Intuitive explanation of even/odd permutation

Question

Given a permutation it can be classified as either even or odd depending on whether it is expressible as a product of even or odd number of transpositions. Is there some geometrical or intuitive meaning for this? Secondly what is the significance of classifying permutations as even or odd?

Thanks

Update: I am looking for a motivation for the classification of permutations as even and odd when a student first encounters them. For example rotations in $D_n$ can be used to motivate $\mathbb{Z}_n$. Likewise I was looking for a general motivation for even/odd permutations. At an advanced level there is the motivation that $A_n$ is simple for $n\ge 5$, but is there any motivation for a novice?

Answer 1

Why "Odd" and "Even"?

The symmetric group $S_n$ maps homomorphically onto the cyclic group of order two $(\{0, 1\}, +_{\operatorname{mod}2})$. By homomorphically, I mean that the map preserves the group structure, in the sense that if we denote the map $f:S_n\rightarrow \{0, 1\}$ then $f(\sigma\tau)=f(\sigma)+_{\operatorname{mod}2}f(\tau)$. Odd permutations map to the non-trivial element $1$ in this map, while even permutations map to the trivial element $0$. This is the reason behind calling them odd and even, because we have the following operations:

• $odd+odd=even$, and $1+_{\operatorname{mod}2}1=0$.
• $even+odd=odd$, and $0+_{\operatorname{mod}2}1=1$.
• $odd+even=odd$, and $1+_{\operatorname{mod}2}0=1$.
• $even+even=even$, and $0+_{\operatorname{mod}2}0=0$.

Therefore, this homomorphism is mimicking addition of odd and even numbers. The even permutations are the kernel of this map (homomorphism), and this means that they form a subgroup (or rather, a normal subgroup, by the first isomorphism theorem), called the alternating group $A_n$.

Applications

For a cool, but, I will admit, slightly advanced application of parity, look up Theorem 11.2.1 of David Joyner's book Adventures in Group Theory. The theorem gives necessary and sufficient conditions for a certain kind of 4-tuple to represent an element of the group underlying the Rubik's cube.

Indeed, I would recommend looking up Joyner's book if you want examples of applications. The book is based around applying group theory to puzzles, and mostly these applications use symmetric group arguments. For example, Section 7.4 of the book proves (using the parity of the permutations) that the 15 puzzle has no solution (this is, of course, fretty's answer). I especially like Section 3.4, which explains a practical application of permutation groups to campanology.

Answer 2

My favourite application of even/odd permutations is in proving that the $15$ puzzle is impossible:

http://www.math.ubc.ca/~cass/courses/m308-02b/projects/grant/fifteen.html

Essentially it boils down to the fact that any sequence of moves getting the empty space back to its original position must define an even permutation...but the puzzle starts off defining an odd permutation.

Answer 3

If you consider orientation in Euclidean space to be intuitive, then the answer is affirmative: a permutation on $n$ symbols is even or odd according to whether it preserves or reverses orientation when acting on the standard basis of $n$-space.