Exercise 2.4.14 ($\mathbb{Q}$ is not finitely generated)

Proof.

(a)

If $G$ is finite, since $G=⟨G⟩$, $G$ is finitely generated.

(b)

Since $\mathbb{Z}=⟨1⟩$, $\mathbb{Z}$ is finitely generated (as every cyclic group).

(c)

Let $H$ a finitely generated subgroup of $\mathbb{Q}$, so that $$H=⟨a_1,a_2,\dots ,a_n⟩,a_1,a_2,\dots ,a_n\in \mathbb{Q}.$$

We write $a_i=\frac{p_i}{q_i}$, where $p_i\in \mathbb{Z},q_i\in \mathbb{Z},q_i>0$, and put $k=q_1{q}_2\cdots q_n$. Then

$$a_i=\frac{p_i}{q_i}=\frac{p_i(k∕q_i)}{k},$$

where $m=p_i(k∕q_i)$ is an integer, so $a_i=m\cdot \frac{1}{k}\in ⟨\frac{1}{k}⟩$. Since the subgroup $⟨\frac{1}{k}⟩$ contains $a_1,a_2,\dots ,a_n$, this shows that $H=⟨a_1,a_2,\dots ,a_n⟩\subseteq ⟨\frac{1}{k}⟩$, so

$$H\le ⟨\frac{1}{k}⟩.$$

We know that a subgroup of a cyclic group is cyclic, so $H$ is cyclic.

Every finitely generated subgroup of the additive group $\mathbb{Q}$ is cyclic.

(d)

If $\mathbb{Q}$ were a finitely generated group, then by part (c) $\mathbb{Q}$ would be cyclic, i.e., $\mathbb{Q}=⟨\frac{p}{q}⟩$, for some integers $p\in \mathbb{Z},q\in \mathbb{Z},q>0$. Then the rational $\frac{1}{2q}\in ⟨\frac{p}{q}⟩$, so $\frac{1}{2q}=n\frac{p}{q}$ for some integer $n$, thus $1=2\mathit{np}$ and $2\mid 1$. Since $2\nmid 1$, this is a contradiction, which proves that $\mathbb{Q}$ is not cyclic. Hence $\mathbb{Q}$ is not finitely generated.