How many cycles $A$ and $B$ can form this cycle How many cycles $A$ and $B$ can form this cycle:
$AB=(axyguimjrcwk)(bvqphsleofzt)(d)(n)$
I can see that $A$ and $B$ must share the cycle $(dn)$, and I believe due to ordering, both $A$ and $B$ must contain the same $24$ cycle each. So there should be only $1$ possible solution for $A$ and $B$. Is this right?
 A: If you have a permutation with a cycle of length $n$, then that cycle can be created by two other permutations $A$ and $B$ with cycles that live within those $n$ elements, as long as those two permutations can be themselves created by composing a particular length-$n$ cyclic permutation $P$ with itself some number of times $a$ and $b$ (including $n$ itself!), so $A=P^a$ and $B=P^b$, and $\gcd (a+b, n)=1$.  Your particular cycle is then $P^{a+b}$.
For instance, $(aie)(bjf)(ckg)(dlh)\times(adgj)(behk)(cfil)=(lkjihgfedcba)$; for $P=(abcdefghijkl)$, this is $P^8P^3=P^{11}\left(=P^{-1}\right)$ 
Similarly, if you have a permutation with $m$ cycles of length $n$, you can create this with two other cycles the same way, using a $P$ of length $mn$, and $\gcd(a+b,mn)=m$.
Finally, two permutation hunks that work on disjoint piles of elements can be combined in a single permutation hunk without a problem.
So, for your case, you have several options here: you can consider the two $12$-cycles as independent (so you're looking at two disjoint $12$-length permutations) or together (so you're looking at a $24$-length permutation); similarly, your two $1$-length permutations can be considered also a single $2$-length permutation, but they don't have to be.
