Let $g$ be a member of a group (or monoid) $G$, then why is it that if $G$ is finite, $g, g^2, g^3, g^4, \dots $, cannot be distinct?

Question

Let $g$ be a member of a group $G$, then why is it that if $G$ is finite, $g, g^2, g^3, g^4, \dots $, cannot be distinct?

Does it have to do with the Pigeon-Hole Principle?

Answer 1

There are infinitely many candidates for powers of $g$ in $G$. But $G$ is finite and all powers of $g$ are in $G$ by closure of the operation. Hence there exists at least two elements, $g^i, g^j$ of $\langle g\rangle\subseteq G$ such that $g^i =g^j$ by the pigeonhole principle since $G$ is finite. Can you continue from here?

Answer 2

Because of closure, $g, g^2, g^3, g^4, ...$ (i.e., $g^n, \;\forall n\in \Bbb N$) must be in $G$.

If $g, g^2, g^3, g^4, ...$ were all distinct, that would be infinitely many elements of $G$,

so $G$ would not be finite.

Answer 3

The map $\mathbb N \to G$ given by $n \mapsto g^n$ cannot be injective because $\mathbb N$ is infinite and $G$ is finite.