F.G. abelian group so that every non-trivial quotient is cyclic

Question

I need to characterize every finitely generated abelian group G that has the following property: $$\frac{G}{S} \text{ is cyclic for every } \lbrace0\rbrace \lneq S\leq G$$ Given the problems before this one, I believe I am supposed to use the structure theorem figure out the underlying structure of the decomposition (for example that the decomposition has only one or two primes and such). I know the question in this post is very similar, but it is not precisely the same and the slight difference in the conditions on the subgroups makes the argument non-applicable, unfortunately.

Answer 1

Suppose that $G$ is not cyclic. This means that in the factorization of $G$ there is a factor $\mathbb Z_{p^a}\times \mathbb Z_{p^b}$. So let's write the group as $\mathbb Z_{p^a}\times \mathbb Z_{p^b}\times H$.

If $H$ is not trivial then $\frac{G}{\{0\}\times\{0\}\times H}$ is a non-cyclic quotient isomorphic to $\mathbb Z_{p^a} \times \mathbb Z_{p^b}$.

If $a>1$ let $C$ be the subgroup of order $p$ of the first factor, then the quotient $\frac{G}{C\times \{0\} \times H}$ will give you a quotient that contains a subgroup isomorphic to $\mathbb Z_p \times \mathbb Z_p$ and is thus not cyclic. The same argument shows $b$ must be $1$.

So the only abelian groups that are not cyclic and fit the bill are those isomorphic to $\mathbb Z_p \times \mathbb Z_p$

Answer 2

First question: how many factors can appear at most in the decomposition?

Second question: in every case you deduced above, which powers can appear? Can you have for example a factor $\mathbb{Z}/2^3\mathbb{Z}$?

Good work!

Answer 3

Let $G=\langle x_1,\dots, x_n\rangle $. Then $G/\langle x_1\rangle\cong \langle x_2,\dots,x_n\rangle $ is cyclic. It follows that $G $ is generated by two elements.

More can be said: it's easy to see that, in case $G $ is not cyclic, we have $G\cong\mathbb Z_p×\mathbb Z_p $, for $p $ prime.