Give an interesting example of a subgroup of the additive group of the rationals.
Solution
The subgroup of dyadic rationals is \[ \ZZ[1/2]=\left\{\frac{m}{2^n}:m\in\ZZ,\ n\ge0\right\}\subseteq(\QQ,+). \]
1. This is a subgroup of \((\QQ,+)\).
It contains \(0\). If \(x=a/2^m\) and \(y=b/2^n\), then after replacing both denominators by \(2^{\max(m,n)}\), the difference \(x-y\) again has denominator a power of \(2\). Hence the subgroup criterion applies.
2. It is not cyclic.
Suppose \(\ZZ[1/2]=\langle q\rangle\) for some nonzero rational \(q=a/b\) in lowest terms. Every integer multiple of \(q\) has reduced denominator dividing \(b\). But \(\ZZ[1/2]\) contains \(1/2^n\) for arbitrarily large \(n\), whose reduced denominators are unbounded. Contradiction.
3. It is a proper subgroup of \(\QQ\).
For example, \(1/3\notin\ZZ[1/2]\), since no power of \(2\) is divisible by \(3\).
Thus \(\ZZ[1/2]\) is a proper, noncyclic subgroup of the additive rationals.