description/proof of that for group action, induced map with fixed group element is bijection
Topics
About: group
About: set
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 action.
- The reader knows a definition of bijection.
Target Context
- The reader will have a description and a proof of the proposition that for any group action, the induced map with any fixed group element is a bijection.
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:
\(G\): \(\in \{ \text{ the groups } \}\)
\(S\): \(\in \{ \text{ the sets } \}\)
\(f\): \(: G \times S \to S\), \(\in \{\text{ the group actions }\}\)
\(g \in G\)
\(f_g\): \(: S \to S, s \mapsto f (g, s)\)
//
Statements:
\(f_g \in \{\text{ the bijections }\}\)
//
2: Natural Language Description
For any group, \(G\), any set, \(S\), any group action, \(f: G \times S \to S\), and any element, \(g \in G\), \(f_g: S \to S, s \mapsto f (g, s)\) is a bijection.
3: Proof
Whole Strategy: Step 1: choose each \(s_1, s_2 \in S\) such that \(s_1 \neq s_2\), suppose that \(f_g (s_1) = f_g (s_2)\), and find a contradiction; Step 2: choose each \(s \in S\) and see that \(f_g (f (g^{-1}, s)) = s\).
Step 1:
Let us see that \(f_g\) is an injection.
Let \(s_1, s_2 \in S\) be any such that \(s_1 \neq s_2\).
Let us suppose that \(f_g (s_1) = f_g (s_2)\).
\(g s_1 = g s_2\). \(s_1 = 1 s_1 = (g^{-1} g) s_1 = g^{-1} (g s_1) = g^{-1} (g s_2) = (g^{-1} g) s_2 = 1 s_2 = s_2\), a contradiction.
So, \(f_g (s_1) \neq f_g (s_2)\), and \(f_g\) is an injection.
Step 2:
Let \(s \in S\) be any.
\(f_g (g^{-1} s) = g (g^{-1} s) = (g g^{-1}) s = 1 s = s\).
So, \(f_g\) is a surjection.
