Parity of a permutation

In mathematics, when "X" is a finite set with at least two elements, the permutations of "X" (i.e. the bijective functions from "X" to "X") fall into two classes of equal size: the even permutations and the odd permutations. If any total ordering of "X" is fixed, the parity (oddness or evenness) of a permutation formula_1 of "X" can be defined as the parity of the number of inversions for "σ", i.e., of pairs of elements "x", "y" of "X" such that and .
The sign, signature, or signum of a permutation "σ" is denoted sgn("σ") and defined as +1 if "σ" is even and −1 if "σ" is odd. The signature defines the alternating character of the symmetric group S"n". Another notation for the sign of a permutation is given by the more general Levi-Civita symbol ("ε""σ"), which is defined for all maps from "X" to "X", and has value zero for non-bijective maps.
The sign of a permutation can be explicitly expressed as
where "N"("σ") is the number of inversions in "σ".
Alternatively, the sign of a permutation "σ" can be defined from its decomposition into the product of transpositions as
where "m" is the number of transpositions in the decomposition. Although such a decomposition is not unique, the parity of the number of transpositions in all decompositions is the same, implying that the sign of a permutation is well-defined.
Example.
Consider the permutation "σ" of the set which turns the initial arrangement 12345 into 34521.
It can be obtained by three transpositions: first exchange the numbers 2 and 4, then exchange 1 and 3, and finally exchange 1 and 5. This shows that the given permutation "σ" is odd. Following the method of the cycle notation article, this could be written, composing from left to right, as
There are many other ways of writing "σ" as a composition of transpositions, for instance
but it is impossible to write it as a product of an even number of transpositions.
Properties.
The identity permutation is an even permutation. An even permutation can be obtained as the composition of an even number and only an even number of exchanges (called transpositions) of two elements, while an odd permutation can be obtained by (only) an odd number of transpositions.
The following rules follow directly from the corresponding rules about addition of integers:
From these it follows that
Considering the symmetric group S"n" of all permutations of the set {1, ..., "n"}, we can conclude that the map
that assigns to every permutation its signature is a group homomorphism.
Furthermore, we see that the even permutations form a subgroup of S"n". This is the alternating group on "n" letters, denoted by A"n". It is the kernel of the homomorphism sgn. The odd permutations cannot form a subgroup, since the composite of two odd permutations is even, but they form a coset of A"n" (in S"n").
If , then there are just as many even permutations in S"n" as there are odd ones; consequently, A"n" contains "n"!/2 permutations. (The reason is that if "σ" is even then is odd, and if "σ" is odd then is even, and these two maps are inverse to each other.)
A cycle is even if and only if its length is odd. This follows from formulas like
In practice, in order to determine whether a given permutation is even or odd, one writes the permutation as a product of disjoint cycles. The permutation is odd if and only if this factorization contains an odd number of even-length cycles.
Another method for determining whether a given permutation is even or odd is to construct the corresponding permutation matrix and compute its determinant. The value of the determinant is the same as the parity of the permutation.
Every permutation of odd order must be even. The permutation in A4 shows that the converse is not true in general.
Equivalence of the two definitions.
This section presents proofs that the parity of a permutation "σ" can be defined in both ways:
Proof 1.
Let "σ" be a permutation on a ranked domain "S". Every permutation can be produced by a sequence of transpositions (2-element exchanges). Let the following be one such decomposition
We want to show that the parity of "k" is equal to the parity of the number of inversions of "σ".
Every transposition can be written as a product of an odd number of transpositions of adjacent elements, e.g.
Generally, we can write the transposition ("i" "i+d") on the set {1...,"i"...,"i+d"...} as the composition of 2"d"−1 adjacent transpositions by recursion on "d":
If we decompose in this way each of the transpositions "T"1 ... "T""k" above, we get the new decomposition:
where all of the "A"1..."Am" are adjacent. Also, the parity of "m" is the same as that of "k".
This is a fact: for all permutation "τ" and adjacent transposition "a," "aτ" either has one less or one more inversion than "τ". In other words, the parity of the number of inversions of a permutation is switched when composed with an adjacent transposition.
Therefore, the parity of the number of inversions of "σ" is precisely the parity of "m", which is also the parity of "k". This is what we set out to prove.
We can thus define the parity of "σ" to be that of its number of constituent transpositions in any decomposition. And this must agree with the parity of the number of inversions under any ordering, as seen above. Therefore, the definitions are indeed well-defined and equivalent.
Proof 2.
An alternative proof uses the Vandermonde polynomial
So for instance in the case , we have
Now for a given permutation "σ" of the numbers {1, ..., "n"}, we define
Since the polynomial formula_7 has the same factors as formula_8 except for their signs, it follows that sgn("σ") is either +1 or −1. Furthermore, if "σ" and "τ" are two permutations, we see that
Since with this definition it is furthermore clear that any transposition of two elements has signature −1, we do indeed recover the signature as defined earlier.
Proof 3.
A third approach uses the presentation of the group S"n" in terms of generators "τ"1, ..., "τ""n"−1 and relations
[Here the generator formula_13 represents the transposition .] All relations keep the length of a word the same or change it by two. Starting with an even-length word will thus always result in an even-length word after using the relations, and similarly for odd-length words. It is therefore unambiguous to call the elements of S"n" represented by even-length words "even", and the elements represented by odd-length words "odd".
Proof 4.
Recall that a pair "x", "y" such that and is called an inversion. We want to show that the count of inversions has the same parity as the count of 2-element swaps. To do that, we can show that every swap changes the parity of the count of inversions, no matter which two elements are being swapped and what permutation has already been applied. 
Suppose we want to swap the "i"th and the "j"th element. Clearly, inversions formed by "i" or "j" with an element outside of will not be affected. 
For the elements within the interval , assume "v""i" of them form inversions with "i" and "v""j" of them form inversions with "j". If "i" and "j" are swapped, those "v""i" inversions with "i" are gone, but inversions are formed. The count of inversions "i" gained is thus , which has the same parity as "n".
Similarly, the count of inversions "j" gained also has the same parity as "n". Therefore, the count of inversions gained by both combined has the same parity as 2"n" or 0. Now if we count the inversions gained (or lost) by swapping the "i"th and the "j"th element, we can see that this swap changes the parity of the count of inversions, since we also add (or subtract) 1 to the number of inversions gained (or lost) for the pair "(i,j)".
Note that initially when no swap is applied, the count of inversions is 0. Now we obtain equivalence of the two definitions of parity of a permutation.
Other definitions and proofs.
The parity of a permutation of formula_14 points is also encoded in its cycle structure.
Let "σ" = ("i"1 "i"2 ... "i""r"+1)("j"1 "j"2 ... "j""s"+1)...("ℓ"1 "ℓ"2 ... "ℓ""u"+1) be the unique decomposition of "σ" into disjoint cycles, which can be composed in any order because they commute. A cycle involving points can always be obtained by composing "k" transpositions (2-cycles):
so call "k" the "size" of the cycle, and observe that, under this definition, transpositions are cycles of size 1. From a decomposition into "m" disjoint cycles we can obtain a decomposition of "σ" into transpositions, where "k""i" is the size of the "i"th cycle. The number is called the discriminant of "σ", and can also be computed as
if we take care to include the fixed points of "σ" as 1-cycles.
Suppose a transposition ("a" "b") is applied after a permutation "σ". When "a" and "b" are in different cycles of "σ" then
and if "a" and "b" are in the same cycle of "σ" then
In either case, it can be seen that , so the parity of "N"(("a" "b")"σ") will be different from the parity of "N"("σ").
If is an arbitrary decomposition of a permutation "σ" into transpositions, by applying the "r" transpositions formula_19 after "t"2 after ... after "t""r" after the identity (whose "N" is zero) observe that "N"("σ") and "r" have the same parity. By defining the parity of "σ" as the parity of "N"("σ"), a permutation that has an even length decomposition is an even permutation and a permutation that has one odd length decomposition is an odd permutation.
Remarks:
Generalizations.
Parity can be generalized to Coxeter groups: one defines a length function ℓ("v"), which depends on a choice of generators (for the symmetric group, adjacent transpositions), and then the function gives a generalized sign map.