Rotation of angle $2k\pi/p$ generates the group of all rotation

Question

In the book Geometric Transformation of Razvan Gelca, there is an argument as follows:

I could understand most of the proof there, however is there any easier explanation for the yellow painted part, i.e: because p is prime then the rotation $2k\pi/p$ generates the group of all rotation of $A_1A_2\cdots A_p$?

As far as I understand there are 2 conditions: $p$ is prime and $A_1A_2\cdots A_p$ is fixed under the rotation $2k\pi/p$. Now the group generated by the rotation $2k\pi/p$ (A) is a cyclic group of order $p$. So my question is:

1- Why it equals to the whole group of all possible rotations (S) if $p$ is prime?

2- So if $p$ is not a prime then $A$ is a proper subset of $S$? Are there any other special properties?

Please help me.

Thanks.

Answer 1

Since $p$ is prime, the group has no non-trivial subgroups. If it did then for some $k$, not all "rotated-permutations" are going to be eventually traversed.

$k$ can be considered the winding number which incidentally is equal to the number of vertices skipped in a rotation by $\frac{2k\pi}{p}$

For illustration see the following figure of polygons and Schläfli symbol $\{n/k\}$ (created by 'Cmglee' for Wikipedia under CC BY-SA 3.0)