Permutation Groups and Multiplication of Permutation

Last Updated : 18 May, 2022

Let G be a non-empty set, then a one-one onto mapping to itself that is as shown below is called a permutation.

The number of elements in finite set G is called the degree of Permutation.
Let G have n elements then P_n is called a set of all permutations of degree n.
P_nis also called the Symmetric group of degree n.
P_n is also denoted by S_n.
The number of elements in P_n or S_ⁿ is $n!$

Examples:

Case1: Let G={ 1 } element then permutation are S_n or P_n = $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$

Case 2: Let G= { 1, 2 } elements then permutations are $\begin{pmatrix} 1 & 2\\ 1 &2 \end{pmatrix} , \begin{pmatrix} 1 & 2\\ 2 &1 \end{pmatrix}$

Case 3: Let G={ 1, 2, 3 } elements then permutation are 3!=6. These are,

$\begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix} , \begin{pmatrix} 1 & 2 & 3\\ 2& 1& 3 \end{pmatrix} , \begin{pmatrix} 1 & 2 & 3\\ 3 & 2 & 1 \end{pmatrix} , \begin{pmatrix} 1 & 2 & 3\\ 1 & 3 & 2 \end{pmatrix} , \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix} , \begin{pmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{pmatrix}$

Reading the Symbol of Permutation

Suppose that a permutation is $\begin{pmatrix} 1 & 2 & 3 &4 &5&6\\ 2&3&1&4&5&6 \end{pmatrix}$

First, we see that in a small bracket there are two rows written, these two rows have numbers. The smallest number is 1 and the largest number is 6.
Starting from the left side of the first row we read as an image of 1 is 2, an image of 1 is 2, an image of 2 is 3, an image of 3 is 1, an image of 4 is 4 (Self image=identical=identity), an image of 5 is 6 and image of 6 is 5.
The above thing can be also read as: Starting from the left side of the first row 1 goes to 2, 2goes to 3, 3goes to,4 goes to 4,5 goes to 6, and 6 goes to 5.

A cycle of length 2 is called a permutation.

Example:

1) $\begin{pmatrix} 4 & 5\\ \end{pmatrix}$

Length is 2, so it is a transposition.

2) $\begin{pmatrix} 1 & 2 & 3\\ \end{pmatrix}$

Length is three, so it is not a transposition.

Multiplication of Permutation

Problem: If $A= \begin{pmatrix} 1 & 2 & 3&4&5\\ 2&3&1&4&5 \end{pmatrix} and \,B= \begin{pmatrix} 1 & 2 & 3 &4&5\\ 1&3&4&5&2 \end{pmatrix}$

Find the product of permutation A.B and B.A

Solution: $A= \begin{pmatrix} 1 & 2 & 3&4&5\\ 2&3&1&4&5 \end{pmatrix} and \,B= \begin{pmatrix} 1 & 2 & 3 &4&5\\ 1&3&4&5&2 \end{pmatrix}$

$A.B= \begin{pmatrix} 1 & 2 & 3&4&5\\ 2&3&1&4&5 \end{pmatrix} . \begin{pmatrix} 1 & 2 & 3 &4&5\\ 1&3&4&5&2 \end{pmatrix}$

$= \begin{pmatrix} 1 & 2 & 3&4&5\\ &&&& \end{pmatrix}$

Here we can see that in first bracket 1 goes to 2 i.e. image of 1 is 2, and in second row 2 goes to 3 i.e. image of 2 is 3.

Hence, we will write 3 under 1 in the bracket shown below,

$= \begin{pmatrix} 1 & 2 & 3&4&5\\ 3&&&& \end{pmatrix}$

Do above step with all elements of first row, answer will be

$A.B= \begin{pmatrix} 1 & 2 & 3&4&5\\ 3&4&1&5&2 \end{pmatrix}$

Similarly,

$B.A= \begin{pmatrix} 1 & 2 & 3&4&5\\ 1&3&4&5&2 \end{pmatrix} . \begin{pmatrix} 1 & 2 & 3 &4&5\\ 2&3&1&4&5 \end{pmatrix}$

$= \begin{pmatrix} 1 & 2 & 3&4&5\\ 2&1&4&5&3 \end{pmatrix}$