Can the group of rational numbers be generated by a finite set of elements?

[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.

 

[Jupiter]

I think I've got a proof.

Let [tex]X=\{\frac{a_1}{b_1},\frac{a_2}{b_2},...,\frac{a_n}{b_n}\}\subset \mathbb{Q}[/tex]. Consider [tex]\langle \frac{1}{b_1b_2...b_n}\rangle[/tex]. Then if [tex]x\in \langle X\rangle [/tex], we know [tex]x=\frac{c_1a_1}{b_1}+\frac{c_2a_2}{b_2}+...+\frac{c_na_n}{b_n}[/tex] for some [tex]c_i\in \mathbb{Z}[/tex]. It follows that [tex]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].
(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.

 

[matt grime]

Is Q a finitely generated what?

 

[Jupiter]

the group of rational numbers is not finitely generated

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.

 

[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.

 

[Jupiter]

One presumes you mean as an additive group.

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

 

[matt grime]

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

 

[NateTG]

Here's an alternative proof:

Let's assume that there is some generating set [tex]\{q_1,q_2...q_n\}[/tex].
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:
[tex]q_j=(-1)^{e_0}\prod_{i=1}^\infty \pi(i)^{e_ij}[/tex]
where [tex]\pi[/tex] is the prime function (i.e. [tex]\pi(1)=2, \pi(2)=3, \pi(3)=5...[/tex]),
[tex]e_i \in \mathbb{Z}[/tex], [tex]e_0 \in {0,1}[/tex] and only a finite number of the exponents are zero.

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

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

If we're considering [tex]Q-\{0\}[/tex] as a multiplicative group, it's easy to see that [tex]\frac{1}{\pi(i_{max}+1)}[/tex] 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