Element of Cyclic Group is not necessarily Generator

Theorem
Let $\gen g = G$ be a cyclic group.

Let $a \in G$

Then it is not necessarily the case that $a$ is also a generator of $G$.

Proof
Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.

Consider the subgroup $\gen 2$ of $\struct {\R_{\ne 0}, \times}$ generated by $2$.

By definition, $\gen 2$ is a cyclic group.

Consider the element $4 \in \struct {\R_{\ne 0}, \times}$.

We have that $4 = 2^2$.

Thus $4 \in \gen 2$.

There exists no $n \in \Z$ such that $4^n = 2$.

But $2 \in \gen 2$.

Thus $\gen 2$ is not generated by $4$.

That is, $4$ is not a generator of $\gen 2$.

Hence the result, by Proof by Counterexample.