Inverse of Permutation Group

Permutation groups are a fundamental concept in abstract algebra, particularly in the study of group theory. They describe the symmetries of objects and the ways elements of a set can be arranged.

What is a Permutation Group?

A permutation group is a set of permutations of a given set that satisfies the group properties such as closure, associativity, the existence of an identity element, and the existence of inverse elements.

• Closure: If you take two permutations from a permutation group and compose them (apply one after the other), the result is a permutation within the same group.
• Associativity: The composition of permutations is associative, meaning that the order in which you compose permutations does not matter.
• Identity Element: There is an identity permutations in the group that leaves every element unchanged when applied.
• Inverse Element: Every permutation in the group has inverse permutations such that when the permutations is composed with its inverse, the result is the identity permutation.

Definition of an Inverse Permutation

The inverse of a permutations σ is a permutation σ^-1 such that when σ is composed with σ^-1 and vice versa, the result is the identity permutation.

Properties of Inverse

• Uniqueness: Every permutations has a unique inverse in the permutation group. This uniqueness follows from the group axioms, which ensure that the inverse is well-defined.
• Order of a Permutation: The order of a permutation (the smallest positive integer k such that σ^k = id) is the same as the order of its inverse. That is, if σ has order k, then σ^-1 also has order k.
• Composition with the Inverse: Composing a permutations with its inverse (in either order) yields the identity permutation:

σ ∘ σ^-1 = σ^-1 ∘ σ = id

This is a fundamental property of group elements.

• Self-Inverse (Involutions): A permutation is called an involution (or self-invers) if applying it twice returns the original arrangement i.e. σ² = id. This means σ = σ^-1.

Inverse of Permutation Group

If the product of two permutations is identical, then each of them is called the inverse of the 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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