let $D =
order of a group
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$\begingroup$
group-theory
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0@Arturo, my answer is in the following. – 2011-04-09
2 Answers
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The order of $D$ is $2n$ - in particular, this presentation gives the dihedral group. See here.
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0Whoops, that should be just n-gon, not 2n. – 2011-04-07
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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.