Non-Trivial Group has Non-Trivial Cyclic Subgroup

Theorem
Let $G$ be a group whose identity element is $e$.

Let $g \in G$.

If $g$ has infinite order, then $\left \langle {g} \right \rangle$ is an infinite cyclic group.

If $\left|{g}\right| = n$, then $\left \langle {g} \right \rangle$ is a cyclic group with $n$ elements.

Thus, every group which is non-trivial has at least one cyclic subgroup which is also non-trivial.

In the case that $G$ is itself cyclic, that cyclic subgroup may of course be itself.

Infinite Order
Suppose that $g$ has infinite order.

We have that $\left \langle {g} \right \rangle$ consists of all possible powers of $g$.

So $\left \langle {g} \right \rangle$ can contain a finite number of elements only if some of these were equal.

Then we would have:
 * $\exists i, j \in \Z, i < j: g^i = g^j$

and so:
 * $g^{j-i} = g^{i-i} = g^0 = e$

which would mean that $g$ was of finite order.

This contradiction leads to the conclusion that $\left \langle {g} \right \rangle$ must be an infinite cyclic group.

Finite Order
If $\left|{g}\right| = n$, we have from Equal Powers of Finite Order Element that there are exactly $n$ different elements of $G$ of the form $g^i$.

Hence $\left \langle {g} \right \rangle$ is a cyclic group with $n$ elements.