# Composition of Permutations is not Commutative

## Theorem

Let $S$ be a set.

Let $\map \Gamma S$ denote the set of permutations on $S$.

Let $\pi, \rho$ be elements of $\map \Gamma S$

Then it is not necessarily the case that:

$\pi \circ \rho = \rho \circ \pi$

where $\circ$ denotes composition.

## Proof

Let $S := \set {1, 2, 3}$.

Let:

 $\displaystyle \pi$ $:=$ $\displaystyle \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1}$ $\displaystyle \rho$ $:=$ $\displaystyle \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2}$

Then:

 $\displaystyle \pi \circ \rho$ $=$ $\displaystyle \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1} \circ \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2}$ $\displaystyle$ $=$ $\displaystyle \dbinom {1 \ 2 \ 3} {3 \ 2 \ 1}$

while:

 $\displaystyle \rho \circ \pi$ $=$ $\displaystyle \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2} \circ \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1}$ $\displaystyle$ $=$ $\displaystyle \dbinom {1 \ 2 \ 3} {2 \ 1 \ 3}$ $\displaystyle$ $\ne$ $\displaystyle \pi \circ \rho$

$\blacksquare$