When Are Two Subgroups of the Rationals Isomorphic?

Abstract

Since most of us have studied fractions in various ways since elementary school, it seems that almost any question about them could be answered with little effort. Consider the following question: two subgroups of the rationals under addition, can we decide whether these subgroups are isomorphic? Given arbitrary groups, it is typically easier to determine that they are not isomorphic than to show that they are. Some of the features at our disposal to show that two groups cannot be isomorphic are element orders, group orders, and certain characteristic subgroups-the center of the group, the commutator subgroup, etc.-which must match up precisely in isomorphic groups. When we consider additive subgroups of the rationals, we lose all of these typical approaches: Every nonzero element has infinite order (a group with this property is said to be torsion-free); since the rationals are commutative under addition, the center of a subgroup is the subgroup itself and commutator subgroups are trivial. Consider, for example, G1, the subgroup consisting of the rationals with square-free denominators, and G2, the subgroup consisting of the rationals with square-free odd denominators. The subgroups G1 and G2 are generated by reciprocals of primes and odd primes, respectively. Each element in these groups is a finite linear combination of the said reciprocals with integer coefficients. We write G1 = (1/2, 1/3, 1/5, .. .), and similarly, G2 = (1/3, 1/5, 1/7, .. .). Since 1/2 , G2, G2 is a proper subgroup of G1 (written G2 < G1). To show that G1 is isomorphic to G2, we will find a mapping between the groups and then prove that the mapping is an isomorphism. To establish this mapping, we look at the solvability of equations in the two groups. For example, the equation 3x = 1 is solvable in both groups, since there is a unique element x in each group that satisfies the equation. (The uniqueness of solutions to equations in torsion-free abelian groups is important and we make repeated use of it throughout the discussion.) This avenue will lead us to the goal of deciding whether subgroups of the rationals are isomorphic, but there is work to be done en route. In particular, for our examples G1 and G2, the equation 2x = 1 is solvable in G1 but not in G2, an observable difference between these subgroups. To see that this does not necessarily mean that the subgroups are nonisomorphic, consider the following: The groups (1/2) and (1/3) are two subgroups of Q that are both infinite cyclic and therefore isomorphic. However, the equation 2x = 1 has a solution in (1/2), but not in (1/3). We conclude that the solvability or nonsolvability of particular equations in isolated instances is not enough to prove that two groups are not isomorphic. In the groups G1 and G2, the only equations of the form mx = 1 that are solvable in G1 and not in G2 are those with m = 21 . j, 374 MATHEMATICS MAGAZINE