Generators of Infinite Cyclic Group

Theorem
Let $\gen g = G$ 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:


 * $\gen g = \set {\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots}$

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

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

Futhermore, $\gen e = \set e \ne G$.

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

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

Then:


 * $\gen {g^n} = \set {\ldots, g^{-2 n}, g^{-n}, e, g^n, g^{2 n}, \ldots}$

But since $\order n > 1$, none of these elements is equal to $g$, because $1 \notin n \Z$.

So:
 * $g \notin \gen {g^n} \implies \gen {g^n} \ne \gen g$

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


 * $\gen g \owns h \mapsto h^n \in \gen {g^n}$