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For a group $G$. let $F(G)$ denote the collection of all subgroups of $G$. Which of the following situation can occur?

A) $G$ is finite but $F(G)$ is infinite.

B)$G$ is infinite but $F(G)$ is finite.

C)$G$ is countable but $F(G)$ is uncountable.

D)$G$ is uncountable but $F(G)$ is countable.

I don't know how i approach this problem. I think any group $G$ of infinite order has an infinite of subgroups as i can take this group generated by power of these elements. But i don't know how i go through this problem. I can sure that A is not true as $G$ is finite the power set of $G$is also finite .

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    $\begingroup$ If $G$ is finite, how many subsets does it have? What does this tell you about A)? $\endgroup$
    – Gerry Myerson
    Nov 16, 2017 at 21:49
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    $\begingroup$ @James: the group you describe is the direct product, not the direct sum, and it is uncountable. The genuine direct sum does work. $\endgroup$
    – Qiaochu Yuan
    Nov 16, 2017 at 23:06
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    $\begingroup$ Yes, that’s why I said the direct sum does work. $\endgroup$
    – Qiaochu Yuan
    Nov 16, 2017 at 23:18
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    $\begingroup$ (D) is impossible: an uncountable group has uncountably many countable subgroups. Because, if there were only countably many, then the union of all of them is a countable set; add another element to that union, and it generates a new countable subgroup. $\endgroup$
    – bof
    Nov 16, 2017 at 23:54
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    $\begingroup$ Let me rework the argument of @bof slightly: if a group has countably many subgroups, then it has countably many cyclic subgroups. Since a group is equal to the union of its cyclic subgroups, and this union is countable, the group is countable. This rewording shows that essentially the same argument (modulo one detail) shows (B) is impossible as well. $\endgroup$
    – Steve D
    Nov 17, 2017 at 5:46

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