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I want to show $$\langle a,b\mid ab^2＝b^3a, ba^2＝a^3\rangle$$ is isomorphic to a trivial group.

My text book proved this like following.

$bab^2a＝a^4b^2a\to ba＝a^4b^2$

$a^4b^2a＝a^3b$ (omitted some process), so $ab^2a＝b$

$b^2a^2＝1, a＝a^4b^3a b^3a^2＝1, b^3＝b^2\to b＝1, a＝1$

This proof is too difficult for me to imitate because there is no oriented strategy of each process. Firstly, I cannot understand why we should start from $bab^2a$.

Other way to prove is also appreciated, I would be happy if you could give me an strategy of deformation.

Shaun
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  • Which textbook are you referring to? – Shaun Jan 07 '20 at 13:34
  • https://math.stackexchange.com/questions/66573/a-particular-two-variable-system-in-a-group does this answer your question? – Math Jan 07 '20 at 13:38
  • This answer is from my teacher's text and not open to everyone, sorry..The second answer is what I wanted, thank you very much. –  Jan 07 '20 at 13:45
  • I am reopening this question as the linked question is different from this one, and is very difficult. Also, the OP reasked this question here. – user1729 Jan 08 '20 at 09:24
  • I also cannot follow the working you have copied from your text book. Is it possible for you to upload a picture of the question+solution? (Just to ensure that there are no typos.) – user1729 Jan 08 '20 at 09:37
  • I’m in the process of proving a similar statement. I have only briefly read the hints and answers since I still want to figure it out by myself. Is any of the following two approaches valid? $(1)$ Let $F_{{a,b}} = F$, and $\phi: F \to G$ be a homomorphism whose kernel $H$ is the normal subgroup generated by the relations that we have. If I can show that $H$ is all of $F$ (i.e. for any elements of $F$, the relations hold), then $F/H$ consists of just one element. And we know $F/H \cong G$, so $G$ must be trivial. – ensbana Mar 16 '21 at 09:01
  • $(2)$ Using Tietze transformations (https://en.wikipedia.org/wiki/Tietze_transformations), I’m hoping to simplify the given presentation into something like $\langle a | a \rangle$. – ensbana Mar 16 '21 at 09:03
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    @ensbana Yes, both those approaches are valid. However, you should be aware that proving triviality is a hard problem in general. The question here is easy, while the question linked in the second comment (this question) is hard. – user1729 Mar 16 '21 at 10:28

1 Answers1

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One way to see that $\langle a,b\mid ab^2＝b^3a, ba^2＝a^3\rangle$ defines the trivial group is to use the second relator as follows: $$\begin{align*} ba^2&=a^3\\ \Rightarrow ba^2\cdot a^{-2}&=a^3\cdot a^{-2}\\ \Rightarrow b&=a \end{align*}$$ Then rewrite $b\rightarrow a$ in the first relator: $$\begin{align*} ab^2&=b^3a\\ \Rightarrow a^3&=a^4\\ \Rightarrow a&=1, b=1\end{align*}$$ Hence, the group is trivial.

Note that there is no general algorithm to decide whether or not a group is trivial, but if a group is trivial then there is a procedure which will prove this. See here for more details.

user1729
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