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 $$H \cong G$$.

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


 * $$n G \ \stackrel {\mathbf {def}} {=\!=} \ \left \langle {a^n} \right \rangle$$

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 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 $$h \in H \Longrightarrow \exists s \in \Z, s > 0: h = g^s$$ 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 a Cyclic Group is Cyclic, as $$G$$ is cyclic, then $$H$$ must also be cyclic.

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