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 $a$).

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.

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 $G$.

Below is a simple conceptual proof that does not employ the high-powered structure theorem.

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 closed under$\rm\ lcm.$

Therefore 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 lcm-closed order set, i.e. with $\rm\: o(X) = $ order of $\rm\: X$

$\rm\quad\quad\quad\quad\ \ X,Y \in G\ \Rightarrow\ \exists\ Z \in G:\ o(Z) = lcm(o(X),o(Y))$

Proof $\ \ $ By induction on $\rm\, o(X)\: o(Y).\ $ If it is $\:1\:$ then trivially $\rm\:Z = 1.\ $ Otherwise

write $\rm\ o(X) = AP,\: \ o(Y) = BP',\ \ P'\!\mid P = p^m > 1,\: $ prime $\rm\: p\:$ coprime to $\rm\: A,B$

Then $\rm\: o(X^P) = A,\ o(Y^{P'}) = B.\ $ By induction there's a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$

so $\rm\ o(X^A\: Z) = P\: lcm(A,B) = lcm(AP,BP') = lcm(o(X),o(Y)).\ \ $ QED