Prove a finite group generated by two involutions is dihedral
Is my following argument correct?
Let $G=\langle x,y\rangle$ be a group generated by involutions $x,y$. Let $n=\mathrm{ord}(xy)$ to get a presentation $G=\langle x,y\mid x^2=y^2=(xy)^n=1\rangle $ so G is dihedral of order $2n$ ?
Further note: I realise now my argument is not sufficient as it remains to show $G$ has no other relations.
I just found an idea from a reference which claims "…So $G$ must have a presentation of the form $G=\langle x,y\mid x^2=y^2=(xy)^m=1\rangle $, then one has to show $m=n$…" in which I do not understand why $G$ has exactly a presentation of such form (the presentation inovlves $m$)? That reference also showed $|\langle x,y\rangle |=2n$ which directly led to the conclusion: $m=n$
Best Answer
If $G$ is finite and has generators $x,y$ of order 2, then the elements of $G$ are $x,xy,xyx,xyxy,xyxyx,\dots$ and $y,yx,yxy,yxyx,yxyxy,\dots$ and as soon as you know the first term in those lists to give you the identity element, you're done. It can't be an element like $xyxyx$, because if that's the identity then you multiply left and right by $x$ to find $yxy$ is the identity, and you multiply left and right by $y$ to find $x$ is the identity. So the defining relation must be $(xy)^m=1$ for some positive integer $m$ (note that $(yx)^m=1$ if and only if $(xy)^m=1$).
So your presentation is $$\langle x,y\mid x^2,y^2,(xy)^m\rangle$$ and you seem happy to accept that as dihedral.