A countable group with uncountably many subgroups

Posted: February 15, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: ,

Problem. Prove that the number of subgroups of (\mathbb{Q},+), the additive group of rational numbers, is uncountable.

Solution. Let P be the set of prime numbers and let S be the set of all non-empty subsets of P. Clearly S is uncountable. The idea is to find a one-to-one map from S to the set of all subgroups of (\mathbb{Q},+), which will obviously prove that the set of subgroups of (\mathbb{Q},+) is uncountable. Now, for every A \in S define

G_A = \{a/b : \ a \in \mathbb{Z}, \ b \in \mathbb{N}, \ \gcd(a,b)=1, \ b=1 \ \text{or every prime divisor of} \ b \ \text{is in} \ A \}.

Clearly G_A \neq \emptyset because A \neq \emptyset (or you can argue that 0 = 0/1 \in G_A). Also if a/b \in G_A, then clearly -a/b \in G_A. Also, if r=a/b \in G_A and s=c/d \in G_A, then r+s = (ad +bc)/(bd). Now, either bd=1 or bd > 1. If bd=1, then by the definition of G_A we have r+s \in G_A. If bd > 1 and p is a prime divisor of bd, then either p \mid b or p \mid d. In either case p \in A. So r+s \in G_A, i.e. G_A is a subgroup of (\mathbb{Q}, +). So, to complete the solution, we only need to show that if A \neq B are in S, then G_A \neq G_B. Well, this is easy to see: since A \neq B, there exists some prime p \in A \setminus B (or you may say that there exists a prime p \in B \setminus A). Now, clearly 1/p \in G_A. But 1/p \notin G_B because p \notin B. So G_A \neq G_B. \ \Box

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Comments
  1. Chandrasekhar says:

    Hi–

    Your blog is excellent. I have a question Can u give example of an Uncountable proper subgroup of R

    Chandrasekhar

    • Yaghoub says:

      Consider R as a vector space over Q, the field of rational numbers. Then, by Zorn’s lemma, there exists a Q-vector subspace G of R such that R = Q + G, where + here is direct. This G clearly satisfies your conditions.

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