# The equivalence of a set and its permutations.

1. Dec 7, 2015

### Odious Suspect

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. Dec 7, 2015

### andrewkirk

I don't think you're missing anything. It is, as you say, a trivial consequence.

3. Dec 7, 2015

### 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.

4. Dec 7, 2015

### Odious Suspect

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.