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In this paper, it is proved in Proposition 10.4.3 that if $G$ is an infinite group such that every nontrivial subgroup of $G$ is of finite index, then $G \cong \mathbb{Z}$. The authors need this fact to prove another theorem. 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.

The proof given in the paper surely won't seem difficult to those who are familiar with the necessary group theoretic background (such as the theorem of Hölder-Burnside-Zassenhaus). I was wondering whether this proposition was known before; and if so, whether it has easier proofs that use less machinery.

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3 Answers 3

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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.

  1. Show the group is torsion-free

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

  1. Show the group has non-trivial center

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

  1. 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.

  1. 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.

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  • $\begingroup$ In your step 4, it is not important that the center be known to be finitely generated since that property is not used. I think step 5 has a gap: we don't know yet that the whole group is abelian, so while the mapping described in step 5 takes values in the center and is injective, it is not immediate that the mapping in step 5 is a homomorphism. $\endgroup$
    – KCd
    Commented Aug 30, 2020 at 4:45
  • $\begingroup$ @KCd: the center being infinite cyclic implies every finite-index subgroup is also infinite cyclic. The mapping in step 5 is the transfer map (and so a homomorphism). $\endgroup$
    – Steve D
    Commented Aug 30, 2020 at 5:33
  • $\begingroup$ Aha, the transfer. That should probably be mentioned within step 5, since superficially the maps in steps 4 and 5 look "similar" but the fact that the transfer map is a homomorphism is a more delicate issue than the mapping in step 4 being a homomorphism. $\endgroup$
    – KCd
    Commented Aug 30, 2020 at 5:42
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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 $G \simeq \mathbb{Z}$.

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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.''

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