Each permutation in permutation group PSL2 consist of fixed points and cycles of equal lengths. Prove or disprove it.

Question

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 $ccdot x_i^d$ or  $ccdot  x_1^k x_i^d$.
How to prove this fact in general (if it is true) or how to find the counterexample (if it is false)?

user10354138 · 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 $(2,m,m,dots,m)$, $mgeq 3$ (hence $mgeq 4$ is even since we can take $m$-th power). WLOG the $2$-cycle is $0toinftyto 0$ and hence the map is $zinmathbb{P}^1(mathbb{F}_q)mapstofrac{lambda}{z}$ for some $lambdainmathbb{F}_q$. Thus the only way to get at least one 2-cycle is if $Ain PSL(2,q)$ order 2 and hence $(2,m,m,dots,m)$ is impossible.

Answered by user10354138 on November 24, 2021

Each permutation in permutation group PSL2 consist of fixed points and cycles of equal lengths. Prove or disprove it.

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