Let $D_n$ be the dihedral group defined by $D_n=$ {$I,R,R^2,...,R^{(n−1)},r,rR,rR^2,...rR^{(n−1)}$}

Theorem. A nontrivial proper subgroup $N$ of $D_n$ is normal in $D_n$ if and only if $N$ is a subgroup of $\langle R \rangle $ or $n$ is even and $N$ is one of $\langle r,R^2 \rangle$ or $\langle rR,R^2 \rangle$ .

How can I prove this theorem?


Here are two very big hints.

First prove $\langle R^k\rangle \unlhd D_n\ \forall\ k\in\mathbb{Z}$, which is relatively easy and solves the case $N\leq \langle R\rangle$.

Second, prove that $N \unlhd D_n$ and $rR^j \in N$ for some $j$ implies $r^2 \in N$, which solves the case "$n$ is even and..."


In this answer to your previous question I explained how to manipulate conjugacy equations in dihedral groups. Figure out what the proper nontrivial subgroups of $D_n$ are (hint: this depends on when $n$ is even or odd), then apply the method I used in that answer to figure out which are invariant under conjugation.

  • $\begingroup$ I know that when $n$ is odd, $⟨r,R^2⟩ and ⟨rR,R^2⟩$ will generate the whole group, so it can't be the proper subgroup, can it? </br> when $n$ is even $⟨r,R^2⟩ and ⟨rR,R^2⟩$ will generate part of the group $D_n$, so it's a proper subgroup of $D_n$. But how does this relate to this theorem. $\endgroup$ – Diane Vanderwaif Apr 12 '13 at 12:12
  • $\begingroup$ @Diane Well... if it generates the whole group, does it meet the definition of a proper subgroup? $\endgroup$ – Douglas B. Staple Apr 16 '13 at 1:21

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