g to the n equals 1 iff the order divides n

Statement

Let $G$ be a group with identity element $e$, and let $g \in G$ be an element of finite order $m$. Let $n$ be an integer. Then $g^n = e$ if and only if $m$ divides $n$.

Remarks

The proof divides $n$ by $m$ with remainder: $n = qm + r$ with $0 \leq r < m$, so that $g^n = (g^m)^q g^r = g^r$. By the minimality of $m$, the power $g^r$ is $e$ only when $r = 0$. A consequence is that $g^a = g^b$ if and only if $a \equiv b \pmod m$.

Explanations

L1 human

An element of order $m$ comes back to the identity exactly after $m$, $2m$, $3m$ steps, and so on. It is never at the identity after a number of steps that is not a multiple of $m$. For example, a rotation by $90$ degrees is the identity after $8$ steps, but not after $6$ steps.