TITLE: Infinite group has infinitely many subgroups, namely cyclic subgroups.
QUESTION [18 upvotes]: If $G$ is an infinite group then $G$ has infinitely many subgroups.
Proof: Let's consider the following set: $C=\{\left \langle g \right \rangle: g\in G \}$ - collection of all cyclic subgroups in $G$ generated by elements of $G$. Two cases are possible:

Exists infinitely many distinct cyclic subgroups $\Rightarrow$ We are done.
Exists finitely many distinct cyclic subgroups for example $C=\{H_1, H_2,\dots, H_n\}$. Then $G=\bigcup \limits_{i=1}^{n}H_i$. Since $G$ is infinite then WLOG suppose that $H_1$ is also infinite, where $H_1=\left \langle g_1 \right \rangle$. Let's consider the following set $\{\left \langle g_1^n \right \rangle: n\in \mathbb{N}\}$ - the collection of all cyclic sugroups of $H_1\subset G.$  Let $K_1=\left \langle g_1 \right \rangle$, $K_2=\left \langle g_1^2 \right \rangle$, $K_3=\left \langle g_1^3 \right \rangle$, $\dots$. It's easy to show that $K_n$ and $K_m$ are distinct for $n\neq m$. Indeed, WLOG take $n<m$ and taking $g_1^n\in K_n$ but $g_1^n\notin K_m$ otherwise $g_1^n=g_1^{ml}$ where $l\in \mathbb{Z}$ $\Rightarrow$ $g_1^{n-ml}=e$ and since $H_1$ is infinite $\Rightarrow$ $n=ml$ which is contradiciton since $m>n$.

Thus, the subgroups $K_n$ for any $n\in \mathbb{N}$ are cyclic subgroups of $H_1$ $\Rightarrow$ cyclic subgroups of $G$.
Is this reasoning correct?

REPLY [25 votes]: I think the contrapositive is much clearer:

If a group has finitely many subgroups, then the group is finite.

Indeed, let $G$ be a group with finitely many subgroups. Then $G$ has finitely many cyclic subgroups. An infinite cyclic group has infinitely many subgroups. Therefore, all cyclic subgroups of $G$ are finite. Finally, $G$ is finite because it is the union of its cyclic subgroups, which is a finite union of finite sets.