multiply two non-disjoint permutations I have two permutations $(2,7,9,4,3)$ and $(2,3,6)$ of $S_9$.
According to this post I want to multiplay these two. Unfortunately multiplication is described there only for disjoint permutations.
The algorithm presented there does not work for me. 
$(2,7,9,4,3)(2,3,6)$
Following that instruction I get $(2,2 ...$
I have started with $2$: 
$$(2,$$
The first cycle, $(2,3,6)$ sends $2$ to $3$. The second cycle, $(2,7,9,4,3)$ sends $3$ again to $2$ so I get: 
$$(2,2 ...$$
And I am stuck in an infinite loop here.
Question: How do I multiply permutations that are not disjoint? Or is it the same way as I should deal with disjoint permutations and pay attention to some additional rules?
 A: Writing both the permutations as product of transpositions, we get 
$(2,7,9,4,3)=(2,3)(2,4)(2,9)(2,7)$
and 
$(2,3,6)=(2,6)(2,3)$. So their multiplication is 
$(2,3)(2,4)(2,9)(2,7)\cdot (2,6)(2,3)$
Now we do it like this :- Start with $2$. So $2\mapsto 3$ due to the rightmost transposition in the product and $3$ is fixed in all the other elements of the product and finally $3$ is mapped to $2$ because of the transposition $(2,3)$. So the overall effect is that $2$ is fixed in this multiplication.
Now let's start with $3$. $3\mapsto 2$ (from the rightmost permutation) which in turn is mapped to $6$ and then $6$ is fixed throughout, so the overall effect is that $3\mapsto 6$. 
Doing the same trick for all the elements we see that 
$(2,3)(2,4)(2,9)(2,7)\cdot (2,6)(2,3)=(3,6,7,9,4)$.  
A: Let $\pi_1=(2\,7\,9\,4\,3)$, $\pi_2=(2\,3\,6)$, $\pi=\pi_1\pi_2$.
You have worked out that $\pi_2(2)=3$ and $\pi_1(3)=2$; hence $\pi(2)=2$, i.e. $\pi$ fixes $2$.
Now do the same for $3$. $\pi_2(3)=6$ and $\pi_1$ fixes $6$, so $\pi(3)=6$. Now look at $6$. $\pi_2(6)=2$ and $\pi_1(2)=7$; therefore $\pi(6)=7$. Continue with $7$. $\pi_2(7)=7$, $\pi_1(7)=9$; therefore $\pi(7)=9$. Carry on, you get $\pi(9)=4$, $\pi(4)=3$ which is what we started with. Hence:
$$(2\,7\,9\,4\,3)\,(2\,3\,6)\ =\ (3\,6\,7\,9\,4)$$
The numbers which don't appear ($1$, $5$, $8$) are fixed.
A: The algorithm works. Start with $2$:
$
(2,7,9,4,3)(2,3,6)=(2)(3,6.7,9,4)
$
Since the any cycle $(k)$ is the same as the identity you can just drop it; in the present case drop $(2)$.
