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Define the order ord$(g)$ of an element $g$ in a finite group $G$ to be $|\{g^0,g^1,g^2,\ldots\}|$.

I want to prove an alternate definition of the order of an element, but I don't know how to prove this implication:

Show for an arbitrary $g\in G$ and $d\in \mathbb{Z}_{>0}$: If $g^d=1$ and $g^{d/t}\neq 1$ for all prime divisors $t$ of $d$, then ord$(g)=d$.

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First notice that $d\geq \mbox{ord}(g)$ since $g^d=1=g^0$.

Assume now that $d> \mbox{ord}(g)$.

So $\exists d'\in \mathbb{Z}_{>0}$ such that $d'

Hence $d\leq \mbox{ord}(g)$ and we are done.