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We know that if $n$ is an odd natural number then the dihedral group $D_{2n}$ is isomorphic to the group $D_n\times \mathbb{Z}_2$ where $D_m$ is the dihedral group of order $2m$, which has presentation $$ D_m:=\langle r, f\rangle =\langle r, f : r^m = f^2 = (rf)^2 =e \rangle \, . $$

My question is: what is the relation between $D_{2n}$ and $D_n$ if $n$ is even? Is there any such thing?

Thank you.

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    $\begingroup$ What are your thoughts on the problem so far? What have you tried? $\endgroup$ – Cameron Buie Dec 16 '15 at 5:56
  • $\begingroup$ Well, if $n$ is odd, then the relation is already there. For even, I think there should be some subgroup of $D_{2n}$ to be isomorphic to $D_n$. But no idea what else should I approach. :-( $\endgroup$ – Anjan3 Dec 16 '15 at 6:01
  • $\begingroup$ It would probably help (us and you) if you were to share the definition(s) of $D_m$ that you have. $\endgroup$ – Cameron Buie Dec 16 '15 at 6:05
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    $\begingroup$ How do you know the relation is there for odd $n$, have you proved it yourself? If you do you'll have a better feeling for whether odd $n$ is essential (and yes, because $n$-gons embed nicely into $2n$-gons, there is at least a subgroup isomorphic to $D_n$ regardless). $\endgroup$ – pjs36 Dec 16 '15 at 6:06
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    $\begingroup$ When $n$ is even the group $D_{2n}$ does not decompose as a direct product, although $D_n$ is isomorphic to a subgroup and a quotient group of $D_{2n}$. $\endgroup$ – Derek Holt Dec 16 '15 at 9:31

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