${ if \ \sigma = \begin{pmatrix}1&2&3&4&5\\ 4&2&5&3&1 \end{pmatrix} , \ \tau = \begin{pmatrix}1&2&3&4&5\\ 3&5&4&2&1 \end{pmatrix} }$
${ Find: (i) O( \sigma\tau) \ \ , \ \ (ii) (\sigma\tau)^{23} }$
i found order ${ O( \sigma\tau)\ }$ = 4
and i don't know law give me
(ii) without multiplication ${( \sigma\tau) }$ 23 time ?
If the order of $(\sigma \tau)$ is $4$, then we know that $(\sigma \tau)^4 = e$. Then $(\sigma \tau)^{4k} = e$ for any $k \in \mathbb{Z}^{+}$. Then $e = (\sigma \tau)^{24} = (\sigma \tau)^{23} \cdot (\sigma \tau)$. Hence, $(\sigma \tau)^{23} = (\sigma \tau)^{-1}$