# Relation between a group's cardinality and number of subgroups

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?

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

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.
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 $S(x)$.