# Writing $2$-cycles as a product of adjacent 2-cycles.

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.

• Are you familiar with the fact that for any permutation $\sigma\in S_n$ you have $$\sigma(a\ b)\sigma^{-1}=(\sigma(a),\sigma(b))?$$ – Servaes Mar 5 '19 at 22:17
• @Servaes I am not. – Ryan Mar 6 '19 at 1:39

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 $$(a\ b)$$.
• So, if we have $$(a\ a+1)(a\ b)(a\ a+1)=(a+1\ b)$$ I guess I am not seeing the big picture and generalization on how this is writing $(a\ b)$ as adjacent 2-cycles. Thanks for your help. – Ryan Mar 6 '19 at 1:43