Let $G=A_{100}$ and assume $G$ acts on set $\Omega$ with $|\Omega|=90$. Prove that $G$ has $90$ orbits of size $1$ on $\Omega$.
My attempts
Using the Orbit-Stabilizer Theorem, for any $x\in \Omega$, we have $$|O_x||G_x| = |G| = \frac{100!}{2},$$ where $O_x$, $G_x$ are the orbit and stabilizer of $x$, respectively.
Also, we know that since $O_x\leq \Omega$, its order must divide $90=1\cdot2\cdot3^2\cdot 5$.
Meanwhile, note that $G_x\leq G$ and that $G$ is simple.
Then we can rule out $|O_x|=2$ because that would imply that $G_x$ is a proper normal subgroup of $G$, contradicting $G$ being simple.
But I can't figure out how to rule out the rest of the possible orders of $O_x$ and deduce that $|O_x|=1$.
Or should I approach from a different angle? I tried using the class equation and the permutation representation of $G$ but neither worked for me.
Any help would be greatly appreciated.