Infinite Group has Infinite Number of Subgroups
Theorem
Let $\struct {G, \circ}$ be an infinite group.
Then $\struct {G, \circ}$ has an infinite number of distinct subgroups.
Proof
There are two cases to consider: either $\struct {G, \circ}$ has an infinite cyclic subgroup, or it does not.
Case 1
Suppose that $\struct {G, \circ}$ has an infinite cyclic subgroup denoted as $H$.
Let $a \in G$ be the element of $G$ such that $\gen a = H$.
Then by Distinct Subgroups of Infinite Cyclic Group, $\sequence {\gen {a^k} }_{k \mathop \in \N^*}$ is a sequence of distinct subgroups of $H$.
So, by Subgroup of Subgroup is Subgroup, it follows that $\struct {G, \circ}$ has an infinite number of subgroups.
Case 2
Suppose that $\struct {G, \circ}$ does not have an infinite cyclic subgroup.
Then for every $x \in G$, we have that $\gen x$ is a finite group.
Let $a, b \in G$.
Let $\sim$ be the equivalence relation on $G$ defined as:
- $a \sim b \iff \gen a = \gen b$
Let $\eqclass a \sim$ denote the equivalence class of $a$ under $\sim$.
Let $x \sim a$.
Then:
- $\gen x = \gen a$
Since $x \in \gen x$, it follows that:
- $x \in \gen a$
It has been shown that:
- $\eqclass a \sim \subseteq \gen a$
By Subset of Finite Set is Finite, every equivalence class under $\sim$ is finite.
By Equivalence Class holds Equivalent Elements:
- $\eqclass a \sim \ne \eqclass b \sim \implies a \not \sim b \implies \gen a \ne \gen b$
So every equivalence class generates a unique subgroup of $\struct {G, \circ}$.
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 a (countably) 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.
$\blacksquare$