order of a group

let $D = <x, y | x^2, y^2, (xy)^n>$. What is the order of $D$? Thank you very much.

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You need to start showing some effort on these problems. Posting so many problems, most (all?) homework, in such a short period, and showing absolutely no work on your part is not going to endear you to the regulars. – Arturo Magidin Apr 7 '11 at 4:14
@Arturo, actually, these are not homework. I do not take the course now. They are some exercises of a quiz in the past. I would like to review these knowledge by myself. I know how to do some parts of it in some way. But I would not like to type so many. I would like to compare my answers with other people. – user Apr 8 '11 at 18:24
@Jianrong: Then you need to say that in your posts. Simply posting so many questions showing so little effort from you is, frankly, abusive on your part. If you are posting questions simply to "compare my answers with other people", you should tell people that. And, frankly, "I don't like to type so much" is a rather bad excuse for your behavior. – Arturo Magidin Apr 8 '11 at 18:28
@Arturo, my answer is in the following. – user Apr 9 '11 at 1:30

The order of $D$ is $2n$ - in particular, this presentation gives the dihedral group. See here.
The number of the elements of $D$ can be counted directly. The elements of $D$ are of the form $x^{\alpha_1}y^{\beta_1}\cdots x^{\alpha_m}y^{\beta_m}$ for some integers $\alpha_i, \beta_i \geq 0$ and $m\geq 0$. By using the relations $x^2=1$, $y^2=1$ and $(xy)^n=1$, the distinct elements of $D$ are \begin{align*} 1, xy, (xy)^2, \cdots, (xy)^{n-1}, \end{align*} \begin{align*} y, x, (xy)^2y=xyx, \cdots, (xy)^{n-1}y=(xy)^{n-2}x. \end{align*} So $D$ has 2n elements.