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?
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.