= Solution
The <order of a group element> $g$ is the least positive integer $m$ for which $g^m=e$, if one exists. In a <finite group>, two terms of $e,g,g^2,\ldots,g^{|G|}$ must agree. If $g^i=g^j$ with $i<j$, cancellation gives $g^{j-i}=e$. Thus such positive integers exist and have a least member: \b[every element of a <finite group> has finite order].
If $h=aga^{-1}$, repeated cancellation shows $h^m=ag^ma^{-1}$ for every positive integer $m$. Consequently $h^m=e$ exactly when $g^m=e$. The least positive exponents are therefore equal:
$$
\boxed{\operatorname{ord}(aga^{-1})=\operatorname{ord}(g).}
$$
This proof also shows that <conjugation> preserves infinite order when the ambient <group> is not finite.
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