description/proof of that for group and finite-order element, order power of element is \(1\) and subgroup generated by element consists of element to non-negative powers smaller than element order
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 order of element of group.
- The reader admits the proposition that for any group, the powers sequence of any element that returns back returns to the element.
Target Context
- The reader will have a description and a proof of the proposition that for any group and its any finite-order element, the order power of the element is \(1\) and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(g\): \(\in G\)
//
Statements:
\(\vert (g) \vert = n \in \mathbb{N} \setminus \{0\}\)
\(\implies\)
\(g^n = 1 \land (g) = \{1, ..., g^{n - 1}\}\)
//
2: Natural Language Description
For any group, \(G\), and any element, \(g \in G\), if the order of \(g\), \(\vert (g) \vert\), is a finite \(n \in \mathbb{N} \setminus \{0\}\), \(g^n = 1 \land (g) = \{1, ..., g^{n - 1}\}\).
3: Proof
Whole Strategy: Step 1: see that \((g) = \{1, g, g^{-1}, g^2, g^{-2}, ...\}\); Step 2: think of \((g, g^2, ...)\) and see that the sequence returns to \(g\); Step 3: see that \((g) = \{1, g, ..., g^{n - 1}\}\).
Step 1:
Let us suppose that \(\vert (g) \vert = n \in \mathbb{N} \setminus \{0\}\).
As \((g)\) consists of all the finite multiplications of \(g\) and \(g^{-1}\) by the definition of subgroup generated by subset of group, \((g) = \{1, g, g^{-1}, g^2, g^{-2}, ...\}\): whether \(k_j \in \mathbb{Z}\) is positive, 0, or negative, \(g^{k_1} ... g^{k_l} = g^{k_1 + ... + k_l}\).
\(\vert (g) \vert = n\) means that \(\{1, g, g^{-1}, g^2, g^{-2}, ...\}\) contains some duplications and has \(n\) distinct elements.
Step 2:
Let us think of \((g, g^2, ...)\).
There is an \(m \in \mathbb{N} \setminus \{0\}\) such that \(\{g, ..., g^m\}\) is distinct and \(g^{m + 1} \in \{g, ..., g^m\}\), because otherwise, \(\{g, g^2 ...\}\) would be distinct, which would mean that \(\vert (g) \vert\) was not finite.
In fact, \(g^{m + 1} = g\), by the proposition that for any group, the powers sequence of any element that returns back returns to the element.
That means that \(g^m = 1\), because \(g^m = g^{m + 1} g^{-1} = g g^{-1} = 1\).
Let us think of \(\{1 = g^0 = g^m, g = g^1, ..., g^{m - 1}\}\), which is distinct.
Step 3:
For each \(k \in \mathbb{N} \setminus \{0\}\) such that \(m \lt k\), \(k = m l + j\), where \(l \in \mathbb{N}\) such that \(0 \le l\) and \(j \in \mathbb{N}\) such that \(0 \le j \lt m\). So, \(g^k = g^{m l + j} = g^{m l} g^j = (g^m)^l g^j = 1^l g^j = 1 g^j = g^j\). So, \(g^k \in \{1, g, ..., g^{m - 1}\}\).
\(g^{-1} = g^{m - 1}\), because \(g g^{m - 1} = g^{m - 1} g = g^m = 1\). \(g^{-m} = 1\), because \(g^{-m} = (g^m)^{-1} = 1^{-1} = 1\).
For each \(k \in \mathbb{N} \setminus \{0\}\) such that \(0 \lt k\), \(k = m l + j\), where \(l \in \mathbb{N}\) such that \(0 \le l\) and \(j \in \mathbb{N}\) such that \(0 \le j \lt m\). So, \(g^{-k} = g^{- m l - j} = (g^{- m})^l g^{- j} = (g^{- m})^l 1 g^{- j} = (g^{- m})^l g^m g^{- j} = 1^l g^{m - j} = 1 g^{m - j} = g^{m - j}\). But \(0 \lt m - j \le m\). When \(m - j = m\), \(g^{-k} = g^m = 1\). So, \(g^{- k} \in \{1, g, ..., g^{m - 1}\}\).
So, \((g) = \{1, g, g^{-1}, g^2, g^{-2}, ...\} = \{1, g, ..., g^{m - 1}\}\).
As \(\vert (g) \vert = n\), in fact, \(m = n\).
So, \((g) = \{1, g, ..., g^{n - 1}\}\).
