I know that Q itself is not cyclic, but what about proper subgroups of Q. All subgroups I can think of so far seem to be cyclic and therefore generated by one fraction. For example if the cyclic group generated by m/n is represented as <m/n>, then every subgroup of this form would be contained in <1/mn>.For example, <2/3> is contained in <1/6> etc.
How about a short direct proof: We can write the generators with a least common denominator D: {Ni/D} for i=1..k Letting F=GCF{Ni}, each Ni/D is a multiple of F/D. So every generator of the subgroup, hence the entire subgroup, is in the cyclic group <F/D>. By Euclid's algorithm, there exist integers Ai so that ΣAiNi=F. ? F/D=ΣAi(Ni/D) Hence F/D, and so all of the cyclic group <F/D>, is in the finitely generated subgroup. Therefore <F/D> is the subgroup in question.
first of all, I assume Q is the rationals, not the reals. (otherwise the rationals are a non-cyclic subgroup). Consider A = {m/n: in lowest terms, there does not exist x such that x^2|n} A is a subgroup of Q. It suffices to show that A is closed under addition. Let m/n, m'/n' be in A. Then there does not exist x such that x^2 | lcm(n,n'). Otherwise X^2 divides n or x^2 divides n'. Consider B = {m/n in lowest terms : 3 does not divide n} also a non-cyclic subgroup. I have no idea how to classify all the subgroups of (Q, +). i think this is a harder problem than you imagined.
