Let $G$ be a group, let $a,b\in G$ with $ab=ba$, and let $|a|=m,~|b|=n$.
Then, $|ab|\mid\textrm{lcm}(m,n)=mn/d$, where $d=\gcd(m,n)$.
I have confusing in the following situation:
Let $\sigma,\tau$ be 'disjoint permutations' in certain symmetric group of orders $m$ and $n$, respectively, where 'disjoint' means that $\sigma(i)\neq i\Rightarrow\tau(i)=i$ and $\tau(j)\neq j\Rightarrow\sigma(j)\neq j$.
Now, suppose that $\sigma\circ\tau=\tau\circ\sigma$.
Then, what is the order of an element $\sigma\circ\tau$? Is it just $|\sigma\circ\tau|=\textrm{lcm}(m,n)$?
I think that $\gcd(m,n)$ need not be equal to $1$.
But, my attempts as follows:
Set $l=\textrm{lcm}(m,n)$. Then, since $\sigma\circ\tau=\tau\circ\sigma$, $(\sigma\circ\tau)^{l}=\textrm{id}$.
Now, assume that $(\sigma\circ\tau)^{l'}=\textrm{id}$ with $0<l'\le l$.
Then, since $\sigma\circ\tau=\tau\circ\sigma$, $\sigma^{l'}=(\tau^{l'})^{-1}$, and which implies that $\sigma^{l'}=(\tau^{l'})^{-1}=\textrm{id}$ since $\sigma$ and $\tau$ are disjoint.
Thus, $m\mid l'$ and $n\mid l'$, furthermore, $l=\textrm{lcm}(m,n)\mid l'$.
Therefore, since $0<l\le l'$, $l=l'$, and hence, the element $\sigma\circ\tau$ is of order $l=\textrm{lcm}(m,n)$.
My attempts seems to be false. Where i made a mistake?
Can someone point me out? Thank you!
There is no mistake in your reasoning. The fact that permutations are disjoint is much stronger than the fact that permutations commute. So, in general for commuting permutations $a, b$ we have $$ \operatorname{ord} (a\cdot b) | \operatorname{lcm} (\operatorname{ord} (a), \operatorname{ord} (b)) $$ But in the case where $a$ and $b$ are disjoint we have $$ \operatorname{ord} (a\cdot b) = \operatorname{lcm} (\operatorname{ord} (a), \operatorname{ord} (b)). $$
Also we can give an exmaple where for commuting but non-disjoint permutations the last equality is not true. We have $$ \operatorname{ord}(a\cdot a^{-1}) = 1 $$ for every permutation $a$, but of course if $a$ has non-identity order equation $ \operatorname{ord} (a\cdot b) = \operatorname{lcm} (\operatorname{ord} (a), \operatorname{ord} (b)) $ is not valid.