Generators of Infinite Cyclic Group

Theorem
Let $\left \langle {g} \right \rangle = G$ be an infinite cyclic group.

Then the only other generator of $G$ is $g^{-1}$.

Proof
By definition, the infinite cyclic group with generator $g$ is:


 * $\left \langle {g} \right \rangle = \left\{{\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots}\right\}$

where $e$ denotes the identity $e = g^0$.

The fact that $g^{-1}$ generates $G$ is shown by Cyclic Group Generator is Not Unique.

Futhermore, $\left \langle {e} \right \rangle = \left\{{e}\right\} \ne G$.

Since $g$ is infinite, we must have $g^i \neq g^j$ for all $i \ne j$, for otherwise $g^r = e$ for some $r \in \Z$, a contradiction.

Let $n \in Z \setminus \left\{{-1, 0, 1}\right\}$.

Then:


 * $\left \langle {g^n} \right \rangle = \left\{{\ldots, g^{-2n}, g^{-n}, e, g^n, g^{2n}, \ldots}\right\}$

But since $\left|{n}\right| > 1$, none of these elements is equal to $g$, because $1 \notin n \Z$.

So:
 * $g \notin \left \langle {g^n} \right \rangle \implies \left \langle {g^n} \right \rangle \ne \left \langle {g} \right \rangle$

Note
While for $n \in Z \setminus \left\{{-1, 0, 1}\right\}$ we have shown that $\left\langle {g^n} \right \rangle$ and $\left \langle {g} \right \rangle$ are different as sets, the two are isomorphic as abstract groups via


 * $ \left \langle {g} \right \rangle \owns h \mapsto h^n \in \left\langle {g^n} \right \rangle$