Prove $(\mathbb Q,+)$ is not free abelian group.

Question

Prove $(\mathbb Q,+)$ is not free abelian group.

My solution:

First of all I prove $(\mathbb Q,+)$ is not finitely generated.

Suppose $\mathbb Q$ is finitely generated then exist $\langle\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},\cdots ,\frac{1}{a_n}\rangle$.

Each $q\in \mathbb Q$ is generated by $\langle\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3}, \cdots, \frac{1}{a_n}\rangle$.

Define $t=\frac{1}{a_{1}} \cdot \frac{1}{a_{2}}\cdot\cdots \cdot\frac{1}{a_{n}} = \frac{1}{a_{1} \cdot a_{2}\cdot \cdots \cdot a_{n}}.$

If we take $x = \frac{1}{t+1} = \frac{1}{(a_{1} \cdot a_{2} \cdot\cdots\cdot a_{n})+1}$ , then $x \notin \langle\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3}, \cdots, \frac{1}{a_n}\rangle.$

$(\mathbb Q,+)$ is not finitely generated.

We know free abelian group has a finite basis that generate the group, there is no finite basis, $(\mathbb Q,+)$ is not free abelian group. (Is it correct ?)

I'd be grateful for your some help!

Answer 1

free abelian group has a finite basis that generate the group, there is no finite basis, $(\mathbb{Q}, +)$ is not free abelian group.

This is incorrect. We can take the free Abelian group on an infinite set of generators.

As for how to solve the problem:

Suppose that $\mathbb{Q}$ is the free group on the set $I$ of generators, with universal map $\iota : I \to \mathbb{Q}$. Consider the map $f : I \to \mathbb{Z}$ given by $f(i) = 1$. Extend this to a group homomorphism $g : \mathbb{Q} \to \mathbb{Z}$ such that $g \circ \iota = f$. Now take some $i \in I$. What is $g(\iota(i) / 2)$?