[Math] Why is this the method to getting transpositions from disjoint cycles

group-theorypermutations

I have the disjoint cycle: $$(156)(2437).$$ Apparently the "method" would get us: $$(1,6)(1,5)(2,7)(2,3)(2,4).$$ Basically you take the first number, and put it as a transposition of the last number and go backwards till you use all of them. I don't understand where this comes form. I don't know why this method works. Does it even work for my case? Oh and the order is 12 with the permutation being odd.

I think I understand $$(1,5)(5,6)(2,4)(4,3)(3,7)$$ much better. Is this a correct way to do transpositions as well?

Best Answer

Assuming we are composing from right to left, the original product of transpositions:

$$ (1,6)(1,5)(2,7)(2,3)(2,4) $$

can be applied to each item to show it agrees with the product of disjoint cycles:

$$ (156)(2437) $$

For example, let's apply the product of transpositions to $1$. Going from right to left, the first three transpositions leave $1$ fixed as it doesn't appear yet. Then $(1,5)$ sends $1$ to $5$, and since $5$ is not "moved" by the last transposition, that's how $1$ is mapped by the permutation as a whole.

Similarly in the product of disjoint cycles, $1$ is not affected by the first (rightmost) cycle, and in the end we see $1$ goes to $5$ by the second cycle $(156)$.

So these agree as functions on the mapping of $1$ to $5$. Showing the equality amounts to showing they agree on all inputs (elements of the domain).

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