Equal Powers of Group Element implies Finite Order

Theorem
Let $\struct {G, \circ}$ be a group.

Let $g \in G$ such that $g^r = g^s$ where $r, s \in \Z: r \ne s$.

Then there exists $m \in \Z_{>0}$ such that:
 * $(1): \quad g^m = e$
 * $(2): \quad 0 \le i < j < m \implies g^i \ne g^j$

Proof
, suppose that $r > s$.

From $g^r = g^s$ it follows from Powers of Group Elements that:
 * $g^{r - s} = g^r g^{-s} = g^s g^{-s} = e$

Thus there exists $t \in \Z_{>0}$ such that $g^t = e$.

Let $m \in \Z_{>0}$ be the smallest such that $g^m = e$.

Suppose $0 \le i < j < m$ such that $g^i = g^j$.

Then $g^{j - i} = e$ and $0 < j - i < m$.

This contradicts the choice of $m$ as the smallest $m \in \Z_{>0}$ such that $g^m = e$.

Hence the result.

Also see

 * Definition:Order of Group Element