Product of Non-Disjoint Cycles

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 $$(1634)(1352)=(14)(2635).$$

$(1352)$ 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.