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I have these two presentations that are supposed to indicate the same group: $$\langle a,b \mid a^3=b^2\rangle$$ $$\langle x,y \mid xyx=yxy\rangle$$ Does anybody know how can set the isomorphism up? Thank you!

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    $\begingroup$ Proper notation is $\langle a,b \mid a^3=b^2\rangle,$ not $<a,b|a^3=b^2>.$ Note \langle, \mid, \rangle. $\qquad$ $\endgroup$ – Michael Hardy Apr 22 '17 at 6:01
  • $\begingroup$ Have you tried drawing the caylies graph? $\endgroup$ – Elad Apr 22 '17 at 6:01
  • $\begingroup$ I didn't do cayley's graphs. Would it be easier? $\endgroup$ – Dac0 Apr 22 '17 at 6:18
  • $\begingroup$ This is really not very difficult. You should try and do it yourself. $\endgroup$ – Derek Holt Apr 22 '17 at 8:24
  • $\begingroup$ You are right, once you've seen it it's really easy $\endgroup$ – Dac0 Apr 22 '17 at 11:19
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Starting from $\langle x,y \mid xyx=yxy\rangle$, we have $xyx=yxy$, so $(xyx)(yxy)=(xyx)(xyx)$, so $(xy)(xy)(xy)=(xyx)(xyx)$, so by letting $a=xy$ and $b=xyx$ we have $a^3=b^2$, so you can deduce the isomorphism wanted : $$\langle a,b \mid a^3=b^2\rangle\longrightarrow \langle x,y \mid xyx=yxy\rangle\\a\mapsto xy\\b\mapsto xyx\\x\leftarrow\!\shortmid a^{-1}b\\y\leftarrow\!\shortmid b^{-1}a^2$$

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  • $\begingroup$ $y$ is actually $ba^{-1}$. $\endgroup$ – Steve D Apr 22 '17 at 8:24
  • $\begingroup$ actually it is $b^{-1}a^2$ $\endgroup$ – Jennifer Apr 22 '17 at 8:36
  • $\begingroup$ umm..those are the same (given the relation in the group) $\endgroup$ – Steve D Apr 22 '17 at 8:37
  • $\begingroup$ my bad sorry ${}{}$ $\endgroup$ – Jennifer Apr 22 '17 at 8:40

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