prove that a group, G, of infinite order has infinitely many distinct subgroups. How would I go about proving the following statement?: 
Suppose $G$ is a group with |$G$| = $\infty$. Prove that $G$ has infinitely many $\underline{distinct}$ subgroups.
 A: There are two cases:


*

*There is $g \in G$ such that $\lvert g \rvert = \infty$. Then $\langle g \rangle$ is isomorphic to $\mathbb{Z}$ which has infinitely many distinct subgroups.

*Every $g \in G$ has finite order $\lvert g \rvert < \infty$. Pick $g_0 \in G$ and define (recursively) $g_n$ to be any element of $G$ such that $g_n \notin \langle g_0 \rangle \cup \dotsb \cup \langle g_{n-1} \rangle$. Since the set on the right is always finite (being a finite union of finite sets) then there is always such $g_n$ because $G$ is infinite. It follows that $\langle g_n \rangle$ give us infinitely many distinct subgroups of $G$.

A: Hint 1: If $G$ has finitely many [distinct] subgroups, then it has finitely many [distinct] cyclic subgroups.
Hint 2: If $G$ has finitely many [distinct] subgroups, then all its cyclic subgroups must have finite order.
Hint 3: Combining the above two hints with the fact that each element of $G$ lies in a cyclic subgroup yields $|G| < \infty$.
A: Hints:


*

*Show that the infinite cyclic group has infinitely many distinct subgroups.

*Let $g_1$ be any non-unit element of $G$. If $G_1 = \langle g\rangle$ is infinite, we're done. If not, choose any $g_2$ in the set $G\backslash G_1$, and so on.
