Verifying that $\mathbb Q=\bigcup_{n\ge 1} H_n$ Let $G=(\mathbb Q,+)$ and $r_1=p_1/q_1, r_2=p_2/q_2\in G$. I want to prove that:


*

*$\langle r_1,r_2\rangle\subseteq \langle\frac{1}{q_1q_2}\rangle$


*If $r_1,r_2,...,r_n\in G$ then $\langle r_1,r_2,...,r_n\rangle$ is cyclic.


*If $H_n=\langle \frac{1}{n!}\rangle$ then every $H_n$ is cyclic, $$H_1\subset H_2\subset H_3\subset...$$ and $\mathbb Q=\bigcup_{n\ge 1} H_n$.

I can see that $r_1\in\langle\frac{1}{q_1q_2}\rangle$ and the same is true for $r_2$. This ends proving $1.$ I guess I should use the first part to see the second part clearly. So, I think I should have $r_i\in \langle\frac{1}{q_1q_2q_3...q_n}\rangle$. Is my guess right because it makes the second proved? For the third part it is obvious that every $H_n$ is cyclic. Assuming that chain of inclusions is correct and that $\bigcup_{n\ge 1} H_n\subseteq\mathbb Q$, just make me an small finial hint. Please verify my way of solution. I will be thankful if you have any other point of views. Thanks
 A: Your intuition seems correct, but your arguments need to be more rigorous. The crucial step in proving all of these statements is knowing that $a \in \langle b \rangle$ means there is $n \in \Bbb Z$ for which $a = nb$. This follows from the definition of cyclic groups in the additive notation. Another important fact is that $a \in \langle b \rangle$ implies $\langle a \rangle \subset \langle b \rangle$. Try to prove this if you aren't already familiar with it.
To get to specifics, 1 follows from the statements above. For 2, it's not sufficient to prove $r_i \in \left\langle \frac{1}{q_1 \cdots q_n} \right\rangle$. This would imply $\langle r_1, \ldots, r_n \rangle \subset \left\langle \frac{1}{q_1 \cdots q_n} \right\rangle$. You need to show $\langle r_1 \cdots r_n \rangle = \langle s \rangle$ for some $s \in \Bbb Q$.
For 3 and 4, use my initial statements and make sure to show containment in both directions in 4!
A: You need to work a bit harder to prove (2). Suppose that $r_1=\frac6{11}$ and $r_2=\frac{10}{13}$; then $\langle r_1,r_2\rangle$ is $\left\langle\frac2{143}\right\rangle$, not $\left\langle\frac1{143}\right\rangle$, though it’s still cyclic. Note that everything in $\langle r_1,r_2\rangle$ has the form
$$\frac{mp_1+np_2}{q_1q_2}\;;$$
what do you know about $\{mp_1+np_2:m,n\in\Bbb Z\}$ that gives you a generator for this group?
Once you’ve done that, you can use induction to extend from $\langle r_1,r_2\rangle$ to $\langle r_1,\dots,r_k\rangle$; the induction step will just use part (2).
Showing that every rational number $\frac{p}q$ is in some $H_n$ is easy: think about the relative sizes of $n$ and $q$.
