description/proof of that for group, subgroup, and element of group, if \(k\) is 1st positive power to which element belongs to subgroup, multiples of \(k\) are only powers to which element belongs to subgroup
Topics
About: group
The table of contents of this article
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, any subgroup, and any element of the group, if \(k\) is the 1st positive power to which the element belongs to the subgroup, the multiples of \(k\) are the only powers to which the element belongs to the subgroup.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(g'\): \(\in G'\)
\(k\): \(\in \mathbb{N} \setminus \{0\}\) such that \(g'^k \in G \land \forall j \in \mathbb{N} \setminus \{0\} \text{ such that } j \lt k (g'^j \notin G)\)
//
Statements:
\(\forall l \in \mathbb{Z}, \forall j \in \mathbb{N} \text{ such that } 0 \le j \lt k (g'^{l k + j} \in G \iff j = 0)\)
//
2: Proof
Whole Strategy: Step 1: let \(m = l k + j\) and suppose that \(g'^m \in G\), and see that \(j = 0\); Step 2: let \(m = l k + j\) and suppose that \(j = 0\), and see that \(g'^m \in G\).
Step 1:
Let \(m = l k + j\).
Note that \(m\) s cover all the integers.
Let us suppose that \(g'^m \in G\).
\(g'^m = g'^{l k + j} = g'^{l k} g'^j = (g'^k)^l g'^j\). So, \(g'^j = (g'^k)^{- l} (g'^k)^l g'^j = (g'^k)^{- l} g'^m\). But as \(g'^k \in G\), \((g'^k)^{- l} \in G\), and also as \(g'^m \in G\), \(g'^j = (g'^k)^{- l} g'^m \in G\).
As \(0 \le j \lt k\), by the supposition of this proposition, \(j = 0\).
Step 2:
Let \(m = l k + j\).
Let us suppose that \(j = 0\).
\(g'^m = g'^{l k + j} = g'^{l k} = (g'^k)^l \in G\) as \(g'^k \in G\).
