# Even and odd permutations

In mathematics, the permutations of a finite set (i.e. the bijective mappings from the set to itself) fall into two classes of equal size: the even permutations and the odd permutations. The parity (oddness or evenness) of a permutation $sigma$ of order $n$ is defined as the parity of the number of inversions, i.e., such pairs of indices $\left(i,j\right)$ that

The sign or signature of a permutation $sigma$ is denoted sgn(&sigma;) and defined as +1 if $sigma$ is even and −1 if $sigma$ is odd. It can be explicitly expressed as:$sgn\left(sigma\right)=\left(-1\right)^\left\{N\left(sigma\right)\right\}$ where $N\left(sigma\right)$ is the number of inversions in $sigma$. Another notation for the sign of a permutation is the Levi-Civita symbol ($epsilon_\left\{ijk\right\} ,$ e.g.), which is defined to be zero for non-bijective maps from the finite set to itself.

Example

Consider the permutation &sigma; of the set {1,2,3,4,5} which turns the initial arrangement 12345 into 34521.It can be obtained by three transpositions: first exchange the places of 1 and 3, then exchange 2 and 4, and finally exchange 1 and 5. This shows that the given permutation &sigma; is odd. Using the notation explained in the permutation article, we can writeThere are (infinitely) many other ways of writing &sigma; as a composition of transpositions, for instance:$sigma=\left(2 3\right) \left(1 2\right) \left(2 4\right) \left(3 5\right) \left(4 5\right);$,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 from the identity permutation by an even number of exchanges (called transpositions) of two elements, while an odd permutation can be obtained by an odd number of transpositions.

* the composition of two even permutations is even
* the composition of two odd permutations is even
* the composition of an odd and an even permutation is oddFrom these it follows that
* the inverse of every even permutation is even
* the inverse of every odd permutation is odd

Considering the symmetric group "Sn" of all permutations of the set {1,...,"n"}, we can conclude that the
$operatorname\left\{sgn\right\} : S_n o \left\{-1,1\right\}$that assigns to every permutation its signature is a group homomorphism.

Furthermore, we see that the even permutations form a subgroup of "Sn". This is the alternating group on "n" letters, denoted by "An". 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 "An" (in "Sn").

If "n">1, then there are just as many even permutations in "Sn" as there are odd ones; consequently, "An" contains "n"!/2 permutations. [The reason: if &sigma; is even, then (12)&sigma; is odd; if &sigma; is odd, then (12)&sigma; is even; the two maps are inverse to each other.]

A cycle is even if and only if its length is odd. This follows from formulas like:("a" "b" "c" "d" "e") = ("a" "e") ("b" "e") ("c" "e") ("d" "e")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.

Every permutation of odd order must be even; the converse is not true in general.

Proof of connection between the sign of a permutation and the number of transpositions

Every permutation can be produced by a sequence of transpositions: with the first transposition we put the first element of the permutation in its proper place, the second transposition puts the second element right etc. Given a permutation &sigma;, we can write it as a product of transpositions in many different ways. We want to show that either all of those decompositions have an even number of transpositions, or all have an odd number.

Suppose we have two such decompositions: :&sigma; = T'1 T'2 ... T'k' :&sigma; = Q'1 Q'2 .... Q'm'. We want to show that k' and m' are either both even, or both odd.

Every transposition can be written as a product of an odd number of transpositions of adjacent elements, e.g.:(2 5) = (2 3)(3 4)(4 5)(4 3)(3 2)If we decompose in this way each of the transpositions T'1...T'k' and Q'1..Q'm' aboveinto an odd number of adjacent transpositions, we get the new decompositions::&sigma; = T1 T2 ... Tk :&sigma; = Q1 Q2 .... Qmwhere all of the T1...Tk Q1...Qk are adjacent, k-k' is even, and m-m' is even.

Now compose the inverse of T1 with &sigma;. T1 is the transposition ("i", "i"+1) of two adjacent numbers, so, compared to &sigma;, the new permutation &sigma;("i", "i"+1) will have exactly one inversion pair less (in case ("i","i"+1) was an inversion pair for &sigma;) or more (in case ("i", "i"+1) was not an inversion pair). Then apply the inverses of T2, T3, ... Tk in the same way, "unraveling" the permutation &sigma;. At the end we get the identity permutation, whose "N" is zero. This means that the original N(&sigma;) less k is even.

We can do the same thing with the other decomposition, Q1... Qm, and it will turn out that the original "N"(&sigma;) less m is even.

Therefore, m - k is even, as we wanted to show.

We can now define the transposition &sigma; to be even if "N"(&sigma;) is an even number, and odd if "N"(&sigma;) is odd. This coincides with the definition given earlier but it is now clear that every permutation is either even or odd.

An alternative proof uses the polynomial :

Since the polynomial $P\left(x_\left\{sigma\left(1\right)\right\},dots,x_\left\{sigma\left(n\right)\right\}\right)$ has the same factors as $P\left(x_1,dots,x_n\right)$ except for their signs, if follows that sgn(&sigma;) is either +1 or −1. Furthermore, if &sigma; and &tau; are two permutations, we see that:$operatorname\left\{sgn\right\}\left(sigma au\right) = frac\left\{P\left(x_\left\{sigma\left( au\left(1\right)\right)\right\},ldots,x_\left\{sigma\left( au\left(n\right)\right)\right\}\right)\right\}\left\{P\left(x_1,ldots,x_n\right)\right\}$::::$= frac\left\{P\left(x_\left\{sigma\left(1\right)\right\},ldots,x_\left\{sigma\left(n\right)\right\}\right)\right\}\left\{P\left(x_1,ldots,x_n\right)\right\} cdot frac\left\{P\left(x_\left\{sigma\left( au\left(1\right)\right)\right\},ldots, x_\left\{sigma\left( au\left(n\right)\right)\right\}\right)\right\}\left\{P\left(x_\left\{sigma\left(1\right)\right\},ldots,x_\left\{sigma\left(n\right)\right\}\right)\right\}$::::$= operatorname\left\{sgn\right\}\left(sigma\right)cdotoperatorname\left\{sgn\right\}\left( au\right)$Since with this definition it is furthermore clear that any transposition of two adjacent elements has signature -1, we do indeed recover the signature as defined earlier.

A third approach uses the presentation of the group "Sn" in terms of generators $au_1,dots, au_\left\{n-1\right\}$ and relations
* $au_i^2 = 1$ for all "i"
* $au_i au_\left\{i+1\right\} au_i = au_\left\{i+1\right\} au_i au_\left\{i+1\right\}$ for all "i" < "n"-1
* $au_i au_j = au_j au_i$ if |"i"-"j"| > 2. [Here the generator $au_i$ represents the transposition ("i", "i"+1).] 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 "Sn" represented by even-length words "even", and the elements represented by odd-length words "odd".

* The fifteen puzzle is a classic application, though it actually involves a groupoid.
* Zolotarev's lemma

References

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