# Torsion-free group with a subgroup of finite index [duplicate]

Possible Duplicate:
Torsion-free virtually-Z is Z

Let $G$ be a torsion-free group with a subgroup $H$ of finite index isomorphic to $\mathbb{Z}$. Is $G$ isomorphic to $\mathbb{Z}$?

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## marked as duplicate by Arturo Magidin, t.b., Asaf Karagila, Sasha, Jonas TeuwenNov 23 '11 at 23:58

If this is a homework question, see meta.math.stackexchange.com/questions/1803/… – Dimitrije Kostic Nov 21 '11 at 16:47
It is not a homework. I just met torsion-free groups and I tried to find some basic properties of it. I was unable to solve this question. – student Nov 21 '11 at 16:51
The question has been asked there: math.stackexchange.com/questions/70497/… – PseudoNeo Nov 21 '11 at 16:58
Okay. I got it. – student Nov 21 '11 at 17:05

Generally, we say $G$ is "virtually $\mathcal{P}$" if it has a subgroup $H$ that has property $\mathcal{P}$ and is of finite index in $G$. When the property $\mathcal{P}$ is inherited to subgroups (if $H$ has property $\mathcal{P}$, and $K$ is a subgroup of $H$, then $K$ has property $\mathcal{P}$), then in fact $G$ is virtually $\mathcal{P}$ if and only if it has a normal subgroup with property $\mathcal{P}$ that is of finite index (since every subgroup of $G$ of finite index contains a normal subgroup of $G$ of finite index).

So your question is whether a torsion-free virtually cyclic group must be cyclic.

The answer is "yes": in fact, a virtually cyclic group must be either finite (every finite group is virtually cyclic), finite by infinite cyclic, or finite by infinite-dihedral. For a proof of the case you are interested in, see this previous question.

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