Question

3.
Show that the set Q* of all non-zero rational numbers forms an
infinite abelian group under the operation of multiplication of
rational numbers.​

Answer 1

Let G be the set of positive rational numbers. G is a group under mutiplications. Let Π be the set of prime numbers. We claim G is a free abelian group with a basis Π.

Let r∈G. there exist positive integers a,b such that r=ab. Since a and b can be written as products of prime numbers, G is generated by Π.

Let p1,…,pr be distict prime numbers. Suppose pn11⋯pnrr=1, where all ni are integers. It suffices to prove that all ni=0. If all ni≥0, clearly all ni=0. Hence we assume not all ni≥0. Without loss of generality, we can assume that n1,…,nk≥0 and nk+1,…nr<0. Then pn11⋯pnkk=p−nk+1k+1⋯p−nrr. But this is a contradiction because Z is a UFD