Infinite Group has Infinite Number of Subgroups

Theorem
Let $\left({G, \circ}\right)$ be an infinite group.

Then $\left({G, \circ}\right)$ has an infinite number of distinct subgroups.

Proof
There are two cases to consider: either $\left({G, \circ}\right)$ has an infinite cyclic subgroup, or it does not.

Case 1
Suppose that $\left({G, \circ}\right)$ has an infinite cyclic subgroup denoted as $H$.

Let $a \in G$ be the element of $G$ such that $\left\langle{a}\right\rangle = H$.

Then by Distinct Subgroups of Infinite Cyclic Group, $\left\langle{\left\langle{a^k}\right\rangle}\right\rangle_{k \mathop \in \N^*}$ is a sequence of distinct subgroups of $H$.

And so by Subgroup of Subgroup is Subgroup, it follows that $\left({G, \circ}\right)$ has an infinite number of subgroups.

Case 2
Suppose that $\left({G, \circ}\right)$ does not have an infinite cyclic subgroup.

Then for every $x \in G$, we have that $\left\langle{x}\right\rangle$ is a finite group.

Let $a, b \in G$.

Let $\sim$ be the equivalence relation on $G$ defined as:
 * $a \sim b \iff \left\langle{a}\right\rangle = \left\langle{b}\right\rangle$

Let $\left[\!\left[{a}\right]\!\right]_\sim$ denote the equivalence class of $a$ under $\sim$.

Let $x \sim a$.

Then:
 * $\left\langle{x}\right\rangle = \left\langle{a}\right\rangle$

Since $x \in \left\langle{x}\right\rangle$, it follows that:
 * $x \in \left\langle{a}\right\rangle$

It has been shown that:
 * $\left[\!\left[{a}\right]\!\right]_\sim \subseteq \left\langle{a}\right\rangle$

By Subset of Finite Set is Finite, every equivalence class under $\sim$ is finite.

By Equivalence Class holds Equivalent Elements:
 * $\left[\!\left[{a}\right]\!\right]_\sim \ne \left[\!\left[{b}\right]\!\right]_\sim \implies a \not \sim b \implies \left\langle{a}\right\rangle \ne \left\langle{b}\right\rangle$

So every equivalence class generates a unique subgroup of $\left({G, \circ}\right)$.

By Union of Equivalence Classes is Whole Set, the union of all equivalence classes under $\sim$ must equal $G$.

From Finite Union of Finite Sets is Finite, it follows that this must be an infinite union.

Thus there must exist an infinite number of equivalence classes, and hence an infinite number of subgroups.

The result then follows from Proof by Cases.