Show that Aut $(C_n)=D_n$

Question

Problem statement: I need to find out Aut $(C_n)$ is isomorphic to $D_n$.

I already know that it is isomorphic, so now all I need to do is to prove it.

Any help is appreciated. I was hoping for a duplicate post, but couldn't find it. Thank you!

Answer 1

Firstly note that any automorphism can be obtained in this way: A given vertex $v$ may be mapped to any of the $n$ vertices available (including itself). As soon as that is done, an adjacent vertex to $v$ has only two choices left: it can either be in the counter clockwise direction to $v$ or in the clockwise direction to $v$. Once that choice is also made, no other choices are required. Hence we get $2n$ automorphisms this way and there can be no others.

Also, it is clear that two kinds of automorphisms suffice to generate this group: rotation, and swapping the notion of clockwise and counter clockwise (assuming we draw the cycle graph as equally spaced points on the unit circle; there is no loss of generality in doing that). But both these automorphisms also generate the dihedral group $D_n$ which also has $2n$ elements. It follows that $A(C_n)=D_n$.

This answer is taken from my blog post.