置换群的逆

排列组的逆-：如果两个排列的乘积是相同的排列，那么它们中的每一个都被称为彼此的逆。

例如-：排列

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

彼此相反，因为它们的乘积是$\begin{pmatrix} a1&a2&a3&........&an\\ a1&a2&a3&........&an \end{pmatrix}$

这是一个相同的排列。

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)