938: For 2 Positive Natural Numbers Whose Greatest Common Divisor Is 1, Integers Modulo Multiplication of Numbers Group Is 'Groups - Homomorphisms' Isomorphic to Direct Product of Integers Modulo 1st Number Group and Integers Modulo 2nd Number Group

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description/proof of that for 2 positive natural numbers whose greatest common divisor is 1, integers modulo multiplication of numbers group is 'groups - homomorphisms' isomorphic to direct product of integers modulo 1st number group and integers modulo 2nd 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: Proof

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Starting Context

• The reader knows a definition of integers modulo natural number group.
• The reader knows a definition of direct product of structures.
• The reader admits the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
• 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 for any 2 positive natural numbers whose greatest common divisor is 1, the integers modulo the multiplication of the numbers group is 'groups - homomorphisms' isomorphic to the direct product of the integers modulo the 1st number group and the integers modulo the 2nd 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:
$\{n_1,n_2\}$: $\subseteq \mathbb{N}\setminus \{0\}$
${\mathbb{Z}}_{n_1{n}_2}$: $=\text{ the integers modulo }{n}_1{n}_2\text{ group }$
${\mathbb{Z}}_{n_1}$: $=\text{ the integers modulo }{n}_1\text{ group }$
${\mathbb{Z}}_{n_2}$: $=\text{ the integers modulo }{n}_2\text{ group }$
${\mathbb{Z}}_{n_1}\times {\mathbb{Z}}_{n_2}$: $=\text{ the direct product }$
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Statements:
$gcd(n_1,n_2)=1$
$⟹$
${\mathbb{Z}}_{n_1{n}_2}{\cong }_{groups}{\mathbb{Z}}_{n_1}\times {\mathbb{Z}}_{n_2}$
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2: Proof

Whole Strategy: Step 1: define a to-be-isomorphism, $f:{\mathbb{Z}}_{n_1}\times {\mathbb{Z}}_{n_2}\to {\mathbb{Z}}_{n_1{n}_2}$; Step 2: see that $f$ is indeed a 'groups - homomorphisms' isomorphism.

Step 1:

Let us define a to-be-isomorphism, $f:{\mathbb{Z}}_{n_1}\times {\mathbb{Z}}_{n_2}\to {\mathbb{Z}}_{n_1{n}_2}$.

To think of it, we have the only option: as $([0],[0])=([0],[n_2])$ needs to be mapped to $[0]$, in order for $f$ to be group homomorphic, $([0],[1])$ needs to be mapped to $[(n_1{n}_2)/n_2]=[n_1]$, and likewise, $([1],[0])$ needs to be mapped to $[(n_1{n}_2)/n_1]=[n_2]$.

Extending it linearly, $([j],[k])$ is mapped to $[j{n}_2+k{n}_1]$.

Let us see that it is indeed well-define.

For $([j],[k])=([j+n_1{l}_1],[k+n_2{l}_2])$, $[(j+n_1{l}_1)n_2+(k+n_2{l}_2)n_1]=[j{n}_2+k{n}_1+n_1{l}_1{n}_2+n_2{l}_2{n}_1]=[j{n}_2+k{n}_1+n_1{n}_2(l_1+l_2)]=[j{n}_2+k{n}_1]$, so, the mapping does not depend on the representatives.

Step 2:

Let us see that $f$ is indeed a 'groups - homomorphisms' isomorphism.

1st, let us see that $f$ is a group homomorphism.

For each $([j],[k]),([j^{\prime }],[k^{\prime }])$, $f(([j],[k])+([j^{\prime }],[k^{\prime }]))=f(([j+j^{\prime }],[k+k^{\prime }]))=[(j+j^{\prime })n_2+(k+k^{\prime })n_1]=[j{n}_2+k{n}_1+j^{\prime }{n}_2+k^{\prime }{n}_1]=[j{n}_2+k{n}_1]+[j^{\prime }{n}_2+k^{\prime }{n}_1]=f(([j],[k]))+f(([j^{\prime }],[k^{\prime }]))$.

So, $f$ is a group homomorphism, by the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.

Let us see that $f$ is injective.

For any $([j],[k]),([j^{\prime }],[k^{\prime }])\in {\mathbb{Z}}_{n_1}\times {\mathbb{Z}}_{n_2}$, let us suppose that $f(([j],[k]))=f(([j^{\prime }],[k^{\prime }]))$.

$[j{n}_2+k{n}_1]=f(([j],[k]))=f(([j^{\prime }],[k^{\prime }]))=[j^{\prime }{n}_2+k^{\prime }{n}_1]$, which means that $j{n}_2+k{n}_1=j^{\prime }{n}_2+k^{\prime }{n}_1+l{n}_1{n}_2$ for an $l\in \mathbb{Z}$.

$(j-j^{\prime })n_2=(k^{\prime }-k)n_1+l{n}_1{n}_2$.

Now, as $gcd(n_1,n_2)=1$, when $n_1$ and $n_2$ are factorized in prime numbers, $n_1$ and $n_2$ do not have any common prime number. As $(k^{\prime }-k)n_1+l{n}_1{n}_2$ is a multiple of $n_1$, $(j-j^{\prime })n_2$ is a multiple of $n_1$, but as $n_2$ has no common prime factor with $n_1$, $j-j^{\prime }$ has to have all the prime factors of $n_1$, which means that $j-j^{\prime }$ is a multiple of $n_1$, which means that $[j]=[j^{\prime }]$. Likewise, $[k]=[k^{\prime }]$.

So, $([j],[k])=([j^{\prime }],[k^{\prime }])$.

So, $f$ is injective.

As $|{\mathbb{Z}}_{n_1}\times {\mathbb{Z}}_{n_2}|=n_1{n}_2=|{\mathbb{Z}}_{n_1{n}_2}|$, injective $f$ is surjective.

So, $f$ 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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