# Infinite abelian group whose all nontrivial subroups have finite index

D.J.S. Robinson, A Course in the Theory of Groups, 2d edition, exercise 4.1.3, p. 98, asks for a proof of the following statement :

Statement 1. If $$G$$ is an infinite abelian group all of whose proper quotient groups (I understand : quotients by nonzero subgroups) are finite, then $$G$$ is (infinite) cyclic.

I can prove it, but only by use of the following theorem, which is proved later in the book (4.2.10, p. 103) :

Statement 2. If an abelian group is finitely generated, it is a direct sum of finitely many cyclic (finite or infinite) subgroups.

(I give a proof of Statement 1 by Statement 2 below.) My question is:

Is it possible to prove Statement 1 without using Statement 2 in a more or less explicit manner?

Here is my proof of Statement 1 by use of Statement 2.

Let $$G$$ be as in Statement 1. Note it additively. For every nonzero element $$x$$ of $$G$$, $$\mathbb{Z}x$$ is a nontrivial subgroup of $$G$$, thus, by hypothesis,

(1) the quotient $$G/ \mathbb{Z}x$$ is finite.

Thus, since $$G$$ is infinite by hypothesis, $$\mathbb{Z}x$$ is infinite. Since this is true for every nonzero element $$x$$ of $$G$$,

(2) $$G$$ is torsion-free.

Choose a nonzero element $$x$$ of $$G$$ (it is possible, since $$G$$ is infinite by hypothesis). By (1), there exists a finite set $$\{a_{1}, \ldots , a_{n} \}$$ of elements of $$G$$ such that $$a_{1} + \mathbb{Z}x, \ldots , a_{n} + \mathbb{Z}x$$ are all the elements of $$G/ \mathbb{Z}x$$. Thus $$G$$ is generated by $$x, a_{1}, \ldots, a_{n}$$, thus $$G$$ is finitely generated, thus, by Statement 2,

(3) $$G$$ is a direct sum $$H_{1} \oplus \cdots \oplus H_{k}$$ of finitely many nontrivial cyclic subgroups.

Since $$G$$ is nontrivial, we have $$k \geq 1$$.

By (2), $$G$$ is torsion-free, thus every $$H_{i}$$ is torsion-free. Thus, since $$H_{i}$$ is nontrivial,

(4) every $$H_{i}$$ is infinite.

Since $$H_{1}$$ is not trivial, $$G / H_{1}$$ is finite (by hypothesis of the exercice). But, by (3), $$G / H_{1}$$ is isomorphic to $$H_{2} \oplus \cdots \oplus H_{k}$$, thus $$H_{2} \oplus \cdots \oplus H_{k}$$ is finite. Thus, in view of (4), $$k = 1$$, thus, by (3), $$G = H_{1}$$ . Thus, by (3), $$G$$ is (infinite) cyclic and we are done.

Edit (January 15, 2022). If I am not wrong, the proof of satement 1 can easily be extracted from Robinson's proof of statement 2. Let $$G$$ be as in statement 1. As noted, $$G$$ is torsion-free. Since $$G$$ is infinite, we can choose a nonzero element $$a$$ of $$G$$. Then $$$$ is a nontrivial subgroup of $$G$$, thus, by hypothesis, $$$$ is of finite index $$m > 0$$ in $$G$$. Then $$x \mapsto mx$$ defines a homomorphism from $$G$$ to $$$$. Since $$G$$ is torsion-free, this homomorphism is injective, thus $$G$$ is isomorphic to a subgroup of $$$$, thus $$G$$ is cyclic.

• If you'd like a study buddy, let me know. I'm reading the same book. Currently, I'm up to the exercises labelled 5.2 (page 138). Commented Nov 25, 2021 at 15:12
• @Shaun How did you solve exerc. 4.1.3 ? Commented Nov 25, 2021 at 15:18
• I'm afraid I don't remember, @Panurge. I didn't write it down. Commented Nov 25, 2021 at 15:31

Let $$x \in G$$ be such that $$G/\mathbb{Z}x$$ has minimal order among all quotients of the form $$G/\mathbb{Z}g$$, $$0 \neq g \in G$$. We claim that $$G = \mathbb{Z}x$$.

If this were not the case, then there would exist $$y \in G - \mathbb{Z}x$$ and $$y + \mathbb{Z}x$$ has finite order $$n > 1$$. We then have $$ny = m x$$ for some $$m \in \mathbb{Z}$$ and $$n$$ and $$m$$ are coprime (since $$n$$ is minimal with $$ny \in \mathbb{Z}x$$). We can thus write $$1 = m a + n b$$ for some $$a,b \in \mathbb{Z}$$. Let us set $$z = ay + bx$$, then we have $$z \in G - \mathbb{Z}x$$ (since $$a$$ and $$n$$ are coprime) and $$n z = n (ay + bx)= a n y + b n x = (m a + n b) x = x$$ so that $$\mathbb{Z}x \subseteq \mathbb{Z}z$$. Actually this is proper containment, as $$z \in G - \mathbb{Z}x$$. It follows that $$G/\mathbb{Z}z$$ has smaller order than $$G/\mathbb{Z}x$$ which is a contradiction to our choice of $$x$$. It follows that $$G = \mathbb{Z}x$$ and we are done.

Suppose $$G$$ is non-cyclic. Lets find a contradiction.

As you have pointed out, $$G$$ contains an infinite cyclic subgroup of finite index. Let's call one such subgroup $$Z$$, and write $$n$$ for its index.

As $$G$$ is non-cyclic but contains a cyclic subgroup of finite index, there exist elements $$x, y\in G$$ such that the subgroup $$\langle x, y\rangle$$ (i.e. the minimal subgroup of $$G$$ containing $$x$$ and $$y$$) is non-cyclic. As $$x^n, y^n\in Z$$, we have that $$x$$ and $$y$$ have a common power, so write $$x^p=y^q$$. By replacing $$x$$ with $$x^{-1}$$ or $$y$$ with $$y^{-1}$$, we may assume $$p, q>0$$. Write $$z:=xy$$. If $$p\geq q$$ then consider the subgroup $$\langle x, z\rangle$$. Otherwise, consider the subgroup $$\langle z, y\rangle$$. Both subgroups are infact equal to $$H$$, but we now have $$x^{p-q}=z^q$$ if $$p\geq q$$, or $$z^p=y^{q-p}$$ if $$q>p$$. Now repeat this process, with $$(x, z)$$ or $$(z, y)$$ in place of $$(x,y)$$, until one of the exponents is $$0$$ (as we have reduced the size of the exponents, this will necessarily happen) - the other one will not be $$0$$, so we have $$H=\langle x_k, y_k\rangle$$ with either $$x_k^{p_k}=1$$, $$p_k>1$$, or $$y_k^{q_k}=1$$, $$q_k>1$$. It follows that $$H=\langle x, y\rangle$$ contains a non-trivial element of finite order, which is our required contradiction.

• I will read these two answers tomorrow and I will then make the difficult choice of which answer to rate as the best. Commented Nov 25, 2021 at 15:29

Here's an answer for arbitrary modules over commutative rings $$A$$: an $$A$$-module $$M$$ is a just-(infinite length) $$A$$-module ($$M$$ is has infinite length, every proper quotient of $$M$$ is has infinite length) if and only if, $$I$$ denoting the annihilator of $$M$$, $$A/I$$ is a Krull noetherian domain of dimension 1 and $$M$$ is isomorphic to a nonzero ideal of $$A/I$$.

Proof: first the assumption implies that $$M$$ is noetherian. It follows that $$A/I$$ is noetherian.

To conclude, we can suppose that $$A$$ is noetherian and $$M$$ is a faithful $$A$$-module. The assumption implies that $$M$$ does not contain the direct sum of two nonzero modules. Hence $$M$$ has a single associated ideal, say $$P$$. This applies to any nonzero cyclic submodule of $$M$$. Hence any nonzero $$m\in M$$ has a an annihilator contained in $$P$$. If by contradiction some $$m\in M$$ as annihilator strictly contained in $$M$$, then $$Pm$$ is a nonzero submodule of infinite index, contradiction. So every nonzero $$m\in M$$ has annihilator $$P$$. Hence $$P=0$$ and $$M$$ is isomorphic to a submodule of the field of fractions of $$A=A/P$$, thus, since it is finitely generated, is isomorphic to an ideal of $$A$$.

Conversely, any nonzero ideal in a noetherian domain of Krull dimension 1 satisfies the condition.

If now $$M$$ is just-infinite ($$M$$ is infinite and every proper quotient of $$M$$ is finite), then $$M$$ is also just-(infinite length), so has the required form: ideal $$J/I$$ in $$A/P$$ with $$A/P$$ noetherian domain of Krull dimension 1. Then for such a module, just-infinite is equivalent to every residual field $$A/M$$ of $$A/P$$ being finite. Indeed, we can suppose $$P=0$$. The condition is sufficient. Conversely, if $$M$$ is a maximal ideal, then $$MJ\neq J$$ by Nakayama's lemma, and we see that $$J/MJ$$ is an infinite proper quotient of $$J$$.