Subgroup of Infinite Cyclic Group is Infinite Cyclic Group

Theorem
Let $G = \left\langle{a}\right\rangle$ be an infinite cyclic group generated by $a$, whose identity is $e$.

Let $g \in G, g \ne e: \exists k \in \Z, k \ne 0: g = a^k$.

Let $H = \left \langle {g} \right \rangle$.

Then $H \le G$ and $H \cong G$.

Thus, all non-trivial subgroups of an infinite cyclic group are themselves infinite cyclic groups.

A subgroup of $G = \left \langle {a} \right \rangle$ is denoted as follows:


 * $n G := \left \langle {a^n} \right \rangle$

This notation is usually used in the context of $\left({\Z, +}\right)$, where $n \Z$ is (informally) understood as the set of integer multiples of $n$.

Proof
The fact that $H \le G$ follows from the definition of group generator.

By Infinite Cyclic Group is Isomorphic to Integers:
 * $G \cong \left({\Z, +}\right)$

Now we show that $H$ is of infinite order.

Suppose $\exists h \in H, h \ne e: \exists r \in \Z, r > 0: h^r = e$.

But:
 * $h \in H \implies \exists s \in \Z, s > 0: h = g^s$

where $g = a^k$.

Thus:
 * $e = h^r = \left({g^s}\right)^r = \left({\left({a^k}\right)^s}\right)^r = a^{k s r}$

and thus $a$ is of finite order.

This would mean that $G$ was also of finite order.

So $H$ must be of infinite order.

From Subgroup of Cyclic Group is Cyclic, as $G$ is cyclic, then $H$ must also be cyclic.

From Infinite Cyclic Group is Isomorphic to Integers:
 * $H \cong \left({\Z, +}\right)$

Therefore, as $G \cong \left({\Z, +}\right)$:
 * $H \cong G$

Comment
The interesting thing to note here is that a non-trivial subgroup of an infinite group is itself isomorphic to the group of which it is a subgroup.

This can be compared with the result Infinite Set Equivalent to Proper Subset.