# Every even permutation is a product of one or more cycles of length 3

Here's the outline of the proof I propose:

Let $\sigma \in S_n$ be any even permutation. Then by definition, $\sigma$ can be written as a product of even number of transpositions in various ways. Write $\sigma$ as a product of transpositions in one such way. Now, pair the first transposition with the second transposition, and the third with the fourth and every $(2n-1)$th transposition with the $(2n)$th transposition. Perform the following operation in every pair; $(xz)(xy)=(xyz)$ where $y\ne z$ and $(xy)(ab)=(ayz)(xab)$ where $x\ne y\ne a\ne b$. Thus, we've shown every even permutation is a product of one or more cycles of length $3$.

Is my proof correct?

Are there any alternative and neater ways to prove this?

• You may want to choose a representation of $\sigma$ as a product of the minimal number of transpositions (i.e. can't have less factors), so that no adjacent transpositions would cancel. Otherwise, everything is fine. – Alexander Burstein Jan 11 '18 at 16:36
• Possible duplicate of math.stackexchange.com/questions/1868836/…. – lhf Jan 12 '18 at 0:09

## 1 Answer

Looks good. You could maybe spend a little more time enumerating cases (what happens if y = z? what if x = z?) but the cases you've glossed over are trivial.

I do not know of an alternate proof.