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I just got these presentations of groups:

$\langle a,b\mid aba^{-1}b^{-1}\rangle$

$\langle a,b\mid aba^{-1}b^{-2},bab^{-1}a^{-2}\rangle$

$\langle a,b\mid abab^{-1}\rangle $

Are any of them trivial? How do you prove it?

Zev Chonoles
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sheldoor
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    A group given by a presentation with more generators than relators must be infinite, never mind trivial! – user1729 Nov 25 '11 at 19:59
  • For the second group, I will mention http://math.stackexchange.com/questions/66573/a-particular-two-variable-system-in-a-group/66632#66632 –  Nov 26 '11 at 21:43

2 Answers2

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Standard techniques that work with the kinds of groups you've given is to compute the abelianization of the group. At least one of your groups is abelian - so you can describe it using the classification of finitely-generated abelian groups.

Sometimes the abelianization gives you further ideas, like trying to prove your group is a semi-direct product. There's something called the Reidemeister-Schreir technique that will help you with this. Your 3rd group appears to be a semi-direct product.

Ryan Budney
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    I assume you actually know the third group very well; it's the fundamental group of the Klein bottle. –  Nov 26 '11 at 21:45
  • @user641: What you assume is irrelevant to this forum. This isn't a forum for anonymous users to solve people's homework problems. – Ryan Budney Jan 12 '17 at 18:27
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For the first and last group, try finding a homomorphism from your group $G$ to an abelian group. If you can find an onto map $G \to A$ where $A$ is nontrivial, then $G$ is non-trivial. If you have a group $G=\langle a,b\mid r\rangle$ and you decide you want to map $G\to A$ by $a\mapsto x$ and $b \mapsto y$, you get a homomorphism if and only if the result of substituting $x$ for $a$ and $y$ for $b$ in $r$ is trivial.

For the middle one, you've got $aba^{-1}=b^2$ and $bab^{-1}=a^2$ (1). Rewriting the first one gives $ab (ba^{-1}b^{-1}) = b$. Work out $ba^{-1}b^{-1}$ from (1) and substitute...

Zev Chonoles
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Matthew Towers
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  • thanks but are there any rules of rewriting like you did? – sheldoor Nov 25 '11 at 20:05
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    No. (Actually it's a theorem that the answer is no: see http://en.wikipedia.org/wiki/Word_problem_for_groups - there's no general algorithm that will decide whether or not a given word is the identity). So every time you can do it, it's a little victory against the universe. – Matthew Towers Nov 25 '11 at 21:02