# Permutation as a Product of $2$ cycles

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

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.

• Your answer looks correct to me. – jwc845 Sep 22 '18 at 23:57
• Why don't you just work out the product of those six $2$-cycles and see what you get? Hint: it's correct. – bof Sep 22 '18 at 23:57

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.