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

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!

• 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? Aug 6 '19 at 22:33

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 $$\langle A \rangle$$.

$$\newcommand{gae}{\newcommand{#1}{\operatorname{#1}}} \gae{gcd} \gae{lcm}\newcommand{gen}{\left\langle{#1}\right\rangle}$$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 $$H=\gen{\frac{\gcd(n,h)}{\lcm(m,k)}}$$.

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