TITLE: ord $(b)|\max\{\text{ord}(g)|g\in G\}$ for all $b\in G\,$ a finite abelian group
QUESTION [6 upvotes]: Let $a$ be an element of maximum order from a finite Abelian group $G$. Prove that for any element $b$, $|b|$ divides $|a|$ (order of $b$ divides order of $a$).

REPLY [17 votes]: 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 $m$.
Now let $a$ be your element of maximum order, and $b$ any other element. Suppose $p$ is a prime dividing $|b|$ to a higher power than $|a|$. Write $|a| = p^i m$ and $|b| = p^j n$, where $j>i$ and $p$ divides neither $m$ nor $n$. Then $a^{p^i}$ and $b^{n}$ have coprime orders, so $a^{p^i} b^n$ has order $p^j m > |a|$, a contradiction.