description/proof of that for group and finite-order element, inverse of element has order of 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 order of element of group.
- The reader admits 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.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any group and any finite-order element, the inverse of the element has the order of 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:
\(G\): \(\in \{\text{ the groups }\}\)
\(g\): \(\in G\), such that \(\vert (g) \vert \lt \infty\)
//
Statements:
\(\vert (g^{-1}) \vert = \vert (g) \vert\)
//
2: Natural Language Description
For any group, \(G\), and any finite-order element, \(g \in G\), \(\vert (g^{-1}) \vert = \vert (g) \vert\).
3: Proof
Whole Strategy: Step 1: let \(n := \vert (g) \vert\) and see that \((g^{-1})^n = 1\); Step 2: see that there is no \(k \in \mathbb{N} \setminus \{0\}\) such that \(k \lt n\) and \((g^{-1})^k = 1\); Step 3: conclude the proposition.
Step 1:
Let \(n := \vert (g) \vert\).
\(g^n = 1\), by 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.
\((g^{-1})^n = g^{- n} = (g^{n})^{-1} = 1^{-1} = 1\).
Step 2:
Let us see that there is no \(k \in \mathbb{N} \setminus \{0\}\) such that \(k \lt n\) and \((g^{-1})^k = 1\).
Let us suppose that there was such a \(k\).
\(g^k = (g^{- k})^{-1} = ((g^{-1})^k)^{-1} = 1^{-1} = 1\), a contradiction against \(\vert (g) \vert = n\), by 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.
Step 3:
\((g^{-1}) = \{1, g^{-1}, ..., (g^{-1})^{n - 1}\}\), by 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.
That means that \(\vert (g^{-1}) \vert = n = \vert (g) \vert\).
