Skip to content
Related Articles

Related Articles

Even and Odd Permutations and their theorems
  • Last Updated : 18 Feb, 2021

Even Permutations:

A permutation is called even if it can be expressed as a product of even number of transpositions.

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 number of transpositions is even, so it is a 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 transposes so it is a even permutation.

Odd Permutations:

A permutation is called even if it can be expressed as a product of odd number of transpositions.



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 a 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 a 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 are both even. 

Let P1 and P2 be the product of 2n and 2m transpositions respectively, where n and m are positive integers. 

Then each of P, P2 and P2 P1 is product of 2n + 2m transpositions, where 2n + 2m is evidently an even integer. 

Hence, P1 P2 and P2P, are even permutations. 

Case II. If P1, P2 are both odd. Let Pi and 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 P2P, 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 P, be an odd and P2 be an even permutation. Also let P, and P2 be the product of (2n + 1) and 2 and transpositions respectively, where n and m are positive integers. 

Then each of P1 P2 and P2P1 is product of (2n + 1) + 2m i.e. [ 2 ( n+ m )+1] transpositions , where 2(n+ m) + 1 is evidently and 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. 

For example 

Hence I is an even permutation. (See definition) 

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.

But P and I are even               (See Theorem 2 above), 

so P-1 is also even                  (See Theorem  1 (a) above)

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, the identity permutation.

But P and I are odd               (See Theorem 2 above),

so P-1 is also odd.                  (See Theorem  1 (b) above)

My Personal Notes arrow_drop_up
Recommended Articles
Page :