description/proof of that for different group left actions, corresponding group right actions are different
Topics
About: set
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of group right action that corresponds to group left action.
Target Context
- The reader will have a description and a proof of the proposition that for any different group left actions, the corresponding group right actions are different.
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:
\(G\): \(\in \{ \text{ the groups } \}\)
\(S\): \(\in \{ \text{ the sets } \}\)
\(f_1\): \(: G \times S \to S\), \(\in \{\text{ the group left actions }\}\)
\(f_2\): \(: G \times S \to S\), \(\in \{\text{ the group left actions }\}\)
\(f'_1\): \(: S \times G \to S\), \(= \text{ the group right action that corresponds to } f_1\}\)
\(f'_2\): \(: S \times G \to S\), \(= \text{ the group right action that corresponds to } f_2\}\)
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Statements:
\(f_1 \neq f_2\)
\(\implies\)
\(f'_1 \neq f'_2\)
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2: Proof
Whole Strategy: Step 1: suppose that \(f'_1 = f'_2\) and see that \(f_1 = f_2\); Step 2: conclude the proposition.
Step 1:
Let us suppose that \(f'_1 = f'_2\).
For each \((s, g) \in S \times G\), \(f'_1 ((s, g)) = f'_2 ((s, g))\).
That means that \(f_1 ((g^{-1}, s)) = f_2 ((g^{-1}, s))\).
As \((s, g^{-1}) \in S \times G\), \(f_1 (({g^{-1}}^{-1}, s)) = f_2 (({g^{-1}}^{-1}, s))\).
That means that \(f_1 ((g, s)) = f_2 ((g, s))\).
So, after all, for each \((g, s) \in G \times S\), \(f_1 ((g, s)) = f_2 ((g, s))\), which means that \(f_1 = f_2\).
Step 2:
As the contraposition, if \(f_1 \neq f_2\), \(f'_1 \neq f'_2\).
