Prove the existence of an element with order = exponent of a Finite Abelian Group Let $G$ be a finite abelian group. If $m=$exponent of $G$, then by definition $m=lcm(|g_1|,...,|g_n|)$ for all $g_i \in G$ . It suffices to show if $g \in G$ has the maximum order, then $\forall g_i \in G$, $|g_i|$ divides $|g|$. But then I'm stuck.
 A: (1). $g^n=1\implies |g|$ divides $n.$ Proof: Otherwise $n=m|g|+r$ for some integers $m,r$ with $1\leq r<|g|.$ But then  $g^r=g^n (g^{|g|})^m=1$ with $1\leq r<|g|$, contrary to the def'n of $|g|.$
(2).. If $n>0$ and $n$ divides $|g|$ then $|g^n|=|g|/n.$ Proof: We have  $(g^n)^{|g|/n}=g^{|g|}=1$,  so by (1), $|g^n|$ divides $|g|/n.$ Therefore  $$|g^n|\leq |g|/n.$$  But  $g^{n|g^n|}=(g^n)^{|g^n|}=1,$ so by (1), $|g|$ divides  $n|g^n|,$ so  $|g|\leq n|g^n|$. Therefore $$|g|/n\leq |g^n|.$$ 
(3).(i).  If $\gcd (|g|,|g'|=1)$ then $|gg'|=|g|\cdot |g'|.$
Proof: $(gg')^{|g|\cdot |g'|}=(g^{|g|})^{|g'|}\cdot (g'^{|g'|})^{|g|}=1$.   So by (1) $|gg'|$ divides $|g|\cdot |g'|.$ $$ \text { Therefore }\bullet \quad |gg'|\leq |g|\cdot |g'|.$$  And we have $g^{|gg'|}=g'^{(-|gg'|)}$ so $$ \quad g^{|gg'|\cdot |g'|}=  
  (g^{|gg'|})^{|g'|}=     (g'^{(-|gg'|)})^{|g'|}=     g'^{(-|gg'|\cdot |g'|)}= (g'^{|g'|})^{(-|gg'|)}=1.$$ So by (1), $|g|$  divides  $|gg'|\cdot |g'|.$  Since $\gcd(|g|,|g'|)=1$ this implies $$|g| \text { divides } |gg'|.$$ Interchanging $g$ with $g'$  we also have $$|g'|\text {  divides } |gg'|.$$ Now since $|g|$ and $|g'|$ both divide $|gg'|$ with  $\gcd (|g|,|g'|)=1,$ we have: $|g|\cdot |g'|$ divides $|gg'|.$ $$\text {Therefore }\bullet \bullet  \quad|g|\cdot |g'|\leq |gg'|.$$ By $\bullet$ and $\bullet \bullet$ we have $|gg'|=|g| \cdot |g'|$.
(3).(ii). If $g_1,...,g_k$ are  members of $G$ such that $\gcd (|g_i|
,|g_j|)=1$ when $i\ne j,$  then $$|\prod_{i=1}^kg_i|=\prod_{i=1}^k|g_i|.$$ 
Proof: $k=1$ is trivial.   If $k>1$ use induction on $k$: Let $g=\prod_{i=1}^{k-1}g_i$ and $g'=g_k.$ If $|g|=\prod_{i=1}^{k-1}|g_i|$ then $\gcd(g,g')=1$ so by (3).(i) we have $|gg'|=|g|\cdot |g'|=\prod_{i=1}^k|g_i|.$   
(4).  Let $\{p_1,...,p_k\}$ be a set of $k$ distinct primes such that  for every $g\in G$ the order of $g$ is $\prod_{i=1}^kp_i^{e_i(g)}$ where each $e_i(g)$ is a non-negative integer. 
For each $i$ let $x_i\in G$ such that $e_i(x_i)=\max \{e_i(x):x\in G\}.$ 
Let $n_i=|x_i|/p_i^{e_i(x_i)}.$ Let $y_i=x_i^{n_i}.$ 
By (2) we have $|y_i|=|x_i|/n_i=p_i^{e_i( x_i)}.$ 
Let $z=\prod_{i=1}^k y_i.$ Then  $|y|=$  $|\prod_{i=1}^ky_i|=$ $\prod_{i=1}^k|y_i|=\prod_{i=1}^kp_i^{e_i(x_i)}$ by (3).(ii). 
For every $g\in G$ we have $e_i(g)\leq e_i(x_i)$ by def'n of $x_i,$ so $|g|\leq |y|.$ In fact $|g|$ divides $|y|.$  
REMARK. In the case where $G$ is the set of non-zero members of  finite field $F$, where the group operation of $G$ is multiplication in $F$: Let $|G|$ be the number of members of $G.$ Let  $y$ be as in (4). Now $|g|$ divides $|y|$ for every $g\in G$. So in F the number of solutions of the polynomial $x^{|y|}-1=0$ is $|G|$, so $|y|\geq |G|.$ But (Lagrange's theorem) $|y|$ divides $|G|.$ So $|y|=|G|$, implying that  $G$ is a cyclic group.
