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In Serre's book "Trees" on page 10 the following exercise is given:

Show that the group defined by the presentation $$x_2x_1x_2^{-1}=x_1^2, \hspace{7pt} x_3x_2x_3^{-1}=x_2^2, \hspace{7pt} x_1x_3x_1^{-1}=x_3^2$$ is trivial.

Comparing to what was done before, clearly, the approach he used to prove that a smilarily defined group is infinite - will fail.
My question is - how would you approach this? just sequentailly substiuting one word into another to show that, say, $x_1=1$? Is there any smarter way that just brute-forcing something that cancels out?

Thanks in advance for any help.

N.B: Being an exercise from "Trees", I would expect it to be asked here before. I did my best trying to find it - but couldn't. So I'm sorry if it turns out to be a duplicate...

Dennis Gulko
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    This also is mentioned in http://math.stackexchange.com/questions/479835/higman-group-with-3-generators-is-trivial . Also http://mathoverflow.net/questions/51718/about-an-exercise-in-serres-trees . Noone bothered to prove the triviality :/ – darij grinberg Jul 03 '15 at 11:52
  • So, judging by the comments there, my approach was correct. But then this exercise seems completely pointless... – Dennis Gulko Jul 03 '15 at 12:00
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    More beating around the bush: http://math.stackexchange.com/questions/661636/does-this-group-presentation-define-a-nontrivial-group – darij grinberg Jul 03 '15 at 12:02
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    As to the guys who voted to close: The question you offer does, indeed, contain an answer, but not the question directly. In fact, in that answer it is said that "It is well known that..." – Dennis Gulko Jul 03 '15 at 14:06
  • Because I didn't find that question before posting my question... – Dennis Gulko Jul 04 '15 at 19:48

1 Answers1

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Observe first that $$ x_2x_1=x_1^2x_2,\quad x_3x_2=x_2^2x_3,\quad x_1x_3=x_3^2x_1 $$ and more generally $$ x_2^jx_1^k = x_1^{2^jk} x_2^j,\qquad x_3^jx_2^k = x_2^{2^jk} x_3^j,\qquad x_1^jx_3^k = x_3^{2^jk} x_1^j $$ for any $j,k\in\mathbb{N}$. Then $$ x_1^2 x_2^2 x_3 \;=\; x_1^2 x_3 x_2 \;=\; x_3^4 x_1^2 x_2 \;=\; x_3^4 x_2 x_1 \;=\; x_2^{16}x_3^4x_1 \;=\; x_2^{16}x_1x_3^2 \;=\; x_1^{2^{16}}x_2^{16}x_3^2, $$ and solving for $x_3$ gives $$ x_3 \,=\, x_2^{-16}x_1^{2-2^{16}}x_2^2. $$ Then $$ x_2^2 \;=\; x_3x_2x_3^{-1} \;=\; x_2^{-16}x_1^{2-2^{16}}x_2 x_1^{2^{16}-2}x_2^{16} \;=\; x_2^{-16}x_1^{2^{16}-2}x_2^{17} $$ and it follows that $x_2 = x_1^{2^{16}-2}$. In particular, $x_2$ and $x_1$ must commute, so the relation $$ x_2x_1 = x_1^2x_2 $$ proves that $x_1 = 1$, and hence $x_2=1$ and $x_3=1$.

Note: The main trick here was the initial string of equations. In general, we you have relations in a group that pairs of elements "almost" commute, e.g. $x_2x_1=x_1^2 x_2$, you can get a lot of mileage from trying to implement the equations $$ abc \;=\; acb \;=\; cab \;=\; cba \;=\; bca \;=\; bac \;=\; abc. $$ For "real" commutation you get the same $a$, $b$, and $c$ at the end, but for "fake" commutation you usually get something slightly different than what you started with, in this case something with $x_3^2$ instead of $x_3$. Once we found an expression for $x_3$ in terms of $x_1$ and $x_2$ the rest was pretty straightforward.

Jim Belk
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  • I found this really interesting, especially your illuminating comment at the end about the main trick. I wonder if you know off the top of your head any other examples where this trick can be applied? – fourier1234 Jul 08 '19 at 09:38
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    @fourier1234 I've seen this trick before, but I don't remember any specific examples. However, I can give some geometric insight. In the Cayley graph of a group, commutation relations are "squares", and the six squares in the string of equations $abc=acb=\cdots = abc$ fit together to make a cube. For "fake" commutation relations, the six squares don't quite fit together, so intuitively you get something like a cube but with one extra face that represents a new relation. Going through the six equations is how you derive this new relation from the almost-squares. – Jim Belk Mar 19 '21 at 00:45