I am trying to find out what the following group is:

$$G = \langle a, b \mid ab^2 = b^2a,\ a^4 = b^3\rangle.$$

Due to the isomorphism problem for groups, there is no algorithmic way to approach questions like this in general. The only technique I know of is to consider the abelianisation of $G$, which, if I'm not mistaken, is the group given by the same generators and relations, together with the additional relation $ab = ba$. In this case, aside from the fact that you can remove the first relation, it doesn't seem be any simpler to determine.

So my questions are as follows:

  1. What is the group $G$?
  2. Without using the answer to the first question (i.e. using only the presentation), what is the abelianisation of $G$?

In addition, any descriptions of other techniques that one can use to get a better understanding of a group from a presentation would be very much appreciated.

  • $\begingroup$ How do you know it is only one group(not a hint but a question.) $\endgroup$ – Jorge Fernández Hidalgo Dec 19 '13 at 2:38
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    $\begingroup$ You are not mistaken about the abelianisation. It is easy to check that $\langle a,b|a^4=b^3,ab=ba\rangle$ is generated by $ab^{-1}$ which is of order ${\rm lcm}(3,4)=12$, so is $C_{12}$. @user4140 Do you understand what a presentation of a group is? | Oops $ab^{-1}$ has infinite order, I am confusing $a^4=b^3$ with $a^4,b^3$. (Re: the below comment by B.S.) $\endgroup$ – anon Dec 19 '13 at 2:47
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    $\begingroup$ @anon: GAP tells that the group is infinite. $\endgroup$ – mrs Dec 19 '13 at 2:50

Firstly, $a$ commutes with $b^2$ by the first relation, while clearly $a$ commutes with $a^4$ so $a$ commutes with $b^3$ by the second relation. This means that the subgroup $H=\langle a, b^2, b^3\rangle$ is abelian, as the generators pairwise commute. However, $b^3b^{-2}=b\in H$ and so $G=H$. Thus, $G$ is an abelian group. As has already been noted, the abelianisation of $G$ is infinite cyclic, and thus we conclude that $G$ is infinite cyclic.

  • $\begingroup$ Curiously, the same answer (but different method) as mine, so not just the cyclicity of the abelianization, but that of the whole group had already been notred. $\endgroup$ – Igor Rivin Dec 19 '13 at 14:07
  • $\begingroup$ @IgorRivin I was unsure if you were concluding that the group was infinite cyclic, or you just thought that it probably was (you say should be). If the former, I was unsure how you were concluding (although I get your point now - saying something like "therefore, $G$ is generated by $b^2$ and by $a$" would make it clear). $\endgroup$ – user1729 Dec 19 '13 at 14:23
  • $\begingroup$ Oh, come on. I show the group is abelian and then note that the abelianization is $\mathbb{Z}.$ How much more needs to be said? $\endgroup$ – Igor Rivin Dec 19 '13 at 14:25
  • $\begingroup$ @IgorRivin You seem to be implying that I somehow "stole" your answer. As you point out, our answers are the same (which is good - there is only one answer) and, moreover, my method is different from yours. I am unsure why you seem to be taking offence over this... $\endgroup$ – user1729 Dec 19 '13 at 14:35

The abelianization is generated by $c=a^{-1}b$ because $c^{3}=a^{-3}b^3=a$ and $c^4=ac=b$. Hence all relations become redundant and we end up with $\mathbb Z$.

  • $\begingroup$ Nicely Done, Hagen. $\endgroup$ – mrs Dec 19 '13 at 2:52
  • $\begingroup$ Could you explain why you chose to look at the element $a^{-1}b$? Also, do you know what the original group $G$ is? $\endgroup$ – Michael Albanese Dec 19 '13 at 2:55
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    $\begingroup$ @MichaelAlbanese The relation $a^4=b^3$ can be written as simply $a^{4}b^{-3}$ (it is standard to suppress $=e$ in presentations), which is $a(ab^{-1})^3$ or $(ab)^4b^{-1}$ in the abelianisation (as $a,b$ commute), which tells us $(ab^{-1})^3=a^{-1}$ and $(ab^{-1})^4=b$, so $a$ and $b$ are obtained from $ab^{-1}$ so it is a generator. (Hagen has it backwards from me, $a^{-1}b$, but same idea.) $\endgroup$ – anon Dec 19 '13 at 3:06
  • $\begingroup$ @anon: Thanks. That is very helpful. $\endgroup$ – Michael Albanese Dec 19 '13 at 3:14

$b^2$ commutes with $a,$ (by the first relation), so every element in the group can be written as $$(b^2)^k \prod (a^{i_j}b^3)^l a^m=b^{2k} a^n,$$ (by the second relation). Given the abelianization, the group should be $\mathbb{Z}.$

  • $\begingroup$ Can you write $b$ in that form? $\endgroup$ – Hagen von Eitzen Dec 19 '13 at 3:04
  • $\begingroup$ @HagenvonEitzen Yes, since $a^4=b^3,$ then $b=b^{-2}a^4$ $\endgroup$ – Igor Rivin Dec 19 '13 at 3:12
  • $\begingroup$ Of course, the original form was written too quickly, this is now corrected (but no substantial change). $\endgroup$ – Igor Rivin Dec 19 '13 at 3:53

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