An elegant proof that the pairity of a permutation is well-defined

Question

I want to explain to non-mathematicians a very nice proof that the 15-puzzle with 14 and 15 replaced is not solvable.

For that, one crucial argument is the fact that:

If we can write the same permutation as a product of transpositions in two different ways, then the pairity of the number of transpositions is the same.

I know a standrad induction proof to this statement, but it's quite boring, technical and can't be properly explained to "ordinary" people, as all other proofs I know.

I'm looking for a proof as elegant as one can find. Maybe a proof using coloring of some kind (it feels to me like there must be one), or visual graph theory, or binary arithmetic, etc.

I'm curious to hear your answers!

You can define it as the determinant of the permutation matrix, but I'm not sure this would speak to non-mathematicians... — Arnaud D., Jun 17 '17 at 11:50
@ArnaudD. But then you have to define the determinant without talking about pairity of permutations, which is possible but not a very fun approach — 35T41, Jun 17 '17 at 11:54
@MaudPieTheRocktorate Thanks, but this is exactly what I have so far. Note that they treat the pairity as something that is already defined, and in particular they assume what I'm trying to find a proof for — 35T41, Jun 17 '17 at 11:56

score 2 · Accepted Answer · answered Jun 17 '17 at 12:15

I took this from https://math.stackexchange.com/a/46476/256345

Basically, $\forall \sigma\in S_n$, write in two rows $$1, 2, \cdots, n$$ $$1, 2,\cdots, n$$ Now connect each $k$ in the top to $\sigma(k)$ in the bottom. Require that each curve intersection is simple, that is, no three curves intersect at one point, and no tangency, and finally, no self-intersection (which changes parity!)

Then define the parity of a permutation as the parity of intersections.

This is very geometric and easy to demonstrate. Now you only need to convince the audience that this is a well-defined parity, that is, it does not depend on how you draw the curves.

Step 1: Convince them that any two ways to draw the curves can be turned to each other through two of the three Reidemeister moves. This should be rather intuitive if you draw two diagrams, one with wavy wavy curves, another with straight curves, and ask them to mentally "taut" the wavy curves to the straight curves.

Step 2: Show each of the two Reidemeister moves does not change parity.

There is a "dynamical" interpretation of this crossing number, at least in the case where your curves admit no horizontal tangents. Take numbers 1, ..., n on the real line and start moving them continuously (actually, smoothly) in such a way that after one unit of time they come back to the original positions, but permuted according to sigma. Assume all collisions to be crossings between pairs of numbers. Then, it is clear that the permutation sigma can be obtained as a composition of transpositions, one for each crossing. — Jairo Bochi, Dec 01 '22 at 17:28

Christian Blatter · Answer 2 · 2017-06-20T17:28:34.367

The essential point is the following: Multiplying a permutation $\sigma\in {\cal S}_n$ with a transposition $(x,y)$ changes the parity of the number of cycles: The permutation $\sigma':=(x,y)\circ\sigma$ has one cycle more or one cycle less less than $\sigma$.

Proof. Assume that $x\ne y$ are elements of $[n]$, and ${\bf a}$, ${\bf b}$ are disjoint words (maybe empty) over the alphabet $[n]$ not containing $x$ or $y$. Then $$(x,y)\circ(x{\bf a}y{\bf b})= (x{\bf a})\circ(y{\bf b})\ .$$ Since $(x,y)^2={\rm id}$ we then also have $$(x,y)\circ\bigl((x{\bf a})\circ(y{\bf b})\bigr)=(x{\bf a}y{\bf b})\ .\qquad\square$$

An elegant proof that the pairity of a permutation is well-defined

2 Answers2