Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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: $m=n$

share|cite|improve this question
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
How do you know $xy$ has finite order? – Gerry Myerson Jun 19 '12 at 3:47
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
@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
up vote 4 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.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.