1
$\begingroup$

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$(g)=d$.

1 Answers 1

1

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' and $g^{d'}=1$. This implies that $d$ is a multiple of $d'$, i.e. $d'\cdot m=d$. Now let $p\in \mathbb{N}$ be any prime divisor of $m$, which is clearly also a prime divisor of $d$, then we have: $g^{d/p}=g^{d'\cdot m / p}=1^{m/p}=1.$ This means that there exists a prime divisor of $d$ (namely $p$) such that $g^{d/p}=1$, which gives us the desired contradiction.

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