What is the Group of Bijections of the circle Ok, this problem is driving me crazy because the problem says:

In the group of bijections of the circle, build a succesion of subgroups $G_i$ such that $G_i$ is a subgroup of $G_{i+1}$, and $|G_i|=i!$.

I understand it like this:
The circle has infinite points, if I have one, the bijection is basically a rotation of $0°$ because it ends up in the same point, if I have $2$ points in the circle, then I have $2!$ permutations that is, $2$ rotations of the circle, if I have $3$ elements with $3!=6$ possible permutations $= 6$ rotations, if I have $4$ elements then there are $4!=24$ permutations or rotations.
That is how I understand it but I don't know how could those rotations be, I mean graphically I can't think of a way to express those permutations. I want to understand these first examples to extend it to any $G_i$ group.
Thanks in advance for your support.
 A: Rotations!
We start with a rotation of order $1$, that means some rotation which when applied once is the identity on the circle. Well, we may take the rotation by $360^°$ for this. Now we are looking for some rotation which when applied twice is the identity on the circle. A rotation by $180^°$ will do. Moreover, applying this rotation twice we end up with  rotation by $360^°$! So we have found our first two subgroups $G_1\subseteq G_2$ both of correct order.
How to generalize this? Well, what we did here, to go from $G_1$ to $G_2$, was dividing $360^°$ by $2$. So the next step should be dividing $180^°$ by $3$. Geometrically, we just evenly divide the arcs of the intermediate points of the rotation, i.e. where some fixed point on the circle stops while applying the rotations. So $G_3$ is generated by a rotation by $180/3=60^°$. You can check that this subgroup has $3!=6$ elements.

 In the same way you get $G_i$ as generated by a rotation of $360/i!$ degrees simply because these rotations have order precisely $i!$ and applying the rotation $i$ times will give you a rotation by $i(360/i!)=360/(i-1)!$ degrees, i.e. the generator of $G_{i-1}$. Hence these are the required subgroups.

