## A countable group with uncountably many subgroups

Posted: February 15, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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 $G_A \neq G_B. \ \Box$

1. Argha Mazumder says:

From gcd(a,b)=1 and gcd(c,d)=1 , how to show that gcd(bd,ad+bc)=1 ??

2. 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.