600: For Group as Direct Sum of Finite Number of Normal Subgroups, Element Is Uniquely Decomposed and Decomposition Is Commutative

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for group as direct sum of finite number of normal subgroups, element is uniquely decomposed and decomposition is commutative

---

Topics

About: group

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

---

Starting Context

• The reader knows a definition of group as direct sum of finite number of normal subgroups.
• The reader admits the proposition that any group as direct sum of finite number of normal subgroups is the group as direct sum of any reordered and combined normal subgroups.
• The reader admits the proposition that for any group as direct sum of finite number of normal subgroups, the product of any subset of the normal subgroups is the group as direct sum of the subset.

---

Target Context

• The reader will have a description and a proof of 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.

---

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.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$G$: $\in \{\text{ the groups }\}$
$\{G_1,...,G_n\}$: $\subseteq \{\text{ the normal subgroups of }G\}$
//

Statements:
$G=\text{ the direct sum of }\{G_1,...,G_n\}$
$⟹$
$\mathrm{\forall }{p}_1...p_n=p_1^{\prime }...p_n^{\prime }\in G\text{ such that }{p}_j,p_j^{\prime }\in G_j(p_j=p_j^{\prime }\wedge \mathrm{\forall }\sigma \in \{\text{ the permutations of }\{1,...,n\}\}(p_1...p_n=p_{{\sigma }_1}...p_{{\sigma }_n}))$.
//

---

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 $\{G_1,...,G_n\}$, for each $p_1...p_n\in G$ such that $p_j\in G_j$, $p_1...p_n$ is the unique decomposition of the element, and for each permutation of $\{1,...,n\}$, $\sigma$, $p_1...p_n=p_{{\sigma }_1}...p_{{\sigma }_n}$.

---

3: Proof

Let us prove it inductively with respect to $n$.

Let $n=2$.

As $p_1{p}_2=p_1^{\prime }{p}_2^{\prime }$, ${p_1}^{-1}{p}_1{p}_2{p_2^{\prime }}^{-1}={p_1}^{-1}{p}_1^{\prime }{p}_2^{\prime }{p_2^{\prime }}^{-1}$, so, $p_2{p_2^{\prime }}^{-1}={p_1}^{-1}{p}_1^{\prime }$, but the left hand side is in $G_2$ and the right hand side is in $G_1$, so, the both side is in $G_1\cap G_2=\{1\}$, so, $p_2{p_2^{\prime }}^{-1}={p_1}^{-1}{p}_1^{\prime }=1$, which implies that $p_1^{\prime }=p_1$ and $p_2^{\prime }=p_2$.

For each $p_2{p}_1\in G$, $p_2{p}_1=p_1^{\prime }{p}_2^{\prime }$ for a $p_1^{\prime }\in G_1$ and a $p_2^{\prime }\in G_2$, because $G=G_1{G}_2$. ${p_2}^{-1}{p}_2{p}_1={p_2}^{-1}{p}_1^{\prime }{p}_2^{\prime }$, so, $p_1={p_2}^{-1}{p}_1^{\prime }{p}_2^{\prime }={p_2}^{-1}{p}_1^{\prime }{p}_2{p_2}^{-1}{p}_2^{\prime }=p_1^{″}{p_2}^{-1}{p}_2^{\prime }$ for a $p_1^{″}\in G_1$, because $G_1$ is a normal subgroup, but as the decomposition is unique, $p_{1}1=p_1^{″}{p_2}^{-1}{p}_2^{\prime }$ implies that ${p_2}^{-1}{p}_2^{\prime }=1$, so, $p_2^{\prime }=p_2$. Likewise, $p_2{p}_1{p_1}^{-1}=p_1^{\prime }{p}_2^{\prime }{p_1}^{-1}$, so, $p_2=p_1^{\prime }{p}_2^{\prime }{p_1}^{-1}=p_1^{\prime }{p_1}^{-1}{p}_1{p}_2^{\prime }{p_1}^{-1}=p_1^{\prime }{p_1}^{-1}{p}_2^{″}$ for a $p_2^{″}\in G_2$, but as the decomposition is unique, $1p_2=p_1^{\prime }{p_1}^{-1}{p}_2^{″}$ implies that $p_1^{\prime }{p_1}^{-1}=1$, so, $p_1^{\prime }=p_1$. So, $p_2{p}_1=p_1{p}_2$.

Let us suppose that the proposition holds through $n=n^{\prime }$.

Let us suppose that $G$ is the direct sum of $\{G_1,...,G_{n^{\prime }+1}\}$.

$G$ is the direct sum of $\{G_1,G_2...G_{n^{\prime }+1}\}$, by the proposition that any group as direct sum of finite number of normal subgroups is the group as direct sum of any reordered and combined normal subgroups.

For each $p=p_1...p_{n^{\prime }+1}=p_1^{\prime }...p_{n^{\prime }+1}^{\prime }$, $p=p_1(p_2...p_{n^{\prime }+1})=p_1^{\prime }(p_2^{\prime }...p_{n^{\prime }+1}^{\prime })$, where $p_2...p_{n^{\prime }+1},p_2^{\prime }...p_{n^{\prime }+1}^{\prime }\in G_2...G_{n^{\prime }+1}$, so, $p_1=p_1^{\prime }$ and $p_2...p_{n^{\prime }+1}=p_2^{\prime }...p_{n^{\prime }+1}^{\prime }$, by applying this proposition for $n=2$.

$G_2...G_{n^{\prime }+1}$ is the direct sum of $\{G_2,...,G_{n^{\prime }+1}\}$, by the proposition that for any group as direct sum of finite number of normal subgroups, the product of any subset of the normal subgroups is the group as direct sum of the subset. So, applying this proposition for $n=n^{\prime }$, $p_j=p_j^{\prime }$ for each $2\le j\le n^{\prime }+1$.

So, $p_j=p_j^{\prime }$ for each $1\le j\le n^{\prime }+1$.

For each $\sigma$, when ${\sigma }_1=1$, $p_1...p_{n^{\prime }+1}=p_{{\sigma }_1}...p_{{\sigma }_{n^{\prime }+1}}$ by applying this proposition for $n=n^{\prime }$; when ${\sigma }_1\ne 1$, ${\sigma }_1=k$ for a $2\le k\le n^{\prime }+1$, and $p_1...p_{n^{\prime }+1}=p_1(p_2...p_{n^{\prime }+1})=(p_2...p_{n^{\prime }+1})p_1=p_2(p_3...p_{n^{\prime }+1}{p}_1)=(p_3...p_{n^{\prime }+1}{p}_1)p_2=...=p_k{p}_{k+1}...p_{n^{\prime }+1}{p}_1...p_{k-1}=p_{{\sigma }_1}{p}_{k+1}...p_{n^{\prime }+1}{p}_1...p_{k-1}=p_{{\sigma }_1}...p_{{\sigma }_{n^{\prime }+1}}$, where the 'until (not including)'-the-last equals are by applying this proposition for $n=2$ and the last equal is by applying this proposition for $n=n^{\prime }$.

So, this proposition holds for each $2\le n$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>