# Thread: Q is not finitely generated

1. ## Q is not finitely generated

Hello all Mathematicians and lover of math:

I need your help in verifying this statement :

Show that every finitely generated subgroup of ( Q , + ) is cyclic, show also thet ( Q , + ) is not isomorphic to the set of positive rational numbers with usual multiplication

What I know is that the generator for any finitely-generated subgroup of Q is the smallest non-zero member of that subgroup

2. Originally Posted by fuzzy topology
Hello all Mathematicians and lover of math:

I need your help in verifying this statement :

Show that every finitely generated subgroup of ( Q , + ) is cyclic, show also thet ( Q , + ) is not isomorphic to the set of positive rational numbers with usual multiplication

What I know is that the generator for any finitely-generated subgroup of Q is the smallest non-zero member of that subgroup
Suppose that $\displaystyle G\subseteq Q$ is finitely generated. Then $\displaystyle G=\left\{n_1\cdot\frac{p_1}{q_1}+\cdots+n_k\cdot\f rac{p_k}{q_k}:n_k\in\mathbb{Z}\right\}\subseteq \left\{\frac{n}{q_1\cdots q_k}:n\in\mathbb{Z}\right\}=\left\langle\frac{1}{q _1\cdots q_k}\right\rangle$. Thus, $\displaystyle G$ is a subgroup of a cyclic group, and thus cyclic.

For the second part, use this.