Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone please help me out on how to use the fundamental theorem of finitely generated abelian groups to prove that $({\mathbb{Q}},+)$ is not finitely generated?

share|cite|improve this question
What are $Q$ and $(Q,+)$? – Chris Eagle Feb 2 '13 at 11:24
I don't see why you need that theorem. It couldn't be finitely generated since, letting $p$ be any prime greater than the product of the denominators of the generators in lowest form, one could not generate $1/p$. – Amit Kumar Gupta Feb 2 '13 at 11:28
im sorry I have just managed to find from Detexify the correct writing of the Rational Numbers.(not so familiar with SE yet) – Faye Feb 2 '13 at 11:28
@Amit, I would like to say that I agree with you as it seems to not finitely generate whenn letting p a prime number. Thank you all for the very quick feedback. – Faye Feb 2 '13 at 11:34
Are you ever going to answer my question? – Chris Eagle Feb 2 '13 at 11:38
up vote 13 down vote accepted

Assume that $(\mathbb Q,+)$ is finitely generated. Then, by the fundamental theorem there exist $m,n\ge 0$ and $d_1\mid\cdots\mid d_m$ with $d_i>1$ such that $\mathbb Q\simeq \mathbb Z/d_1\mathbb Z\oplus\cdots\oplus\mathbb Z/d_m\mathbb Z\oplus \mathbb Z^n$. If $m\ge 1$, then there exists $x\in\mathbb Q$, $x\neq 0$, such that $d_1x=0$, a contradiction. Thus we get $m=0$. Then $\mathbb Q\simeq \mathbb Z^n$. If $n\ge 2,$ then there exist $x_1,x_2\in\mathbb Q$ which are linearly independent over $\mathbb Z$. But $x_1=a_1/b_1$ and $x_2=a_2/b_2$ give $(b_1a_2)x_1+(-b_2a_1)x_2=0$, a contradiction. So we must have $n=1$, that is, $\mathbb Q$ is cyclic. Assume that it is generated by $a/b$ with $b\ge 1$. Then $\frac{1}{b+1}$ can not be written as $\frac{ka}{b}$, and again we reached to a contradiction.

(Of course, there are simpler and much more natural arguments to show that $(\mathbb Q,+)$ is not finitely generated.)

Edit. In particular, this shows that the additive group of any field of characteristic $0$ is not finitely generated. (However the property holds for any infinite field.)

share|cite|improve this answer
This argument gives the nice bonus that $\Bbb{Q}$ isn't free (as an abelian group) at all, whether on finitely many or infinitely many generators. – Chris Eagle Feb 2 '13 at 11:49
@ChrisEagle Actually I've borrowed the argument for the free case from the general one showing that $\mathbb Q$ is not free. – user26857 Feb 2 '13 at 11:52
Thank you both for your feedback! – Faye Feb 2 '13 at 11:55
It is as you have used the exact Theorem that I wanted used, but I did not know how to. But thank you for your help! – Faye Feb 2 '13 at 12:03
+1 Very nice and complete approach. – S. Snape Feb 2 '13 at 18:45

If $\mathbb Q$ wants to be finitely generated, then it can't be divisible group. But it is.

share|cite|improve this answer
Nice hint/answer there. +1 – DonAntonio Feb 2 '13 at 11:28
Thank you for your feedback! – Faye Feb 2 '13 at 11:42
Very nice hint! +1 – amWhy Feb 2 '13 at 14:48

As per the OP's request, here's an explanation of my comment:

I don't see why you need that theorem (FToFGAG). It couldn't be finitely generated since, letting p be any prime greater than the product of the denominators of the generators in lowest form, one could not generate 1/p.

Suppose $(\mathbb{Q}, +)$ were finitely generated. Let $$\left\{\frac{n_1}{d_1}, \dots, \frac{n_k}{d_k}\right\}$$ be a generating set. Let $p$ be any prime that doesn't divide $\prod_{i=1}^k d_i$. Then clearly one cannot generate $\frac 1p$ by adding and subtracting (whole number multiples of) the elements of the generating set, contradicting the fact that it's supposed to be a generating set.

To be explicit, let's consider an arbitrary ($\mathbb{Z}$-linear) combination of the elements in the generating set:

$$m_1\frac{n_1}{d_1}+ \cdots + m_k\frac{n_k}{d_k}=\frac{\mathrm{long\ expression}}{\prod_i^kd_i}$$

There's no way to reduce such a fraction to $\frac 1p$ if $p$ doesn't divide the denominator.

share|cite|improve this answer

In a finitely generated $0$ characteristic Abelian group you cannot divide arbitrary many times by $2$, say.

share|cite|improve this answer
Thank you for your feedback! – Faye Feb 2 '13 at 11:42
This is false: in $\Bbb{Z}/3\Bbb{Z}$, for example, you can divide by $2$ as many times as you like. – Chris Eagle Feb 2 '13 at 11:43
You're right. $0$ characteristic is also needed for this arguement. – Berci Feb 3 '13 at 13:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.