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# Understand the problem

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This 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.
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• We 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.
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• We 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}.
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• Now 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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