Non-Trivial Group has Non-Trivial Cyclic Subgroup

Theorem
Let $$G$$ be a group, and 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 not trivial has at least one cyclic subgroup which is also not 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.