Understand the problem

A countable group can have only countably many distinct subgroups.
Source of the problem
TIFR GS 2018 Part A Problem 21
Group Theory
Difficulty Level
Suggested Book
Abstract Algebra, Dummit and Foote

Start with hints

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

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