598: Normal Subgroup of Group Is Normal Subgroup of Subgroup of Group Multiplied by Normal Subgroup

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description/proof of that normal subgroup of group is normal subgroup of subgroup of group multiplied 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: Proof

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Starting Context

• The reader knows a definition of normal subgroup of group.
• The reader admits the proposition that for any group, any subgroup of the group multiplied by any normal subgroup of the group is a subgroup of the group.
• The reader admits the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it.

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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 of the group is a normal subgroup of any subgroup of the group multiplied by the 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:
$G$: $\in \{\text{ the groups }\}$
$G_1$: $\in \{\text{ the subgroups of }G\}$
$G_2$: $\in \{\text{ the normal subgroups of }G\}$
//

$G_2\in \{\text{ the normal subgroups of }{G}_1{G}_2\}$
$\wedge$
$G_2\in \{\text{ the normal subgroups of }{G}_2{G}_2\}$
//

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2: Natural Language Description

For any group, $G$, any subgroup, $G_1$, of $G$, and any normal subgroup, $G_2$, of $G$, $G_2$ is a normal subgroup of $G_1{G}_2$ and $G_2$ is a normal subgroup of $G_2{G}_1$.

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3: Proof

$G_1{G}_2\subseteq G$ and $G_2{G}_1\subseteq G$ are indeed some subgroups of $G$, by the proposition that for any group, any subgroup of the group multiplied by any normal subgroup of the group is a subgroup of the group.

1st let us think of being a normal subgroup of $G_1{G}_2$.

$G_2\subseteq G_1{G}_2$ is obvious. So, $G_2$ is a subgroup of $G_1{G}_2$.

For any $p\in G_1{G}_2$, $p{G}_2{p}^{-1}\subseteq G_2$, because $p\in G$. So, $G_2$ is a normal subgroup of $G_1{G}_2$, by the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it.

Let us think of being a normal subgroup of $G_2{G}_1$.

$G_2\subseteq G_2{G}_1$ is obvious. So, $G_2$ is a subgroup of $G_2{G}_1$.

For any $p\in G_2{G}_1$, $p{G}_2{p}^{-1}\subseteq G_2$, because $p\in G$. So, $G_2$ is a normal subgroup of $G_2{G}_1$, by the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it.

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References

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