Suppose that a group G acts on a set X, the number of elements in G is 9 and the number of elements in X is 27. Prove that if there is some orbit with exactly one element then there are at least two orbits with exactly one element.
I have no clue what to do here. I was thinking Burnsides or the Orbit Stabilizer theorem but I don't know how to apply them when all I know is the orders of the group and set.
On
If you add the size of all the orbits you get $27$, which is a multiple of $3$.
On the other hand, by the orbit stabilier theorem each orbit has size $1,3$ or $9$.
This means that all orbits that have more than one element has a number of elements that is a multiple of $3$.
We conclude that the total number of elements that are alone in their own orbit is also a multiple of $3$, hence it cannot be exactly $1$.
Hints:
The number of elements in each orbit is a divisor of $\;9\;$ , so if there is an orbit with one element , since the sum over all elements in all orbits must be $\;27\;$, we must have another two orbits with one element (do the sums, taking into account you already have one $\;1\;$ and all the other numbers must be divisors of $\;9\;$ ...)