Hello there. I have this two cycle permutation problems to prove,

1. $\displaystyle \left ( a b \right )^{-1} = \left ( a b \right )$

2. $\displaystyle (1,2,3, .... ,n)^{-1} = (n,n-1, ... ,2,1)$

Any idea?

Printable View

- Mar 13th 2013, 11:31 AMDonnieDarkoCycle permutation problem
Hello there. I have this two cycle permutation problems to prove,

1. $\displaystyle \left ( a b \right )^{-1} = \left ( a b \right )$

2. $\displaystyle (1,2,3, .... ,n)^{-1} = (n,n-1, ... ,2,1)$

Any idea? - Mar 13th 2013, 05:22 PMNehushtanRe: Cycle permutation problem
#1. What is $\displaystyle (a\ b)(a\ b)$?

#2. Use #1 and induction. (Note: $\displaystyle (1\ 2\ \cdots\ n\ n+1)=(1\ n+1)(1\ 2\ \cdots\ n)$.) - Mar 14th 2013, 11:31 AMDonnieDarkoRe: Cycle permutation problem
1. It is decomposition into disjoint cycle?

2. I tried to solve it like this.

$\displaystyle \tau = (a_{1} a_{2} ... a_{n} )$

$\displaystyle \tau(a_{i})=a_{i+1}$

$\displaystyle \tau^{-1}(a_{i+1})=a_{i}$

$\displaystyle \tau: a_{1} \mapsto a_{2} \mapsto ...\mapsto a_{n}$

$\displaystyle \tau^{-1}: a_{n} \mapsto a_{n-1} \mapsto ...\mapsto a_{1}$

- Its obvious that tau^(-1), is just tau written in reverse.