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July 25, 2017

TIFR-2013 problem 1 | Countable group-Uncountable Subgroup

Try this problem from TIFR-2013 problem 1 based on the Countable group-Uncountable Subgroup.

Question: TIFR-2013 problem 1


Every countable group \(G\) has only countable number of distinct subgroups.


How many subgroups does \(\mathbb{Q}\) have?


A very natural way to go from countably infinite set to uncountable set is to take its power set (set of all subsets).

Let \(X=\left \{p_1,p_2,...\right \}\) be the set of all primes. Since we can count the primes as "first prime is 2, second is 3, third is 5,..." this set is a countably infinite set. (Other way of seeing this is \(X\) is an infinite subset of \(\mathbb{N}\).)

The power set of X, \(\mathbb{P}(X)\) is therefore an uncountable set.

The natural way subgroups come to the mind is cyclic groups generated by \(1/p_i\). But if we only take this collection we will end up with as many groups as there are primes, which will not be helpful.

But let us take subgroup generated by a collection of \(1/p_i\)'s. That will take us to the set of subsets of X, which will give what we desire.

Let us make this more formal:

Consider a subset \(A\) of \(X\). \(G_A\) be the subgroup generated by \(\left \{1/p_\alpha | p_\alpha \in A\right \}\)

The only obstacle that can occur is that these \(G_A\)s are not distinct for distinct \(A\)s.  In other words, the function from \(\mathbb{P}(X)\) to subgroups of \(\mathbb{Q}\) which takes a set \(A\) to \(G_A\) must be one-to-one.

Let \(G_A=G_B\) We wish to show that \(A\subseteq B\) and \(B\subseteq A\).

Let \(p\in A\). Then \(1/p\in G_A\). Then \(1/p\in G_B\). Now if \(p\notin B\) then there is no way that \(1/p\) comes in \(G_B\), because any linear combination of \(1/p_\alpha\)s where \(p_\alpha\) is different from p, will never give p in the denominator (because of the lcm process in addition of two rationals). Therefore, \(p\in B\). This proves that \(A\subseteq B\). Since there is nothing special about \(A\) and \(B\), we can interchange the roles of these two, and finally conclude that \(A=B\).

Therefore for each subset of \(X\), we get a different subgroup of \(\mathbb{Q}\), and since there are uncountable number of subsets of X, we get an uncountable number of subgroups of \(\mathbb{Q}\).


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