Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is my problem:

Let $G=\langle a,b|a^l=b^3=1,(ab)^3=(a^{-1}b)^3 \rangle$. Find the order of $\frac{G}{G'}$ and then verify that if $G$ is metabelian.

What I have done: I added the relation $[a,b]=1$ to $G$'s relations and found that:

  1. $\frac{G}{G'}=\langle a,b|a=b^3=1,[a,b]=1 \rangle\cong\mathbb Z_3$

  2. $\frac{G}{G'}=\langle a,b|a^2=b^3=1,[a,b]=1 \rangle\cong\mathbb Z_6$

  3. $\frac{G}{G'}=\langle a,b|a^3=b^3=1,[a,b]=1 \rangle\cong\mathbb Z_3\times\mathbb Z_3$

  4. $\frac{G}{G'}=\langle a,b|a^6=b^3=1,[a,b]=1 \rangle\cong\mathbb Z_3\times\mathbb Z_6$

for the last part of the question, I have to probe if $G'$ is abelian or not. This is my idea:

For 1. ; Since $a=1$ is a relation of the quotient group so, $a\in G'$ and also $b^3\in G'$. This means that any conjugations of $a$ or $b^3$ would be in the derived subgroup as well. For example $b^{-1}ab$ or $a^{-1}b^3a$. Now I think to use such these conjugations and the relations of $G$ simultaneously, to find a presentation of $G'$ and see that if this subgroup is abelian.

This would be really a long harsh road :) solving the problem just in the case 1. Honestly, I don't know any way but Todd-Coxeter Algorithm for such these problems. May I ask to note me some useful hint? Thanks

share|cite|improve this question
A simple computer calculation shows that it is not metabelian for $l=2$, and $G$ is infinite in that case. It is metabelian of order $81$ for $l=3$. For $l=5$, it is finite of order 375 and not metabelian. Are you sure you have the presentation correct? – Derek Holt Oct 4 '12 at 8:43
@DerekHolt: This problem was given to me as you see. What you noted make me doubt if it is printed correct. I didn't know about your calculation. – Babak S. Oct 4 '12 at 8:56
@DerekHolt: Your conclusions show this problem leads some strange results so it can't be a routine and practical problem. Thanks for your time prof. Holt. – Babak S. Oct 4 '12 at 9:07
For $l$ odd the group is apparently finite of order $3\cdot l^3$, $l<40$ – i. m. soloveichik Oct 4 '12 at 15:45
@i.m.soloveichik: Thanks for your attempt here. Maybe the presentation was printed wrong. It took more time of me in a dark air. :( – Babak S. Oct 4 '12 at 16:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.