What can be said about the group generated by the union of the two dihedral group $D_n$ and $D_m$, for all natural numbers $n, m$? Is it $D_{lcm(n,m)}$? After some heuristics it does seem to me the union of the isometries of the rectangle and that of the square yields the isometries of the dodecagon.


Edit: the dihedral groups are seen as sub groups of $O(2,R)$


closed as unclear what you're asking by user228113, Derek Holt, hardmath, 3SAT, user223391 Feb 20 '16 at 16:40

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  • $\begingroup$ Generated how? Within some larger group? $\endgroup$ – DonAntonio Feb 18 '16 at 21:19
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    $\begingroup$ You need to be specific -- you can't just take the 'union of groups.' You can join them if they are subgroups of a third group, but how they are subgroups affects the outcome. $\endgroup$ – Thomas Andrews Feb 18 '16 at 21:21
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    $\begingroup$ What does $D_n$ mean to you, exactly? (Not everyone uses the same notation for dihedral groups.) What are the isometries of the square and rectangle, what is their "union", and how is it related to the isometries of the dodecagon? I'm sure you're not claiming that $4\times8=24$, so I must be misunderstanding something you wrote. If you actually showed your work it might be easier to understand. $\endgroup$ – David K Feb 18 '16 at 21:23
  • $\begingroup$ I get you consider all the dihedral groups as subgroups of $O(2,\Bbb R)$. Is it true? Or are you considering them as subgroups of $S_{\Bbb N}$ through the natural maps $D_n\hookrightarrow S_{n}\hookrightarrow S_{\Bbb N}$ ? $\endgroup$ – user228113 Feb 18 '16 at 21:24
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    $\begingroup$ Please edit your question so that it's clear. Users shouldn't have to read a bunch of comments to understand what the question is. $\endgroup$ – user147263 Feb 18 '16 at 22:32

The subgroup of $O(2)$ generated by $D_n$ and $D_m$, regarded as subgroups of $O(2)$ in the usual way, is $D_{\text{lcm}(m, n)}$.

But it's important to note that when people say "$D_n$" by default they mean $D_n$ as an abstract group, independent of any particular embedding into a larger group such as $O(2)$. In that setting there's no meaningful way to take the "union" of two groups. There is a meaningful coproduct, but it's usually very large.

  • $\begingroup$ @Raphael: I don't know, but it's not a very difficult exercise to just do. $\endgroup$ – Qiaochu Yuan Feb 19 '16 at 15:37