# Is there a better description of this class of discrete groups?

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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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
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

$\cal C$ is the class of finite extensions of finitely generated groups having an infinite cyclic free factor.
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 $A\cap(B*\langle c\rangle_\infty)$.