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