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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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    Also related: http://math.stackexchange.com/questions/487128/prove-that-a-group-generated-by-two-elements-of-order-2-x-and-y-is-isomo2014-01-16

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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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    Really elegant!2018-04-02
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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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define generators: $a = xyx$, $b = xy$

Clearly a is an involution since it is conjugate with $y$. Further, this generates the group since it generates $x$ and $y$.

Next, we note that $aba = xyxxyxyx = xyyxyx = xxyx = yx$ which is the inverse of $b$. Since $a$ is an involution and $a,b$ generate the group, this means that conjugating any element of the group with $a$ will yield its inverse since this is true for all group generators.

Now, consider any word sequence of $a,b,b^{-1}$. Using the relation we just proved, we can always move the factors of $a$ and reduce this to an sequence of the form $a b^k$ if the number of $a$ factors is odd, or $b^k$ if the number of factors is even.

If $b$ has finite order, then clearly the group is dihedral. If it has infinite order, then it is the infinite dihedral group.