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

A group $G$ is said to be boundedly generated if (it is finitely generated and) there exists a finite family of cyclic subgroups (not necessarily normal or distinct) $\lbrace C_i \rbrace_{i =1, \ldots, k}$ such that $G = C_1 \cdots C_k$.

Simplest examples of boundedly generated groups are finite groups and polycyclic groups. Free groups are not boundedly generated, and $SL_2(\mathbb{Z})$ also isn't (since passing to finite index subgroups does not change this property).

A priori, the order in the product matters. Assume $G = C_1 \cdots C_k$ is a boundedly generated group so that $G = C_{\sigma(1)} \cdots C_{\sigma(k)}$ for any permutation $\sigma \in Sym_k$. In particular, one may then assume the $C_i$ are distinct. Except virtually polycyclic groups, which group has this property?

share|improve this question
1  
"boundedly generated group where the order of the product does not matter": I don't understand what you mean by this. –  Yves Cornulier Jul 9 '13 at 18:30
    
oh sorry, that was pretty unclear indeed... –  Antoine Jul 9 '13 at 19:07
    
ok, it's clearly defined now. –  Yves Cornulier Jul 9 '13 at 19:29
    
Actually, are there constructions of groups with bgp, which are not lattices in Lie/alberaic groups? (Polycyclic groups and groups of integer points in higher rank algebraic groups are the only examples I am aware of.) –  Misha Jul 10 '13 at 15:30
2  
@Misha: there are Muranov's construction of boundedly generated infinite f.g. groups that are simple. –  Yves Cornulier Jul 10 '13 at 15:42
show 1 more comment

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.