Write in array form, product of disjoint cycles, product of 2-cycles... In symmetric group $S_7$, let $A= (2 3 5)(2 7 5 4)$ and $B= (3 7)(3 6)(1 2 5)(1 5)$.
Write $A^{-1}$, $AB$, and $BA$ in the following ways:
(i) Array Form
(ii) Product of Disjoint Cycles
(iii) Product of $2$-Cycles  
Also are any of $A^{-1}, AB$, or $BA$ in $A_7$?

I won’t do the problem, but I will answer the same questions for the permutation $A$; see if you can use that as a model. I assume throughout that cycles are applied from left to right.
$A=(235)(2754)=(2345)(2754)(1)(6)$; that means that $A$ sends $1$ to $1$, $2$ to $3$, $3$ to $5$ to $4$, $4$ to $2$, $5$ to $2$ to $7$; $6$ to $6$, and $7$ to $5$. Thus, its two-line or array representation must be $$\binom{1234567}{1342765}\;.$$
Now that we have this, it’s easy to find the disjoint cycles. Start with $1$; it goes to $1$, closing off the cycle $(1)$. The next available input is $2$; it goes to $3$, which goes to $4$, which goes to $2$, giving us the cycle $(234)$. The next available input is $5$; it goes to $7$, which goes right back to $5$, and we have the cycle $(57)$. Finally, $(6)$ is another cycle of length $1$. Thus, $A=(1)(234)(57)(6)$ or, if you’re supposed to ignore cycles of length $1$, simply $(234)(57)$.
Finally, you should know that a cycle $(a_1a_2\dots a_n)$ can be written as a product of $2$-cycles in the following way:
$$(a_1a_2\dots a_n)=(a_1a_2)(a_1a_3)\dots(a_1a_n)\;.$$
Thus, $A=(23)(24)(56)$.
One further hint: one easy way to find $A^{-1}$ is to turn the two-line representation of $A$ upside-down (and then shuffle the columns so that the top line is in numerical order).
