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Let $H$ be the subgroup of all rotations in $D_n$ and let $\phi$ be an automorphism of $D_n$. Prove that $\phi(H) = H$. In words, an automorphism of $D_n$ carries rotations to rotations.

I understand that if $\phi$ is an automorphism, then it is an isomorphism from the group onto itself. Further, I understand (at least conceptually) why the subgroup $H$ of rotations of $D_n$ ought to map back to rotations-- that at least makes sense to me when I think about it.

What I am a bit unclear about is the sketch of the proof when it comes to the properties of $D_n$. I know I need to show the following (as far as the isomorphism is concerned):

  1. I need to show that $\phi$ is 1-1, i.e., assume $\phi(h) = \phi(h')$ and then prove $h = h'$.
  2. I need to show $\phi$ is onto, i.e., for any element $h$ in $H$, there's a $h'$ in $H$ such that $\phi(h') = h$.
  3. I need to show $\phi$ is operation-preserving, i.e., $\phi(hh') = \phi(h)\phi(h')$ in $H$.

How do I apply these standard properties specifically to $D_n$ in the problem? Also, when I say "I know," I realize I probably do not "know," and am open to being corrected.

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On property 2. I put $h$ and $h'$ both in $H$ rather than putting $h'$ in some other group, say $H'$, because this is an automorphism, not an isomorphism. I'm wondering if that is proper. –  GaMbiT Mar 20 '13 at 9:29
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You can already assume a lot of the things you mention that you have to prove because you already know $\phi$ is an automorphism. Your basic goal is to show that for any $h \in H$, $\phi(h) \in H$. –  muzzlator Mar 20 '13 at 9:34
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This is not true if you include the "degenerate" dihedral group of 4 elements! In the Klein 4-group, no subgroup of order 2 is characteristic. –  yatima2975 Mar 20 '13 at 11:36
    
Well, it's the Fitting subgroup, so it's characteristic. Does that count? :) –  Alexander Gruber Mar 20 '13 at 20:00
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2 Answers

Consider the element $g$ of order $n$ that generates all rotations. Then $\phi(g^m) = \phi(g)^m$. Show that $\phi(g)$ must have the same order as $g$ and so it must be a rotation.

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Another possibility: $H$ is the only abelian subgroup of index two in $D_n$ (I supposed $n > 4$).

In fact, you can classify the maximal abelian subgroups of $D_n$. If $n$ is odd, $D_n$ has $n+1$ maximal abelian subgroups: $n$ cyclic subgroups of order two (each one generated by a reflection) and one cyclic subgroup of order $n$ (generated by a rotation). If $n$ is even, $D_n$ has $1+n/2$ maximal abelian subgroups: $n/2$ subgroups isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ (generated by two perpendicular reflections) and one cyclic subgroup of order $n$ (generated by a rotation).

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