869: Integers Modulo Natural Number Group

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definition of integers modulo natural number group

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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: Note

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Starting Context

• The reader knows a definition of group.
• The reader knows a definition of quotient set.

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Target Context

• The reader will have a definition of integers modulo natural number group.

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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:
$\mathbb{Z}$: $=\text{ the integers set }$
$n$: $\in \mathbb{N}$
$\sim$: $\in \{\text{ the equivalence relations on }\mathbb{Z}\}$, such that $\mathrm{\forall }{z}_1,z_2\in \mathbb{Z}(\mathrm{\exists }l\in \mathbb{Z}(z_1=z_2+ln)⟺z_1\sim z_2)$
$\ast \mathbb{Z}/n$: $=\mathbb{Z}/\sim$ as the quotient set with the group operation specified below
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Conditions:
$\mathrm{\forall }[z_1],[z_2]\in \mathbb{Z}/n([z_1]+[z_2]=[z_1+z_2])$
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2: Note

The operation is usually denoted as $+$ because it is based on $+$ on $\mathbb{Z}$.

Let us see that the operation is well-defined.

That is about that $[z_1+z_2]$ does not depend on the representatives, $z_1,z_2$.

Let $z_1^{\prime },z_2^{\prime }\in \mathbb{Z}$ be such that $[z_1]=[z_1^{\prime }]$ and $[z_2]=[z_2^{\prime }]$. That means that $z_1^{\prime }=z_1+l_{1}n$ and $z_2^{\prime }=z_2+l_{2}n$. $[z_1^{\prime }+z_2^{\prime }]=[z_1+l_{1}n+z_2+l_{2}n]=[z_1+z_2+(l_1+l_2)n]=[z_1+z_2]$.

Let us see that $\mathbb{Z}/n$ is indeed a Abelian group.

The operation is closed, because $[z_1+z_2]\in \mathbb{Z}/n$.

$[0]$ is the identity element: $[0]+[z]=[0+z]=[z]$ and $[z]+[0]=[z+0]=[z]$.

For each $[z]$, $[-z]$ is the inverse: $[z]+[-z]=[z-z]=[0]$ and $[-z]+[z]=[-z+z]=[0]$.

The operation is associative: for each $[z_1],[z_2],[z_3]\in \mathbb{Z}/n$, $([z_1]+[z_2])+[z_3]=[z_1+z_2]+[z_3]=[z_1+z_2+z_3]=[z_1]+[z_2+z_3]=[z_1]+([z_2]+[z_3])$.

So, $\mathbb{Z}/n$ is a group.

$\mathbb{Z}/n$ is Abelian: for each $[z_1],[z_2]\in \mathbb{Z}/n$, $[z_1]+[z_2]=[z_1+z_2]=[z_2+z_1]=[z_2]+[z_1]$.

Obviously, $\mathbb{Z}/n=\{[0],...,[n-1]\}$.

In fact, $\mathbb{Z}/n=\mathbb{Z}/(n\mathbb{Z})$, the quotient group by the normal subgroup: $n\mathbb{Z}$ is a subgroup: for each $n{z}_1,n{z}_2\in n\mathbb{Z}$, $n{z}_1+n{z}_2=n(z_1+z_2)\in n\mathbb{Z}$, $0=n0\in n\mathbb{Z}$, and $-n{z}_1=n(-z_1)\in n\mathbb{Z}$, and is a normal subgroup, because $\mathbb{Z}$ is Abelian.

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References

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