description/proof of that for group, normal subgroup, and quotient group, representatives set multiplied by element is representatives set
Topics
About: group
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of quotient group of group by normal subgroup.
- The reader knows a definition of representatives set of quotient set.
Target Context
- The reader will have a description and a proof of the proposition that for any group, any normal subgroup, and the quotient group, any representatives set multiplied by any element is a representatives set.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the normal subgroups of } G'\}\)
\(G' / G\): \(= \text{ the quotient group of } G' \text{ by } G\)
\(f\): \(: G' / G \to G'\), \(\in \{\text{ the representatives maps for } G' / G\}\)
\(\overline{G' / G - f}\): \(= \text{ the representatives set }\)
\(p\): \(\in G'\)
//
Statements:
\(p \overline{G' / G - f} \in \{\text{ the representatives sets }\}\)
\(\land\)
\(\overline{G' / G - f} p \in \{\text{ the representatives sets }\}\)
//
2: Natural Language Description
For any group, \(G'\), any normal subgroup, \(G \subseteq G'\), the quotient group, \(G' / G\), any representatives map, \(f: G' / G \to G'\), the corresponding representatives set, \(\overline{G' / G - f}\), and any element, \(p \in G'\), \(p \overline{G' / G - f}\) or \(\overline{G' / G - f} p\) is a representatives set.
3: Proof
Whole Strategy: Step 1: express \(\overline{G' / G - f}\) as \(\{p_\alpha \vert \alpha \in A\}\) and \(G' / G\) as \(\{p_\alpha G \vert \alpha \in A\}\) and define \(\{p p_\alpha G \vert \alpha \in A\}\) and \(\{p_\alpha p G \vert \alpha \in A\}\); Step 2: see that for each element of \(\{p_\alpha G \vert \alpha \in A\}\), there is a corresponding element in \(\{p p_\alpha G \vert \alpha \in A\}\); Step 3: see that there is no duplication in \(\{p p_\alpha G \vert \alpha \in A\}\); Step 4: see that for each element of \(\{p_\alpha G \vert \alpha \in A\}\), there is a corresponding element in \(\{p_\alpha p G \vert \alpha \in A\}\); Step 5: see that there is no duplication in \(\{p_\alpha p G \vert \alpha \in A\}\).
Hereafter, we frequently use the fact that for each \(p \in G'\), \(p G = G p\), which is because the definition of normal subgroup implies that \(p G p^{-1} = G\), which implies that \(p G = p G p^{-1} p = G p\).
Step 1:
Let us express \(\overline{G' / G - f}\) as \(\{p_\alpha \vert \alpha \in A\}\) for a possibly uncountable index set, \(A\).
\(G' / G = \{p_\alpha G \vert \alpha \in A\}\).
\(p \overline{G' / G - f} = \{p p_\alpha \vert \alpha \in A\}\) and \(\overline{G' / G - f} p = \{p_\alpha p \vert \alpha \in A\}\). They do not have any duplication, because \(p p_\alpha = p p_\beta\) or \(p_\alpha p = p_\beta p\) implies that \(p_\alpha = p_\beta\), which implies that \(\alpha = \beta\).
Let us define \(\{p p_\alpha G \vert \alpha \in A\}\) and \(\{p_\alpha p G \vert \alpha \in A\}\).
If \(\{p p_\alpha G \vert \alpha \in A\}\) and \(\{p_\alpha p G \vert \alpha \in A\}\) do not have any duplication and \(G' / G = \{p p_\alpha G \vert \alpha \in A\} = \{p_\alpha p G \vert \alpha \in A\}\), this proposition is true. So, let us prove it.
Step 2:
For each \(p_\beta G \in \{p_\alpha G \vert \alpha \in A\}\), let us find a \(p p_\gamma G \in \{p p_\alpha G \vert \alpha \in A\}\) such that \(p_\beta G = p p_\gamma G\).
\(p^{-1} p_\beta \in p_\gamma G\) for a \(\gamma\). That means that \(p^{-1} p_\beta = p_\gamma p'\) for a \(p' \in G\). So, \(p_\gamma = p^{-1} p_\beta {p'}^{-1}\). Then, \(p p_\gamma G = p p^{-1} p_\beta {p'}^{-1} G = p_\beta {p'}^{-1} G = p_\beta G\), because \({p'}^{-1} \in G\).
That means that \(G' / G \subseteq \{p p_\alpha G \vert \alpha \in A\}\), and as \(\{p p_\alpha G \vert \alpha \in A\} \subseteq G' / G\) is obvious, \(G' / G = \{p p_\alpha G \vert \alpha \in A\}\).
Step 3:
Let us see that there is no duplication in \(\{p p_\alpha G \vert \alpha \in A\}\).
Let us suppose that \(\alpha \neq \beta\) and \(p p_\alpha G = p p_\beta G\). That would mean that \(p_\alpha \neq p_\beta\) and \(p_\alpha G \neq p_\beta G\).
\(p_\alpha G = p^{-1} p p_\alpha G = p^{-1} p p_\beta G = p_\beta G\), a contradiction.
Step 4:
For each \(p_\beta G \in \{p_\alpha G \vert \alpha \in A\}\), let us find a \(p_\gamma p G \in \{p_\alpha p G \vert \alpha \in A\}\) such that \(p_\beta G = p_\gamma p G\).
\(p_\beta p^{-1} \in p_\gamma G = G p_\gamma\) for a \(\gamma\). That means that \(p_\beta p^{-1} = p' p_\gamma\) for a \(p' \in G\). So, \(p_\gamma = {p'}^{-1} p_\beta p^{-1}\). Then, \(p_\gamma p G = {p'}^{-1} p_\beta p^{-1} p G = ({p'}^{-1} p_\beta) G = G ({p'}^{-1} p_\beta) = (G {p'}^{-1}) p_\beta = G p_\beta\), because \({p'}^{-1} \in G\),\(= p_\beta G\).
Step 5:
Let us see that there is no duplication in \(\{p_\alpha p G \vert \alpha \in A\}\).
Let us suppose that \(\alpha \neq \beta\) and \(p_\alpha p G = p_\beta p G\). That would mean that \(p_\alpha \neq p_\beta\) and \(p_\alpha G \neq p_\beta G\).
\(p_\alpha G = G p_\alpha = G (p_\alpha p) p^{-1} = (p_\alpha p G) p^{-1} = (p_\beta p G) p^{-1} = ((p_\beta p) G) p^{-1} = G p_\beta p p^{-1} = G p_\beta = p_\beta G\), a contradiction.
