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In relation to my previous question, I am curious about what exactly are the normal subgroups of a dihedral group $D_n$ of order $2n$.

It is easy to see that cyclic subgroups of $D_n$ is normal. But I suspect that case analysis is needed to decide whether dihedral subgroups of $D_n$ is normal.

A little bit of Internet search suggests the use of semidirect product $(\mathbb Z/n\mathbb Z) \rtimes (\mathbb Z/2\mathbb Z) \cong D_n$, but I do not know the condition for subgroups of a semidirect product to be normal.

I would be grateful if you could suggest a way to enumerate the normal subgroups of $D_n$ that does not resort to too much of case analysis.

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1 Answer 1

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Here is a nice answer: the dihedral group is generated by a rotation $R$ and a reflection $F$ subject to the relations $R^n=F^2=1$ and $(RF)^2=1$. For $n$ odd the normal subgroups are given by $D_n$ and $\langle R^d \rangle$ for all divisors $d\mid n$. If $n$ is even, there are two more normal subgroups, i.e., $\langle R^2,F \rangle$ and $\langle R^2,RF \rangle$.

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    $\begingroup$ You have lost track of the group $D_n$ itself. $\endgroup$
    – Alex M.
    Jun 12, 2016 at 10:37
  • $\begingroup$ Yes, you are right. Of course, the group itself should be included. $\endgroup$ Jun 12, 2016 at 11:39
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    $\begingroup$ By saying $\langle{R^d}\rangle$ is a normal subgroup for all divisors $d\ \mid\ {n}$, you've actually already included {1}, because $n\ \mid\ {n}$. $\endgroup$
    – Rasputin
    Oct 31, 2016 at 20:22
  • $\begingroup$ @ Dietrich Burde : Is there any other normal subgroups of $D_{2n}$ besides these normal subgroups $\endgroup$
    – user120386
    May 22, 2017 at 10:08
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    $\begingroup$ @user120386 No, this is a complete classification of normal subgroups, see Keith Conrad's article on dihedral groups. $\endgroup$ May 22, 2017 at 13:51

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