Subgroups of the rationals is either generated by one element or infinitely generated

Question

The following situation is given: Consider $(\mathbb{Q},+)$ as an ordered abelian group: $x,y\in \mathbb{Q}, \; x\le y\; :\Leftrightarrow y-x\ge 0$. Let $\mathbb{Q}_+=\{c\in\mathbb{Q}\mid c\ge 0\}.$

Let $U\neq 0$ be an abelian group such and there exists an embedding $\iota:U\to \mathbb{Q}$, with $\iota(1_U)=1_\mathbb{Q}$. Suppose that $\iota(U)\cap \mathbb{Q}_+$ contains no minimal element other than $1_\mathbb{Q}$.

I don't know why the following 2 claims are correct:

1) Suppose $\iota(U)\cong \mathbb{Z}$, then $U$ is generated by $1_U$.

2) Suppose $\iota(U)\ncong \mathbb{Z}$, then it's infinitely generated.

For 1): I know that $1_\mathbb{Z}$ generates $\mathbb{Z}$ as an infinite cyclic group. If I identify $1_\mathbb{Q}$ with $1_\mathbb{Z}$, and since $\iota$ is an isomorphism with $\iota(1_U)=1_\mathbb{Z}$, then $1_U$ must be a generator. Is this argumentation ok?

How to argue for 2?

I appreciate any help.

Answer 1

If $H=\iota(U)$ is finitely generated, with generators $a_1/b_1,\ldots, a_k/b_k$, prove that (i) $H\subseteq\frac1N\Bbb Z$ for some $N$, (ii) $H=\frac MN\Bbb Z$ for some $M$.

Answer 2

Hint 1: a subgroup of $(\mathbb{Q},+)$ is finitely generated if and only if it has a smallest positive element. Hint 2: if a subgroup of $(\mathbb{Q},+)$ is finitely generated, then it is generated by its smallest positive element.

For example: subgroup of $(\mathbb{Q},+)$ generated by 2 and 7 has smallest positive element $1$; subgroup generated by $\frac{1}{2}$ and $\frac{4}{7}$ has smallest positive element $\frac{1}{14}$.