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What are examples of groups such that every finitely generated subgroup is isomorphic to $\mathbb{Z}$?

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2 Answers 2

up vote 11 down vote accepted

Suppose any non-trivial finitely generated subgroup of $G$ is isomorphic to $\Bbb{Z}$, and that $G$ itself is nontrivial. Choose some nonzero $e \in G$. For any $g \in G$, the subgroup $\langle e, g\rangle$ is isomorphic to $\Bbb{Z}$, so in particular $ne=mg$ for some integers $m,n$. Define a map $f:G \to \Bbb{Q}$ by $g \mapsto n/m$. Then $f$ is a well-defined group homomorphism; since $G$ is torsion-free and $e$ nonzero, it is straightforward to check that $f$ is injective. So $G$ is isomorphic to a nontrivial subgroup of $\Bbb{Q}$, which are classified, e.g., in this paper.

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There are no such groups, because the trivial subgroup of any group is finitely generated and not isomorphic to $\mathbb{Z}$.

However, if we restrict our attention to non-trivial subgroups, then the group $\mathbb{Z}$ under addition is an example of such a group, as any non-trivial subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ for some $n\geq1$, hence isomorphic to $\mathbb{Z}$ via the map $f:\mathbb{Z}\to n\mathbb{Z}$, $f(x)=nx$.

A less obvious example is the group $\mathbb{Q}$ under addition.

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Why did someone find this downvote-worthy? –  Zev Chonoles Jul 19 '12 at 18:31

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