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Let $\cal C$ be the smallest class of finitely generated discrete countable groups such that

  1. $\cal C$ contains all infinite cyclic groups,
  2. if $G$ is any finitely generated countable discrete group and $H\in \cal C$ then $G\ast H \in \cal C$,
  3. If $G$ is any countable discrete group and $H$ is a finite index subgroup then $G\in {\cal C} \iff H\in \cal C$.

In 1. I write "all infinite cyclic groups" to avoid adding that $\cal C$ is closed under isomorphism, but feel free to replace it with your favorite infinite cyclic group and adding that $\cal C$ is closed under isomorphisms.

Question. Is there a better ("more intrinsic") description of the class $\cal C$?

My motivation: I have been studying when for a given group $G$ is there a finite set $M$ and a subshift of finite type $S\subset M^G$ such that $G$ acts on $S$ freely (not essentially freely, topologically freely, etc, but freely). It's not difficult to see groups in class $\cal C$ don't admit free subshifts of finite type, and together with a colleague we can prove that some groups (e.g. polycyclic, but also some with infinitely many ends) not in $\cal C$ admit free subshifts of finite type. It seems natural to conjecture $\cal C$ is the class of groups which don't admit a free subshift of finite type, so I'd like to understand it.

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1  
Finitely generated and virtually free. –  Misha Aug 5 '13 at 12:21
    
@Misha, in pt. 2 $G$ is any countable group, not necessarily $G\in \cal C$ –  Łukasz Grabowski Aug 5 '13 at 12:34
3  
By Kurosh's theorem, these are finite extensions of groups having a free factor $\mathbb{Z}$. –  Mark Sapir Aug 5 '13 at 13:20
    
Thanks Mark, I changed the question, is the answer now the same? –  Łukasz Grabowski Aug 5 '13 at 13:21
    
Now, the answer is all countable groups, because the trivial group $H$ belongs to the class by (1); so, all countable groups $G$ lie in the class by (2). –  Anton Klyachko Aug 5 '13 at 13:27

1 Answer 1

up vote 4 down vote accepted

$\cal C$ is the class of finite extensions of finitely generated groups having an infinite cyclic free factor.

(as in Mark's comment concerning another version of the question).

To prove this it suffice to note that the class of such finite extensions is closed with respect to finite-index subgroup. Indeed, a finite-index subgroup of a finitely generated group is finitely generated. Now,
if $A$ is a finite-index subgroup of a finite extension of $B*\langle c\rangle_\infty$, then $A\cap \langle c\rangle_\infty$ is a finite-index subgroup of $\langle c\rangle_\infty$ and, hence, is infinite cyclic. By Kurosh's subgroup theorem, $A\cap \langle c\rangle_\infty$ is a free factor of $A\cap(B*\langle c\rangle_\infty)$ (and $A$ is a finite extension of $A\cap(B*\langle c\rangle_\infty)$.

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I DeTeXified. Hope that is ok. –  Benjamin Steinberg Aug 25 '13 at 3:26
    
Almost. I replaced "set" with "class" in the deTeXified sentence. –  Anton Klyachko Aug 25 '13 at 7:37

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