Show that a group \(G\) is isomorphic to a subgroup of the additive group \(\mathbb Q\) if and only if \(G\) is countable and every finite subset of \(G\) is contained in an infinite cyclic subgroup of \(G\).
The trivial group \(G=\{0\}\) is a subgroup of \((\QQ,+)\), but its finite subset \(\{0\}\) is not contained in any infinite cyclic subgroup of \(G\). The equivalence holds for every nontrivial group \(G\).
Solution
1. Every nontrivial subgroup \(H\leq(\QQ,+)\) is countable and everyfinite subset of \(H\) is contained in an infinite cyclic subgroup of \(H\).
Countability follows from \(H\subseteq\QQ\). Let \[ S=\{q_1,\ldots,q_r\}\subseteq H \] be finite. If every \(q_i=0\), choose any nonzero \(h\in H\); then \(S\subseteq\langle h\rangle\).
Otherwise write the nonzero \(q_i\) with a common positive denominator: \[ q_i=\frac{a_i}{N}. \] Let \(d\) be the greatest common divisor of the nonzero integers \(a_i\). By Bezout’s identity, \(d/N\) is an integer linear combination of the \(q_i\), so \[ \frac dN\in H. \] Each \(q_i\) is an integer multiple of \(d/N\). Hence \[ S\subseteq\left\langle\frac dN\right\rangle, \] and this cyclic subgroup is infinite.
2. Suppose \(G\) is countable and every finite subset of \(G\) is containedin an infinite cyclic subgroup. Then there is an increasing sequence\[ C_1\subseteq C_2\subseteq\cdots \]of infinite cyclic subgroups whose union is \(G\).
Enumerate \[ G=\{g_1,g_2,\ldots\}. \] Choose an infinite cyclic subgroup \(C_1\) containing \(g_1\). Inductively, suppose \[ C_n=\langle c_n\rangle \] has been chosen and contains \(g_1,\ldots,g_n\). By hypothesis, the finite set \(\{c_n,g_{n+1}\}\) is contained in an infinite cyclic subgroup \(C_{n+1}\). Then \(C_n=\langle c_n\rangle\subseteq C_{n+1}\) and \(g_{n+1}\in C_{n+1}\). Thus the construction continues, and every \(g_n\) belongs to \(C_n\), so the union is \(G\).
3. There are injective homomorphisms\[ \phi_n:C_n\longrightarrow\QQ \]such that\[ \phi_{n+1}|_{C_n}=\phi_n. \]
Choose a generator \(c_n\) of each \(C_n\). Since \(C_n\subseteq C_{n+1}\), there is a nonzero integer \(m_n\) such that \[ c_n=m_n c_{n+1}. \] Define recursively nonzero rational numbers \(r_n\) by \[ r_1=1, \qquad r_{n+1}=\frac{r_n}{m_n}, \] and define \[ \phi_n(kc_n)=kr_n \qquad(k\in\ZZ). \] Each \(\phi_n\) is an injective homomorphism. Moreover, \[ \phi_{n+1}(c_n) = \phi_{n+1}(m_nc_{n+1}) = m_nr_{n+1} = r_n = \phi_n(c_n), \] so the two homomorphisms agree on the generator of \(C_n\) and therefore on all of \(C_n\).
4. The compatible maps in step 3 define an injective homomorphism\[ \phi:G\longrightarrow\QQ. \]
For \(g\in G\), choose \(n\) with \(g\in C_n\) and set \[ \phi(g)=\phi_n(g). \] Compatibility in step 3 makes this independent of the choice of \(n\). Given \(g,h\in G\), both lie in some common \(C_N\), so \[ \phi(g+h)=\phi_N(g+h)=\phi_N(g)+\phi_N(h)=\phi(g)+\phi(h). \] If \(\phi(g)=0\), choose \(n\) with \(g\in C_n\). Then \(\phi_n(g)=0\), and injectivity of \(\phi_n\) gives \(g=0\). Thus \(\phi\) is injective.
5. The group \(G\) is isomorphic to a subgroup of \((\QQ,+)\).
By step 4, \(\phi\) identifies \(G\) with the subgroup \(\phi(G)\) of \((\QQ,+)\).
6. Q.E.D.
For nontrivial groups, step 1 proves the forward implication, and steps <1>2–<1>5 prove the converse.