The elements of the alternating group, called even permutations, are the products of even numbers of identity map, an empty product of no transpositions, is an even permutation since zero is even; it is the identity element of the group.
22.
The elements of the alternating group, called even permutations, are the products of even numbers of identity map, an empty product of no transpositions, is an even permutation since zero is even; it is the identity element of the group.
23.
Note that the Vandermonde determinant is " alternating " in the entries, meaning that permuting the \ alpha _ i by an odd permutation changes the sign, while permuting them by an even permutation does not change the value of the determinant.
24.
Because this method generates permutations that alternate between being even and odd, it may easily be modified to generate only the even permutations or only the odd permutations : to generate the next permutation of the same parity from a given permutation, simply apply the same procedure twice.
25.
That is, the sum of all even permutations of ( " a ", ( " b ", " c " ) ) must be zero ( where the permutation is performed by leaving the parentheses fixed and interchanging letters an even number of times ).
26.
:: : If you've got the even ones, you can get the odd ones by picking any single odd permutation ( for example swapping the first and second element ) and applying it once to each even permutation in the list . talk ) 01 : 11, 25 July 2010 ( UTC)
27.
If > 1, then there are just as many even permutations in " S " as there are odd ones; consequently, " A " contains ! / 2 permutations . [ The reason : if ? is even, then is odd; if ? is odd, then is even; the two maps are inverse to each other .]
28.
Studied the analogue of the 15 puzzle on arbitrary finite connected and non-separable graphs . ( A graph is called separable if removing a vertex increases the number of components . ) He showed that, except for polygons, and one exceptional graph on 7 vertices, it is possible to obtain all permutations unless the graph is bipartite, in which case exactly the even permutations can be obtained.
29.
The identity holds provided that for any two vertices " A " and " B " of the graph, the number of odd Eulerian paths from " A " to " B " is the same as the number of even ones . ( Here a path is called odd or even depending on whether its edges taken in order give an odd or even permutation of the 2 " n " edges . ) Swan showed that this was the case provided the number of edges in the graph is at least 2 " n ", thus proving the Amitsur Levitzki theorem.
