Prove a group generated by two involutions is dihedral 
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: $m=n$
 A: 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. 
A: 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.
A: 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.
A: To me, the dihedral group $D_{2n}$ is the group $$\langle r,s:r^n=s^2=1,sr=r^{-1}s \rangle$$
If we define $r=xy,s=x$ then notice how $G$ is also generated by $r$ and $s$ (subject to the relations in $G$).
In addition to $r^n=s^2=1$ we have that $(sr)^2=1$ i.e. $sr=r^{-1}s^{-1}=r^{-1}s$. Therefore, $G$ must be a quotient of $D_{2n}$. Because $$G=\langle r,s:r^n=s^2=1,sr=r^{-1}s,\text{and any other relations in $G$}\rangle$$
$$\approx\frac{\langle r,s:r^n=s^2=1,sr=r^{-1}s\rangle}{\langle\text{other relations in $G$}\rangle}$$
If we can show $|G|=2n$ then there must be no other relations and $G\approx D_{2n}$. 
Now $G=\{s^ir^j:0\leq i<2,0\leq j<n\}$ and $s^ir^j=1$ forces $i=j=0$ (Gerry Myerson explained this bit in his answer). This proves that $|G|=2n$.
A: 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.
