718: For Group, Normal Subgroup, and Quotient Group, Representatives Set Multiplied by Element Is Representatives Set

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description/proof of that for group, normal subgroup, and quotient group, representatives set multiplied by element is representatives set

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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

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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.

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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 the quotient group, any representatives set multiplied by any element is a representatives set.

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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:
$G^{\prime }$: $\in \{\text{ the groups }\}$
$G$: $\in \{\text{ the normal subgroups of }{G}^{\prime }\}$
$G^{\prime }/G$: $=\text{ the quotient group of }{G}^{\prime }\text{ by }G$
$f$: $:G^{\prime }/G\to G^{\prime }$, $\in \{\text{ the representatives maps for }{G}^{\prime }/G\}$
$\overline{G^{\prime }/G-f}$: $=\text{ the representatives set }$
$p$: $\in G^{\prime }$
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Statements:
$p\overline{G^{\prime }/G-f}\in \{\text{ the representatives sets }\}$
$\wedge$
$\overline{G^{\prime }/G-f}p\in \{\text{ the representatives sets }\}$
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2: Natural Language Description

For any group, $G^{\prime }$, any normal subgroup, $G\subseteq G^{\prime }$, the quotient group, $G^{\prime }/G$, any representatives map, $f:G^{\prime }/G\to G^{\prime }$, the corresponding representatives set, $\overline{G^{\prime }/G-f}$, and any element, $p\in G^{\prime }$, $p\overline{G^{\prime }/G-f}$ or $\overline{G^{\prime }/G-f}p$ is a representatives set.

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3: Proof

Whole Strategy: Step 1: express $\overline{G^{\prime }/G-f}$ as $\{p_{\alpha }|\alpha \in A\}$ and $G^{\prime }/G$ as $\{p_{\alpha }G|\alpha \in A\}$ and define $\{p{p}_{\alpha }G|\alpha \in A\}$ and $\{p_{\alpha }pG|\alpha \in A\}$; Step 2: see that for each element of $\{p_{\alpha }G|\alpha \in A\}$, there is a corresponding element in $\{p{p}_{\alpha }G|\alpha \in A\}$; Step 3: see that there is no duplication in $\{p{p}_{\alpha }G|\alpha \in A\}$; Step 4: see that for each element of $\{p_{\alpha }G|\alpha \in A\}$, there is a corresponding element in $\{p_{\alpha }pG|\alpha \in A\}$; Step 5: see that there is no duplication in $\{p_{\alpha }pG|\alpha \in A\}$.

Hereafter, we frequently use the fact that for each $p\in G^{\prime }$, $pG=Gp$, which is because the definition of normal subgroup implies that $pG{p}^{-1}=G$, which implies that $pG=pG{p}^{-1}p=Gp$.

Step 1:

Let us express $\overline{G^{\prime }/G-f}$ as $\{p_{\alpha }|\alpha \in A\}$ for a possibly uncountable index set, $A$.

$G^{\prime }/G=\{p_{\alpha }G|\alpha \in A\}$.

$p\overline{G^{\prime }/G-f}=\{p{p}_{\alpha }|\alpha \in A\}$ and $\overline{G^{\prime }/G-f}p=\{p_{\alpha }p|\alpha \in A\}$. 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 $\{p{p}_{\alpha }G|\alpha \in A\}$ and $\{p_{\alpha }pG|\alpha \in A\}$.

If $\{p{p}_{\alpha }G|\alpha \in A\}$ and $\{p_{\alpha }pG|\alpha \in A\}$ do not have any duplication and $G^{\prime }/G=\{p{p}_{\alpha }G|\alpha \in A\}=\{p_{\alpha }pG|\alpha \in A\}$, this proposition is true. So, let us prove it.

Step 2:

For each $p_{\beta }G\in \{p_{\alpha }G|\alpha \in A\}$, let us find a $p{p}_{\gamma }G\in \{p{p}_{\alpha }G|\alpha \in A\}$ such that $p_{\beta }G=p{p}_{\gamma }G$.

$p^{-1}{p}_{\beta }\in p_{\gamma }G$ for a $\gamma$. That means that $p^{-1}{p}_{\beta }=p_{\gamma }{p}^{\prime }$ for a $p^{\prime }\in G$. So, $p_{\gamma }=p^{-1}{p}_{\beta }{p^{\prime }}^{-1}$. Then, $p{p}_{\gamma }G=p{p}^{-1}{p}_{\beta }{p^{\prime }}^{-1}G=p_{\beta }{p^{\prime }}^{-1}G=p_{\beta }G$, because ${p^{\prime }}^{-1}\in G$.

That means that $G^{\prime }/G\subseteq \{p{p}_{\alpha }G|\alpha \in A\}$, and as $\{p{p}_{\alpha }G|\alpha \in A\}\subseteq G^{\prime }/G$ is obvious, $G^{\prime }/G=\{p{p}_{\alpha }G|\alpha \in A\}$.

Step 3:

Let us see that there is no duplication in $\{p{p}_{\alpha }G|\alpha \in A\}$.

Let us suppose that $\alpha \ne \beta$ and $p{p}_{\alpha }G=p{p}_{\beta }G$. That would mean that $p_{\alpha }\ne p_{\beta }$ and $p_{\alpha }G\ne p_{\beta }G$.

$p_{\alpha }G=p^{-1}p{p}_{\alpha }G=p^{-1}p{p}_{\beta }G=p_{\beta }G$, a contradiction.

Step 4:

For each $p_{\beta }G\in \{p_{\alpha }G|\alpha \in A\}$, let us find a $p_{\gamma }pG\in \{p_{\alpha }pG|\alpha \in A\}$ such that $p_{\beta }G=p_{\gamma }pG$.

$p_{\beta }{p}^{-1}\in p_{\gamma }G=G{p}_{\gamma }$ for a $\gamma$. That means that $p_{\beta }{p}^{-1}=p^{\prime }{p}_{\gamma }$ for a $p^{\prime }\in G$. So, $p_{\gamma }={p^{\prime }}^{-1}{p}_{\beta }{p}^{-1}$. Then, $p_{\gamma }pG={p^{\prime }}^{-1}{p}_{\beta }{p}^{-1}pG=({p^{\prime }}^{-1}{p}_{\beta })G=G({p^{\prime }}^{-1}{p}_{\beta })=(G{p^{\prime }}^{-1})p_{\beta }=G{p}_{\beta }$, because ${p^{\prime }}^{-1}\in G$,$=p_{\beta }G$.

Step 5:

Let us see that there is no duplication in $\{p_{\alpha }pG|\alpha \in A\}$.

Let us suppose that $\alpha \ne \beta$ and $p_{\alpha }pG=p_{\beta }pG$. That would mean that $p_{\alpha }\ne p_{\beta }$ and $p_{\alpha }G\ne p_{\beta }G$.

$p_{\alpha }G=G{p}_{\alpha }=G(p_{\alpha }p)p^{-1}=(p_{\alpha }pG)p^{-1}=(p_{\beta }pG)p^{-1}=((p_{\beta }p)G)p^{-1}=G{p}_{\beta }p{p}^{-1}=G{p}_{\beta }=p_{\beta }G$, a contradiction.

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References

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