- #1

- 317

- 26

## Homework Statement

Determine ##\phi(R_{180})##, if ##\phi:D_n\to D_n## is an automorphism where ##n## is even so let ##n=2k##.

The solutions manual showed that since the center of ##D_n## is ##\{R_0, R_{180}\}## and ##R_{180}## is not the identity then it can only be that ##\phi(R_{180})=R_{180}##. This made some sense since to preserve structure ##\phi## must map centers to centers. My solution below is indeed sloppy. Although, I want to know if the procedure is acceptable in the general case, for instance when we want to know the image of some rotation not in the center of ##D_n##. Thanks!

**2. Relevant propositions**

($) Let ##H## be the subgroup of all rotations in ##D_n## and ##\phi:D_n\to D_n## be an automorphism, then ##\phi(H)=H##

## The Attempt at a Solution

Let ##n=2k## such that ##k\in\Bbb{N}##. By ($), if ##H## is a subgroup of rotations, ##\phi(H)=H##; hence ##\phi(R_{180})\in H##. Also, note that if ##R_{360}=R_0=R_2k##, then ##R_{180}=R_k##. Picking an integer ##m\in\Bbb{Z}_{2k}## where ##gcd(m,2k)=1## so that ##\Bbb{Z}_{2k}=\langle m\rangle##; that is ##\forall s\in\Bbb{Z}_{2k}\exists t\in\Bbb{Z}## such that ##s\equiv mt \pmod {2k}##. Suppose ##k\equiv mv\pmod {2k}##. Then ##\phi(R_{180})=\phi(R_k)=\phi(R_{mv})=\phi(R_{m}^{v})=\phi(R_m)^v##.