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?
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 $S(x)$.
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.
