Computing the permutation group of $((1 2 3 4)(4 2 7)^{-1}(7 1 5 6))^{125}$

Question

$((1 2 3 4)(4 2 7)^{-1}(7 1 5 6))^{125} = ((1 2 3 4)(7 2 4)(7 1 5 6))^{125}=(1 2 3 4)^{125}(7 2 4)^{1 2 5}(7 1 5 6)^{125}$

$125 \equiv 1 (mod 4) $ 4 is the length of the cycle $(1 2 3 4)$

$125 \equiv 2 (mod 3) $ 3 is the length of the cycle $(7 2 4)$

$125 \equiv 1 (mod 4) $ 4 is the length of the cycle $(7 1 5 6)$

so I can rewrite it as $(1 2 3 4)(7 2 4)^{2}(7 1 5 6)$

$(7 2 4)^2=(7 4 2)$

so I have to compute $(1 2 3 4)(7 4 2)(7 1 5 6)$ I think that the solution is $(1)(2 3)(4 5 6 7)$

Answer 1

You can't say that $((1 2 3 4)(7 2 4)(7 1 5 6))^{125}=(1 2 3 4)^{125}(7 2 4)^{1 2 5}(7 1 5 6)^{125}$ because these cycles do not necessarily commute since they are not disjoint.

First write $(1 2 3 4)(4 2 7)^{-1}(7 1 5 6)$ as a product of disjoint cycles.

Then reduce $125$ mod the length of each cycle.