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Let $$G = \langle a,b: ab^2 = b^3, ba^3 = a^2b\rangle.$$ Show that

  1. $a^2b^4a^{-2}= b^9$

  2. $a^3b^4a^{-3}= b^9$ (Hint: Use $ba^3 = a^2b$ and 1.)

  3. $|G| = 1$. (Hint: Use 1 and 2.)

I have successfully shown 1 but I am currently struggling on how to show 2 using the provided hints and don't know how to approach 3. I would appreciate any help in furthering my understanding of this.

Edit: My solution for 1.

$a^2b^4a^{-2} = aab^2b^2a^{-2} = a(ab^2)b^2a^{-2} = ab^3ab^2a^{-2} = ab^3(ab^2)a^{-2} = ab^3b^3aa^{-2} = ab^6a^{-1} = ab^2b^4a^{-1} = b^3ab^4a^{-1} = b^3(ab^2)b^2a^{-1} = b^3b^3ab^2a^{-1} = b^6(ab^2)a^{-1} = b^6b^3aa^{-1}=b^9$

waterr
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    Please provide more context. For example, share your solution 1. – Shaun Mar 14 '24 at 15:49
  • Compare with this duplicate, interchanging $a$ and $b$. No need to compute $b^9$. – Dietrich Burde Mar 14 '24 at 15:51
  • That is no duplicate, @DietrichBurde. This problem concludes that $G$ is trivial, whereas that answer uses that $G$ is trivial. – Shaun Mar 14 '24 at 15:55
  • @shaun No, it doesn't. It shows: 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: So it is shown that the group is trivial. – Dietrich Burde Mar 14 '24 at 15:56
  • My mistake: I didn't read it properly, I guess, @DietrichBurde. Sorry! – Shaun Mar 14 '24 at 15:58
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    Alright, then let me close it, before the OP tries to work with this unnecessarily complicated computation. – Dietrich Burde Mar 14 '24 at 15:59
  • @DietrichBurde but the question specifically wants me to compute for $b^9$ and then use the fact that since those elements equal $b^9$ to show that the group is trivial – waterr Mar 14 '24 at 16:04
  • Don't just go by what the questions wants. The question can be solved much better, so why to follow this. Your teacher will be delighted when you come up with a better solution. – Dietrich Burde Mar 14 '24 at 16:08

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