# The orbits of the action of ${\rm Aut}(D_4)$ on $D_4$.

Consider the action of $${\rm Aut}(D_4)$$ on $$D_4$$ by $$\psi \cdot g = \psi(g)$$, where $$D_4$$ is the dihedral group on $$4$$ elements (so a square). I want to find the orbits of this action.

Instead of finding the elements of $${\rm Aut}(D_4)$$, I thought that maybe I could use the fact that $${\rm Aut}(D_4) \cong D_4$$.

I also started out by noting that for any $$\psi \in{\rm Aut}(D_4)$$, $$\psi(r) = r$$ or $$r^3$$ since the order must be the same, and the remaining generator $$s$$ maps to $$sr, sr^2, sr^3$$ or , $$sr^4$$, but not $$r^2$$ because then it can't generate $$D_4$$. But I am stuck and would appreciate any insight.

Thanks!

• Well, you are done ! You just have to notice that $id$, and $r^2$ are fixed by any automorphisms. You then get $4$ orbits $\{id\}, \{r^2\}, \{ r,r^3\}, \{s,sr,sr^2,sr^3\}$. – GreginGre Nov 24 '19 at 10:43

You are practically there. You just need to compile all the information together. By the way you have defined the group action, two elements $$a$$ and $$b$$ are in the same orbit if and only if there exists an automorphism $$\phi$$ such that $$\phi(a) = b$$.
• Note that the identity $$e$$ will always get sent to itself. Thus, the orbit of the identity is just $$\{e\}$$
• Next, you have noted that $$r$$ can only be sent to $$r$$ and $$r^3$$ by an automorphism. Thus, the orbit of $$r$$ is just $$\{r,r^3\}$$ There is nothing else in this orbit, otherwise this would imply $$r$$ could be sent to another element besides $$r$$ and $$r^3$$.
• Similarly, we have that the orbit of $$s$$ is just $$\{s, sr, sr^2, sr^3 \}$$ since you have noted that these are the only elements $$s$$ can be sent to.
• Lastly, we must have that the orbit of $$r^2$$ is just $$\{r^2\}$$ You have already noted that $$r^2$$ cannot be the image of $$r$$ and $$s$$ implying it cannot be in their orbits. Also, it is not in the orbit of the identity. Thus, the only option is for $$r^2$$ to be in an orbit by itself. You can check that any automorphism does indeed fix $$r^2$$.