Permutation and cycles I need to determine the missing number to fulfill the following reproduction:
$$\pi=\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&1&2&6&7&8}$$
And here is the representation of $pi$ as the product of transpositions:
$$(1 i) (1 3) (2 5) (9 8) (8 7) (7 6) (3 6) = \pi$$
The result for $i$ must be 2 but I have no clue how to get there.
I tried to start at $(3 6)$ and then check on which transposition it gets pictured on but it doesn't seem to work for me. Any ideas on how to generally solve these kind of problems?
-Freddy

According to the comments, you’re multiplying from right to left. If you multiply the transpositions that you know completely, you get
$$\pmatrix{6&3&7&8&9&5&2&1&4\\1&9&6&7&8&2&5&3&4}\;,$$
which we rearrange into standard form as
$$\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&2&1&6&7&8}\;.\tag{1}$$
We want to choose $i$ so that 
$$(1i)\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&2&1&6&7&8}=\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&1&2&6&7&8}\;.\tag{2}$$
There are several ways to think about this. 


*

*You can notice that $(1)$ and $\pi$ differ only in what they do to $5$ and $6$, and it’s pretty clear that setting $i=2$ will fix that discrepancy. 

*You can notice that if you carry out the multiplication on the lefthand side of $(2)$, you’ll get $$\pmatrix{1&2&3&4&5&6&7&8&9\\\cdot&\cdot&\cdot&\cdot&\cdot&i&\cdot&\cdot&\cdot}\;,$$ so if this product is to be $\pi$, $i$ must be $2$. (Of course you should then finish the multiplication to check.)

*You can use the fact that a transposition is its own inverse: $(2)$ holds if and only if $$\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&2&1&6&7&8}=(1i)\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&1&2&6&7&8}\;,$$ and the product on the right is going to have a column $\binom5i$, so again $i$ must be $2$.
