Describe the isomorphism classes of subgroups of the additive group of rational numbers \((\mathbb{Q}, +)\) (rank 1 torsion-free abelian groups). (1) How many isomorphism classes are there (what is the exact cardinality)? (2) Are the subrings \(\mathbb{Z}[1/p \mid p \in S]\) associated to different sets of primes \(S \subseteq \mathbb{P}\) distinct and non-isomorphic? (3) Do these exhaust all subgroups of \(\mathbb{Q}\)? (Explain Baer’s classification via height vectors and types).
Solution
Let \(G\le\QQ\) be nonzero and fix \(0\ne x\in G\). For each prime \(p\), define \[ h_p(x)=\sup\{m\ge0:x\in p^mG\}\in\NN\cup\{\infty\}. \] The sequence \((h_p(x))_p\) is called a characteristic of \(G\).
1. Changing the nonzero element changes its characteristic only at finitely many primes, and never changes which coordinates are infinite.
If \(0\ne y\in G\), then \(by=ax\) for nonzero integers \(a,b\). In the torsion-free group \(G\le\QQ\), \(h_p(nz)=h_p(z)+v_p(n)\) for every nonzero integer \(n\) and \(0\neq z\in G\), since \(nz\in p^mG\) if and only if \(z\in p^{m-v_p(n)}G\) for \(m\ge v_p(n)\). Hence for every prime \(p\), \[ h_p(y)=h_p(x)+v_p(a)-v_p(b) \] with the convention \(\infty+c=\infty\). Only primes dividing \(ab\) can change, and an infinite height remains infinite. Thus the equivalence class of the characteristic is independent of the chosen nonzero element.
Two characteristics \((a_p)\) and \((b_p)\) are called equivalent when they agree at all but finitely many primes and \[ a_p=\infty\iff b_p=\infty \] for every \(p\). Their equivalence class is the type of \(G\).
2. Two nonzero subgroups of \(\QQ\) are isomorphic if and only if they have the same type.
1. Conversely, equivalent height data differ by such a finite rational rescaling, which gives an isomorphism.
3. There are exactly \(2^{\aleph_0}\) isomorphism classes.
There are at most \(2^{\aleph_0}\) subgroups because \(\QQ\) is countable.
For each set of primes \(S\), let \[ G_S=\ZZ[1/p:p\in S]. \] For \(1\in G_S\), \[ h_p(1)=\begin{cases} \infty,&p\in S,\\ 0,&p\notin S. \end{cases} \] Thus \(G_S\cong G_T\) implies \(S=T\). Since the set of primes is countable, there are \(2^{\aleph_0}\) choices of \(S\). Hence the number of isomorphism classes is exactly \(2^{\aleph_0}\).
4. The groups \(G_S\) do not exhaust all subgroups of \(\QQ\).
Choose finite numbers \(e_p\ge0\) with \(e_p>0\) for infinitely many primes \(p\) (for example \(e_p=1\) for all \(p\)), and set \[ G=\left\langle p^{-j}:p\text{ prime},\ 0\le j\le e_p\right\rangle\le\QQ. \] Then \(h_p(1)=e_p\). Such a type can have infinitely many positive finite coordinates, whereas every \(G_S\) has only heights \(0\) or \(\infty\). Hence it is not of localization type.
Therefore Baer’s types classify all nonzero subgroups of \(\QQ\), and there are exactly continuum many isomorphism classes.