description/proof of that for group, normal subgroup, and subgroup, subsets of quotient group that contain cosets of subgroup are same or disjoint
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
- 4: Note
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any group, any normal subgroup, and any subgroup, the subsets of the quotient group that contain any 2 cosets of the subgroup are same or disjoint, and all the such subsets are bijective.
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:
\(G''\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the normal subgroups of } G''\}\)
\(G'\): \(\in \{\text{ the subgroups of } G''\}\)
\(G'' / G\): \(= \text{ the quotient group }\), \(= \{g''_\gamma G \vert \gamma \in C\}\)
\(g''_\alpha G'\): \(= \text{ the left coset }\), \(\alpha \in A\)
\(g''_\beta G'\): \(= \text{ the left coset }\), \(\beta \in A\)
\(S_\alpha\): \(= \{g''_\gamma G \in G'' / G \vert g''_\gamma G \cap g''_\alpha G' \neq \emptyset\}\)
\(S_\beta\): \(= \{g''_\gamma G \in G'' / G \vert g''_\gamma G \cap g''_\beta G' \neq \emptyset\}\)
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Statements:
\(S_\alpha = S_\beta \lor S_\alpha \cap S_\beta = \emptyset\)
\(\land\)
\(S_\alpha \cong S_\beta\), where \(\cong\) denotes being 'sets - maps' isomorphic: bijective.
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2: Natural Language Description
For any group, \(G''\), any normal subgroup of \(G''\), \(G\), any subgroup of \(G''\), \(G'\), the quotient group, \(G'' / G = \{g''_\gamma G \vert \gamma \in C\}\), the left cosets, \(g''_\alpha G'\) where \(\alpha \in A\) and \(g''_\beta G'\) where \(\beta \in A\), and the subsets of \(G'' / G\), \(S_\alpha := \{g''_\gamma G \in G'' / G \vert g''_\gamma G \cap g''_\alpha G' \neq \emptyset\}\) and \(S_\beta := \{g''_\gamma G \in G'' / G \vert g''_\gamma G \cap g''_\beta G' \neq \emptyset\}\), \(S_\alpha = S_\beta\) or \(S_\alpha \cap S_\beta = \emptyset\). Furthermore, \(S_\alpha \cong S_\beta\), where \(\cong\) denotes being 'sets - maps' isomorphic: bijective.
3: Proof
Whole Strategy: Step 1: take \(S_1 := \{g'_\gamma G \in G'' / G \vert g'_\gamma G \cap 1 G' \neq \emptyset\}\) with \(g'_\gamma \in G'\); Step 2: see that \(S_\alpha = g''_\alpha S_1\) and \(S_\beta = g''_\beta S_1\); Step 3: see that if \(S_\alpha \cap S_\beta \neq \emptyset\), \(S_\alpha = S_\beta\); Step 4: conclude the proposition.
Step 1:
For the left coset, \(1 G'\), Let us take the corresponding \(S_1 := \{g''_\gamma G \in G'' / G \vert g''_\gamma G \cap 1 G' \neq \emptyset\}\).
Let us denote the set of such \(\gamma\) s as \(C' \subseteq C\).
As an element of \(G'\), \(g'_\gamma\), is contained in \(g''_\gamma G\), \(g''_\gamma\) can be taken to be \(g'_\gamma\). So, we suppose that \(g'_\gamma \in G'\) for each \(\gamma \in C'\) and \(S_1 = \{g'_\gamma G \vert \gamma \in C'\}\).
Step 2:
Let us see that \(S_\alpha = g''_\alpha S_1 = \{g''_\alpha g'_\gamma G \vert \gamma \in C'\}\).
As \(G' \subseteq \cup S_1\), \(g''_\alpha G' \subseteq g''_\alpha \cup S_1 = g''_\alpha \cup_{\gamma \in C'} g'_\gamma G = \cup_{\gamma \in c'} g''_\alpha g'_\gamma G = \cup_{\gamma \in c'} g''_\alpha S_1\).
For each \(g''_\alpha g'_\gamma G\), let us see that \(g''_\alpha g'_\gamma G \cap g''_\alpha G' \neq \emptyset\). \(g''_\alpha g'_\gamma 1 = g''_\alpha g'_\gamma \in g''_\alpha g'_\gamma G \cap g''_\alpha G'\).
That means that \(S_\alpha = g''_\alpha S_1\).
The element of \(g''_\alpha S_1\) are distinct, because if \(g''_\alpha g'_\gamma G = g''_\alpha g'_{\gamma'} G\), \(g''_\alpha g'_\gamma = g''_\alpha g'_{\gamma'} g\) for a \(g \in G\), so, \(g'_\gamma = g'_{\gamma'} g\), which means that \(g'_\gamma \in g'_{\gamma'} G\), which means that \(\gamma = \gamma'\).
Likewise, \(S_\beta = g''_\beta S_1\).
That implies that \(S_\alpha \cong S_1 \cong S_\beta\).
Step 3: let us see that if \(S_\alpha \cap S_\beta \neq \emptyset\), \(S_\alpha = S_\beta\).
\(S_\alpha = \{g''_\alpha g'_\gamma G \vert \gamma \in C'\}\) and \(S_\beta = \{g''_\beta g'_\gamma G \vert \gamma \in C'\}\).
Let us suppose that \(S_\alpha \cap S_\beta \neq \emptyset\).
\(g''_\alpha g'_\gamma G = g''_\beta g'_{\gamma'} G = G g''_\beta g'_{\gamma'}\) (because \(G\) is a normal subgroup: for each \(g'' \in G''\), \(g''^{-1} G g'' = G\), so, \(G g'' = g'' G\)), which means that \(g''_\alpha g'_\gamma = g g''_\beta g'_{\gamma'}\) for a \(g \in G\), so, \(g''_\alpha = g g''_\beta g'_{\gamma'} {g'_\gamma}^{-1}\). So, \(g''_\alpha g'_{\gamma''} G = G g''_\alpha g'_{\gamma''} = G g g''_\beta g'_{\gamma'} {g'_\gamma}^{-1} g'_{\gamma''} = G g''_\beta g'_{\gamma'} {g'_\gamma}^{-1} g'_{\gamma''} = g''_\beta g'_{\gamma'} {g'_\gamma}^{-1} g'_{\gamma''} G\), but \(g'_{\gamma'} {g'_\gamma}^{-1} g'_{\gamma''} \in G'\), which means that \(g'_{\gamma'} {g'_\gamma}^{-1} g'_{\gamma''} G = g'_{\gamma'''} G\) for a \(\gamma''' \in C'\). So, \(g''_\alpha g'_{\gamma''} G = g''_\beta g'_{\gamma'''} G\), which means that \(g''_\alpha g'_{\gamma''} G \in S_\beta\), so, \(S_\alpha \subseteq S_\beta\).
By the symmetry, \(S_\beta \subseteq S_\alpha\), and so, \(S_\alpha = S_\beta\).
Step 4:
By the proposition that any proposition 1 or any proposition 2 if and only if if not the proposition 2, the proposition 1, \(S_\alpha = S_\beta\) or \(S_\alpha \cap S_\beta = \emptyset\).
4: Note
As a corollary, when the order of \(G'' / G\) is finite, the order of \(S_\alpha\) divides the order of \(G'' / G\): \(\vert S_\alpha \vert \vert \vert \vert G'' / G \vert\). That is because otherwise, the cosets of \(G'\) could not cover \(G''\): for example, when \(G'' / G = \{1 G, g'_2 , g'_3 G, g''_4 G, g''_5 G\}\) and \(S_1 = \{1 G, g'_2 G, g'_3 G\}\), \(g''_4 G, g''_5 G\) could not be covered, because like \(S_\alpha = \{1 G, g'_4 G, g'_5 G\}\) could not happen, because \(S_1 \cap S_\alpha \neq \emptyset\).
