597: Finite Product of Normal Subgroups Is Commutative and Is Normal Subgroup

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description/proof of that finite product of normal subgroups is commutative and is 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, the product of any finite number of subgroups is associative.
• 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, the product of any finite number of normal subgroups is commutative and is a 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:
$G$: $\in \{\text{ the groups }\}$
$\{G_1,...,G_n\}$: $\subseteq \{\text{ the normal subgroups of }G\}$
$G_1...G_n$:
$G_{{\sigma }_1}...G_{{\sigma }_n}$: $\sigma$ is any permutation of $(1,...,n)$
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Statements:
$G_1...G_n=G_{{\sigma }_1}...G_{{\sigma }_n}$
$\wedge$
$G_1...G_n\in \{\text{ the normal subgroups of }G\}$
//

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2: Natural Language Description

For any group, $G$, and any normal subgroups, $G_1,...,G_n$, of $G$, $G_1...G_n=G_{{\sigma }_1}...G_{{\sigma }_n}$ where $\sigma$ is any permutation of $(1,...,n)$, and $G_1...G_n$ is a normal subgroup of $G$.

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3: Proof

Let us prove the commutativity inductively.

Let $n=2$. $G_1{G}_2=G_2{G}_1$? For each $g_1{g}_2\in G_1{G}_2$, $g_1{g}_2=g_1{g}_2{g_1}^{-1}{g}_1$, but $g_2^{\prime }:=g_1{g}_2{g_1}^{-1}\in G_2$, because $G_2$ is a normal subgroup; so, $g_1{g}_2=g_2^{\prime }{g}_1\in G_2{G}_1$. So, $G_1{G}_2\subseteq G_2{G}_1$. Likewise, $G_2{G}_1\subseteq G_1{G}_2$. So, $G_1{G}_2=G_2{G}_1$.

Let us suppose that the commutativity holds through $n=n^{\prime }$. $G_1...G_{n^{\prime }+1}=G_{{\sigma }_1}...G_{{\sigma }_{n^{\prime }+1}}$? $G_{{\sigma }_{n^{\prime }+1}}=G_k$. $G_1...\widehat{G_k}...G_{n^{\prime }+1}=G_{{\sigma }_1}...G_{{\sigma }_{n^{\prime }}}$ by the supposition. $G_1...\widehat{G_k}...G_{n^{\prime }+1}{G}_k=G_{{\sigma }_1}...G_{{\sigma }_{n^{\prime }}}{G}_{{\sigma }_{n^{\prime }+1}}$. As the multiplications are associative, by the proposition that for any group, the product of any finite number of subgroups is associative, $G_1...\widehat{G_k}...G_{n^{\prime }}{G}_{n^{\prime }+1}{G}_k=G_1...\widehat{G_k}...G_{n^{\prime }}(G_{n^{\prime }+1}{G}_k)=G_1...\widehat{G_k}...G_{n^{\prime }}(G_k{G}_{n^{\prime }+1})=G_1...\widehat{G_k}...(G_{n^{\prime }}{G}_k)G_{n^{\prime }+1}=G_1...\widehat{G_k}...(G_k{G}_{n^{\prime }})G_{n^{\prime }+1}=...=G_1...G_k...G_{n^{\prime }+1}$. So, $G_1...G_{n^{\prime }+1}=G_{{\sigma }_1}...G_{{\sigma }_{n^{\prime }+1}}$.

So, $G_1...G_n=G_{{\sigma }_1}...G_{{\sigma }_n}$ for each $2\le n$.

Let us prove that $G_1...G_n$ is a normal subgroup inductively.

Let $n=2$. Is $G_1{G}_2$ a normal subgroup? $G_1{G}_2$ is a subgroup 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. For each $g\in G$, $g{G}_1{G}_2{g}^{-1}=G_1{G}_2$? For each $g{g}_1{g}_2{g}^{-1}\in g{G}_1{G}_2{g}^{-1}$, $g{g}_1{g}^{-1}g{g}_2{g}^{-1}$, but $g{g}_1{g}^{-1}\in G_1$ and $g{g}_2{g}^{-1}\in G_2$, so, $g{g}_1{g}_2{g}^{-1}\in G_1{G}_2$. So, $g{G}_1{G}_2{g}^{-1}\subseteq G_1{G}_2$. So, $g{G}_1{G}_2{g}^{-1}=G_1{G}_2$, 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 suppose that the claim holds through $n=n^{\prime }$. $G_1...G_{n^{\prime }+1}=(G_1...G_{n^{\prime }})G_{n^{\prime }+1}$, but $(G_1...G_{n^{\prime }})$ is a normal subgroup by the supposition, and $(G_1...G_{n^{\prime }})G_{n^{\prime }+1}$ is a normal subgroup.

So, $G_1...G_n$ is a normal subgroup for each $2\le n$.

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References

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