math 150 modern algebra Find the number of even and odd per

math 150 ; modern algebra

Find the number of even and odd permutations in Sn for all n. Present the following permutation as a product of non-intersecting cycles for all k: (1 2 3)k =

Solution

A. If n > 1, 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: if is even, then (12) is odd; if is odd, then (12) is even; the two maps are inverse to each other.

B. (1 2 3) k