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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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    What is your definition of the infinite dihedral group?2012-05-29
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    I know it is just generated by two elements of order 2 and no relation is there at the presentation of it.2012-05-29
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    If you know that the finite dihedral group $D_{2n}$ is isomorphic to the semidirect product of the cyclic group of order $n$ and the cyclic group of order 2, then it should be clear that your $D_\infty$ is the semidirect product of $\mathbb{Z}$ and the cyclic group of order 2, which is (one of) the definition(s) of the infinite dihedral group.2012-05-29
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    Yes, I got it. Thanks Chris. Thanks Turgeon.2012-05-29
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    I just forget that I can take $k=1$, the generator of Z.2012-05-29
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    The first matrix in Chris's comment has infinite order, not $\,2$ .2012-05-29
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    @DonAntonio: Yes, I meant of course to write the two matrices you give in your answer. I'll delete my comment.2012-05-29
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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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    $\LaTeX$ possibility: rather than align the $0$'s manually, you can use `$$\left(\begin{array}{rc}...\end{array}\right)$$` and have the column with the minus signs aligned on the right, instead of in the center. Of course, you have to manually put in the parentheses, so you may prefer to use multiple thin spaces and do it manually anyway.2012-05-29
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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