**Problem**. Prove that the number of subgroups of the additive group of rational numbers, is uncountable.

**Solution**. Let be the set of prime numbers and let be the set of all non-empty subsets of Clearly is uncountable. The idea is to find a one-to-one map from to the set of all subgroups of which will obviously prove that the set of subgroups of is uncountable. Now, for every define

Clearly because (or you can argue that ). Also if then clearly Also, if and then Now, either or If then by the definition of we have If and is a prime divisor of then either or In either case So i.e. is a subgroup of So, to complete the solution, we only need to show that if are in then Well, this is easy to see: since there exists some prime (or you may say that there exists a prime ). Now, clearly But because So

From gcd(a,b)=1 and gcd(c,d)=1 , how to show that gcd(bd,ad+bc)=1 ??

Hi–

Your blog is excellent. I have a question Can u give example of an Uncountable proper subgroup of R

Chandrasekhar

Consider R as a vector space over Q, the field of rational numbers. Then, by Zorn’s lemma, there exists a Q-vector subspace G of R such that R = Q + G, where + here is direct. This G clearly satisfies your conditions.