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Let $$G_1=\langle a,b,c:\hspace{0.2cm}abc^{-1}a^{-1}b^{-1}c=1\rangle,$$ $$G_2=\langle d,e:\hspace{0.2cm}de d^{-1}e^{-1}=1\rangle.$$ How can I construct (if possible) a group isomorphism between $G_1$ and $G_2$?

I try to think of this as identifying the edges of a square and a hexagon both representing the torus, but I get stuck.

Thanks for the comments and help.

Dann
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    Why are you are asking for help with proving something that is not true? Doing that wastes people's time. – Derek Holt Feb 27 '22 at 07:43
  • @DerekHolt: The OP did not know if it was true or not. There are similar questions on MO and here which suggest that it is true. https://math.stackexchange.com/questions/1978562/what-surface-do-we-get-by-joining-the-opposite-edges-of-a-hexagon – markvs Feb 27 '22 at 17:05
  • See also https://mathoverflow.net/questions/128974/why-isnt-langle-x-y-zxyzx-1y-1z-1-rangle-a-hyperbolic-surface-grou – markvs Feb 27 '22 at 17:13

1 Answers1

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These groups are not isomorphic because the abelianization of the first group is $\Bbb Z^3$ and the abelianization of the second group is $\Bbb Z^2$. In fact the first group is the free product $\Bbb Z*\Bbb Z^2=\langle b, ab, c^{-1}b\rangle$, where $(ab)$ commutes with $(c^{-1}b)$.

markvs
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