# Understand the problem

A countable group can have only countably many distinct subgroups.

##### Source of the problem

TIFR GS 2018 Part A Problem 21

##### Topic

Group Theory

##### Difficulty Level

Easy

##### Suggested Book

Abstract Algebra, Dummit and Foote

# Start with hints

Do you really need a hint? Try it first!

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.

**Exercise**

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

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