description/proof of that for group and element, if there is positive natural number to power of which element is 1 and there is no smaller such, integers of which powers to which element are 1 are only multiples of number
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
- 4: Note
Starting Context
- The reader knows a definition of group.
Target Context
- The reader will have a description and a proof of the proposition that for any group and any element, if there is a positive natural number to power of which the element is 1 and there is no smaller such, the integers of which powers to which the element are 1 are only the multiples of the number.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(g\): \(\in G\)
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Statements:
\(\exists n \in \mathbb{N} \setminus \{0\} (g^n = 1 \land \lnot \exists n' \in \mathbb{N} \setminus \{0\} (n' \lt n \land g^{n'} = 1))\)
\(\implies\)
\(n z\) s where \(z \in \mathbb{Z}\) are the only integers such that \(g^{n z} = 1\)
//
2: Natural Language Description
For any group, \(G\), and any element, \(g \in G\), if there is an \(n \in \mathbb{N} \setminus \{0\}\) such that \(g^n = 1\) and there is no \(n' \in \mathbb{N} \setminus \{0\}\) such that \(n' \lt n \land g^{n'} = 1\), \(n z\) s where \(z \in \mathbb{Z}\) are the only integers such that \(g^{n z} = 1\).
3: Proof
Whole Strategy: Step 1: suppose that \(g^{n z + k} = 1\) where \(0 \le k \lt n\) and see that \(k = 0\).
Step 1:
Let us suppose that \(\exists n \in \mathbb{N} \setminus \{0\} (g^n = 1 \land \lnot \exists n' \in \mathbb{N} \setminus \{0\} (n' \lt n \land g^{n'} = 1))\).
Let us suppose that \(g^{n z + k} = 1\) where \(0 \le k \lt n\).
\(g^{n z + k} = g^{n z} g^k = (g^n)^z g^k = 1^z g^k = g^k = 1\).
By the supposition, \(k = 0\).
4: Note
As an immediate corollary, when \(g \neq 1\) and \(g^p = 1\) where \(p \in \{\text{ the prime numbers }\}\), there is no \(n \in \mathbb{N} \setminus \{0\}\) such that \(n \lt p \land g^n = 1\), because otherwise, there would be the smallest \(n' \in \mathbb{N} \setminus \{0\}\) such that \(n' \le n \land g^{n'} = 1\) (\(n'\) might be \(n\)), and \(n' z\) would be the only integers such that \(g^{n' z} = 1\), which contradicts \(g^p = 1\), because \(p\) could not be \(n' z\): \(n' \neq 1\).
