# Isomorphism between two group presentation

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!

• Proper notation is $\langle a,b \mid a^3=b^2\rangle,$ not $<a,b|a^3=b^2>.$ Note \langle, \mid, \rangle. $\qquad$ – Michael Hardy Apr 22 '17 at 6:01
• Have you tried drawing the caylies graph? – Elad Apr 22 '17 at 6:01
• I didn't do cayley's graphs. Would it be easier? – Dac0 Apr 22 '17 at 6:18
• This is really not very difficult. You should try and do it yourself. – Derek Holt Apr 22 '17 at 8:24
• You are right, once you've seen it it's really easy – 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 : $$\langle a,b \mid a^3=b^2\rangle\longrightarrow \langle x,y \mid xyx=yxy\rangle\\a\mapsto xy\\b\mapsto xyx\\x\leftarrow\!\shortmid a^{-1}b\\y\leftarrow\!\shortmid b^{-1}a^2$$
• $y$ is actually $ba^{-1}$. – Steve D Apr 22 '17 at 8:24
• actually it is $b^{-1}a^2$ – Jennifer Apr 22 '17 at 8:36
• my bad sorry ${}{}$ – Jennifer Apr 22 '17 at 8:40