Mathematics Asked by Slepecky Mamut on November 24, 2021
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
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)?
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
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