# $H=\langle a,b|a^2=b^3=1,(ab)^n=(ab^{-1}ab)^k\rangle$

Let $G=\langle a,b|a^2=b^3=1,(ab)^n=(ab^{-1}ab)^k \rangle$. Prove that $G$ can be generated with $ab$ and $ab^{-1}ab$. And from there, $\langle(ab)^n\rangle\subset Z(G)$.

Problem wants $H=\langle ab,ab^{-1}ab \rangle$ to be $G$. Clearly, $H\leqslant G$ and after doing some handy calculation which takes time I've got:

1. $ab^{-1}=(ab^{-1}ab)(ab)^{-1}\in H$

2. $b=b^{-2}=(ab)^{-1}ab^{-1}\in H;\\ a=(ab)b^{-1}\in H$

So $G\leqslant H$ and therefore $G=H=\langle ab,ab^{-1}ab\rangle$. For the second part, I should prove that $N=\langle(ab)^n\rangle\leqslant Z(G)$. May I ask to help me? Thanks.

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It is clear that $(ab)^n$ commutes with $ab$; since you proved that $G=H$, $G$ is generated by $ab$ and by $ab^{-1}ab$, so to show that $(ab)^n$ is in $Z(G)$ you are left with showing that $(ab)^n$ also commutes with $ab^{-1}ab$. Can you do that? –  Mariano Suárez-Alvarez Sep 6 '12 at 6:01
@MarianoSuárez-Alvarez: Oh yes. I see what you wanted. (ab)^n(ab)=(ab)(ab)^n is clear to me now. Thanks Mariano. I am doing the second. –  Babak S. Sep 6 '12 at 6:09
but...$(ab)^n = (ab^{-1}ab)^k$ which clearly commutes with $ab^{-1}ab$ –  David Wheeler Sep 6 '12 at 6:20
@DavidWheeler: I was on a wrong road. Thanks. Now it is obvious. I neglected (ab)^n=(ab^-1ab)^k. Thanks. –  Babak S. Sep 6 '12 at 6:28
@AmirHoseinSadeghiManesh: My dear brother, in that time I was working on some problems including presentation of groups. I think this problem had some defects in the body, so it was left unsolved here. Eltemase 2a daram. ;-) –  Babak S. Jan 9 '13 at 11:14