Problem: Consider the arbitrary 2-cycle $(a\ b)$ from $S_n$. Find a way to write this permutation as a product of adjacent 2-cycles.
What I do know:
A transposition is a single cycle of length 2. An adjacent transposition is of the form $(i\ i+1)$. For example, $(3\ 7)$ is a non-adjacent transposition, but $(3,4)$ is an adjacent transposition. It turns out that the set of transpositions for $S_n$ is a generating set for $S_n$.
So, to write the arbitrary 2-cycle $(a\ b)$ from $S_n$ as a product of 2 adjacent cycles, would it look something like this:
I can start at some arbitrary values, say, $(1\ 2)$. Then,
$(1\ 2)(a\ a+1)(a+1\ a+2)\cdots(b-2\ b-1)(b-1\ b)$ and eventually, I will get to $(a\ b)$?
I guess my question is, would this be a valid answer or would I need to generalize it more? And, is it ok to start at some arbitrary values such as the one I chose? Thanks for your help.
Best Answer
In general the product of adjacent transpositions $$(1\ 2)(a\ a+1)(a+1\ a+2)\cdots(b-2\ b-1)(b-1\ b),$$ does not eventually reach $(a\ b)$. For example, if $(a\ b)=(3\ 4)$ then your product is $$(1\ 2)(3\ 4)\neq(3\ 4).$$ A less degenerate example would be $(a\ b)=(4\ 8)$. Then your product is $$(1\ 2)(4\ 5)(5\ 6)(6\ 7)(7\ 8)=(1\ 2)(4\ 5\ 6\ 7\ 8)\neq(4\ 8).$$
In stead, use the fact that $$(c\ c+1)(a\ c)(c\ c+1)=(a\ c+1).$$ In this way, starting from $c=a+1$ we get $$(a+1\ a+2)(a\ a+1)(a+1\ a+2)=(a\ a+2),$$ then again with $c=a+2$ to get $$(a+2\ a+3)(a+1\ a+2)(a\ a+1)(a+1\ a+2)(a+2\ a+3)=(a\ a+3),$$ and you can continue this way all the way to $c=b-1$ to get $(a\ b)$.