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

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?

• How would $\frac14$ be generated? – Mastrem Aug 13 at 8:33
• As for infinitely generated: For any finite set of fractions, each fraction that they generate has a denominator dividing the product of the denominators of the fractions in that original set finite set. This cannot be the entirety of $\mathbb{Q}$ – Mastrem Aug 13 at 8:37
• Well after trying, I can't find a combination to make $\frac{1}{4}$, so I think it isn't possible, but how to prove this properly? – Belgium_Physics Aug 13 at 8:39
• In general, $\frac{1}{l^n}$ where $l$ is prime and $n>1$ is not in the subgroup generated by $\{\frac{1}{p} | p\text{ is prime}\}$. – Kenneth Yeo Aug 13 at 8:39
• Possible duplicate of Showing $\mathbb{Q}$ is a locally cyclic group – Crostul Aug 13 at 8:40

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.

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.

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 $$\frac{1}{\mathrm{lcm}(q_1,\dots, q_n)}$$