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, subgroup generated by element consists of element to non-negative powers smaller than 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
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 subgroup generated by the element consists of the element to the non-negative powers smaller than 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:
\(G\): \(\in \{\text{ the groups }\}\)
\(g\): \(\in G\)
//
Statements:
\(\exists n \in \mathbb{N} \setminus \{0\} (g^n = 1 \land \lnot \exists n' \in \mathbb{N} \setminus \{0\} (n' \lt n \land g^{n'} = 1))\)
\(\implies\)
\((g) = \{1, g, ..., g^{n - 1}\}\)
//
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 \(n' \lt n \land g^{n'} = 1\), \((g) = \{1, g, ..., g^{n - 1}\}\).
3: Proof
Whole Strategy: Step 1: see that \((g) = \{1, g, g^{-1}, g^2, g^{-2}, ...\}\); Step 2: see that \(\{1, g, ..., g^{n - 1}\}\) is distinct; Step 3: see that \((g) = \{1, g, ..., g^{n - 1}\}\).
Step 1:
Let us suppose that \(\exists n \in \mathbb{N} \setminus \{0\} (g^n = 1 \land \lnot \exists n' \in \mathbb{N} \setminus \{0\} (n' \lt n \land g^{n'} = 1))\).
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, \(g^{k_1} ... g^{k_l} = g^{k_1 + ... + k_l}\).
Step 2:
Let us see that \(\{1, g, ..., g^{n - 1}\}\) is distinct.
Let us suppose that \(\{1, g, ..., g^{n - 1}\}\) was not distinct, which would mean that there were some \(k, l \in \mathbb{N}\) such that \(0 \le k \lt l \le n - 1\) and \(g^k = g^l\).
\(1 = g^k g^{- k} = g^l g^{- k} = g^{l - k}\), where \(0 \lt l -k \le n - 1\), a contradiction.
So, \(\{1, g, ..., g^{n - 1}\}\) is distinct.
Step 3:
For each \(k \in \mathbb{N} \setminus \{0\}\) such that \(n \lt k\), \(k = n l + j\), where \(l \in \mathbb{N}\) such that \(1 \le l\) and \(j \in \mathbb{N}\) such that \(0 \le j \lt n\). So, \(g^k = g^{n l + j} = g^{n l} g^j = (g^n)^l g^j = 1^l g^j = 1 g^j = g^j\). So, \(g^k \in \{1, g, ..., g^{n - 1}\}\).
\(g^{-1} = g^{n - 1}\), because \(g g^{n - 1} = g^{n - 1} g = g^n = 1\). \(g^{-n} = 1\), because \(g^{-n} = (g^n)^{-1} = 1^{-1} = 1\).
For each \(k \in \mathbb{N} \setminus \{0\}\), \(k = n l + j\), where \(l \in \mathbb{N}\) such that \(0 \le l\) and \(j \in \mathbb{N}\) such that \(0 \le j \lt n\). So, \(g^{-k} = g^{- n l - j} = (g^{- n})^l g^{- j} = (g^{- n})^l 1 g^{- j} = (g^{- n})^l g^n g^{- j} = 1^l g^{n - j} = 1 g^{n - j} = g^{n - j}\). But \(0 \lt n - j \le n\). When \(n - j = n\), \(g^{-k} = g^n = 1\). So, \(g^{- k} \in \{1, g, ..., g^{n - 1}\}\).
So, \((g) = \{1, g, g^{-1}, g^2, g^{-2}, ...\} = \{1, g, ..., g^{n - 1}\}\).
