Showing $H=langle a,b|a^2=b^3=1,(ab)^n=(ab^{-1}ab)^krangle$.
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, $langle(ab)^nranglesubset Z(G)$.
Problem wants $H=langle ab,ab^{-1}ab rangle$ to be $G$. Clearly, $Hleqslant G$ and after doing some handy calculation which takes time I've got:
$ab^{-1}=(ab^{-1}ab)(ab)^{-1}in H$
$b=b^{-2}=(ab)^{-1}ab^{-1}in H$
$a=(ab)b^{-1}in H$
So $Gleqslant H$ and therefore $G=H=langle ab,ab^{-1}abrangle$.
For the second part, I should prove that $N=langle(ab)^nrangleleqslant Z(G)$.
Please help me.
Thanks.
group-theory group-presentation combinatorial-group-theory
edited Nov 30 '18 at 19:57
Shaun
8,820113681
asked Sep 6 '12 at 5:58
mrsmrs
1
|
show 1 more comment
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, $langle(ab)^nranglesubset Z(G)$.
Problem wants $H=langle ab,ab^{-1}ab rangle$ to be $G$. Clearly, $Hleqslant G$ and after doing some handy calculation which takes time I've got:
$ab^{-1}=(ab^{-1}ab)(ab)^{-1}in H$
$b=b^{-2}=(ab)^{-1}ab^{-1}in H$
$a=(ab)b^{-1}in H$
So $Gleqslant H$ and therefore $G=H=langle ab,ab^{-1}abrangle$.
For the second part, I should prove that $N=langle(ab)^nrangleleqslant Z(G)$.
Please help me.
Thanks.
group-theory group-presentation combinatorial-group-theory
edited Nov 30 '18 at 19:57
Shaun
8,820113681
asked Sep 6 '12 at 5:58
mrsmrs
1
1
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-Álvarez
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.
– mrs
Sep 6 '12 at 6:09
1
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.
– mrs
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. ;-)
– mrs
Jan 9 '13 at 11:14
|
show 1 more comment
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, $langle(ab)^nranglesubset Z(G)$.
Problem wants $H=langle ab,ab^{-1}ab rangle$ to be $G$. Clearly, $Hleqslant G$ and after doing some handy calculation which takes time I've got:
$ab^{-1}=(ab^{-1}ab)(ab)^{-1}in H$
$b=b^{-2}=(ab)^{-1}ab^{-1}in H$
$a=(ab)b^{-1}in H$
So $Gleqslant H$ and therefore $G=H=langle ab,ab^{-1}abrangle$.
For the second part, I should prove that $N=langle(ab)^nrangleleqslant Z(G)$.
Please help me.
Thanks.
group-theory group-presentation combinatorial-group-theory
edited Nov 30 '18 at 19:57
Shaun
8,820113681
asked Sep 6 '12 at 5:58
mrsmrs
1
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, $langle(ab)^nranglesubset Z(G)$.
Problem wants $H=langle ab,ab^{-1}ab rangle$ to be $G$. Clearly, $Hleqslant G$ and after doing some handy calculation which takes time I've got:
$ab^{-1}=(ab^{-1}ab)(ab)^{-1}in H$
$b=b^{-2}=(ab)^{-1}ab^{-1}in H$
$a=(ab)b^{-1}in H$
So $Gleqslant H$ and therefore $G=H=langle ab,ab^{-1}abrangle$.
For the second part, I should prove that $N=langle(ab)^nrangleleqslant Z(G)$.
Please help me.
Thanks.
group-theory group-presentation combinatorial-group-theory
group-theory group-presentation combinatorial-group-theory
edited Nov 30 '18 at 19:57
Shaun
8,820113681
asked Sep 6 '12 at 5:58
mrsmrs
1
edited Nov 30 '18 at 19:57
Shaun
8,820113681
asked Sep 6 '12 at 5:58
mrsmrs
1
edited Nov 30 '18 at 19:57
Shaun
8,820113681
edited Nov 30 '18 at 19:57
Shaun
8,820113681
edited Nov 30 '18 at 19:57
Shaun
8,820113681
8,820113681
asked Sep 6 '12 at 5:58
mrsmrs
1
asked Sep 6 '12 at 5:58
mrsmrs
1
asked Sep 6 '12 at 5:58
mrsmrs
1
1
1
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-Álvarez
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.
– mrs
Sep 6 '12 at 6:09
1
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.
– mrs
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. ;-)
– mrs
Jan 9 '13 at 11:14
|
show 1 more comment
1
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-Álvarez
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.
– mrs
Sep 6 '12 at 6:09
1
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.
– mrs
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. ;-)
– mrs
Jan 9 '13 at 11:14
1
1
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-Álvarez
Sep 6 '12 at 6:01
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-Álvarez
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.
– mrs
Sep 6 '12 at 6:09
@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.
– mrs
Sep 6 '12 at 6:09
1
1
but...$(ab)^n = (ab^{-1}ab)^k$ which clearly commutes with $ab^{-1}ab$
– David Wheeler
Sep 6 '12 at 6:20
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.
– mrs
Sep 6 '12 at 6:28
@DavidWheeler: I was on a wrong road. Thanks. Now it is obvious. I neglected (ab)^n=(ab^-1ab)^k. Thanks.
– mrs
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. ;-)
– mrs
Jan 9 '13 at 11:14
@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. ;-)
– mrs
Jan 9 '13 at 11:14
|
show 1 more comment
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1
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-Álvarez
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.
– mrs
Sep 6 '12 at 6:09
1
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.
– mrs
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. ;-)
– mrs
Jan 9 '13 at 11:14