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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?2012-12-12
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    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](https://projecteuclid.org/euclid.ijm/1256044634) (groups with a single defining relator are very natural).2012-12-14
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    (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](http://mathoverflow.net/questions/28999/is-there-a-universal-countable-group-a-countable-group-containing-every-countab).)2012-12-14
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    @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 MR8401242018-12-11

2 Answers 2

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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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    +1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow.2012-12-12
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    not getting in head, could it be elaborated a little?2012-12-12
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    @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 $\mathbb N$?2012-12-12
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    well, why it is countably infinite group as you say $\mathbb{N}$ has uncountably many subsets?2012-12-12
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    Isn't that an answer to a different question? Wouldn't some free product instead be the answer? My knowledge of free products is rudimentary, but I find this question interesting.2012-12-12
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    @gnometorule: Which different question do you think this answers? You can _also_ construct a counterexample using free groups, but that doesn't mean the one I give here fails to work.2012-12-12
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    I think OP means to find an uncountable subgroup. Your (nice) example has a finite (not uncountably) infinite sub group. Maybe I'm just not getting it. :)2012-12-12
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    @gnometorule: To me, the question clearly speaks of the _number of subgroups_, not the size of any particular subgroup.2012-12-12
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    You are right, and I need a refresher course in proper reading.2012-12-12
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    @Henning Makholm how each infinite subset produces different subgroup? Do you mean We have to take all singleton sets containing every element of infinite subset and all possible combinations of them?2018-11-13
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    @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).2018-11-13
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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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    I like this example. In addition it shoulws that a countable ring can have uncountably many subrings.2012-12-14
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    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.2014-08-17
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    @Bryan: Hmm, is it obvious that these subgroups are non-isomorphic?2014-08-17
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    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.2014-08-17
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    @Bryan: Indeed, thank you.2014-08-17
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    @Robert I'm not sure I follow. What does $p$-divisible mean here?2018-02-05
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    @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, $py=x$.2018-02-05
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    @NateEldredge WHy sets of prime are uncountable ? Please can you explain2018-11-27
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    @Shubham: There are infinitely many primes, and the power set of any infinite set is uncountable (Cantor's theorem).2018-11-27
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    Thanks a lot I don't know about that2018-11-27
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    @Shubham: See https://en.wikipedia.org/wiki/Cantor%27s_theorem. It's the same result that tells you that $\mathbb{R}$ is uncountable.2018-11-27