无偏差排列群的轨道

David Ellis, Scott Harper
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引用次数: 0

摘要

让 $G$ 是一个度数为 $n$ 的非难有限置换群。等价地,一个有限群从来不是一个原子群的共轭的联合。如果 $G$ 是不传递的,那么 $G$ 就可能没有衍生,即使 $G$ 只有两个轨道,而且两个轨道的大小都是$(1/2+o(1))n$,这种情况也可能发生。然而,我们猜想,如果$G$有两个大小恰好为$n/2$的轨道,那么$G$确实有一个 derangement,而且当$G$至少原始地作用于其中一个轨道时,我们证明了这一猜想。等价地,我们猜想一个有限群从来不是两个同阶原子群共轭的联合,当至少有一个子群是最大群时,我们证明了这一猜想。我们证明了这一猜想的其他情况,并强调了我们的结果与互变交集族和多项式模数根的联系。同时,我们还证明了伊斯贝尔猜想的一个线性变体,它与素幂级数的变化有关。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Orbits of permutation groups with no derangements
Let $G$ be a nontrivial finite permutation group of degree $n$. If $G$ is transitive, then a theorem of Jordan states that $G$ has a derangement. Equivalently, a finite group is never the union of conjugates of a proper subgroup. If $G$ is intransitive, then $G$ may fail to have a derangement, and this can happen even if $G$ has only two orbits, both of which have size $(1/2+o(1))n$. However, we conjecture that if $G$ has two orbits of size exactly $n/2$ then $G$ does have a derangement, and we prove this conjecture when $G$ acts primitively on at least one of the orbits. Equivalently, we conjecture that a finite group is never the union of conjugates of two proper subgroups of the same order, and we prove this conjecture when at least one of the subgroups is maximal. We prove other cases of the conjecture, and we highlight connections our results have with intersecting families of permutations and roots of polynomials modulo primes. Along the way, we also prove a linear variant on Isbell's conjecture regarding derangements of prime-power order.
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