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

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.

• Observation: $xyxy^{-1}$ is equivalent to $[x,y]$ since $x^2=1$. – Shaun Apr 19 at 15:26

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.
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.
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.