Inverse of Permutation Group

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)

Last Updated : 22 Feb, 2021

Inverse of Permutation Group

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