# Finitely generated abelian group with certain properties

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.

• You appear to have assumed that $G$ is finite - but it is easy to show that $G$ cannot be infinite. – Derek Holt Jul 2 '15 at 21:37
• Thanks for the remark. If $G$ was infinite, then it contains at least one copy of $\mathbb Z$, but this also means it contains $2\mathbb Z$ and $3\mathbb Z$ and neither of these two groups contain the other. – user156441 Jul 2 '15 at 22:16

One of them is the following: when $G=\mathbb Z/p^{\alpha}\mathbb Z\times\mathbb Z/p^{\beta}\mathbb Z$ you said that $\mathbb Z/p\mathbb Z\subset G$. This is right up to isomorphism: we identify $\mathbb Z/p\mathbb Z$ with the subgroup $p^{\alpha-1}\mathbb Z/p^{\alpha}\mathbb Z\times \{0\}$ of $G$.
If you know this, then should be easy to conclude: when $G=\mathbb Z/p^{\beta}\mathbb Z$ the subgroups of $G$ are $p^i\mathbb Z/p^{\beta}\mathbb Z$ with $i=0,1,\dots,\beta$, and therefore $\beta=3$.