For the standard permutation representation of the finite projective special linear group $\text{PSL}_2$, I have noticed the fact that every permutation contains the cycles of one length, except for fixed points. For instance, $\text{PSL}_2[16]$ contains
- 1088 permutations with 15-cycle (and two fixed points)
- 1920 permutations with 17-cycle
- 544 permutations with three 5-cycles (and two fixed points)
- 272 permutations with five 3-cycles (and two fixed points)
- 255 permutations with eight 2-cycles (and one fixed point)
and one identity. A similar result is true for any PSL2 over the finite field with less than 100 elements (computational proof).
As a consequence, each monomial of the cycle index polynomial has the form $c\cdot x_i^d$ or $c\cdot x_1^k x_i^d$.
How to prove this fact in general (if it is true) or how to find the counterexample (if it is false)?
Best Answer
Recall an automorphism of $\mathbb{P}^1$ fixing three points is the identity. So every nonidentity element can only fix at most 2 points. Apply this also to their powers, the only remaining case to rule out is cycle type $(2,m,m,\dots,m)$, $m\geq 3$ (hence $m\geq 4$ is even since we can take $m$-th power). WLOG the $2$-cycle is $0\to\infty\to 0$ and hence the map is $z\in\mathbb{P}^1(\mathbb{F}_q)\mapsto\frac{\lambda}{z}$ for some $\lambda\in\mathbb{F}_q$. Thus the only way to get at least one 2-cycle is if $A\in PSL(2,q)$ order 2 and hence $(2,m,m,\dots,m)$ is impossible.