Let $|G|=21$,with G acting on set X of size 23. There is no element of X fixed by every element of the group G. Determine the number of orbits.

Question

Question is:

Let G be a group of order 21,that acts on set X of size 23. Suppose that there is no element of X fixed by every element of the group G. Determine the number of orbits in this action.

Does this imply,that by orbit stabilizer theorem, $$|G|=|orb(i)|*|stab(i)|=|orb(i)|*1$$

Does this mean the orbit of every element in the set X under this action is 21?

Any help solving this task in general would be appreciated.

Answer 1

By your condition no orbit is of size $1$. The size of every orbit must divide $21$. So it is either $3,7$ or $21$. The sum of sizes must be 23. This rules out size $21$ (because $21+3>23$). There must be an orbit of size 7 since $3\not|23$. There cannot be only one such orbit and the number of them cannot be 3 or more (again because $21+3>23$ and $3\not| 23-7=16$). So the number of orbits of size $7$ is 2, and the number of orbits of size $3$ is $3$. Altogether there are 5 orbits.