Group of order 275 act on set of size 18, what is the minimum number of orbit of lenght 1?

Question

Let $G$ be a group of order 275 acting on set of size 18, what is the minimum number of orbit of length 1?

I think it is 2 because we then have $1+1+5+11$ all of the numbers in the sum are divisors of by $275=5^2*11$.

Answer 1

You have shown that there is a solution which involves 2 orbits of length one, but not yet that you cannot do with fewer.

For that, note that you cannot write $17=18-1$ as a sum of 5s and 11s (clearly no divisor $>18$ will be relevant and 18 cannot be written as a sum of 5s and 11s).

Answer 2

By orbit stabilizer theorem $|G/Stab(a)|=|O(a)|$ With this we know that orbit must divide f order of group So only possibility is 1 . So there are 18 orbits

Answer 3

Since, order of orbit always divides the order of group and sum of orders of orbits is equal to order of set, which is 18.So, possible orders for orbits are 1,5 or 11.

So, only possible class equations of the action are

18=1+1+5+11

18=1+1+1+5+5+5

18=1+1+...+1(8 times)+5+5

18=1+1+...+1(13 times)+5

number of orbits of order 1 is equal to number 1's in the class equation.

so, least number number of orbits of length 1 is 2.