Writing an arbitrary two-cycle, say $(a,b)\in S_n$ as a product of adjacent transpositions.

Question

I need to express $(a,b)$, an arbitrary transposition, as a product of adjacent transpositions. I have figured out a method for the form $(a,a+1)$ and $(a,a+2)$ but the general form has me stumped.

Answer 1

The conjugate of $(a,a+2)$ by $(a+2,a+3)$ is $(a,a+3)$, that is $$(a+2,a+3)(a,a+2)(a+2,a+3)=(a,a+3)$$ so if you can express $(a,a+2)$ in terms of adjacent transpositions, you can do that for $(a,a+3)$. In general, $$(b,b+1)(a,b)(b,b+1)=(a,b+1)$$ so if you can do $(a,b)$ then you can do $(a,b+1)$.

Answer 2

For $(a,b)$, with $a\lt b$, write $(b, b-1)(b-1, b-2)\cdots (a+2, a+1)(a+1, a)(a+2,a+1)\cdots(b, b-1)$

Answer 3

Assume many persons are standing in a line. Our aim is to make the first person in the line to go to the last position and vice versa. We also want everyone else to be in their own position. Allowed moves are two persons standing next to each other can swap their positions.

Call the persons A,B,C,D,E in that order.

Move 1: BACDE (first two swap)

Move 2: BCADE (2nd and third swap their positions)

Move 3: BCDAE (3rd and 4th swap)

Move 4: BCDEA (4th and 5th swap)

Now let us start reverse movement

Move 6: BCEDA (3rd and 4th swap)

Move 7: BECDA (2nd and 3rd swap)

Move 8: EBCDA (first two swap)

Objective achieved.