589: Quotient Group of Group by Normal Subgroup

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definition of quotient group of group by normal subgroup

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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: Note

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Starting Context

• The reader knows a definition of normal subgroup of group.

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Target Context

• The reader will have a definition of quotient group of group by normal subgroup.

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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:
$G^{\prime }$: $\in \{\text{ the groups }\}$
$G$: $\in \{\text{ the normal subgroups of }{G}^{\prime }\}$
$\ast G^{\prime }/G$: $=\{[g^{\prime }]\subseteq G^{\prime }|g^{\prime }\in G^{\prime }\wedge \mathrm{\forall }{g}^{″}\in G^{\prime }(g^{″}\in g^{\prime }G⟺g^{″}\in [g^{\prime }])\}$ with the group operation
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Conditions:
$[g^{\prime }][g^{″}]=[g^{\prime }{g}^{″}]$
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2: Natural Language Description

For any group, $G^{\prime }$, and any normal subgroup, $G$, of $G^{\prime }$, $G^{\prime }/G:=\{[g^{\prime }]\subseteq G^{\prime }|g^{\prime }\in G^{\prime }\wedge \mathrm{\forall }{g}^{″}\in G^{\prime }(g^{″}\in g^{\prime }G⟺g^{″}\in [g^{\prime }])\}$ with the group operation, $[g^{\prime }][g^{″}]=[g^{\prime }{g}^{″}]$

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3: Note

$G^{\prime }/G$ is indeed a set of equivalent classes: each $g^{\prime }\in G^{\prime }$ belongs to $[g^{\prime }]$, because $g^{\prime }=g^{\prime }1\in g^{\prime }G$ while $1\in G$; if $g^{\prime }\in [g^{″}]$ (which implies that $g^{\prime }=g^{″}{g}_1$ for a $g_1\in G$), $[g^{″}]=[g^{\prime }]$, because for each $g^{‴}\in [g^{″}]$, $g^{‴}=g^{″}{g}_2$, where $g_2\in G$, $=g^{\prime }{g_1}^{-1}{g}_2\in g^{\prime }G$, and for each $g^{‴}\in [g^{\prime }]$, $g^{‴}=g^{\prime }{g}_2$, where $g_2\in G$, $=g^{″}{g}_1{g}_2\in g^{″}G$; so, each element of $G^{\prime }$ belongs to the unique class.

$[g^{\prime }][g^{″}]=[g^{\prime }{g}^{″}]$ is indeed well-defined: for each $g^{\prime }{g}_1\in [g^{\prime }]$ and $g^{″}{g}_2\in [g^{″}]$, $g^{\prime }{g}_1{g}^{″}{g}_2=g^{\prime }{g}^{″}{g}^{″-1}{g}_1{g}^{″}{g}_2$, but $g^{″-1}{g}_1{g}^{″}\in G$ because $G$ is a normal subgroup (in fact, for this, $G$ is required to be a normal subgroup), so, $g^{\prime }{g}_1{g}^{″}{g}_2\in [g^{\prime }{g}^{″}]$ and so $[g^{\prime }{g}_1{g}^{″}{g}_2]=[g^{\prime }{g}^{″}]$, which mean that the definition does not depend on the representations of the classes.

$G^{\prime }/G$ is indeed a group: $[1]$ is the identity element, because $[1][g^{\prime }]=[1g^{\prime }]=[g^{\prime }]=[g^{\prime }1]=[g^{\prime }][1]$; $[g^{\prime }][g^{″}]=[g^{\prime }{g}^{″}]\in G^{\prime }/G$; $[g^{\prime -1}][g^{\prime }]=[g^{\prime -1}{g}^{\prime }]=[1]=[g^{\prime }{g}^{\prime -1}]=[g^{\prime }][g^{\prime -1}]$; $([g^{\prime }][g^{″}])[g^{‴}]=[g^{\prime }{g}^{″}][g^{‴}]=[g^{\prime }{g}^{″}{g}^{‴}]=[g^{\prime }][g^{″}{g}^{‴}]=[g^{\prime }]([g^{″}][g^{‴}])$.

While this definition uses $g^{\prime }G$, using $G{g}^{\prime }$ instead does not make any difference, because $g^{\prime }G=G{g}^{\prime }$: for each $g^{″}\in g^{\prime }G$, $g^{″}=g^{\prime }{g}_1$ for a $g_1\in G$, but $=g^{\prime }{g}_1{g}^{\prime -1}{g}^{\prime }$ and $g^{\prime }{g}_1{g}^{\prime -1}\in G$ because $G$ is a normal subgroup, which implies that $g^{″}\in G{g}^{\prime }$; for each $g^{″}\in G{g}^{\prime }$, $g^{″}=g_1{g}^{\prime }$ for a $g_1\in G$, but $=g^{\prime }{g}^{\prime -1}{g}_1{g}^{\prime }$ and $g^{\prime -1}{g}_1{g}^{\prime }\in G$ because $G$ is a normal subgroup, which implies that $g^{″}\in g^{\prime }G$.

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References

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