Find a formula for number of orbits under action of $D_{4}$

We colour each side of a square with $$k \geq 1$$ colours. Find a formula for the number of orbits under the action of $$D_{4}=\{ e , r,r^{2},r^{3},s,sr,sr^{2},sr^{3} \}$$ on the set of colours.

Now as far as I know an orbit of an element $$a \in A$$ under the action of group $$G$$ is defined as the set $$G \cdot a = \{ g \cdot a : g \in G\}$$. In our case $$A$$ is the colouring of the square.

My attempt: Now if we number the sides of the square we can represent them as a set $$A=\{1,2,3,4\}$$.

Now the generators of $$G$$ are $$\langle r \rangle$$ and $$\langle s \rangle$$. No matter the original state of the square $$\langle r \rangle$$ can only rotate it (by $$90$$ degrees), and thus the orbit then contains $$4$$ elements (since $$r^{4}=e)$$. Now this is also dependent on the amount of colours used (if $$k=1$$, then the orbit just contains the original square).

As you probably can tell, I don't really grasp the problem yet. Any suggestions are welcome.

• This kind of problems screams for the use of the Cauchy-Frobenius lemma. – Andreas Caranti May 21 at 15:36
• What exactly is meant by "We colour each side of a square with $k \geq 1$ colours"? Do they mean that we give each side a single color and that we have $k \geq 1$ colors to choose from? – Charles Hudgins May 21 at 15:37
• @CharlesHudgins: I interpret that sentence the way you do. There is no furher elaboration on this part. But since each side is coloured by one of the $k\geq 1$ colours, I assume there are $k$ colours to choose from. – Mathbeginner May 21 at 16:03
• Then I don't see a way to avoid breaking into cases. What happens when $k = 1$ (trivial)? What happens when $k = 2$? Here I think there are $3$ cases not covered by the previous case. What happens when $k = 3$? I think there are $2$ distinct cases here not covered by the previous cases. What happens when $k = 4$? I think there's just one new case to consider here. Finally, when $k \geq 5$, we get no new cases. – Charles Hudgins May 21 at 16:07
• I just realized it's possible I misinterpreted what they meant by side. By "side," do they mean "edge" or "face"? – Charles Hudgins May 21 at 16:08