩ is not the fundamentl group of the "hexagon" in the linked question: the vertices are not all the same point, but 3 are P and 3 are Q≠P."

plshelp
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  • Do $f(a)$ and $f(b)$ commute? What have you tried? – Dietrich Burde Jun 27 '23 at 08:09
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    They are definitely not isomorphic, as they have different abelianization. The group $\langle x,y,z,|, xyz = zyx\rangle$ is not the fundamentl group of the "hexagon" in the linked question: the vertices are not all the same point, but 3 are $P$ and 3 are $Q\neq P$. – Pomponazzo Jun 27 '23 at 08:52

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The group $G=\langle x,y,z\mid xyz=zyx\rangle$ is non-abelian, hence it cannot be isomorphic to the group $\langle a,b\mid ab=ba\rangle \cong \Bbb Z\times \Bbb Z$.

Dietrich Burde
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    (+1) Before reading your answer I was about to write exactly the same sentence as a comment, but doesn't it need a proof? Can we by luck find three $2\times2$ noncommuting matrices $X,Y,Z$ satistying $XYZ=ZYX$? – Anne Bauval Jun 27 '23 at 09:33
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    Yes we can: $X=Y$ s.t. $X^2=I$ and $Z$ not commuting with $X.$ – Anne Bauval Jun 27 '23 at 09:37
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    @AnneBauval Thank you. I agree, my "answer" needs a bit more of an argument. – Dietrich Burde Jun 27 '23 at 09:49
  • Thanks for the answer, a proof is not required; my confusion was about the spaces being homeomorphic, but @Desperados resolved that. – plshelp Jun 27 '23 at 10:43