604: Group as Direct Sum of Finite Number of Normal Subgroups Is 'Groups - Homomorphisms' Isomorphic to Direct Product of Subgroups

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description/proof of that group as direct sum of finite number of normal subgroups is 'groups - homomorphisms' isomorphic to direct product of subgroups

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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
• 4: Proof

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Starting Context

• The reader knows a definition of group as direct sum of finite number of normal subgroups.
• The reader knows a definition of direct product of structures.
• The reader knows a definition of %category name% isomorphism.
• The reader admits the proposition that for any group as the direct sum of any finite number of normal subgroups, each element is uniquely decomposed and the decomposition is commutative.
• The reader admits the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that any group as the direct sum of any finite number of normal subgroups is 'groups - homomorphisms' isomorphic to the direct product of the subgroups.

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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:
$G$: $\in \{\text{ the groups }\}$
$\{G_1,...,G_n\}$: $\subseteq \{\text{ the normal subgroups of }G\}$
$f$: $:G\to G_1\times ...\times G_n,p_1...p_n\mapsto (p_1,...,p_n)$
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Statements:
$G=\text{ the direct sum of }\{G_1,...,G_n\}$
$⟹$
$G\cong G_1\times ...\times G_n$ by $f$, where $\cong$ denotes a 'groups - homomorphisms' isomorphism and $\times$ denotes the direct product.
//

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2: Natural Language Description

For any group, $G$, and any normal subgroups, $G_1,...,G_n$, of $G$, such that $G$ is the direct sum of $\{G_1,...,G_n\}$, $G$ is 'groups - homomorphisms' isomorphic to the direct product, $G_1\times ...\times G_n$.

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3: Note

$G$ is of course not the same with $G_1\times ...\times G_n$, because an element of $G$ is $p_1...p_n$, while any element of $G_1\times ...\times G_n$ has the form of $(p_1,...,p_n)$.

$G$ does not need to be Abelian, because the proposition that for any group as the direct sum of any finite number of normal subgroups, each element is uniquely decomposed and the decomposition is commutative does not require so.

But when $G$ is Abelian, the direct product is also called 'direct sum of modules': any Abelian group is a module.

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4: Proof

Each element of $G$ can be uniquely decomposed as $p_1...p_n$ for a $p_j\in G_j$, by the proposition that for any group as the direct sum of any finite number of normal subgroups, each element is uniquely decomposed and the decomposition is commutative. So, $f$ is well-defined.

$f$ is bijective, because for each $p_1...p_n\ne p_1^{\prime }...p_n^{\prime }\in G$, $p_j\ne p_j^{\prime }$ for a $j$, which implies that $(p_1,...,p_n)\ne (p_1^{\prime },...,p_n^{\prime })$; for each $(p_1,...,p_n)\in G_1\times ...\times G_n$, there is $p_1...p_n$, which is mapped to $(p_1,...,p_n)$.

Let us prove that $f$ is a group homomorphism. $f(1...1)=(1,...,1)$, which is the identity element of $G_1\times ...\times G_n$. These transformations profusely use the commutativity by the proposition that for any group as the direct sum of any finite number of normal subgroups, each element is uniquely decomposed and the decomposition is commutative without further mentioning it: move $p_1^{\prime }$ by $f(p_1...p_n{p}_1^{\prime }...p_n^{\prime })=f(p_1(p_2...p_n{p}_1^{\prime })p_2^{\prime }...p_n^{\prime })=f(p_1(p_1^{\prime }{p}_2...p_n)p_2^{\prime }...p_n^{\prime })$; then, move $p_2^{\prime }$ likewise as $=f(p_1{p}_1^{\prime }{p}_2{p}_2^{\prime }{p}_3...p_n{p}_3^{\prime }...p_n^{\prime })$; and so on, after all, $=f(p_1{p}_1^{\prime }...p_n{p}_n^{\prime })$. So, $=(p_1{p}_1^{\prime },...,p_n{p}_n^{\prime })=(p_1,...,p_n)(p_1^{\prime },...,p_n^{\prime })=f(p_1...p_n)f(p_1^{\prime }...p_n^{\prime })$. $f((p_1...p_n{)}^{-1})=f(p_n^{-1}...p_1^{-1})=f(p_1^{-1}...p_n^{-1})$ (by the proposition that for any group as the direct sum of any finite number of normal subgroups, each element is uniquely decomposed and the decomposition is commutative) $=(p_1^{-1},...,p_n^{-1})=f(p_1...p_n{)}^{-1}$. So, $f$ is a group homomorphism.

$f$ is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.

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References

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