Order of a cyclic subgroup equals the order of its generator

Statement

Let $G$ be a group and let $g \in G$. If $g$ has finite order $n$, then $\langle g \rangle = \{e, g, g^2, \ldots, g^{n-1}\}$, these $n$ elements are distinct, and thus $|\langle g \rangle| = n$. If $g$ has infinite order, then the powers $g^k$ for $k \in \mathbb{Z}$ are distinct, and $\langle g \rangle$ is infinite.

Remarks

Here $\langle g \rangle$ is the cyclic subgroup generated by $g$, and $|\langle g \rangle|$ is its order. The proof uses the divisibility criterion for powers: $g^a = g^b$ exactly when $n$ divides $a - b$. With Lagrange's theorem, this fact shows that the order of an element of a finite group divides the order of the group.

Explanations

L1 human

The two meanings of the word "order" agree. The order of an element is the number of distinct powers that it has, and these powers form the subgroup that it generates. For example, a rotation by $120$ degrees has order $3$, and its powers are the three rotations by $0$, $120$ and $240$ degrees.