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I am working on a problem saying that:

If $D_{2n}=\left\{\begin{pmatrix}\epsilon & k \\ 0 & 1\end{pmatrix}|\epsilon=±1,k\in\mathbb Z_n \right \}$ then for any $n\in\mathbb N$, $D_{2n}$ is a quotient group of $D_\infty=\left\{\begin{pmatrix}\epsilon & k \\ 0 & 1\end{pmatrix}|\epsilon=±1,k\in\mathbb Z \right \}$ May I ask if this later group is infinite dihedral group as we have already known? May I ask how it can be? Thanks.

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    @Chris, that's what I thought. No need to erase the comment, though. Thanks.2012-05-29

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An idea: it seems reasonable easy to show your group is generated by $\,\alpha:=\begin{pmatrix}-1&0\\\,\,\,\,\,0&1\end{pmatrix}\,,\,\beta:=\begin{pmatrix}-1&1\\\,\,\,\,\,0&1\end{pmatrix}$ and also $\,\,\alpha^2=\beta^2=1\,$...and that's all, so in fact this group is the free product $\,\,C_2*C_2\,$ which, as

it happens, is the infinite dihedral group.

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    Thanx a bunch. I tried to do the {rc} thing with the begin{pmatrix} command but it did nothing, so I assumed the LaTeX version supported by this site doesn't have that command, or that it doesn't work now.2012-05-29