ord $(b)|\max\{\text{ord}(g)|g\in G\}$ for all $b\in G\,$ a finite abelian group 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 $a$). 
 A: Here's a simple proof that avoids the high-power structure theorem in the lone prior answer, and highlights the arithmetical essence (the set of all orders of elements of $\rm\,G\,$ is $\color{#c00}{\text{closed under lcm}}$).
Theorem $\rm\ \ \ maxord(G)\ =\ expt(G)\ $ for a finite abelian group $\rm\: G,\ $ i.e.
$\rm\ \ \ max\ \{ ord(g) : \: g \in G\}\ =\  min\ \{ n>0 : \: g^n = 1\ \:\forall\ g \in G\}$
Proof $\ $ By  lemma below, $\rm\: S = \{ ord(g) : \:g \in G \}$  is a finite set of naturals $\color{#c00}{\text{closed under lcm}}$.
Hence every $\rm\ s \in S\:$  is a
divisor of the max elt $\rm\, m\ $  [else  $\rm\, lcm(s,m) > m$],$\ $ so $\rm\ m = expt(G).$
Lemma $\ $ A finite abelian group $\rm\,G\,$ has an $\color{#c00}{\text{lcm-closed}}$ order set, i.e. with $\rm\: o(X) = $ order of $\rm\: X$
$\rm\qquad\qquad\ X,Y \in G\ \Rightarrow\ \exists\ Z \in G:\  o(Z) = lcm(o(X),o(Y))$
Proof $\ $ See this Oct 2010 answer for a short simple proof.
A: 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.
A: By the structure theorem for finite(ly-generated) abelian groups, there exist $d_1, \cdots, d_k \in \mathbb{Z}^+$ such that
$$G \cong \dfrac{\mathbb{Z}}{d_1 \mathbb{Z}} \oplus \cdots \oplus \dfrac{\mathbb{Z}}{d_k \mathbb{Z}}$$
and $d_i$ divides $d_{i+1}$ for each $1 \le i < k$.
But then each element has order dividing $d_k$, and $d_k$ is the maximum order of any element of $G$.
