Here $a,b$...etc are decomposed cycle and they form permutation.
Let $a$,$b$ be elements of finite orders $m$ and $n$ respectively in a group $G$. Suppose that $ba=ab$ and that $\langle a\rangle\cap \langle b\rangle=\{e\}$. Then the order of $ab$ is the l.c.m. of $m$ and $n$.
Proof:
Let $l=mp=nq$ be l.c.m. of $m$ and $n$. Then $(ab)^l=a^lb^l=a^{mp}b^{nq}=(a^m)^p(b^n)^q=e^pe^q=e$. (Here $e$ is identity) so the order of $ab$ is at most $l$. Now let $k>0$ and suppose that $(ab)^k=e$. Then $a^kb^k=e$ so $a^k=b^{-k}$. But then $a^k\in\langle a\rangle\cap\langle b\rangle$ so $a^k=e$. Similarly, $b^k=e$. $k$ is common multiple of $m$ and $n$ so $k\ge l$. Thuse, the order of $ab$ is at least $l$. As we have already seen that it is at most $l$, it must be $l$
Why would $a^k\in\langle a\rangle\cap\langle b\rangle$? Was it a suppose statement? Another problem is why is $k\ge l$? Why shouldn't it be $k\le l$?