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.

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

Let:

$\ds \pi$

$:=$

$\ds \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1}$

$\ds \rho$

$:=$

$\ds \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2}$

Then:

$\ds \pi \circ \rho$

$=$

$\ds \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1} \circ \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2}$

$\ds $

$=$

$\ds \dbinom {1 \ 2 \ 3} {3 \ 2 \ 1}$

while:

$\ds \rho \circ \pi$

$=$

$\ds \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2} \circ \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1}$

$\ds $

$=$

$\ds \dbinom {1 \ 2 \ 3} {2 \ 1 \ 3}$

$\ds $

$\ne$

$\ds \pi \circ \rho$

$\blacksquare$

1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.2$. Commutative and associative operations: Example $62$