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?

Thanks in advance for the answers.
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 $<a>$ is a nontrivial subgroup of $G$, thus, by hypothesis, $<a>$ is of finite index $m > 0$ in $G$. Then $x \mapsto mx$ defines a homomorphism from $G$ to $<a>$. Since $G$ is torsion-free, this homomorphism is injective, thus $G$ is isomorphic to a subgroup of $<a>$, thus $G$ is cyclic.
 A: 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.
A: 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.
A: 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 $J$.
