order of an element in symmetric group .

Question

let $f=(9~6\ 3 \ 5 \ 1 \ 4)(2\ 3 \ 4 \ 5 \ 7 \ 8 )$, $g=(4 \ 6 \ 7 \ 3 \ 1)(1 \ 4 \ 2 \ 6 \ 5 \ 7 \ 8)$ and $h=(1 \ 3 \ 5 \ 2)(2 \ 3 \ 6 \ 7)$ such that $f,g,h \in S_{9}$.

find order of $g^{-2}h^{-2}fh^{2}g^{2}$.

we know that in any group for any $x,a \in G$ we have $|x^{-1}.a .x^{}|=|a|$. so we must find order of $f$. we have $f=(9~6\ 3 \ 5 \ 1 \ 4)(2\ 3 \ 4 \ 5 \ 7 \ 8 )=(1 \ 5 )(2 \ 3 \ 7 \ 8 )(4 \ 9 )$ so $|f|=4$ then $|g^{-2}h^{-2}fh^{2}g^{2}|=4$

Is there an easier way to solve it?

Answer 1

You forgot about the $6$. The correct factorization into disjoint cycles is $$f=(1,5)(2,3,7,8)(4,9,6)$$ So $o(f)=\text{lcd}(2,4,3)=12$