- #1
Odious Suspect
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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##."
"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##."