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Is there a name for (infinite) groups such that every non-trivial, proper subgroup has finite index (e.g. $\mathbb{Z}$)?

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I think the only such group is $\mathbb{Z}$. If every nontrivial subgroup of $G$ is of finite index and $G$ is infinite, then $G$ is torsionfree and contains a subgroup of finite index isomorphic to $\mathbb{Z}$. The only group with this property is $\mathbb{Z}$. – Mikko Korhonen May 2 '13 at 15:54
up vote 8 down vote accepted

The only group with this property is the infinite cyclic group $\mathbb{Z}$.

Let $G$ be an infinite group where every nontrivial subgroup has finite index. Then $G$ must be torsionfree, so $G$ contains a subgroup $H \cong \mathbb{Z}$ of finite index. Then $H$ will contain a normal subgroup $N$ of $G$ such that $N \cong \mathbb{Z}$ and $G/N$ is finite, so $G$ is finitely generated.

Let $g_1, \ldots, g_n$ be a set of generators for $G$, where each $g_i \neq 1$. Now each $C_G(g_i)$ has finite index, so $\cap C_G(g_i) = Z(G)$ has finite index. By Schur's theorem, the commutator subgroup of $G$ is finite. Thus it must be trivial and $G$ is abelian. Classification of finitely generated abelian groups shows that $G \cong \mathbb{Z}$.

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Fantastic! This is an interesting application Schur's theorem. – Marshal Kurosh Aug 3 '13 at 5:57

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