Even And Odd Permutations Article Index for
Even
Website Links For
Even
 

Information About

Even And Odd Permutations




The sign or '''signature ''' of a permutation, with the notation '''sgn(σ)''', is defined as +1 if the permutation is even and -1 if it is odd. Another notation for it is the Levi-Civita Symbol , which is also defined for non-bijective maps from the finite set to itself, with the value zero.


EXAMPLE

Consider the permutation σ 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 σ is odd. Using the notation explained in the Permutation article, we can write
\sigma=\begin{bmatrix}1&2&3&4&5\
3&4&5&2&1\end{bmatrix} = (1 5) (2 4) (1 3)
There are (infinitely) many other ways of writing σ as a Composition of transpositions, for instance
:\sigma=(2 3) (1 2) (2 4) (3 5) (4 5)\;,
but it is impossible to write it as a product of an even number of transpositions.


FACTS

The identity permutation is an even permutation since it can be written as (1 2)(1 2).

The following rules follow directly from the corresponding rules about addition of integers:
  • 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 odd

  • From 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 map
:\operatorname{sgn} : S_n o \{-1,1\}
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 odd ones; consequently, ''An'' contains ''n''! /2 permutations. reason: if σ is even, then (12)σ is odd; if σ is odd, then (12)σ 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'' ''b'') (''a'' ''c'') (''a'' ''d'') (''a'' ''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.


PROOFS THAT EVERY PERMUTATION IS EITHER EVEN OR ODD


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.

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 σ is a given permutation, we define an ''inversion pair'' for σ to be a pair of indices (''i'',''j'') such that ''i''<''j'' and σ(''i'')>σ(''j''). Let ''N''(σ) be the number of inversion pairs of σ. Now if we compose σ with the transposition (''i'', ''i''+1) of two adjacent numbers, then, compared to σ, the new permutation σ(''i'', ''i''+1) will have exactly one inversion pair less (in case (''i'',''i''+1) was an inversion pair for σ) or more (in case (''i'', ''i''+1) was not an inversion pair). So any product of an odd number of transpositions of adjacent elements will have an odd value of ''N'', and any product of an even number of transpositions of adjacent elements will have an even value of ''N''. We can now define σ to be even if ''N''(σ) is even, and odd if ''N''(σ) 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
:P(x_1,\ldots,x_n)=\prod_{i
So for instance in the case ''n'' = 3, we have
:P(x_1, x_2, x_3) = (x_1 - x_2)(x_2 - x_3)(x_1 - x_3)\;
Now for a given permutation σ of the numbers {1,...,''n''}, we define
:\operatorname{sgn}(\sigma)= rac{P(x_{\sigma(1)},\ldots,x_{\sigma(n)})}{P(x_1,\ldots,x_n)}
Since the polynomial ''P''(''x''σ(1),...,''x''σ(''n''))
has the same factors as P(''x''1,...,''x''''n'') except for their signs, if follows that
sgn(σ) is either +1 or −1. Furthermore, if σ and τ are two permutations, we see that
:\operatorname{sgn}(\sigma au) = rac{P(x_{\sigma( au(1))},\ldots,x_{\sigma( au(n))})}{P(x_1,\ldots,x_n)}
:::: = rac{P(x_{\sigma(1)},\ldots,x_{\sigma(n)})}{P(x_1,\ldots,x_n)} \cdot rac{P(x_{\sigma( au(1))},\ldots, x_{\sigma( au(n))})}{P(x_{\sigma(1)},\ldots,x_{\sigma(n)})}
:::: = \operatorname{sgn}(\sigma)\cdot\operatorname{sgn}( au)
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 τ1,...,τ''n''-1 and relations
  • τ''i''2 = 1   for all ''i''

  • τ''i''τ''i''+1τ''i'' = τ''i''+1τ''i''τ''i''+1   for all ''i'' < ''n''-1