Is the additive rational group $\mathbb{Q},+$ generated by $\frac{1}{p}$ where p is a prime?

Question

So it is known that the additive group of rationals numbers $\mathbb{Q},+$ is generated by $\frac{1}{n}$ with $n \in \mathbb{N_0}$ so that: $$\mathbb{Q},+ =grp\{\frac{1}{n} | n \in \mathbb{N}\}$$

Now I was wondering if the group can be generated by $\frac{1}{p}$ where $p$ is a prime. That would give: $$\mathbb{Q},+ =grp\{\frac{1}{p} | \text{p is prime}\}$$

Therefore every $\frac{1}{n}$ should be written as a sum of $\frac{1}{p}$ but I don't know if this is possible.

Whether or not the statement is true, can we conclude from it that $\mathbb{Q},+$ is infinitely generated?

Thanks in advance

Answer 1

The answer is no.

Indeed, when you add fractions of the form $\frac1p$ with $p$ prime you obtain numbers where the denominator is square-free.

For instance, you never get $\frac14$.

To see that $\Bbb Q$ is not finitely generated as a group you can reason in a similar way: if $$ \frac{a_1}{b_1},\frac{a_2}{b_2},...,\frac{a_r}{b_r} $$ is a finite set of reduced (i.e. ${\rm gcd}(a_i,b_i)=1$) rational numbers the subgroup generated by them cannot contain $\frac1p$ where $p$ is a prime number not dividing any of the $b_i$'s.

Answer 2

First of all, the statement is not true, since $\frac14$ is not an element of the group, generated by $\frac1p$ where $p$ is prime.

Second of all, even if the set you describe did generate $\mathbb Q$, you could not, from that, conclude that $\mathbb Q$ is infinitely generated. For example, $\mathbb Z$ is generated by $\mathbb Z$, but is it infinitely generated? Of course not. It is also generated by $\{1\}$, and is clearly finitely generated.

Remember: The existence of an infinite generating set does not exclude the possibility of a different finite generating set.

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To actually prove that $\mathbb Q$ is not finitely generated, you should use a proof by contradiction, and also try to show the fact that $\{\frac{p_1}{q_1}\dots \frac{p_n}{q_n}\}$ (assuming here that $\gcd(p_i, q_i)=1$) generates the same group as $$\frac{1}{\mathrm{lcm}(q_1,\dots, q_n)}$$