Order of element equal to least common multiple 
Let $G$ be a group, and $a,b\in G$. Suppose $\operatorname{ord}(a)=m, \operatorname{ord}(b)=n$, and that $ab=ba$. Prove that there is an element $c\in G$ such that $\operatorname{ord}(c)=\operatorname{lcm}(m,n)$.

Let $A=\operatorname{lcm}(m,n)$. I consider $(ab)^A=a^Ab^A=1$, so the order of $ab$ divides $A$. What can we do to find an element with order exactly $A$? 
 A: 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
A: 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 $(m,n)$.
A: Since $a$ and $b$ commute, we know that $(ab)^k = a^kb^k$ for any integer $k$.
As you pointed out, letting $A = \text{lcm}(m,n)$, we know that the order of $ab$ is at most $A$ (and divides $A$).
So, since we don't have any group elements running around besides products of $a$'s and $b$'s, let's try showing that we must have ord($ab$)=$A$. That is, no smaller divisor of $A$ will do the trick. Raising $ab$ to a power that's a multiple of $m$, we see
$(ab)^{mk} = ((ab)^m)^k = b^k$ = 1 if and only if $n$ divides $k$. 
A similar argument should tell you about a relationship between $m$ and $k$, and this should give you want you want. Think about it, and I can provide more details if necessary.
