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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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    @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

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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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    @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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    @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