1

I am learning about permutations right now. This is the definition in my textbook and one that is also similarly on Wikipedia.

A permutation of a set $X$ is a bijection $p: X \rightarrow X$ on that set

Every definition of a permutation I have seen claims that permutations on a set X is bijective. I am trying to reason this out formally, but I think I am not doing it properly. For a permutation to be a bijection on a set $X$, the function must be one-to-one and onto.

Let us define a set with $n$ elements as $X_n = \{1,2,\dots n\}$. The permutations of $X_n$ total $n!$. Each element in $X_n$ will have $1!, 2!, \dots n!$ permutations. Now it looks pretty clear to me that this is one-to-one and onto, but how do I state this more formally to make the conclusion more obvious to others?

Evan Kim
  • 2,399
  • 5
    I think you mean $p : X \rightarrow X$ in your highlight? – Stuart Green Apr 07 '19 at 16:11
  • 2
    If you are struggling with this then ask yourself how you define a permutation. For example, what is a permutation on ${1,2,3,4,5}$ by your definition? – John Douma Apr 07 '19 at 16:15
  • It's not clear to me exactly what your question is, but https://math.stackexchange.com/questions/1399781/why-do-we-not-have-to-prove-definitions may be relevant. – Eric Wofsey Apr 07 '19 at 16:27
  • @JohnDouma Before reading the answers, I didn't really know. After reading the answers, I would say a permutation is the # of all the ways you can reorder the elements ${1,2,3,4,5}$, which would be 5! – Evan Kim Apr 07 '19 at 17:59
  • @EricWofsey I will take a look, it looks like an interesting post – Evan Kim Apr 07 '19 at 18:00
  • Be careful. Each of the $5!$ ways is a permutation, i.e. there are $5!$ permutations. The total number of ways to rearrange the set is not a single permutation. – John Douma Apr 07 '19 at 18:01

3 Answers3

2

With permutations we are counting all the ways to rearrange the elements into a set of $n$ elements. Clearly we use every element and the elements of the codomain are the same size, so naturally the function is bijective.

Derek Luna
  • 2,732
  • 8
  • 19
2

Image of Explanation

One way to visualize this concept is to think of the cartesian coordinate plane. If my function fails the horizontal line test, then it is not injective. If every y does not have a corresponding x input that maps to it, then it is not surjective.

Neel Sandell
  • 306
  • 1
  • 10
1

It follows from the definition of a permutation really. In basic terms, if you have a line of n objects, a permutation is a reordering of those objects. You are not extending the number of positions available or reducing the number of objects, so all positions in the new line are filled by the original objects. It is onto. Since we cannot have two objects occupy the same position in the line, it is also one to one. Hence it is bijective.

Chris Moorhead
  • 187