716: For Group, Normal Subgroup, and Subgroup, Subsets of Quotient Group That Contain Cosets of Subgroup Are Same or Disjoint

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description/proof of that for group, normal subgroup, and subgroup, subsets of quotient group that contain cosets of subgroup are same or disjoint

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Topics

About: group

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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

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Starting Context

• The reader knows a definition of quotient group of group by normal subgroup.
• The reader admits the proposition that any proposition 1 or any proposition 2 if and only if if not the proposition 2, the proposition 1.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$G^{″}$: $\in \{\text{ the groups }\}$
$G$: $\in \{\text{ the normal subgroups of }{G}^{″}\}$
$G^{\prime }$: $\in \{\text{ the subgroups of }{G}^{″}\}$
$G^{″}/G$: $=\text{ the quotient group }$, $=\{g_{\gamma }^{″}G|\gamma \in C\}$
$g_{\alpha }^{″}{G}^{\prime }$: $=\text{ the left coset }$, $\alpha \in A$
$g_{\beta }^{″}{G}^{\prime }$: $=\text{ the left coset }$, $\beta \in A$
$S_{\alpha }$: $=\{g_{\gamma }^{″}G\in G^{″}/G|g_{\gamma }^{″}G\cap g_{\alpha }^{″}{G}^{\prime }\ne \mathrm{\varnothing }\}$
$S_{\beta }$: $=\{g_{\gamma }^{″}G\in G^{″}/G|g_{\gamma }^{″}G\cap g_{\beta }^{″}{G}^{\prime }\ne \mathrm{\varnothing }\}$
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Statements:
$S_{\alpha }=S_{\beta }\vee S_{\alpha }\cap S_{\beta }=\mathrm{\varnothing }$
$\wedge$
$S_{\alpha }\cong S_{\beta }$, 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 $G^{″}$, $G^{\prime }$, the quotient group, $G^{″}/G=\{g_{\gamma }^{″}G|\gamma \in C\}$, the left cosets, $g_{\alpha }^{″}{G}^{\prime }$ where $\alpha \in A$ and $g_{\beta }^{″}{G}^{\prime }$ where $\beta \in A$, and the subsets of $G^{″}/G$, $S_{\alpha }:=\{g_{\gamma }^{″}G\in G^{″}/G|g_{\gamma }^{″}G\cap g_{\alpha }^{″}{G}^{\prime }\ne \mathrm{\varnothing }\}$ and $S_{\beta }:=\{g_{\gamma }^{″}G\in G^{″}/G|g_{\gamma }^{″}G\cap g_{\beta }^{″}{G}^{\prime }\ne \mathrm{\varnothing }\}$, $S_{\alpha }=S_{\beta }$ or $S_{\alpha }\cap S_{\beta }=\mathrm{\varnothing }$. Furthermore, $S_{\alpha }\cong S_{\beta }$, where $\cong$ denotes being 'sets - maps' isomorphic: bijective.

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3: Proof

Whole Strategy: Step 1: take $S_1:=\{g_{\gamma }^{\prime }G\in G^{″}/G|g_{\gamma }^{\prime }G\cap 1G^{\prime }\ne \mathrm{\varnothing }\}$ with $g_{\gamma }^{\prime }\in G^{\prime }$; Step 2: see that $S_{\alpha }=g_{\alpha }^{″}{S}_1$ and $S_{\beta }=g_{\beta }^{″}{S}_1$; Step 3: see that if $S_{\alpha }\cap S_{\beta }\ne \mathrm{\varnothing }$, $S_{\alpha }=S_{\beta }$; Step 4: conclude the proposition.

Step 1:

For the left coset, $1G^{\prime }$, Let us take the corresponding $S_1:=\{g_{\gamma }^{″}G\in G^{″}/G|g_{\gamma }^{″}G\cap 1G^{\prime }\ne \mathrm{\varnothing }\}$.

Let us denote the set of such $\gamma$ s as $C^{\prime }\subseteq C$.

As an element of $G^{\prime }$, $g_{\gamma }^{\prime }$, is contained in $g_{\gamma }^{″}G$, $g_{\gamma }^{″}$ can be taken to be $g_{\gamma }^{\prime }$. So, we suppose that $g_{\gamma }^{\prime }\in G^{\prime }$ for each $\gamma \in C^{\prime }$ and $S_1=\{g_{\gamma }^{\prime }G|\gamma \in C^{\prime }\}$.

Step 2:

Let us see that $S_{\alpha }=g_{\alpha }^{″}{S}_1=\{g_{\alpha }^{″}{g}_{\gamma }^{\prime }G|\gamma \in C^{\prime }\}$.

As $G^{\prime }\subseteq \cup S_1$, $g_{\alpha }^{″}{G}^{\prime }\subseteq g_{\alpha }^{″}\cup S_1=g_{\alpha }^{″}{\cup }_{\gamma \in C^{\prime }}{g}_{\gamma }^{\prime }G={\cup }_{\gamma \in c^{\prime }}{g}_{\alpha }^{″}{g}_{\gamma }^{\prime }G={\cup }_{\gamma \in c^{\prime }}{g}_{\alpha }^{″}{S}_1$.

For each $g_{\alpha }^{″}{g}_{\gamma }^{\prime }G$, let us see that $g_{\alpha }^{″}{g}_{\gamma }^{\prime }G\cap g_{\alpha }^{″}{G}^{\prime }\ne \mathrm{\varnothing }$. $g_{\alpha }^{″}{g}_{\gamma }^{\prime }1=g_{\alpha }^{″}{g}_{\gamma }^{\prime }\in g_{\alpha }^{″}{g}_{\gamma }^{\prime }G\cap g_{\alpha }^{″}{G}^{\prime }$.

That means that $S_{\alpha }=g_{\alpha }^{″}{S}_1$.

The element of $g_{\alpha }^{″}{S}_1$ are distinct, because if $g_{\alpha }^{″}{g}_{\gamma }^{\prime }G=g_{\alpha }^{″}{g}_{{\gamma }^{\prime }}^{\prime }G$, $g_{\alpha }^{″}{g}_{\gamma }^{\prime }=g_{\alpha }^{″}{g}_{{\gamma }^{\prime }}^{\prime }g$ for a $g\in G$, so, $g_{\gamma }^{\prime }=g_{{\gamma }^{\prime }}^{\prime }g$, which means that $g_{\gamma }^{\prime }\in g_{{\gamma }^{\prime }}^{\prime }G$, which means that $\gamma ={\gamma }^{\prime }$.

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 $S_{\alpha }\cap S_{\beta }\ne \mathrm{\varnothing }$, $S_{\alpha }=S_{\beta }$.

$S_{\alpha }=\{g_{\alpha }^{″}{g}_{\gamma }^{\prime }G|\gamma \in C^{\prime }\}$ and $S_{\beta }=\{g_{\beta }^{″}{g}_{\gamma }^{\prime }G|\gamma \in C^{\prime }\}$.

Let us suppose that $S_{\alpha }\cap S_{\beta }\ne \mathrm{\varnothing }$.

$g_{\alpha }^{″}{g}_{\gamma }^{\prime }G=g_{\beta }^{″}{g}_{{\gamma }^{\prime }}^{\prime }G=G{g}_{\beta }^{″}{g}_{{\gamma }^{\prime }}^{\prime }$ (because $G$ is a normal subgroup: for each $g^{″}\in G^{″}$, $g^{″-1}G{g}^{″}=G$, so, $G{g}^{″}=g^{″}G$), which means that $g_{\alpha }^{″}{g}_{\gamma }^{\prime }=g{g}_{\beta }^{″}{g}_{{\gamma }^{\prime }}^{\prime }$ for a $g\in G$, so, $g_{\alpha }^{″}=g{g}_{\beta }^{″}{g}_{{\gamma }^{\prime }}^{\prime }{g_{\gamma }^{\prime }}^{-1}$. So, $g_{\alpha }^{″}{g}_{{\gamma }^{″}}^{\prime }G=G{g}_{\alpha }^{″}{g}_{{\gamma }^{″}}^{\prime }=Gg{g}_{\beta }^{″}{g}_{{\gamma }^{\prime }}^{\prime }{g_{\gamma }^{\prime }}^{-1}{g}_{{\gamma }^{″}}^{\prime }=G{g}_{\beta }^{″}{g}_{{\gamma }^{\prime }}^{\prime }{g_{\gamma }^{\prime }}^{-1}{g}_{{\gamma }^{″}}^{\prime }=g_{\beta }^{″}{g}_{{\gamma }^{\prime }}^{\prime }{g_{\gamma }^{\prime }}^{-1}{g}_{{\gamma }^{″}}^{\prime }G$, but $g_{{\gamma }^{\prime }}^{\prime }{g_{\gamma }^{\prime }}^{-1}{g}_{{\gamma }^{″}}^{\prime }\in G^{\prime }$, which means that $g_{{\gamma }^{\prime }}^{\prime }{g_{\gamma }^{\prime }}^{-1}{g}_{{\gamma }^{″}}^{\prime }G=g_{{\gamma }^{‴}}^{\prime }G$ for a ${\gamma }^{‴}\in C^{\prime }$. So, $g_{\alpha }^{″}{g}_{{\gamma }^{″}}^{\prime }G=g_{\beta }^{″}{g}_{{\gamma }^{‴}}^{\prime }G$, which means that $g_{\alpha }^{″}{g}_{{\gamma }^{″}}^{\prime }G\in S_{\beta }$, 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 $S_{\alpha }\cap S_{\beta }=\mathrm{\varnothing }$.

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4: Note

As a corollary, when the order of $G^{″}/G$ is finite, the order of $S_{\alpha }$ divides the order of $G^{″}/G$: $|S_{\alpha }||||G^{″}/G|$. That is because otherwise, the cosets of $G^{\prime }$ could not cover $G^{″}$: for example, when $G^{″}/G=\{1G,g_2^{\prime },g_3^{\prime }G,g_4^{″}G,g_5^{″}G\}$ and $S_1=\{1G,g_2^{\prime }G,g_3^{\prime }G\}$, $g_4^{″}G,g_5^{″}G$ could not be covered, because like $S_{\alpha }=\{1G,g_4^{\prime }G,g_5^{\prime }G\}$ could not happen, because $S_1\cap S_{\alpha }\ne \mathrm{\varnothing }$.

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References

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