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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!

Shaun
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    "We know free abelian group has a finite bases". That's false. A free abelian group has a basis. You can have free abelian groups of infinite rank. For example, $\oplus_{n\in\mathbb{N}}\mathbb{Z}$ is a free abelian group that does not have a finite basis. – Arturo Magidin Oct 08 '21 at 19:25
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    Not all free groups are finitely generated as free groups. – lulu Oct 08 '21 at 19:25
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    See this solution. – Arturo Magidin Oct 08 '21 at 19:26
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    @ArturoMagidin Not sure this should have been closed. The person was asking for comment on their specific solution, not asking how to solve the problem in general. Furthermore, interpreting the linked solution requires understanding enough about modules to know that they're equivalent to Abelian groups and being able to translate module arguments to group arguments - probably not useful to someone taking a first course in abstract algebra. – Mark Saving Oct 08 '21 at 19:33
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    @MarkSaving: Yes, I don't normally close "solution-verification" questions. But the comments on the proposed solution had been given, and you posted an independent solution rather than simply addressing the proposed solution. If we are going to add independent answers that do not progress from the proposed solution whose verification is being asked about, then it's a duplicate. – Arturo Magidin Oct 08 '21 at 19:36
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    @ArturoMagidin Thanks, I shall keep this in mind. – Mark Saving Oct 08 '21 at 19:37

1 Answers1

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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 $g(\iota(i) / 2)$?

Mark Saving
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