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Showing $\mathbb{Q}$ is a locally cyclic group

A group $G$ is locally cyclic if every finitely generated subgroup is cyclic. Prove that $(\mathbb{Q},+)$ is a locally cyclic group.

My strategy was to take a finite set of generators in $\mathbb{Q}$ to form a subgroup $H$ and then trying to find a monomorphism from $H \rightarrow \mathbb{Z}$ so that $H$ is isomorphic to a subgroup of $\mathbb{Z}$ and is thus cyclic.

Is this a good strategy, can somebody help me give me some insight into what the monomorphism might be? Thanks!

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    $\begingroup$ It could be a good strategy...... but the question is how to find that pesky monomorophism? A question for you: have you tried any examples? Say, pick your favorite random triple of rational numbers and see what they generate? $\endgroup$
    – Lee Mosher
    Aug 6 '19 at 22:33
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Let me suggest the following strategy:

Consider a non-empty finite set $A = \lbrace \frac{a_1}{b_1},\dots,\frac{a_r}{b_r} \rbrace \subset \mathbb{Q}$. Then we have $$\langle A \rangle = \left\lbrace n_1\frac{a_1}{b_1}+ \dots + n_r\frac{a_r}{b_r} \mid n_1,\dots,n_r \in \mathbb{Z} \right\rbrace,$$ which means that $\langle A \rangle$ is a subgroup of $\langle \frac{1}{b_1\cdots b_r} \rangle$. As subgroups of cyclic groups are cyclic, we get that $\langle A \rangle$ is cyclic.

This also shows how you could find a monomorphism into the integers. There is certainly an isomorphism $\langle \frac{1}{b_1\cdots b_r} \rangle \rightarrow \mathbb{Z}$, given by sending the generator $\frac{1}{b_1\cdots b_r}$ to $1 \in \mathbb{Z}$. Now just restrict that isomorphism to $\langle A \rangle$.

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$\newcommand{gae}[1]{\newcommand{#1}{\operatorname{#1}}} \gae{gcd} \gae{lcm}\newcommand{gen}[1]{\left\langle{#1}\right\rangle}$The lemma needs only be proved for subgroups generated by two elements. Let $H=\gen{\frac nm,\frac hk}$ with $\gcd(n,m)=\gcd(h,k)=1$. There are some $x,y\in\Bbb Z$ such that $x\frac{k}{\gcd(k,m)}n+y\frac{m}{\gcd(k,m)}h=\gcd\left(\frac{k}{\gcd(k,m)}n,\frac{m}{\gcd(k,m)}h\right)$. Now, since $$\gcd\left(\frac{k}{\gcd(k,m)},\frac{m}{\gcd(k,m)}\right)=1\\\gcd\left(\frac{k}{\gcd(k,m)},h\right)\mid\gcd(k,h)=1\\ \gcd\left(\frac{m}{\gcd(k,m)},n\right)\mid\gcd(m,n)=1,$$

necessarily $\gcd\left(\frac{k}{\gcd(k,m)}n,\frac{m}{\gcd(k,m)}h\right)=\gcd(n,h)$. Thus $H\ni x\frac{n}{m}+y\frac hk=\frac{x\frac{k}{\gcd(k,m)}n+y\frac{m}{\gcd(k,m)}h}{\lcm(n,k)}=\frac{\gcd(n,h)}{\lcm(m,k)}$

Now, it is clear that both a $\frac nm$ and $\frac hk$ are integer multiples of $\frac{\gcd(n,h)}{\lcm(m,k)}$, therefore $H=\gen{\frac{\gcd(n,h)}{\lcm(m,k)}}$.

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  • $\begingroup$ Ah so you use induction on the number of generators? Brilliant!! $\endgroup$ Aug 7 '19 at 0:53
  • $\begingroup$ @MathematicalMushroom yes, that's the idea. $\endgroup$
    – user239203
    Aug 7 '19 at 16:41

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