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Inverse of Permutation Group
  • Last Updated : 22 Feb, 2021

Inverse of Permutation Group-: If the product of two permutations is the identical permutation then each of them is called inverse of each other.

For Example-: The permutations 

\begin{pmatrix} a1 &a2&a3&........&an\\ b1 & b2 & b3&........&bn \end{pmatrix} and \begin{pmatrix} b1 & b2 & b3&........&bn\\ a1 &a2&a3&........&an \end{pmatrix}

are inverse of each other since their product is \begin{pmatrix} a1&a2&a3&........&an\\ a1&a2&a3&........&an \end{pmatrix}

which is an identical permutation.



Example 1-: Find the inverse of permutation \begin{pmatrix} 1 & 2 & 3&4\\ 1 & 3 & 4&2 \end{pmatrix}

 Solution-: Let the inverse of permutation be  \begin{pmatrix} 1 & 2 & 3&4\\ a & b & c&d \end{pmatrix} \

where a, b, c and d are to be calculated.

Then According to definition of Inverse of Permutation

\begin{pmatrix} 1 & 2 & 3&4\\ 1 & 3 & 4&2 \end{pmatrix}\begin{pmatrix} 1 & 2 & 3&4\\ a & b & c&d \end{pmatrix}= \begin{pmatrix} 1 & 2 & 3&4\\ 1 & 2 & 3&4 \end{pmatrix}

or \begin{pmatrix} 1 & 2 & 3&4\\ a & c & d&b \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3&4\\ 1 & 2 & 3&4 \end{pmatrix}

∴ b=4 , c=2 , a=1 , d=3

∴ Required inverse is \begin{pmatrix} 1 & 2 & 3&4\\ 1 & 4 & 2&3 \end{pmatrix}



Example 2-: Calculate A-1 if A=\begin{pmatrix} 1 & 2 & 3&4&5\\ 2&3&1&5&4 \end{pmatrix}

Solution-: Let the inverse of A be \begin{pmatrix} 1 & 2 & 3&4&5\\ a&b&c&d&e \end{pmatrix}

where a, b, c, d and e are to be calculated.

Then According to definition of Inverse of Permutation

 \begin{pmatrix} 1 & 2 & 3&4&5\\ 2&3&1&5&4 \end{pmatrix}\begin{pmatrix} 1 & 2 & 3&4&5\\ a&b&c&d&e \end{pmatrix} =\begin{pmatrix} 1 & 2 & 3&4&5\\ 1&2&3&4&5 \end{pmatrix}

or \begin{pmatrix} 1 & 2 & 3&4&5\\ b&c&a&d&e \end{pmatrix}= \begin{pmatrix} 1 & 2 & 3&4&5\\ 1&2&3&4&5 \end{pmatrix}

∴ b=1 , c=2 , a=3 , e=4 , d=5

∴ We have A-1\begin{pmatrix} 1 & 2 & 3&4&5\\ 3&1&2&5&4 \end{pmatrix}

Example 3-:  If f=\begin {pmatrix} 1 & 2 & 3&4&5\\ 1 & 5& 3&2&4 \end{pmatrix} and \,\,\, g=\begin{pmatrix} 1 & 2 & 3&4&5\\ 2& 3& 1&5&4 \end{pmatrix}

then compute f-1o g-1.

Solution-: 

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

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

f-1o g-1= \begin{pmatrix} 1 & 2 & 3&4&5\\ 1 & 4 & 3&5&2 \end{pmatrix} o\begin{pmatrix} 1 & 2 & 3&4&5\\ 3 & 1 & 2&5&4 \end{pmatrix}

f-1o g-1=\begin{pmatrix} 1 & 2 & 3&4&5\\ 3 & 5 & 2&4&1 \end{pmatrix}

Example 4-: If P1=\begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix} , P2= \begin{pmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{pmatrix} ,P3=\begin{pmatrix} 1 & 2 & 3\\ 3 & 2 & 1 \end{pmatrix}

Find (P1 o P2)-1  and (P2 o P3)-1.

Solution-: P1 o P2= \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix}o\begin{pmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix}

                P2 o P3= \begin{pmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{pmatrix}o\begin{pmatrix} 1 & 2 & 3\\ 3 & 2 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3\\ 2 & 1 & 3 \end{pmatrix}

Also, we know that if P-1 be the inverse of permutation P, then P-1 o P = I .

∴ (P1 o P2)-1 = inverse of \begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix}  

∴ (P2 o P3)-1 = inverse of \begin{pmatrix} 1 & 2 & 3\\ 2 & 1 & 3 \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3\\ 2 & 1 & 3 \end{pmatrix}

Example 5-: Prove that (1  2  3  …….  n )-1 = ( n  n-1  n-3 …..  2  1)

Solution-: ( 1  2  3  …..  n)= \begin{pmatrix} 1 & 2 & 3&.......&n-1&n\\ 2 & 3 & 4&.......&n&1 \end{pmatrix}

                 =\begin{pmatrix} 1 & 2 & 3&.......&n-1&n\\ 2 & 3 & 4&.......&n&1 \end{pmatrix}o\begin{pmatrix} n & n-1 &.......&3&2&1\\ n-1 & n-2&.......&2&1&n \end{pmatrix}

                 =\begin{pmatrix} 1 & 2 & 3&.......&n-1&n\\ 2 & 3 & 4&.......&n&1 \end{pmatrix}o\begin{pmatrix} 2 & 3 &.......&n&n-1\\ 1 & 2 &.......&n-1&n \end{pmatrix}

                 =\begin{pmatrix} 1 & 2 &.......&n-1&n\\ 1 & 2 &.......&n-1&n \end{pmatrix} =I

Hence, (1  2  3  …….  n )-1 = ( n  n-1  n-3 …..  2  1)

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