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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.

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  • $\begingroup$ This kind of problems screams for the use of the Cauchy-Frobenius lemma. $\endgroup$ – Andreas Caranti May 21 at 15:36
  • $\begingroup$ 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? $\endgroup$ – Charles Hudgins May 21 at 15:37
  • $\begingroup$ @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. $\endgroup$ – Mathbeginner May 21 at 16:03
  • $\begingroup$ 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. $\endgroup$ – Charles Hudgins May 21 at 16:07
  • $\begingroup$ I just realized it's possible I misinterpreted what they meant by side. By "side," do they mean "edge" or "face"? $\endgroup$ – Charles Hudgins May 21 at 16:08

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