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I encountered this problem in Sims' "Computation with Finitely Presented Groups".

Show that $\langle x,y\mid x^2, y^3, xyxy^{-1}, (xy)^7\rangle$ is trivial.

The book uses coset enumeration or something similar but I haven't got up to that point yet - it's on page 231 after all - so I don't quite understand the derivation of the fact that it's trivial.

I was introduced to coset enumeration in Johnson's "Presentation$\color{red}{s}$ of Groups (Old Version)". I didn't quite understand it then either.

For my own contribution to this post, I have run the presentation through GAP, like so:

gap> F:=FreeGroup(2);
<free group on the generators [ f1, f2 ]> 
gap> gens:=[(F.1)^2, (F.2)^3, (F.1)*(F.2)*(F.1)*(F.2)^(-1), ((F.1)*(F.2))^7];
[ f1^2, f2^3, f1*f2*f1*f2^-1, (f1*f2)^7 ] 
gap> G:=F/gens;
<fp group on the generators [ f1, f2 ]> 
gap> Size(G);
1

As you can see, this is not very edifying.

I want to understand it, not just compute it.

Please help :)

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    $\begingroup$ Observation: $xyxy^{-1}$ is equivalent to $[x,y]$ since $x^2=1$. $\endgroup$ – Shaun Apr 19 at 15:26
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It follows from your observation that $[x,y]=1$ that the group is abelian. It is also finitely generated so by the structure theorem of finitely generated abelian groups it has to take the form $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$ where $n,m\in\mathbb{N}$ ($m,n$ could be zero). From the fact that $x^2=y^3=1$ we have that the group is either trivial or is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$ (additive) as these are the only such groups with generators of order $2$ and $3$. But the latter doesn't satisfy that $(xy)^7=1$ (in additive writting $7(x+y)=0$) hence the group must be trivial.

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  • $\begingroup$ I've posted an answer of my own :) $\endgroup$ – Shaun Apr 19 at 18:39
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    $\begingroup$ @Shaun Looks good. To be honest, you pretty much solved half the question with your observation right from the beginning. $\endgroup$ – Yanko Apr 19 at 19:13
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It should be clear that $\langle \,x,y\mid x^2, y^3, xyxy^{-1}\,\rangle$ is $\Bbb Z/6\Bbb Z$ (and we can choose the isomorphism so that $x=3+6\Bbb Z$ and $y=2+6\Bbb Z$), hence our $G$ is a quotient thereof. Note that the final condition means that $(xy)^7$, or $-1+6\Bbb Z$, is in the kernel, hence that quotient is trivial.

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  • $\begingroup$ I've posted an answer of my own :) $\endgroup$ – Shaun Apr 19 at 18:39
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Since $[x, y]$ is a relator, the group is abelian. Thus $$\begin{align}1&=(xy)^7\\ &=x^7y^7\\ &=(x^2)^3x(y^3)^2y\\ &=xy, \end{align}$$ so $x=y^{-1}$, meaning $x^2=x^3$, so both $x$ and $y$ are $e$. Hence the group is trivial.

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