Q:

Writing permutations as products of disjoint cycles

How can I write these permutations as  products of disjoint cycles?
i.$\;\;(1234)(513)$
ii.$\;\;(13526)(53)(46215)$
iii.$\;(13)(12)(32)(143)$

A:

I'll spell out how to do it for (i.) and call that a hint for the other two.
For (i.), think about where each element ends up.
$1 \rightarrow 2$ in the first cycle, and $2\rightarrow 2$ in the second cycle.  So in the product, $1\rightarrow 2\rightarrow 2$.  In short, $1$ goes to $2$.
$2\rightarrow 3$ in the first cycle, and $3\rightarrow 5$ in the second cycle.  So in the product $2\rightarrow 3\rightarrow 5$.  In short, $2$ goes to $5$.
Since we left at $5$ in the last step, we start at $5$ here.
$5\rightarrow 5$ in the first cycle and $5\rightarrow 1$ in the second.  In short, $5$ goes to $1$.  We've gotten back to where we started, so the first cycle is $(125)$.
Now, we begin again with the first element we haven't seen (in that cycle) which is $3$.
$3\rightarrow 4$ in the first cycle and $4\rightarrow 4$ in the second, so $3$ goes to $4$.
$4\rightarrow 1$ in the first cycle and $1\rightarrow 3$ in the second, so $4$ goes to $3$.  This closed the cycle and we have $(34)$.  We've now seen all the elements, so we are done.
The disjoint cycle notation is: $(125)(34)$

