description/proof of that for group, powers sequence of element that returns back returns to element
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
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, the powers sequence of any element that returns back returns to the element.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(p\): \(\in G\)
\((p, p^2, ...)\):
//
Statements:
\(\exists n \in \mathbb{N} \setminus \{0\} (\{p, ..., p^n\} \text{ is distinct } \land p^{n + 1} \in \{p, ..., p^n\})\)
\(\implies\)
\(p^{n + 1} = p\)
//
2: Natural Language Description
For any group, \(G\), any element, \(p \in G\), and the powers sequence, \((p, p^2, ...)\), if there is an \(n \in \mathbb{N} \setminus \{0\}\) such that \(\{p, ..., p^n\}\) is distinct and \(p^{n + 1} \in \{p, ..., p^n\}\), \(p^{n + 1} = p\).
3: Proof
Whole Strategy: Step 1: let \(p^{n + 1} = p^k\) where \(1 \le k \le n\) and see that \(k = 1\).
Step 1:
There is a \(k \in \mathbb{N}\) such that \(1 \le k \le n\) such that \(p^{n + 1} = p^k\).
\(0 \le n - k\), so, \(2 \le n - k + 2 \le n - 1 + 2 = n + 1\).
As \(p^{n + 1} = p^k\), \(p^{n - k + 2} = p^{n + 1 - (k - 1)} = p^{k - (k - 1)} = p\). As \(2 \le n - k + 2 \le n + 1\), \(n - k + 2 = n + 1\), because otherwise, \(\{p, ..., p^n\}\) would not be distinct. That implies that \(k = 1\).
