Order of element equal to least common multiple [duplicate]

Theorem $\ $ A finite abelian group $\rm\:G\:$ has an lcm-closed order set, i.e. with $\rm\: o(X) = $ order of $\rm\: X$

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

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

Let $m=\prod_{i=1}^t p_i^{m_i}$ and $n=\prod_{i=1}^t p_i^{n_i}$ where $p_1,p_2,\cdots,p_t$ are distinct primes and $m_i,n_i \geq 0$.

Furthermore, may assume that $m_i < n_i$ if $1 \leq i \leq l$ and $m_i \geq n_i$ if $l+1 \leq i \leq t$.

Let $c=\prod_{i=1}^l p_i^{m_i},d=\prod_{i=l+1}^t p_i^{n_i}$.

We can conclude that $o(a^c)=\prod_{i=l+1}^t p_i^{m_i}$ and $o(b^d)=\prod_{i=1}^l p_i^{n_i}$, clearly that $(o(a^c),o(b^d))=1$ and $a^cb^d=b^da^c$. Hence the order of $a^cb^d$ is $\prod_{i=1}^l p_i^{n_i} \cdot \prod_{i=l+1}^t p_i^{m_i}=[m,n]$, which completes the proof.

Remark: The method above can also be used to prove the existence of an element in $G$ with order $(m,n)$.