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
Topics
About: group
The table of contents of this article
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.
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.
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:
\(\{n_1, n_2\}\): \(\subseteq \mathbb{N} \setminus \{0\}\)
\(\mathbb{Z}_{n_1 n_2}\): \(= \text{ the integers modulo } n_1 n_2 \text{ group }\)
\(\mathbb{Z}_{n_1}\): \(= \text{ the integers modulo } n_1 \text{ group }\)
\(\mathbb{Z}_{n_2}\): \(= \text{ the integers modulo } n_2 \text{ group }\)
\(\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2}\): \(= \text{ the direct product }\)
//
Statements:
\(gcd (n_1, n_2) = 1\)
\(\implies\)
\(\mathbb{Z}_{n_1 n_2} \cong_{groups} \mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2}\)
//
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 \(([j], [k]), ([j'], [k'])\), \(f (([j], [k]) + ([j'], [k'])) = f (([j + j'], [k + k'])) = [(j + j') n_2 + (k + k') n_1] = [j n_2 + k n_1 + j' n_2 + k' n_1] = [j n_2 + k n_1] + [j' n_2 + k' n_1] = f (([j], [k])) + f (([j'], [k']))\).
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 \(([j], [k]), ([j'], [k']) \in \mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2}\), let us suppose that \(f (([j], [k])) = f (([j'], [k']))\).
\([j n_2 + k n_1] = f (([j], [k])) = f (([j'], [k'])) = [j' n_2 + k' n_1]\), which means that \(j n_2 + k n_1 = j' n_2 + k' n_1 + l n_1 n_2\) for an \(l \in \mathbb{Z}\).
\((j - j') n_2 = (k' - k) n_1 + l n_1 n_2\).
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 \((k' - k) n_1 + l n_1 n_2\) is a multiple of \(n_1\), \((j - j') n_2\) is a multiple of \(n_1\), but as \(n_2\) has no common prime factor with \(n_1\), \(j - j'\) has to have all the prime factors of \(n_1\), which means that \(j - j'\) is a multiple of \(n_1\), which means that \([j] = [j']\). Likewise, \([k] = [k']\).
So, \(([j], [k]) = ([j'], [k'])\).
So, \(f\) is injective.
As \(\vert \mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \vert = n_1 n_2 = \vert \mathbb{Z}_{n_1 n_2} \vert\), injective \(f\) is surjective.
So, \(f\) is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
