Let $α ∈ S_9$ be the permutation given by $$α(1) = 6;α(2) = 8;α(3) = 7;α(4) = 9;$$ $$α(5) = 3;α(6) = 4;α(7) = 5;α(8) = 2;α(9) = 1.$$ Decompose the permutations α first as a product of disjoint cycles and then as a product of transpositions. Find $α^{−1}$ and $α^2$ . What are the order and sign of α? Find $α^{122}$ .
I have found that $a=(1 6 4 9),(2 8),(3 7 5)$ and $o(a)=LCM(4,2,3)=12$.
The sign of $a=(=1)*(-1)*(1)=1$
$a=(1 9)(1 4)(1 6) (2 8)(3 5)(3 7)$
$a^{-1}=(1 9 4 6)(2 8)(3 5 7)$ and $a^2=$ $(1 2 3 4 5 6 7 8 9 / 4 2 5 1 7 9 3 8 6)$ so, $a^2$ is $(1 4)(3 5 7)(6 9)$
How can I find $a^{122}$?
In any group, if the order of an element $g$ is $n$, we have $$ g^N=g^{N\bmod n}. $$
This is very often at work when calculating a huge power of a number modulo a prime $p\,$: it suffices to apply Fermat's theorem to know the order of the (non divisible by $p$) number, which asserts the order of the number mod. $p$ is a divisor of $p-1$, hence $$x^N\equiv x^{N\bmod p-1}\mod p.$$