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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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  • $\begingroup$ Have you considered a countable direct product of countable groups? $\endgroup$ – user108903 Dec 12 '12 at 16:23
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    $\begingroup$ 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). $\endgroup$ – user1729 Dec 14 '12 at 12:10
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    $\begingroup$ (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.) $\endgroup$ – user1729 Dec 14 '12 at 12:13
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    $\begingroup$ @Shaun The full citation is G. Baumslag and C. F. Miller, A remark on the subgroups of finitely generated groups with one defining relation, Illinois J. Math., Volume 30, Issue 2 (1986), 255-257 MR840124 $\endgroup$ – user1729 Dec 11 '18 at 16:17
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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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  • $\begingroup$ +1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow. $\endgroup$ – Hagen von Eitzen Dec 12 '12 at 16:26
  • $\begingroup$ not getting in head, could it be elaborated a little? $\endgroup$ – Marso Dec 12 '12 at 16:26
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    $\begingroup$ @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 $\mathbb N$? $\endgroup$ – Hagen von Eitzen Dec 12 '12 at 16:27
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    $\begingroup$ You are right, and I need a refresher course in proper reading. $\endgroup$ – gnometorule Dec 12 '12 at 17:10
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    $\begingroup$ @ramanujan: How would they produce identical subgroups? If $A\ne B$ then there is an $x$ that is either in $A\setminus B$ or in $B\setminus A$, so $\{x\}$ would only be in one of the subgroups (and therefore in particular they are not the same subgroup). $\endgroup$ – Henning Makholm Nov 13 '18 at 21:24
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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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    $\begingroup$ I like this example. In addition it shoulws that a countable ring can have uncountably many subrings. $\endgroup$ – Marc van Leeuwen Dec 14 '12 at 13:18
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    $\begingroup$ 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. $\endgroup$ – Robert Wolfe Aug 17 '14 at 20:10
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    $\begingroup$ 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. $\endgroup$ – Robert Wolfe Aug 17 '14 at 20:26
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    $\begingroup$ @Robert I'm not sure I follow. What does $p$-divisible mean here? $\endgroup$ – Anu Feb 5 '18 at 12:21
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    $\begingroup$ @Anu for every element $x$ of $G_A$ there is an element $y$ of $G_A$ such that $y^p=x$, or, in this case, $py=x$. $\endgroup$ – Robert Wolfe Feb 5 '18 at 14:04
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There is another one: $F_\infty$, the free group on countably many letters $x_1,x_2, \ldots$.

It is countable because each of its elements is a finite string of symbols from a countable alphabet. At the same time, for every subset $A\subset \mathbb{N}$, there is a subgroup $H_A$ generated by the set $\{x_i | i \in A\}$. These subgroups are distinct, so we get uncountably many of them.

Also, $F_\infty$ embeds into $F_2$, so we can get these uncountably many subgroups within a finitely presented group!

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