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The equivalence of a set and its permutations.

  1. Dec 7, 2015 #1
    The following is from an introduction to groups. It is not clear to me why the authors bothered to introduce the subset ##\mathfrak{Q}\subseteq \mathfrak{R}## and a subset ##\mathfrak{K}\subseteq \mathfrak{S}^{\mathfrak{R}}## into the discussion. (3) seems to follow trivially from the one-to-one and onto properties of ##\sigma##. Am I missing something here?

    "Let ##\mathfrak{R}## be a set, which we shall now call a space in order to distinguish it from other sets to be considered later; and correspondingly, its elements ##P,Q,R,\ldots## will be called points. Let ##\mathfrak{S}^{\mathfrak{R}}## be the set of permutations on ##\mathfrak{R}## : that is, the set of one-to-one mappings of ##\mathfrak{R}## onto itself. If ##\sigma \in \mathfrak{S}^{\mathfrak{R}}## we denote by ##P\sigma## the image of the point ##P\in \mathfrak{R}## under the mapping ##\sigma##. Then ##\sigma## has the following properties:


    $$\text{(1)} P\sigma \in \mathfrak{R} \text{ for all } P\in \mathfrak{R}$$

    $$\text{(2)} P_1\sigma =P_2\sigma \text{ implies } P_1=P_2 $$

    More generally, if for a subset ##\mathfrak{Q}\subseteq \mathfrak{R}## and a subset ##\mathfrak{K}\subseteq \mathfrak{S}^{\mathfrak{R}}## we denote by ##\mathfrak{Q}\mathfrak{R}## the set of elements ## P\sigma, P\in \mathfrak{Q}, \sigma \in \mathfrak{K}## then the fact that ##\sigma## is a mapping onto ##\mathfrak{R}## is equivalent to

    $$\text{(3)} \mathfrak{R}\sigma =\mathfrak{R}\text{.}$$"


    There is a footnote as follows: "No distinction is made here between an element and the set containing it as the sole member. Thus ##\sigma## in (3) in fact represents {##\sigma##}, the set consisting only of ##\sigma##."
     
  2. jcsd
  3. Dec 7, 2015 #2

    andrewkirk

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    I don't think you're missing anything. It is, as you say, a trivial consequence.
     
  4. Dec 7, 2015 #3

    fresh_42

    Staff: Mentor

    (1) is closure or into, (2) one-to-one or injective and (3) onto or surjective. I think the author only wanted to introduce his notations and summarize what "one-to-one onto" means.
     
  5. Dec 7, 2015 #4
    Indeed. Before introducing ##\mathfrak{Q}\mathfrak{R}##, I didn't have permission to write (3). I also now realize that the "space" being introduced is not the set forming the group being exhibited. That set is ##\mathfrak{S}^{\mathfrak{R}}##, and the operation is the product of permutations.

    My source is: https://mitpress.mit.edu/books/fundamentals-mathematics-0 Volume 1. IIRC, their definitions for some of the structures in modern algebra are non-standard. It's sure no page-turner.
     
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