# Q not finitely generated

1. Apr 28, 2004

### Jupiter

Can someone guide me through the proof (or point me to where I can find the proof) that the group of rational numbers is not finitely generated?
I know that it helps to break it into steps, the first of which you show that any finitely generated subgroup of Q is contained in a cyclic subgroup (and hence is cyclic), and in the second step you show that Q itself is not cyclic.

2. Apr 28, 2004

### Jupiter

I think I've got a proof.

Let $$X=\{\frac{a_1}{b_1},\frac{a_2}{b_2},...,\frac{a_n}{b_n}\}\subset \mathbb{Q}$$. Consider $$\langle \frac{1}{b_1b_2...b_n}\rangle$$. Then if $$x\in \langle X\rangle$$, we know $$x=\frac{c_1a_1}{b_1}+\frac{c_2a_2}{b_2}+...+\frac{c_na_n}{b_n}$$ for some $$c_i\in \mathbb{Z}$$. It follows that $$x=\frac{c_1b_2b_3...b_na_1+...+b_1b_2...b_{i-1}c_ib_{i+1}...b_na_i+...+b_1b_2...b_{n-1}c_na_n}{b_1...b_n} \in \langle \frac{1}{b_1b_2...b_n}\rangle$$.
(tex not showing up - meant to show containment of x in the cyclic subgroup we're considering)
Thus any finitely generated subgroup of Q is cyclic.
Now if Q were cyclic then Q would be isomporphic to Z. If f were such an isomorphism, then certainly f(1) is not 0, so f(n)=nf(1) for all n means that f(1)/2 has no preimage so f certainly cannot be onto. Hence Q is not cyclic and Q is not finitely generated.

I'd still be interested in alternative proofs, if anyone knows any.

Last edited: Apr 28, 2004
3. Apr 28, 2004

### matt grime

Is Q a finitely generated what?

4. Apr 28, 2004

### Jupiter

finitely generated group
To be honest, I don't know of any other kind of "finitely generated" so I took it for granted that I meant groups.

5. Apr 28, 2004

### matt grime

It is context dependent. Q is a finitely generated field oveer Q, it is finitely generated as a module over its centre for instance. One presumes you mean as an additive group.

6. Apr 28, 2004

### Jupiter

Isn't this the only kind of group I could possibly mean?

7. Apr 29, 2004

### matt grime

Yes. But best to check. I think your proof is correct, but you don't need to consider Z explicitly

8. Apr 29, 2004

### NateTG

Here's an alternative proof:

Let's assume that there is some generating set $$\{q_1,q_2...q_n\}$$.
Since the group is non-trivial, we can freely eliminate the identity element from any generating set.
Now each of the elments in that set can be uniquily represented:
$$q_j=(-1)^{e_0}\prod_{i=1}^\infty \pi(i)^{e_ij}$$
where $$\pi$$ is the prime function (i.e. $$\pi(1)=2, \pi(2)=3, \pi(3)=5...$$),
$$e_i \in \mathbb{Z}$$, $$e_0 \in {0,1}$$ and only a finite number of the exponents are zero.

Then let $$E_j$$ be the set of all $$i$$ such that $$e_{ij} \neq 0$$. And let
$$E=E_1 \cup E_2 ... \cup E_n$$. Clearly $$E$$ is finite since it is a finite union of finite sets. Therefore it has some maximum element $$i_{max}$$.

Now, if we're consdiering $$Q$$ as an additive group, then the allegedly generating set cannot generate $$\frac{1}{\pi(i_{max}+1)}$$ since addition never contributes new negative powers.

If we're considering $$Q-\{0\}$$ as a multiplicative group, it's easy to see that $$\frac{1}{\pi(i_{max}+1)}$$ cannot be generated by the set because multiplication, or inversion of two numbers with zero exponents for a prime will never generate non-zero exponents

Last edited: Apr 29, 2004