1

How to prove that this permutation:
p=(1,10,9,7,6)(2,8,4)(3,5)
is odd.

Thanks much!

ABCD
  • 63

3 Answers3

2

Express the permutation as the product of transpositions, and count the number of transpositions:

  • a permutation is odd if and only if it can be written as the product of odd number of transpositions.
  • Likewise, for even permutations.

One way to express your permutation as the product of transpositions is as follows:

$$(1, 10, 9, 7, 6)(2, 8, 4)(3, 5) = \underbrace{(1, 10)(10, 9)(9, 7)(7, 6)(2, 8)(8, 4)(3, 5)}_{\large 7 \text{ transpositions}}$$

Hence the permutation is odd.

There are many ways to write a permutation as the product of transpositions; but the number of transpositions required to do so will always be even, or always be odd. So one can prove "odd-ness" (respectively "even-ness") by expressing the permutation as the product of transpositions, using your favorite method for doing so, and showing that the number of transpositions is odd (respectively even).

amWhy
  • 209,954
1

Note that a cycle of length $n$ can always be written as $nāˆ’1$ transpositions. So once you decompose the permutation into cycles, you have its parity as the parity of the number of even length cycles. You've got one even length cycle, so the permutation is odd.

vonbrand
  • 27,812
1

Just as small points, cause other answers covered it completely, you also have:

  • If $$\pi=(a_1~~a_2~~...~~a_r)\in S_n$$ be a permutation of length $r$, then we can always have $$\pi=(a_1~~a_2)(a_1~~a_3)(a_1~~a_4)...(a_1~~a_r)$$

  • Considering $\pi$ as a permutation; if the length of $\pi$ is an even number, it is an odd permutation and if the length of $\pi$ is an odd number so it is an even permutation.

Now by two facts above check your permutation again.

Mikasa
  • 67,374