On the complexity of the cogrowth sequence

Pub Date : 2018-05-21 DOI:10.4171/jca/39
J. Bell, M. Mishna
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引用次数: 5

Abstract

Given a finitely generated group with generating set $S$, we study the cogrowth sequence, which is the number of words of length $n$ over the alphabet $S$ that are equal to one. This is related to the probability of return for walks in a Cayley graph with steps from $S$. We prove that the cogrowth sequence is not P-recursive when $G$ is an amenable group of superpolynomial growth, answering a question of Garrabant and Pak. In addition, we compute the cogrowth for certain infinite families of free products of finite groups and free groups, and prove that a gap theorem holds: if $S$ is a finite symmetric generating set for a group $G$ and if $a_n$ denotes the number of words of length $n$ over the alphabet $S$ that are equal to $1$ then either $\limsup_n a_n^{1/n} \le 2$ or $\limsup_n a_n^{1/n} \ge 2\sqrt{2}$.
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关于共生序列的复杂性
给定一个具有生成集$S$的有限生成群,我们研究了协生长序列,它是字母表$S$上长度为$n$且等于1的单词的数目。这与从$S$开始的凯利图中行走的返回概率有关。我们证明了当$G$是一个可服从的超多项式群时,协生长序列是不p递归的,回答了Garrabant和Pak的一个问题。此外,我们计算了有限群和自由群的无限族的自由积的协生长,并证明了一个间隙定理成立:如果$S$是群$G$的有限对称生成集,如果$a_n$表示在字母表$S$上等于$1$的长度为$n$的单词数,则$\limsup_n a_n^{1/n} \le 2$或$\limsup_n a_n^{1/n} \ge 2\sqrt{2}$。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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