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Can anyone please help me out on how to use the fundamental theorem of finitely generated abelian groups to prove the above question?

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What are $Q$ and $(Q,+)$? –  Chris Eagle Feb 2 '13 at 11:24
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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
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4 Answers

up vote 9 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.)

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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. –  B. S. Feb 2 '13 at 18:45
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If $\mathbb Q$ wants to be finitely generated, then it can't be divisible group. But it is.

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

$$m_1\left(\frac{n_1}{d_1}\right) + \dots + m_k\left(\frac{n_k}{d_k}\right) = \frac{\mathrm{long\ expression}}{\prod_i^kd_i}$$

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

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

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Thank you for your feedback! –  Faye Feb 2 '13 at 11:42
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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
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You're right. $0$ characteristic is also needed for this arguement. –  Berci Feb 3 '13 at 13:21
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