Infinite groups whose non-trivial subgroups are of finite index

HINTS:

1. Show the group is finitely generated

Any element generates a finite index subgroup; now take that element, together with a representative for each coset.

2. Show the group is torsion-free

An element of finite order would generate a non-trivial, infinite-index subgroup.

3. Show the group has non-trivial center

Consider the intersection of all (infinite cyclic) subgroups generated by the (finitely many) generators.

4. Show the center is cyclic

The center is a finitely generated abelian group, which has a cyclic subgroup of finite index (just take a subgroup generated by any element). Suppose this subgroup has index $n$: the map $z\rightarrow z^n$ is an injective homomorphism to an infinite cyclic group; thus the center is infinite cyclic.

5. Conclude

If the center has index $m$ in the group, then the transfer map into the center $g\rightarrow g^m$ is an injective group homomorphism. As a subgroup of the infinite cyclic group, it is itself infinite cyclic.

Another possibility is to use a theorem of Schur:

Theorem: Let $G$ be a group. If the center $Z(G)$ is a finite-index subgroup of $G$, then the commutator subgroup $D(G)$ is finite.

This is not a difficult result, see for example here for a proof based on Dixon's book Problems in group theory.

Now, let $G$ be an infinite group whose non-trivial subgroups are of finite-index. If $g_0 \in G - \{ 1 \}$, we know that $\langle g_0 \rangle$ is a finite-index subgroup of $G$; let $\langle g_0 \rangle, g_1 \langle g_0 \rangle, \ldots, g_r \langle g_0 \rangle$ be the set of its left-cosets (suppose $g_i \neq 1$). Then $g_0, g_1, \ldots, g_r$ generate $G$, hence $$Z(G)= \bigcap\limits_{i=0}^r C(g_i),$$ where $C(g_i)$ denotes the centraliser of $g_i$. Because $Z(G)$ is an intersection of finitely-many subgroups of finite-index, we conclude that $Z(G)$ is a finite-index subgroup. Now, Schur's theorem implies that $D(G)$ is finite. Because $G$ is infinite, this implies that $D(G)$ is trivial, ie., $G$ is abelian.

Finally, $G$ is a finitely-generated torsion-free abelian group which is virtually $\mathbb{Z}$. Thanks to the classification of finitely-generated abelian groups, we know that the only possibility is $G \simeq \mathbb{Z}$.

They remark at the end of $\S 1.4$ that the proof is non-trivial and they do not know if this was previously known.

See:

Fedorov, Yu. G.,
On infinite groups of which all nontrivial subgroups have a finite index.
Uspehi Matem. Nauk (N.S.) 6, (1951). no. 1(41), 187-189.

The MathSciNet Review, by Richard Albert Good, starts with this sentence:
``The only group satisfying the conditions in the title of the paper is the infinite cyclic group.''