# Group generated by $x$,$y$|$x^2=y^2=1$

Classify groups generated by two elements $x$, $y$ of order 2.

First, please help clarify: Is it necessary for $x$ and $y$ to be distinct elements? (above is the whole question as seen in the book).

Here's what I have so far:

Let $xy=z$, and let $|z|=n$. Then G would be isomorphic to dihedral group $D_n$

If $x=y$ then $G$ is just $\langle x\rangle$

Would $\langle xy \rangle$ be such a group?

Thanks for any help.

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You can make the $\langle$ and $\rangle$ symbols by using \langle and \rangle. –  JavaMan Nov 20 '12 at 4:25
It is not necessary for $x$ and $y$ to be distinct elements, but if they are equal then the group has order 2 (or possibly 1 if you are allowing $x=y=1$). –  Derek Holt Nov 20 '12 at 9:19

The group $\langle x,y\mid x^2,y^2\rangle$ is the infinite dihedral group $D_\infty$.
In this group, $x$ and $y$ are distinct elements. One way to see this is to notice that there is a group homomorphism $\phi:D_\infty\to E$ where $E$ is the group of euclidean motions of the line, such that $$\phi(x):t\in\mathbb R\mapsto -t\in\mathbb R$$ and $$\phi(y):t\in\mathbb R\mapsto 1-t\in\mathbb R.$$ Indeed, once you show that $\phi$ exists, then obviously $\phi(x)\neq\phi(y)$, so $x\neq y$.
If $G$ is a group generated by two elements of order two, then there exists a surjection $D_\infty\to G$, so what you want to do is to classify all normal subgroups of $D_\infty$. To do this, it is useful to observe that the element $z=xy$ has infinite order in $D_\infty$ (this is obvious if you see what $\phi(z)$ is) and then show that a non-trivial normal subgroup must contain a power of $z$.
Every element of the group must be a finite word in $x$ and $y$, but we don't need words with $xx$ or $yy$ in them because $xx=yy=1$, so you have a very simple description of what all the elements look like. Then think about finding the order of $xy$.