How to prove product of even permutations is even How to prove product of even permutations is even? I've found if the even permutations are disjoint, then it's easy because an even number times another even number is even...so the product of two disjoint even permutations is even...but how about not disjoint even permutations? How to prove then?
Thanks in advance!
(An even permutation of $f$ of $\left \{ 1, 2,\cdots ,n \right \}$ is one with an even number of inversions, that is, pairs $(i,j)$ for which $i<j$ and $f(i)>f(j)$.)
 A: One nice trick is to consider the product
$$P(f)=\prod_{i<j} \frac{f(i)-f(j)}{i-j}$$
From your definition of even, it is relatively straightforward to see that
$$P(f)=1$$ when $f$ is even, and $P(f)=-1$ otherwise.
Now to prove your product property, compute $P(fg)$ given $P(f)=P(g)=1$.
A: Write your permutation as a product of transpositions. Then look at the permutation matrix representation of each transposition. Each one has determinant -1. So if you have an even number of transpositions(inversions), then the determinant is 1 and -1 otherwise. So product of any number of even permutations will have permutation matrix determinant 1 implying its even.
A: Each even permutation can b written as a product of even transposition and then multiply both even permitaion and get product of even permutation necause sum of two even numbers are even...
A: Let $p_1,...,p_m$ be even permutations. Let $P_1,...,P_m$ be the corresponding permutation matrices. Then $$\det(\prod_{i=1}^m P_i) = \prod_{i=1}^{m}\det(P_i) = \prod_{i=1}^{m}(1) = 1$$
Therefore, the product of permutations $\prod_{i=1}^{m} p_i$, which the product of permutation matrices $\prod_{i=1}^m P_i$ corresponds is even.
