Product of Non-Disjoint Cycles I am trying to learn how to find the product of non-disjoint cycles, as you may have guessed from the title. I have the basic idea, but I do not understand it entirely.
I am trying to find $(1352)(256)$ and $(1634)(1352)$, multiplying from right to left.
Please give me some hints/suggestions.
Thank you.
 A: $(1352)$ and $(256)$ are specific permutations: write them out in whatever form it is convenient for you (for example, as a table) The product $(1352)(256)$ is the composition of those functions: compute it.
A: Here's how I do it: take
$$(1352)(256)$$
First, let's see what happens to $1$. Under $(256)$, $1$ stays $1$; then under $(1352)$ it maps to $3$. So $1\mapsto 3$. So the product will start $(13\ldots$. Next, look what happens to $3$.
Fixed under $(256)$, maps to $5$ under $(1352)$. So $3\mapsto 5$, and the product will be $(135\ldots$. Now look at what happens to $5$.
First, $5$ goes to $6$ under $(256)$; then $6$ is fixed by $(1352)$. So $5\mapsto 6$, the product is $(1356\ldots$. Next see what happens to $6$.
Under $(256)$, $6$ maps to $2$. Then $2$ maps to $1$ under $(1352)$, so $6\mapsto 1$, which closes the cycle we have. So the product will be $(1356)\cdots$
Now take the first number that hasn't appeared yet: $2$, and start over. First $2$ maps to $5$, then $5$ maps to $2$. So $2$ is fixed. Thus, the product is 
$$(1352)(256)=(1356)(2) = (1356).$$
Proceeding the same way with $(1634)(1352)$, We have $1$ to $3$ to $4$, so the product starts $(14\ldots$. Then $4$ is fixed and then mapped to $1$, so that closes the cycle. $(14)\cdots$. Now cheking $2$, $2$ goes to $1$, then the second cycle maps it to $6$; so the product is $(14)(26\ldots$. Then $6$ is fixed, then mapped to $3$; then looking at $3$, we have $3$ going to $5$ and then $5$ is fixed. Finally, $5$ is mapped to $2$ and $2$ is fixed, which closes the cycle. So the product is 
$$(1634)(1352)=(14)(2635).$$ 
