# Understand the problem

A countable group can have only countably many distinct subgroups.# Start with hints

Hint 1This is a really artistic problem.

**Pre-Solution Thoughts:**

**Lemma:** A group is finite iff the number of subgroups of a group is finite.(Check!)

The lemma is a way to understand that an infinite group will have infinite number of subgroups.

Hint 2We are considering here set of all subsets of a countable group G, which is a subset of the Power Set of G.Given G is countable the Power Set of G is uncountable.Now we know that \(2^d\), where d is the set of natural numbers denotes the cardinality of the Power Set of G which is an uncountable set as it is bijective to the Real Numbers [Here d is not finite by the way, d is countably infinite].So it is kinda intuitive that it may be uncountable.First I took the group (Z,+), but we all know that the subgroups of Z are nZ only. This doesn’t solve our purpose.So naturally the next choice was the group (Q.+) whose subgroup is (Z,+).While understanding the subgroups of (Q.+), the question is solved.Hint 3We need to understand the subgroups of (Q.+).Consider any rational number q and the subgroup qZ of (Q.+) generated by q.What if we take two rational numbers?For simplicity check the subgroup generated by {1/2 , 1/3}.Prove that the subgroup generated by {1/2 , 1/3} is (1/2.3)Z.{Observe that co-prime property of 2 and 3 is playing an important role}.Hint 4Now what if we take three mutually coprime natural numbers say a,b,c and see the subgroup generated by { 1/a , 1/b ,1/c}. In fact for simplicity take a,b,c to be primes.Observe that it is of the form (1/a.b.c)Z.Hence for every finite subset of primes, we generate a distinct subgroup of (Q,+).Hence the total number of subgroups contained all the subsets of primes, which has a bijection with the Real Numbers as mentioned above.So the answer is False.**Exercise**

- Find all the subgroups of (Q,+) with proof.

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