A permutation is a rearrangement of an ordered set S such that each element is mapped uniquely to itself in a one-to-one correspondence. the permutations of a set X = 1, 2, . . . , n form a group under composition. This group is called the symmetric group Sn of degree n.
Let X be a finite set with at least two elements. The permutations of X can be classified into two equal-sized categories: Even permutation and Odd permutation.
Even Permutations
A permutation is called even if it can be expressed as a product of even number of transpositions.
In other words, a permutation is considered even if it can be obtained by performing an even number of swaps within a set. The sign of an even permutation is always +1.
For a set of n elements where n> 2, the total number of permutations is n!. Since permutations are evenly divided into even and odd types, the number of even permutations is \frac{n!}{2}.
For example, if n = 1, 2, 3, 4, 5,..., the number of even permutations is:
- n = 1 → 0 even permutations
- n = 2 → 1 even permutation
- n = 3 → 3 even permutations
- n = 4 → 12 even permutations
- n = 5 → 60 even permutations
and so on…
Example 1 : \begin{pmatrix} 1 & 2 & 3\\ \end{pmatrix}
=\begin{pmatrix} 1 & 2\\ \end{pmatrix} o \begin{pmatrix} 1 & 3\\ \end{pmatrix} = \begin{pmatrix} 1 & 3\\ \end{pmatrix} o \begin{pmatrix} 2 & 3\\ \end{pmatrix}
=\begin{pmatrix} 1 & 3\\ \end{pmatrix} o \begin{pmatrix} 1 & 2\\ \end{pmatrix} o \begin{pmatrix} 1 & 3\\ \end{pmatrix}o \begin{pmatrix} 1 & 2\\ \end{pmatrix}
Here we can see that the permutation ( 1 2 3 ) has been expressed as a product of transpositions in three ways and in each of them the number of transpositions is even, so it is an even permutation.
Example 2:
\begin{pmatrix} 1 & 2 & 3&4\\ 2 & 4 & 3&1 \end{pmatrix}
=\begin{pmatrix} 1 & 2 & 3&4\\ 2 & 4 & 3&1 \end{pmatrix} =\begin{pmatrix} 1 & 2 & 4\\ \end{pmatrix}o \begin{pmatrix} 3\\ \end{pmatrix}
=\begin{pmatrix} 1 & 2 & 4\\ \end{pmatrix}= \begin{pmatrix} 1&2\\ \end{pmatrix}o \begin{pmatrix} 1&4\\ \end{pmatrix}
The given permutation is the product of two transpositions so it is an even permutation.
Odd Permutations
A permutation is called odd if it can be expressed as a product of odd number of transpositions.
In other words, an odd permutation is a permutation that can be expressed as a product of an odd number of two-element swaps (transpositions). It is denoted by the permutation sign -1.
For a set of n elements where n>2, the total number of permutations is n!. Since permutations are equally divided into even and odd types, the number of odd permutations is \frac{n!}{2}.
For example, if n =1, 2, 3, 4, 5,… the number of odd permutations is:
- n = 1 → 0 odd permutations
- n = 2 → 1 odd permutation
- n = 3 → 3 odd permutations
- n = 4 → 12 odd permutations
- n = 5 → 60 odd permutations
and so on…
Example 1: \begin{pmatrix} 3 & 4& 5&6\\ \end{pmatrix}
=\begin{pmatrix} 3 & 4& 5&6\\ \end{pmatrix}= \begin{pmatrix} 3 & 4\\ \end{pmatrix}o \begin{pmatrix} 3 & 5\\ \end{pmatrix} o \begin{pmatrix} 3 & 6\\ \end{pmatrix}
= \begin{pmatrix} 3 & 4\\ \end{pmatrix}o \begin{pmatrix} 4 & 5\\ \end{pmatrix}o \begin{pmatrix} 5 & 6\\ \end{pmatrix}o \begin{pmatrix} 6 & 4\\ \end{pmatrix}o \begin{pmatrix} 3 & 5\\ \end{pmatrix}
Here we can see that the permutation ( 3 4 5 6 ) has been expressed as a product of transpositions in two ways and in each of them number of transpositions is odd, so it is an odd permutation.
Example 2: \begin{pmatrix} 1&2\\ \end{pmatrix}o \begin{pmatrix} 5&4&3\\ \end{pmatrix}o \begin{pmatrix} 6&7&8\\ \end{pmatrix}
=\begin{pmatrix} 1&2\\ \end{pmatrix}o \begin{pmatrix} 5&4\\ \end{pmatrix}o \begin{pmatrix} 5&3\\ \end{pmatrix} o\begin{pmatrix} 6&7\\ \end{pmatrix}o \begin{pmatrix} 6&8\\ \end{pmatrix}
The given permutation is the product of five transposes so it is an odd permutation.
Theorems on Even and Odd Permutations
Theorem 1
If P1 and P2 are permutations, then
- (a) P1 P2 is even provided P1 and P2 are either both even or both odd.
- (b) P1 P2 is odd provided one of P1 and P2 is odd and the other even.
Proof: (a)
Case I. If P1, P2 both are even.
Let P1 and P2 be the product of 2n and 2m transpositions respectively, where n and m are positive integers.
Then each of P1 P2 and P2 P1 is product of 2n + 2m transpositions, where 2n + 2m is evidently an even integer.
Hence, P1 P2 and P2 P1 are even permutations.
Case II. If P1, P2 , both are odd. Let P1 P2 be the product of (2n + 1) and (2m + 1) transpositions respectively, where n and m are positive integers.
Then each of P1 P2 and P2 P1 is the product (2n + 1) + (2m + 1) i.e., 2 (n + m + 1) transpositions, where 2(n + m + 1) is evidently an even integer.
Hence, P1 P2 and P2 P1 are even permutations.
Proof : (b)
Let P1 be an odd and P2 be an even permutation. Also let P1 and P2 be the product of (2n + 1) and 2m transpositions respectively, where n and m are positive integers.
Then each of P1 P2 and P2 P1 is the product of (2n + 1) + 2m i.e. [ 2 ( n+ m )+1] transpositions , where 2(n+ m) + 1 is evidently an odd integer.
Hence P1 P2 and P2 P1 are odd permutations.
Theorem 2
The Identity permutation is an even permutation.
Proof-: The identity permutation l can always be expressed as the product of two (i.e., even) transpositions.
Since the identity permutation does not swap any elements, it can be considered as the product of zero transpositions—which is an even number.
Example: The identity permutation I =(1) (2) (3)... (n) can be written as the product of zero transpositions, which is an even count.
Hence I is an even permutation.
Theorem 3
The inverse of an even permutation is an even permutation.
Proof-: If P be an even permutation and P-1 be its inverse, then PP-1= I, the identity permutation.
From Theorem 2, we know that the identity permutation is an even permutation. Since the product of two permutations retains the parity (even or odd), and P is already even, it follows that P−1 must also be even.
Thus, the inverse of an even permutation is always an even permutation.
Theorem 4
The inverse of an odd permutation is an odd permutation.
Proof-: If P be an odd permutation and P-1 be its inverse, then PP-1= I,
where I is the identity permutation, which is known to be an even permutation (from Theorem 2).
Since the product of two permutations retains their parity, we analyze:
For the equation PP-1= I to hold, P-1 must also be odd (since an odd permutation multiplied by another odd permutation results in an even permutation).
Thus, the inverse of an odd permutation is always an odd permutation.
Practice Questions on Even and Odd Permutations
1. Is the permutation σ=(1, 2, 3, 4) even or odd?
2. Show that the permutation (1, 3, 5, 4, 2) is an odd permutation by expressing it as a product of transpositions.
3. Consider the permutation (1, 2, 3, 4, 5). Is it an even or odd permutation?
4. Consider the permutation (1, 2)(3, 5)(4, 6). Is this an even or odd permutation?
5. Prove that the permutation (1, 3, 5, 4, 2) is odd by expressing it as a product of transpositions.
Can an even permutation become odd after a swap?
Yes, if you perform a transposition on an even permutation, it becomes odd. This happens because applying a transposition adds one more swap, changing the parity from even to odd.
What is the alternating group An?
The alternating group An is the group of all even permutations of n elements. It plays a crucial role in group theory and is a normal subgroup of the symmetric group Sn. The size of An is half the size of Sn.
Why is the parity of a permutation important?
The parity of a permutation is important in many mathematical areas, such as group theory, algebra, and combinatorics. It helps in understanding the structure of permutations, determining the solvability of certain problems (e.g., Rubik’s cube), and defining properties of groups like the alternating group An .
How do even and odd permutations relate to group theory?
In group theory, even and odd permutations are key in defining alternating groups, which help us study symmetries and structure in mathematical objects.
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