Exercise 2.4.13 (The multiplicative group $\mathbb{Q}^+$ is generated by $\{ \frac{1}{p} \mid p ext{ is a prime} \}$)

Question

Prove that the multiplicative group of positive rational numbers is generated by the set $\{ \frac{1}{p} \mid p 	ext{ is a prime} \}$.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

As given in ``Frequently Used Notation'', $\Q^+$ is the multiplicative group of positive rational numbers.

\begin{proof} Let $\mathbb{P} = \{2, 3, 5, 7,\ldots\}$ be the set of (positive) prime numbers, and
$$A = \left\{ \frac{1}{p} \mid p \in \mathbb{P} ight\}.$$
Then the subgroup $\langle A angle$ of $\Q^+$ contains is closed under inverses, thus $\langle A angle$ contains all the prime numbers, and their powers.
Let $x = \frac{a}{b} >0$ be any element of $\Q^+$, where $a \in \Z^+$ and $b \in \Z^+$, and write
$$a = p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k},\qquad   b = q_1^{b_1}q_2^{b_2}\cdots q_l^{b_l}\qquad (a_i, b_i \in \mathbb{N})$$
the decomposition of $a,\, b$ in prime factors. Then
$$x = \frac{a}{b} = \left(\frac{1}{p_1}ight)^{-a_1}\left(\frac{1}{p_2}ight)^{-a_2}\cdots \left(\frac{1}{p_k}ight)^{-a_k}  \left(\frac{1}{q_1}ight)^{b_1}\left(\frac{1}{p_2}ight)^{b_2}\cdots \left(\frac{1}{q_l}ight)^{b_l} \in \langle A angle .$$

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This shows that
$$\mathbb{Q}^+ =\left \langle  \left\{ \frac{1}{p} \mid p \in \mathbb{P} ight\} ight angle.$$
(and also $\Q^+ = \langle  \{ p \mid p \in \mathbb{P} \} angle = \langle \mathbb{P} angle$.)

\end{proof}

Exercise 2.4.13 (The multiplicative group $\mathbb{Q}^+$ is generated by $\{ \frac{1}{p} \mid p \text{ is a prime} \}$)

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Exercise 2.4.13 (The multiplicative group $\mathbb{Q}^+$ is generated by $\{ \frac{1}{p} \mid p \text{ is a prime} \}$)

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