1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Alexander Burstein Jan 11 '18 at 16:36
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/1868836/…. $\endgroup$ – lhf Jan 12 '18 at 0:09
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.