Question

Express the following permutations as products of transpositions, and determine whether they are even or odd. (a) \(\left(\begin{array}{lllll}1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 2 & 1 & 5\end{array}\right)\), (b) \(\left(\begin{array}{llllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 4 & 1 & 7 & 8 & 3 & 6 & 5 & 2\end{array}\right)\), (c) \(\left(\begin{array}{llllll}1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 4 & 5 & 3 & 2 & 1\end{array}\right)\), (d) \(\left(\begin{array}{lllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 6 & 7 & 2 & 4 & 1 & 5 & 3\end{array}\right)\).

Short answer

The permutation (a) is odd and the permutations (b), (c), and (d) are even.

Step-by-step solution

Identify the swaps

For each permutation, identify the swaps needed to get from the top sequence to the bottom sequence. For example, for the permutation (1 2 3 4 5, 3 4 2 1 5), the swaps are: (1, 3), (2,4), (1,2).

Determine the parity

Count the number of swaps. If the number of swaps is even, the permutation is even. If the number of swaps is odd, the permutation is odd. For example, for the permutation (1 2 3 4 5, 3 4 2 1 5), there are 3 swaps, so the permutation is odd.

Apply Steps 1 and 2 to each permutation

(a) (1 2 3 4 5, 3 4 2 1 5) -> (1, 3), (2,4), (1,2) -> odd permutation (b) (1 2 3 4 5 6 7 8, 4 1 7 8 3 6 5 2) -> (1,4), (2,1), (3,7), (5,3), (6,5), (8,2) -> even permutation (c) (1 2 3 4 5 6, 6 4 5 3 2 1) -> (1,6), (2,4), (3,5), (2,3) -> even permutation (d) (1 2 3 4 5 6 7, 6 7 2 4 1 5 3) -> (1,6), (2,7), (4,1), (5,4) -> even permutation

Key concepts

Transpositions

A transposition in mathematics is the swapping of two elements in a sequence. In the context of permutations, which represent the rearrangement of a sequence into a new order, a transposition is a specific kind of permutation that only exchanges two elements.

For example, if we consider the permutation from the exercise \(a\), \(1 2 3 4 5, 3 4 2 1 5\), one of the transpositions required to achieve this rearrangement is swapping element 1 with element 3, denoted as \(1, 3\). Similarly, we need to swap element 2 with element 4, \(2, 4\), and then element 1 with element 2, \(1, 2\).

Transpositions are essential because any permutation can be broken down into a series of transpositions. This property is utilized in the step-by-step solution to express complex permutations as simpler, understandable actions. Identifying transpositions is the first step to determining the parity of a permutation.

Even and Odd Permutations

In permutation theory, each permutation is either even or odd based on the number of transpositions required to achieve it. An even permutation is one that can be created through an even number of transpositions, while an odd permutation requires an odd number of transpositions.

Referring to the exercise, we can ascertain the parity of each permutation by counting the number of transpositions identified in the first step. For instance, permutation \(a\) involves three transpositions; since three is an odd number, we classify this permutation as odd. Conversely, permutation \(b\) has six transpositions, making it an even permutation because six is an even number.

The concept of even and odd permutations is vital since it is linked to important properties in algebra, like the determinant of a matrix being positive or negative when rows are exchanged—an action related to permutations of the row indices.

Parity of Permutations

The term 'parity' in permutations refers to the classification of permutations as even or odd based on the number of transpositions needed to express the permutation. It is a fundamental aspect of permutation groups in mathematical group theory.

The parity of a permutation remains unchanged no matter how it is decomposed into transpositions, which is a property known as the 'parity invariance'. This invariance means that if one representation of a permutation is even, all other representations involving different transpositions will also be even, and similarly for odd permutations.

In the provided exercise, the step-by-step solution uses the concept of parity to determine the nature of each given permutation. Understanding parity helps students grasp more complex concepts in mathematics, such as the behavior of polynomial equations depending on the sign of the permutation of their roots.