Problem
Characterize all finitely generated abelian $G$ such that every proper subgroup of $G$ is cyclic, $G$ contains exactly two proper subgroups, and for each pair of subgroups $S$,$T$ in $G$ we have $S \subset T$ or $T \subset G$.
My idea was the following:
First notice that $G$ has at most two primes in its factorization, if this is not the case, by Cauchy theorem we would have at least the proper subgroups. So the order of $G$ is of the form $ |G|=p^{\alpha}q^{\beta}$ with $p$ and $q$ primes.
By the structure theorem for f.g. abelian groups, we have $$G \cong \mathbb Z / \langle p^{\alpha_1} \rangle \oplus \cdots \oplus \mathbb Z / \langle p^{\alpha_n} \rangle \oplus \mathbb Z / \langle q^{\beta_1} \rangle \oplus \cdots \oplus \mathbb Z / \langle q^{\beta_m} \rangle$$
where $\sum_{i=1}^n \alpha_i=\alpha$ and $\sum_{j=1}^m \beta_j=\beta$ with $\alpha_i,\beta_j$ all natural numbers. Now suppose there is more than one invariant factor associated to each prime, without loss of generality, we can assume there are at least two invariant factors associated to $p$. Then $\mathbb Z_p \oplus \mathbb Z_p$ is a subgroup of $G$ which is not cyclic. So we must have $$G \cong \mathbb Z / \langle p^{\alpha}\rangle \oplus \mathbb Z / \langle q^{\beta}\rangle.$$
But now $\mathbb Z_p, \mathbb Z_q \subset G$ and clearly $\mathbb Z_p \not \subset \mathbb Z_q$ and $\mathbb Z_q \not \subset \mathbb Z_p$.
In conclusion, $G \cong \mathbb Z / \langle p^{\beta}\rangle$.
I got stuck at this point, I see that $\alpha \geq 3$, but could it be greater than this number? Is it correct what I've done so far? Any help would be appreciated.