Q:

Writing the permutation as the product of disjoint cycles? 

In my notes, it says that $\sigma = (1 6)(2 4 5 6)(1 4 5)(2 3)(7 8 9) = (1 3 2 5 6 4)(7 8 9)$.
In my answer, I found that $1 \rightarrow 6$ then $6 \rightarrow 2$ then $2 \rightarrow 3$ and then 3 is fixed.  So I got as far as (1 3 ...). 
Then I found that $2 \rightarrow 4$ then $4 \rightarrow 5$ then 5 is fixed.  So I got as far as (1 3 2 5 ...). 
The next part is where my answer differs from the solution given.  I found that $4 \rightarrow 5$ then $5 \rightarrow 1$ then 1 is fixed.  So I got (1 3 2 5 4) for the first cycle, rather than (1 3 2 5 6 4). 
Could someone please tell me what I'm doing wrong?  By the way, we have been taught in this course to read permutations from left to right :)

A:

After you get $(1 \; 3 \cdots)$ you should start with the last number you have so far, $3$, and see where you go (e.g., $3 \to 2$) to obtain $(1 \; 3 \; 2 \cdots)$. [Your mistake was to ignore the $3$, and for some reason start with $2$ in your next step.  You need to show $3 \to 2$ if you want to write down $2$ after $3$; you just got lucky this time.] 
Then you look at the last number $2$, and continue; you have already shown $2 \to 5$ you have $(1 \; 3 \; 2 \; 5 \cdots)$.
Then you start from $5$ and note $5 \to 6$ to obtain $(1\; 3 \; 2 \; 5 \; 6 \cdots)$.
Then you show $6 \to 1 \to 4$ to get $(1 \; 3 \; 2 \; 5 \; 6 \; 4 \cdots)$.
Finally, you showed $4 \to 5 \to 1$ so the final cycle is $(1\; 3 \; 2 \; 5 \; 6 \; 4)$.

