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Every countable group has only countably many distinct subgroups.

The above statement is false. How to show it? One counterexample may be sufficient, but I am blind to find it out. I have considered some counterexample only like $(\mathbb{Z}, +)$, $(\mathbb{Q}, +)$ and $(\mathbb{R}, +)$.

Is there any relationship between number of elements in a group and number of its subgroup? I do not know. Please discuss a little. What will be if the group be uncountable?

Thank you for your help.

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    $\begingroup$ Well, $(\Bbb R,+)$ won't get you anywhere, since it is uncountable. $\endgroup$ – Cameron Buie Nov 7 '13 at 15:43
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    $\begingroup$ It certainly does. Each non-negative $\alpha\in\Bbb R$ generates a distinct cyclic subgroup $\langle\alpha\rangle,$ for example, and there are uncountably-many such $\alpha$. $\endgroup$ – Cameron Buie Nov 7 '13 at 16:10
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The countably infinite sum $S$ of copies of the group $G=\mathbb Z/2\mathbb Z$ (cyclic group of order two) indexed by $I$ is countable. Every subset $A$ of the index set $I$ corresponds to a subgroup of $S$ consisting of elements with nonzero components only in the copies of $G$ corresponding to the subset $A\subset I$. This gives uncountably many subgroups because the set of subsets of $I$ is uncountable.

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  • $\begingroup$ Thank you for your answer but I am not strong enough in algebra. Please expand the answer such that I can grasp it. "sum of copies of the group $\frac{\mathbb{Z}}{2\mathbb{Z}}" - not clear. $\endgroup$ – Dutta Nov 7 '13 at 16:13
  • $\begingroup$ Are you familiar with the notion of the sum of finitely many groups? $\endgroup$ – Mikhail Katz Nov 9 '13 at 20:03
  • $\begingroup$ It is clear now. I have studied recently direct product of groups a little. $\endgroup$ – Dutta Nov 10 '13 at 2:58
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I think $\mathbb Q$ is also a counterexample. There are uncountably many subcollections $S$ of the collection of prime numbers. For every such $S$, consider the subgroup $G_S$ of $\mathbb Q$ consisting of numbers $m/n$ where the denominator $n$ is divisible only by prime numbers from $S$. This does the trick.

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  • $\begingroup$ such $S$ has to be finite no? can it be infinite? $\endgroup$ – GA316 Nov 7 '13 at 16:53
  • $\begingroup$ With finite $S$'s you won't get an uncountable collection of subgroups, so you have to allow $S$ to be infinite. $\endgroup$ – Mikhail Katz Nov 7 '13 at 17:26
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This wont answer to your question, but still it is interesting. consider $S_{\mathbb{N}}$ , the set of all bijections from $\mathbb{N}$ to $\mathbb{N}$ which is uncountable in cardinality. Now for any non empty subset $A$ of $\mathbb{N}$ we can find a subgroup of $S_{\mathbb{N}}$ which is isomorphic to $S_A$ namely the collection of all bijections in $S_{\mathbb{N}}$ which fixes all the elements of $\mathbb{N} - A$. Hence $S_{\mathbb{N}}$ has uncountably many subgroups.

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  • $\begingroup$ Thank you for your answer. This is one more example of uncountable group whose collection of subgroups is uncountable. $\endgroup$ – Dutta Nov 8 '13 at 3:13
  • $\begingroup$ Where are you studying? $\endgroup$ – Dutta Nov 8 '13 at 3:14
  • $\begingroup$ @Samprity IMSc, Chennai $\endgroup$ – GA316 Nov 8 '13 at 4:37
  • $\begingroup$ Grate! So you are very meritorious and intelligent. $\endgroup$ – Dutta Nov 8 '13 at 10:20
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Take $\bigoplus_{i \in \Bbb N}\Bbb Z_i $ direct sum of countable many copies of $(\Bbb Z,+)$. It's free group with countable base. $Card(\bigoplus_{i \in \Bbb N}\Bbb Z_i)=Card(\Bbb N^*)$. It's easy to see that set of subgroups of $\bigoplus_{i \in \Bbb N}\Bbb Z_i $, $\{\bigoplus_{i \in A}\Bbb Z_i \mid A\in \mathcal P(\Bbb N)\}$ is uncountable.

$\bigoplus_{i \in \Bbb N}\Bbb Z_i $ is a subgroup of $\prod_{i \in \Bbb N}\Bbb Z_i$ consisting of those sequences which are $0$ everywhere but finite number of places. Thats why its countable. Now for different $A,B\subset \Bbb N$ we get different subgroups $\bigoplus_{i \in A}\Bbb Z_i $, $\bigoplus_{i \in B}\Bbb Z_i $. There is continuum different subset of $\Bbb N$ and we are done.

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  • $\begingroup$ Thank you for your answer. I do not know free group. Please expand the answer which will be easy to understand. $\endgroup$ – Dutta Nov 7 '13 at 17:16

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