Product of consecutive transpositions.

Question

I was interested if this equality holds for arbitrary $n$:

$$(1 \space 2)(2\space3)(3\space 4)...(n-1\space n) (n\space 1)=(1)(2 \space 3 \space 4... \space n-1 \space n)$$

(considering multiplications are done in $S_n$)

It looks like it is true (I've done calculations for $n=2, 3, 4$ and the formula does work). But it's hard for me to prove this in general.

The RHS has a typo, doesn’t it? Why would $1$ have a fixed point? Especially for $n=2$. — Erick Wong, Jun 08 '21 at 23:44
Ahhh ok that makes more sense. Have you tried just tracing the orbit of each element? It probably helps to separate into cases for $1$, $n$, and elements in between. — Erick Wong, Jun 08 '21 at 23:54
Yeah, I've tried. It is especially easy for $1, n$, but it is not that obvious for other elements — Easy Cheat, Jun 08 '21 at 23:59
$1 \rightarrow n \rightarrow n-1 \rightarrow n-2 \rightarrow ... \rightarrow 1$, so $1$ remains untouched. — Easy Cheat, Jun 09 '21 at 00:00

score 1 · Accepted Answer · answered Jun 09 '21 at 00:07

We assume that permutations act from the left, so we apply the transpositions from right-to-left.

As the OP mentions in the comments, $1$ is a fixed point of the permutation in question.

The element $n$ follows the trajectory $n \mapsto 1 \mapsto 2$.

For any other $k$ with $1 < k < n$, element $k$ is untouched until it encounters the transposition $(k\;k+1)$, at which point it moves to $k+1$ which puts it out of reach of all other transpositions.

Hence $k \mapsto k+1$ for all such $k$. Combining this with $n \mapsto 2$ gives the desired cycle structure.

Product of consecutive transpositions.

1 Answers1