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!

  • 1
    $\begingroup$ 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

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 : $$\langle a,b \mid a^3=b^2\rangle\longrightarrow \langle x,y \mid xyx=yxy\rangle\\a\mapsto xy\\b\mapsto xyx\\x\leftarrow\!\shortmid a^{-1}b\\y\leftarrow\!\shortmid b^{-1}a^2$$

  • $\begingroup$ $y$ 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

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.