definition of integers modulo natural number group
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of quotient set.
Target Context
- The reader will have a definition of integers modulo natural 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:
\( \mathbb{Z}\): \(= \text{ the integers set }\)
\( n\): \(\in \mathbb{N}\)
\( \sim\): \(\in \{\text{ the equivalence relations on } \mathbb{Z}\}\), such that \(\forall z_1, z_2 \in \mathbb{Z} (\exists l \in \mathbb{Z} (z_1 = z_2 + l n) \iff z_1 \sim z_2)\)
\(*\mathbb{Z} / n\): \(= \mathbb{Z} / \sim\) as the quotient set with the group operation specified below
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Conditions:
\(\forall [z_1], [z_2] \in \mathbb{Z} / n ([z_1] + [z_2] = [z_1 + z_2])\)
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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 \(z'_1, z'_2 \in \mathbb{Z}\) be such that \([z_1] = [z'_1]\) and \([z_2] = [z'_2]\). That means that \(z'_1 = z_1 + l_1 n\) and \(z'_2 = z_2 + l_2 n\). \([z'_1 + z'_2] = [z_1 + l_1 n + z_2 + l_2 n] = [z_1 + z_2 + (l_1 + l_2) n] = [z_1 + z_2]\).
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 \(n z_1, n z_2 \in n \mathbb{Z}\), \(n z_1 + n z_2 = n (z_1 + z_2) \in n \mathbb{Z}\), \(0 = n 0 \in n \mathbb{Z}\), and \(- n z_1 = n (- z_1) \in n \mathbb{Z}\), and is a normal subgroup, because \(\mathbb{Z}\) is Abelian.
