# Show $G=(\mathbb{Q},+)$ is not finitely generated

I have been reading a proof for this question and I do not understand the final contradiction that the proof arrives at.

Show $G=(\mathbb{Q},+)$ is not finitely generated. (i.e. not generated by a finite set of elements)

Suppose $S=\{x_1,...,x_n\}$ is a finite generating set. Since each $x_i$ is a rational number, we can write $x_i=a_i/b_i$ where $a_i,b_i\in\mathbb{Z}$.

But then $x_1,...,x_n$ are all integer multiples of $1/b_1b_2...b_n$, and so $\langle S\rangle\leqslant\langle 1/b_1b_2...b_n\rangle$, which is a proper subgroup of $(\mathbb{Q},+)$ since it doesn't contain $1/2b_1b_2...b_n$, for example.

How is this a contradiction, is it that since it doesn't contain $1/2b_1b_2...b_n$ it is not a subgroup? Which subgroup property is it breaking?

Also why cant $\langle 1/b_1b_2...b_n\rangle$ contain $1/2b_1b_2...b_n$ is it because multiplying by $1/2$ is not something that can be done since $1/2$ is not an element of $G$ and $1/2b_1b_2...b_n$ cannot be reached from addition of elements and their inverses in $\langle 1/b_1b_2...b_n\rangle$

• You assume that the subgroup $\langle S\rangle$ contains all of $\mathbb{Q}$, thus is should also contain $\frac{1}{2b_1 b_2 ... b_n}$ Dec 16 '14 at 10:43
• It's a contradiction with the fact that the picked set generates $\Bbb Q$. Whatever finite set you pick, it generates a proper subgroup of $\Bbb Q$. Dec 16 '14 at 10:45

The subgroup $S = \langle (b_1 b_2 \cdots b_n)^{-1} \rangle$ doesn't contain $(2b_1b_2 \cdots b_n)^{-1}$ since the subgroup under addition only contains integer multiples of $(b_1b_2 \cdots b_n)^{-1}$. This contradicts the fact that $S$ generates all of $(\Bbb Q, +)$
It is a contradiction because the assumption was that the set $S$ generated the entire group $\mathbb{Q}$, which does contain $\frac{1}{2b_1b_3\cdots b_n}$.
That the element $\frac{1}{2b_1b_2\cdots b_n}$ is not in $\langle S\rangle$ is because any sum of elements from $S$ will have a denominator that is a product of distinct elements from $S$ (by picking a reduced expression for the fraction).
Suppose on contrary that a finite subset $S$ of $\mathbb{Q}$ generates the group ($\mathbb{Q}, +$). Let $P$ be the set of primes appearing in the denominators of the members of $S$. Then $P$ is finite. Take a rational number $r$ of which denominator is a prime not in $P$. Obviously, $r$ does not in the group $<S>$ generated by S. This yield a contradiction of the assumption that S generates $\mathbb{Q}$. Therefore no finite set of $\mathbb{Q}$ can generate $\mathbb{Q}$ Therefore $\mathbb{Q}$ is not finitely generated.
Here is how I would frame the argument. Firstly, I assume the result that $\mathbb{Q}$ is not cyclic. You can find a proof of that here. Now, working along the setup of your argument, if $\mathbb{Q}$ is finitely generated, say $\mathbb{Q}$=$\langle{a_1/b_1,...,a_n/b_n}\rangle$ where each $a_i,b_i$ is a nonzero integer and each fraction is in lowest terms, then each $a_i/b_i$ is an integer multiple of $1/(b_1...b_n)$ since $$a_i/b_i=a_ib_1...b_{i-1}b_{i+1}...b_n(1/b_1...b_{i-1}b_{i}b_{i+1}...b_n)$$ Hence, $\langle{a_1/b_1,...,a_n/b_n}\rangle\subset\langle{1/b_1...b_n}\rangle$ and $\langle{1/b_1...b_n}\rangle$ also generates $\mathbb{Q}$, contradicting the fact that $\mathbb{Q}$ is not cyclic.