Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is it true that every finitely generated infinite simple group has exponential (word-)growth?

Remark: As Mark Sapir has pointed out, the question whether every finitely generated group of subexponential growth is even residually finite has been answered in the negative in

Anna Erschler. Not residually finite groups of intermediate growth, commensurability and non-geometricity, J. Algebra 272 (2004), no. 1, 154–172, http://www.sciencedirect.com/science/article/pii/S0021869303006410.

share|improve this question
7  
Not every finitely generated group of subexponential growth isresidually finite (Erschler). The answer to the first question is not known. –  Mark Sapir Feb 18 '13 at 17:15
    
Thanks for the hint on Erschler's result! –  Stefan Kohl Feb 18 '13 at 17:33
3  
@Stefan: you could maybe edit the question accordingly. On the other hand, it remains an open question whether there exists a nontrivial f.g. group of subexponential growth with no nontrivial finite quotient. (The derived subgroups of full topological groups of infinite minimal subshifts, which are infinite, f.g., amenable and simple, have free subsemigroups by Matui-2011 so have an exponential growth). –  YCor Feb 18 '13 at 20:08
    
@Ives: I have edited the question, providing the reference to Erschler's result. –  Stefan Kohl Feb 18 '13 at 20:49
5  
Btw here's part of the argument for Erschler's result: take for granted that there exists a f.g. group $G$ with a central subgroup $Z$ which is an infinite-dimensional vector space over the field on $p$ elements, such that $G/Z$ has subexponential growth (actually in the example, $p=2$ and $G$ is the 1st Grigorchuk group). If $Z'$ is a hyperplane of $Z$ s.t. $G/Z'$ is residually finite, then there exists a normal subgroup of finite index $N$ such that $Z'=N\cap Z$. There are countably many such $N$ and hence such $Z'$. On the other hand, there are uncountably many $Z'$, so one $G/Z'$ isn't RF. –  YCor Feb 18 '13 at 21:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.