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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: $m=n$

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You know your group is a quotient of the presentation you give, but you have not established that $G$ satisfies no further relations that are not a consequence of these three. So you cannot simply claim that $G$ is given by that presentation just from knowing the order of $xy$. –  Arturo Magidin Jun 19 '12 at 3:46
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How do you know $xy$ has finite order? –  Gerry Myerson Jun 19 '12 at 3:47
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P.S. What is your definition of dihedral? –  Arturo Magidin Jun 19 '12 at 3:48
    
I mean a finite group. Yes Vondyck's thm implies G is a quotient of a Dyhedral group of order $2n$, but how to show it does not have further relations? Does it work if I replace the $n$ in the proposed presentation of $G$ by $m$ and then show $m=n$?(I just found this idea from a text but do not understand the reason) –  user31899 Jun 19 '12 at 8:54
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@user31899: If you mean "finite", then you are incorrect; there are groups generated by two involutions that are not finite, e.g., the free product of two copies of the cyclic group of order $2$, $C_2*C_2$. –  Arturo Magidin Jun 19 '12 at 14:11

3 Answers 3

up vote 3 down vote accepted

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.

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More algebraicaly: If $xy$ has order $n$, then note that $\langle xy \rangle \lhd \langle x, y \rangle $ in this case. Now neither $x$ nor $y$ is in $\langle xy \rangle ,$ (if one is, the other is, and then $\langle x, y \rangle$ is cyclic, forcing $x = y,$ a contradiction. Clearly we have $\langle x,y \rangle = \langle x \rangle \langle xy \rangle,$ so we have $|\langle x,y \rangle| = 2n.$ Since $\langle x,y \rangle$ is a homomorphic image of a dihedral group with $2n$ elements, it is itself dihedral with $2n$ elements.

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One geometric way to do this is to let $X=\langle x\rangle$ and $Y=\langle y\rangle$ be subgroups of $G=\langle x,y\rangle$, so that $|X|=|Y|=2$. We can then form a graph $\Gamma$ (a Tits geometry) where:

  • the vertices are the right cosets of $X$ and $Y$;
  • there is an edge between $Xg_1$ and $Yg_2$ precisely when $Xg_1\cap Yg_2\neq\emptyset$.

You can then check the following properties easily:

  • $\Gamma$ is connected, because $X$ and $Y$ generate $G$;
  • every vertex of $\Gamma$ has valence $2$, because $|X|=|Y|=2$.

Now $\Gamma$ is finite if and only if $G$ is finite. If $\Gamma$ is finite, then it is a polygon with $|G|$ sides, and $G$ acts on this polygon (by right translation). You can check $xy$ acts by a rotation, while $x$ (and also $y$) act by a reflection. [Even if you don't check this, $G$ is acting on a polygon by plane isometries,so...] $G$ is thus dihedral (see below).

If $\Gamma$ is infinite, it then looks like a copy of the real line; again $G$ acts on this space, in such a way that it is infinite dihedral.

Note: For the finite case, the polygon you get is two times too big. This can be remedied by alternately coloring the edges red and blue; $G$ will then always send red edges to red edges, etc., and so the action is really on the "appropriately" sized polygon.

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