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Why are these following situations not possible?

A. An infinite group has finite number of subgroups

B. An uncountable group has countable number of subgroups.

Any infinite group that I can think of now has infinite number of subgroups.But what is the logic behind it ?And why the number has to be uncountable if the group is uncountable?

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    $\begingroup$ For the uncountable case, note that any element of the group generates a (cyclic) subgroup which is finite or countably infinite. $\endgroup$ – André Nicolas Aug 2 '15 at 20:40
  • $\begingroup$ In fact if $G$ is infinite, there are precisely $|G|$ cyclic subgroups. $\endgroup$ – whacka Aug 2 '15 at 20:45
  • $\begingroup$ Question A. is old - see here. Question B is similar to A. $\endgroup$ – Dietrich Burde Aug 2 '15 at 20:47
  • $\begingroup$ math.stackexchange.com/questions/1909385 $\endgroup$ – Watson Oct 24 '16 at 17:43
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Cyclic groups to the rescue!

If an infinite group has an element of infinite order (that is, a subgroup isomorphic to $\Bbb Z$) then it has an infinite number of subgroups (because $\Bbb Z$ does).

Otherwise, every element is of finite order, in which case, if we only have a finite number of subgroups $G = \bigcup\limits_k \langle g_{i_k}\rangle$, which is finite, contradiction.

The uncountable case is similar, a countable union of countable sets is countable.

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Say $G$ is infinite. Let $S(x)$ denote the subgroup generated by $x\in G$. If there exists $x$ such that $S(x)$ is infinite then $G$ has infinitely many subgroups, since an infinite cyclic group has infinitely many subgroups. On the other hand if every $S(x)$ is finite then there must be infinitely many distinct subgroups of the form $S(x)$, since $G=\bigcup_{x\in G}S(x)$.

If $G$ is uncountable a simpler argument works; every $S(x)$ is at most countable, so there must be uncountably many distinct $S(x)$.

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