# Permutation

• Aug 15th 2007, 08:23 AM
r7iris
Permutation
Show that if pi and sigma are permutations such that (pi*sigma)^2 =(pi^2)(sigma^2) then (pi*sigma)=(sigma*pi)
• Aug 15th 2007, 08:26 AM
ThePerfectHacker
Quote:

Originally Posted by r7iris
Show that if pi and sigma are permutations such that (pi*sigma)^2 =(pi^2)(sigma^2) then (pi*sigma)=(sigma*pi)

Just expand.

$(ab)^2 = a^2b^2$
$abab=aabb$
$ab=ba$
• Aug 15th 2007, 08:27 AM
topsquark
Quote:

Originally Posted by r7iris
Show that if pi and sigma are permutations such that (pi*sigma)^2 =(pi^2)(sigma^2) then (pi*sigma)=(sigma*pi)

Simple. We know the inverse permutations (on the left and right) exist, so:
$\displaystyle ( \pi \sigma )^2 = \pi ^2 \sigma ^2$

$\displaystyle \pi \sigma \pi \sigma = \pi ^2 \sigma ^2$

$\displaystyle \pi \sigma \pi \sigma \sigma ^{-1} = \pi ^2 \sigma ^2 \sigma ^{-1}$

$\displaystyle \pi \sigma \pi = \pi ^2 \sigma$

$\displaystyle \pi ^{-1} \pi \sigma \pi = \pi^{-1} \pi ^2 \sigma$

$\displaystyle \sigma \pi = \pi \sigma$

-Dan