description/proof of that for \(2\) maps from group into linearly-ordered set, former is equal to or smaller than or is smaller than latter iff left or right translations of them by element satisfy same relation
Topics
About: group
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of left or right translation of map from group by element.
- The reader knows a definition of linearly-ordered set.
Target Context
- The reader will have a description and a proof of the proposition that for any \(2\) maps from any group into any linearly-ordered set, the former is equal to or smaller than or is smaller than the latter if and only if the left or right translations of them by any element satisfy the same relation.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(S\): \(\in \{\text{ the linearly-ordered sets }\}\)
\(f_1\): \(: G \to S\)
\(f_2\): \(: G \to S\)
\(g\): \(\in G\)
\(_gf_1\): \(= \text{ the left translation }\)
\(_gf_2\): \(= \text{ the left translation }\)
\({f_1}_g\): \(= \text{ the right translation }\)
\({f_2}_g\): \(= \text{ the right translation }\)
//
Statements:
(
\(f_1 \le f_2\)
\(\iff\)
\(_gf_1 \le _gf_2\)
)
\(\land\)
(
\(f_1 \le f_2\)
\(\iff\)
\({f_1}_g \le {f_2}_g\)
)
\(\land\)
(
\(f_1 \lt f_2\)
\(\iff\)
\(_gf_1 \lt _gf_2\)
)
\(\land\)
(
\(f_1 \lt f_2\)
\(\iff\)
\({f_1}_g \lt {f_2}_g\)
)
//
2: Proof
Whole Strategy: Step 1: suppose that \(f_1 \le f_2\); Step 2: see that \(_gf_1 \le _gf_2\) and \({f_1}_g \le {f_2}_g\); Step 3: suppose that \(f_1 \lt f_2\); Step 4: see that \(_gf_1 \lt _gf_2\) and \({f_1}_g \lt {f_2}_g\); Step 5: see that \(_{g^{-1}}(_gf_j) = f_j\) and \(({f_j}_g)_{g^{-1}} = f_j\); Step 6: conclude the proposition.
Step 1:
Let us suppose that \(f_1 \le f_2\).
Step 2:
For each \(g' \in G\), \(_gf_1 (g') = f_1 (g^{-1} g') \le f_2 (g^{-1} g') = _gf_2 (g')\).
So, \(_gf_1 \le _gf_2\).
For each \(g' \in G\), \({f_1}_g (g') = f_1 (g' g^{-1}) \le f_2 (g' g^{-1}) = {f_2}_g (g')\).
So, \({f_1}_g \le {f_2}_g\).
Step 3:
Let us suppose that \(f_1 \lt f_2\).
Step 4:
For each \(g' \in G\), \(_gf_1 (g') = f_1 (g^{-1} g') \lt f_2 (g^{-1} g') = _gf_2 (g')\).
So, \(_gf_1 \lt _gf_2\).
For each \(g' \in G\), \({f_1}_g (g') = f_1 (g' g^{-1}) \lt f_2 (g' g^{-1}) = {f_2}_g (g')\).
So, \({f_1}_g \lt {f_2}_g\).
Step 5:
For each \(g' \in G\), \(_{g^{-1}}(_gf_j) (g') = _gf_j (g g') = f_j (g^{-1} g g') = f_j (g')\).
So, \(_{g^{-1}}(_gf_j) = f_j\).
For each \(g' \in G\), \(({f_j}_g)_{g^{-1}} (g') = {f_j}_g (g' g) = f_j (g' g g^{-1}) = f_j (g')\).
So, \(({f_j}_g)_{g^{-1}} = f_j\).
Step 6:
\(_gf_1 \le _gf_2\) implies that \(_{g^{-1}}(_gf_1) \le _{g^{-1}}(_gf_2)\), by Step 2, which implies that \(f_1 \le f_2\), by Step 5.
\({f_1}_g \le {f_2}_g\) implies that \(({f_1}_g)_{g^{-1}} \le ({f_2}_g)_{g^{-1}}\), by Step 2, which implies that \(f_1 \le f_2\), by Step 5.
\(_gf_1 \lt _gf_2\) implies that \(_{g^{-1}}(_gf_1) \lt _{g^{-1}}(_gf_2)\), by Step 4, which implies that \(f_1 \lt f_2\), by Step 5.
\({f_1}_g \lt {f_2}_g\) implies that \(({f_1}_g)_{g^{-1}} \lt ({f_2}_g)_{g^{-1}}\), by Step 4, which implies that \(f_1 \lt f_2\), by Step 5.
So, the proposition holds.
