Prove that a permutation and its inverse have the same type (cf. Exercise 28).

Name: Problem 33, Chapter 14: Introductory Combinatorics
Uploaded: 2023-09-15T18:11:07-08:00
Duration: 7 min 7 s
Channel: William Mead
Description: Prove that a permutation and its inverse have the same type (cf. Exercise 28).

Prove that a permutation and its inverse have the same type...

Key Concepts

Permutation

A permutation is a bijective function from a set to itself, rearranging the elements in a one-to-one correspondence. In group theory, permutations are the elements of the symmetric group and serve as the fundamental examples of group operations, providing a framework for studying symmetry and structure in mathematical objects.

Cycle Decomposition

Cycle decomposition is the process of expressing a permutation as a product of disjoint cycles, where each cycle indicates a closed sequence of mappings among selected elements. This technique simplifies the analysis of permutations by breaking down complex mappings into more manageable cyclic structures, and it reveals key properties like cycle lengths that are intrinsic to the permutation.

Inverse Permutation

The inverse of a permutation is the function that reverses the mapping of the original permutation, sending every output back to its original input. Within a cycle, the inverse is obtained by reversing the order of the elements. This concept is essential in group theory as it guarantees that every permutation has an inverse, making the symmetric group a proper group under composition.

Cycle Type

The cycle type of a permutation is a description of its structure that lists the lengths of the cycles in its cycle decomposition, disregarding the order of the cycles. This classification is important because it is invariant under conjugation in the symmetric group and determines many properties of the permutation, such as its order. In particular, proving that a permutation and its inverse share the same cycle type shows that they have the same overall structure, reinforcing the idea that the inverse operation preserves key combinatorial features.

Transcript

00:01 So we have a linear function f going from a to b we're claiming that it has an inverse if and only if it is both one to one and onto so first, let's assume that f has an inverse has an inverse has an inverse an inverse, call it f minus one such that f composed f inverse equals, let's see f inverse has to be from b to a let's make a better b here this is from b to a back to b.

00:41 So this is the identity on b and f inverse composed f is the identity on a now what you see here is that in order for this to be true f has to be a bijection on just as a set function, right? f is injective because if f of x equals f of y then we can apply f inverse to it right and we get x equals y.

01:17 There we go and we have y is some element of b and f inverse of y has to be some element of a in a and f of f inverse of y equals y of course in particular then y is in the image of f.

01:35 So f is onto so that part's relatively easy, right? we don't even need linearity if f has an inverse for a set with any type of structure vector space or otherwise then f has to be both one to one and onto something something forgetful functor the in the converse here is gonna be a little bit trickier let's say that f is one to one and onto then it has an inverse as a set function and define f inverse from b to a as as just a set function which is to say we know that f has an inverse function we don't know that f inverse is a linear function and that's what that's what we need to prove now we claim that f inverse is linear to this point first, we'll show that it commutes with addition let b1 and b2 be elements of b such that f of a1 is equal to b1 and f of a2 is equal to b2 right, so then f inverse of b1 plus b2 see, what is that? well we have to this we're gonna consider f of a1 plus a2 right.

03:10 This is equal to f of a1 plus f of a2 by the linearity of f which is equal to b1 plus b2 right, this is all true by the linearity of f.

03:30 Now if we apply f inverse to all of this which we can do right f inverse of f is equal to the identity, so this just becomes a1 plus a2 this is all now f inverse of we can skip over here b1 plus b2 right? we're just skipping from the middle part over to this last inequality and so f of b1 plus b2 is equal to simply a1 plus a2 but since f inverse is the in fact the inverse of f a2 has to equal to f inverse of b2 and a1 has to equal f inverse of a1 because if we apply f to both sides then it'll cancel out the f inverse and that was how we defined...

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