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I need to know whether the following statement is true or false?

Every countable group $G$ has only countably many distinct subgroups.

I have not gotten any counter example to disprove the statement but an vague idea to disprove like: if it has uncountably many distinct subgroup then It must have uncountable number of element?

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Have you considered a countable direct product of countable groups? – user108903 Dec 12 '12 at 16:23
The examples given in the answers prove distinct but not non-isomorphic. This is still true, and is true for certain groups with a single defining relator, by a paper of G. Baumslag and Miller (groups with a single defining relator are very natural). – user1729 Dec 14 '12 at 12:10
(On the other hand, it is not true that a countable group can have uncountably many finitely generated subgroups. You might be interested in this MO question.) – user1729 Dec 14 '12 at 12:13
up vote 16 down vote accepted

One example is the group consisting of all finite subsets of $\mathbb N$, with the group operation being symmetric difference. The group is countably infinite, but for each finite or infinite $A\subseteq \mathbb N$ there's a subgroup consisting of the finite subsets of $A$.

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+1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow. – Hagen von Eitzen Dec 12 '12 at 16:26
not getting in head, could it be elaborated a little? – Un Chien Andalou Dec 12 '12 at 16:26
@Kuttus Which part gets stuck? That the main group is countable? Or that each subset produces a different subgroup? Or that there are uncountably many subsets of $\mathbb N$? – Hagen von Eitzen Dec 12 '12 at 16:27
well, why it is countably infinite group as you say $\mathbb{N}$ has uncountably many subsets? – Un Chien Andalou Dec 12 '12 at 16:37
You are right, and I need a refresher course in proper reading. – gnometorule Dec 12 '12 at 17:10

Let $(\mathbb{Q},+)$ be the group of the rational numbers under addition. For any set $A$ of primes, let $G_A$ be the set of all rationals $a/b$ (in lowest terms) such that every prime factor of the denominator $b$ is in $A$. It is clear that $G_A$ is a subgroup of $\mathbb{Q}$, and that $G_A = G_{A'}$ iff $A = A'$. Since there are uncountably many sets of primes, this produces uncountably many distinct subgroups of the countable group $\mathbb{Q}$.

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I like this example. In addition it shoulws that a countable ring can have uncountably many subrings. – Marc van Leeuwen Dec 14 '12 at 13:18
I like that this also shows that a countable group can have uncountably many distinct and non-isomorphic subgroups. Interestingly this extends to show that we can put continuum-many non-isomorphic group operations on a countable set. – Bryan Aug 17 '14 at 20:10
@Bryan: Hmm, is it obvious that these subgroups are non-isomorphic? – Nate Eldredge Aug 17 '14 at 20:21
Given two distinct collections of primes $A$ and $A'$, let's say $A$ contains $p$ while $A'$ does not. Then $G_A$ is $p$-divisible while $G_{A'}$ is not. – Bryan Aug 17 '14 at 20:26
@Bryan: Indeed, thank you. – Nate Eldredge Aug 17 '14 at 20:30

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