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

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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 🙂

Best Answer

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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