Permutations product Can anyone please help me compute 
$513642798 \times 971265384$
attempt: I start with the right permutation, 
So $9 \to 7 \to 9 = (97)$
however, after that do i go to $7 \to 1 \to 3$ ...? 
I am confused about the order.
Thank you for your help
 A: You are on the right track. What your first step has shown is that $9\rightarrow 9$ not $(97)$, which is shorthand for $9 \rightarrow 7$ and $7 \rightarrow 9$.
Continuing on, you have $7\rightarrow 1 \rightarrow 3$ so that just becomes $7\rightarrow 3$ in the final product. You should now do $3$ so that you can continue writing down the permutation. We get $3\rightarrow 8\rightarrow 5$. So we are building the final answer permutation of $(7 3 8 ...)$. Continue until finished. Those numbers that are like $9$ that just map to themselves are left out of the final answer to indicate that the permutation has not effect on them.
A: For $(513642798)(971265384).$
Start on the right with $1$. We have $1\rightarrow 2$, then on the left, $2 \rightarrow 7$ so we begin by writing $(17\ldots$. Back to the right, $7 \rightarrow 1$ then on the left $1 \rightarrow 3$, and so we fill in $(173\ldots$. Going to $3$ on the right, and continuing in this manner, you will get the closed cycle $(17356)$. 
Now pick up the next smallest number not yet addressed, it being $2$. On the right $2 \rightarrow 6$ then on the left $6\rightarrow 4$. Going back to the right, $4 \rightarrow 9$ then on the left $9 \rightarrow 8$. Try it one more time, and you will see the end of that cycle. We can write
$$(513642798)(971265384)=(17356)(248).$$
Note that $9$ is a one-cycle, $(9)$. It is customary to assume all elements not included are one-cycles, but you could certainly append $(9)$ to the beginning, middle, or end of your permutation product. I would append it to the end, or not write it at all if it is clear that we got the product from permutations on nine elements.
