1. ## cyclic group

Need some help on the following 2 related problems:
Show that any finitely generated subgroup of the additive group of rationals (Q,+,0) is cyclic.
Prove that this group is not isomorphic to the direct product of two copies of it.

2. Originally Posted by MathBird
Need some help on the following 2 related problems:
Show that any finitely generated subgroup of the additive group of rationals (Q,+,0) is cyclic.
Remember some facts about the structure of $\mathbb{Z}$. For $x_1,...,x_l\in \mathbb{Z}$ define $(x_1,...,x_l)$ to be the ideal generated these elements, i.e. $\{\Sigma_i a_ix_i | a_i \in \mathbb{Z} \}$. Since ideals are principal it means $(x_1,...,x_l) = (d)$, where $d\geq 0$. We call $d$ the greatest common divisor of $x_1,...,x_l$.

Now given, a finitely generated subgroup $H$ of $\mathbb{Q}^+$. This means, by definition, $H = \left< \frac{p_1}{q_1},...,\frac{p_2}{q_2} \right>$. The subgroup generated by these elements is precisely $\sum_i a_i \frac{p_i}{q_i}$ where $a_i\in\mathbb{Z}$. Define, $q=\Pi_i q_i$ and $Q_i = \frac{q}{q_i}$. Thus, $\sum_i a_i\frac{p_i}{q_i} = \frac{1}{q}\sum_i a_ip_iQ_i$. Define $d$ to be the greatest common divisor of $p_1Q_1,...,p_nQ_n$. Then, as seen in the first paragraph, $H=\left< \frac{d}{q}\right>$. Thus, we see that $H$ must be cyclic.

Prove that this group is not isomorphic to the direct product of two copies of it.
Since it is an infinite cyclic group (assuming $d>0$) it is isomorphic to $\mathbb{Z}$.
It is trivial to show that $\mathbb{Z}$ is not isomorphic to $\mathbb{Z}\times \mathbb{Z}$.

3. This is great!