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$\operatorname{ord}(b)|\max\set{\operatorname{ord}(g)|g∈G}$ for all $b∈G$
MSE
Lemma. If the orders $|x|,|y|$ of $x,y\in G$ are coprime, then the order of $xy$ is $|x||y|$.
Proof. If $(xy)^m = 1$, then $x^{m|y|} = 1$, so $|x|$ divides $m|y|$. Since $|x|$ and $|y|$ are coprime, this implies $|x|$ divides $m$. Similarly, $|y|$ divides $m$, so by coprimality, their product divides $m$. $_\blacksquare$
Let $a$ be an element of maximum order, and $b\neq a$. Suppose $|b|\nmid |a|$. Then there exists a prime $p$ such that $|a|=p^im$, $|b|=p^jn$, $j>i$, $p\nmid m,n$. Then $|a^{p^i}|=m$, $|b^n|=p^j$ are coprime, so $a^{p^i}b^n$ has order $p^jm>|a|$, a contradiction. Therefore, $|b|\mid |a|$ for all $b\in G$. |
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