Permutation as a Product of $2$ cycles

Question

\begin{bmatrix}1&2&3&4&5&6&7&8\\2&3&4&5&1&7&8&6\end{bmatrix}

I have already written this permutation as disjoint cycles: (12345)(678)

My attempt at a product of two cycles: (12)(23)(34)(45)(67)(78), but I don't really think this is right, and if it is right, I'm not sure why. It's just an honest guess.

Answer 1

Indeed, your product of disjoint cycles correctly represent the given permutation. And also your transpositions from the disjoint cycles are spot on.

See also this answer for a more thorough discussion of ways to write the product of transpositions for disjoint permutations.

Answer 2

You are correct. In general we have $(x_1, x_2, x_3,...,x_k)=(x_1, x_2)(x_2, x_3)(x_3, x_4)...(x_{k-1}, x_k)$. Very easy to see why this is correct, just check where does each of the two permutations send each element.