# cycle

• March 15th 2009, 08:15 AM
Sally_Math
cycle
I have to show that if $sigma$ is a cycle of odd length then $sigma^2$ is a cycle.

$wlog$, assume that
$sigma=(1,2,3,......,m) where m is odd.$
Because m is odd, I can compute that
$sigma^2=(1,2,3,....,m)(1,2,3,....,m)(1,3,5,....,m, 2,4,6,...m-1)$
which is again a cycle.
(Headbang) Is this a suitable proof, and what else can I add to it.

Thank You(Bow)
• March 15th 2009, 10:38 AM
ThePerfectHacker
Quote:

Originally Posted by Sally_Math
I have to show that if $sigma$ is a cycle of odd length then $sigma^2$ is a cycle.

$wlog$, assume that
$sigma=(1,2,3,......,m) where m is odd.$
$sigma^2=(1,2,3,....,m)(1,2,3,....,m)=(1,3,5,....,m ,2,4,6,...m-1)$
I consider this proof to be fine. You can add by showing the the LHS and the RHS agree for all values of $1,2,...,m$ in other words evaluating the LHS for any $1,2,...,m$ will give the same result as RHS and so the two functions (bijections) are the same.