Different argument to part c of this problem
Problem says:A group $H$ is called finitely generated if there is a finite set $A$ s.t. $H=\langle A \rangle$.
(a) Prove that every finite group is finitely generated.
(b) Prove that $\mathbb{Z}$ is finitely generated
(c) Prove that every finitely generated subgroup of the additive group $\mathbb{Q}$ is cyclic [If $H$ is a finitely generated subgroup of $\mathbb{Q}$, show that $H \leq \left \langle \frac{1}{k} \right \rangle$, where $k$ is the product of all the denominators which appear in a set of generators for $H$
(d)Prove that $\mathbb{Q}$ is not finitely generated.
So far for part (c) I have the following:
Let $\langle x_1,\,\dots,\,x_n\rangle = H, x_1=\frac{a_1}{b_1}, x_2=\frac{a_2}{b_2}, \dots, x_{n}=\frac{a_n}{b_n}. k=b_1b_2b_3\dots b_n.$ I know $H\leq \mathbb{Q}$ and $\langle \frac{1}{k}\rangle=\{1, \frac{1}{k},\frac{2}{k},\dots, \frac{n}{k}\}.$ [I'm wondering if this set supposed to be finite and if I've defined the set $\langle \frac{1}{k}\rangle$ correctly?]
Now I want to show $H\leq \langle \frac{1}{k} \rangle$. As $1\in \langle \frac{1}{k} \rangle$ so $1 \in H$ therefore $H\neq \varnothing$. Let (WLOG) $\frac{m_1}{b_1}, \frac{m_2}{b_2}\in H$ and $\frac{m_1}{b_1}, \frac{m_2}{b_2}\in \langle \frac{1}{k}\rangle$. As $H$ is a group, then $-\frac{m_2}{b_2}\in H$. I want to show $\frac{m_1b_2-m_2b_1}{b_1b_2}\in \langle \frac{1}{k}\rangle$. This is the case as $\frac{m_1b_2-m_2b_1}{b_1b_2}=\frac{j}{b_1b_2\dots b_n}$ for some positive integer $j$. i.e. $m_1b_2-m_2b_1=j(b_3\dots b_n)^{-1}.$
I feel like my argument derails at the end but I appreciate any general feedback.
I don't know how I could make the last line work besides saying that $m_1b_2-m_2b_1=j(b_3\dots b_n)^{-1}$ is just some integer.
You're definitely on the right track! A few notes:
$\langle \frac{1}{k} \rangle \cong \mathbb Z$, so it is not finite. Can you prove that $f(n) = \frac{n}{k} : \mathbb{Z} \to \langle \frac{1}{k} \rangle$ is an isomorphism? You'll probably find the first isomorphism theorem helpful.
Knowing that $1 \in \langle \frac{1}{k} \rangle$ has nothing to do with whether $1 \in H$. I'm not sure if this is the mistake you're making, but it's worth remembering that $0$ is the identity of this group. Not $1$.
It seems like you're trying to show $H \leq \langle \frac{1}{k} \rangle$ by using some version of the subgroup criterion (correct me if I'm wrong). This will not actually show what you want. Instead, you want to use the following result:
$$\langle x_i \rangle \leq H \iff \text{ each } x_i \in H$$
This is immediate from the definition of $\langle x_i \rangle$ given on page 62 of my edition of D+F (in case yours is different, it's the start of section 2.4). It's probably a worthwhile exercise for you to prove it, if it isn't obvious to you.
Knowing this, we see how to show that $H \leq \langle \frac{1}{k} \rangle$. It suffices to show that each $x_i (= \frac{a_i}{b_i}) \in \langle \frac{1}{k} \rangle$.
So it remains to show that $\frac{a_i}{b_i} = \frac{m}{k} = \frac{m}{b_1 b_2 \ldots b_n}$ for some $m \in \mathbb{Z}$. Can you find an $m$ that works?
I hope this helps ^_^