Skip to content
Related Articles

Related Articles

Improve Article

Permutation Groups and Multiplication of Permutation

  • Last Updated : 17 Feb, 2021

Permutation: Let G be a non-empty set, then a one-one onto mapping to itself that is 

is called a permutation.

  • The number of elements in finite set G is called the degree of Permutation.
  • Let G have n elements then Pn is called set of all permutation of degree n.
  • Pn  is also called the Symmetric group of degree n.
  • Pn is also denoted by Sn.
  • The number of elements in Pn or Sn is n!

Examples:

Case1: Let G={ 1 } element then permutation are Sn or Pn\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&2&5 \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&5&4 \end{pmatrix}

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

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.




My Personal Notes arrow_drop_up
Recommended Articles
Page :