description/proof of that for group and subgroup, left and right cosets of subgroup by element in complement of subgroup are contained in complement of subgroup
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of left or right coset of subgroup by element of group.
Target Context
- The reader will have a description and a proof of the proposition that for any group and any subgroup, the left and right cosets of the subgroup by any element in the complement of the subgroup are contained in the complement of the subgroup.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(g'\): \(\in G' \setminus G\)
//
Statements:
\(g' G \subseteq G' \setminus G\)
\(\land\)
\(G g' \subseteq G' \setminus G\)
//
2: Proof
Whole Strategy: Step 1: suppose that there was a \(g' g \in G\) and find a contradiction; Step 2: suppose that there was a \(g g' \in G\) and find a contradiction.
Step 1:
Let us see that \(g' G \subseteq G' \setminus G\).
Let us suppose that there was a \(g \in G\) such that \(g' g \in G\).
\(\widetilde{g} := g' g \in G\).
\(g' = \widetilde{g} g^{-1} \in G\), because \(G\) was a group, a contradiction against \(g' \in G' \setminus G\).
So, there is no such \(g\).
That means that \(g' G \subseteq G' \setminus G\).
Step 2:
Let us see that \(G g' \subseteq G' \setminus G\).
Let us suppose that there was a \(g \in G\) such that \(g g' \in G\).
\(\widetilde{g} := g g' \in G\).
\(g' = g^{-1} \widetilde{g} \in G\), because \(G\) was a group, a contradiction against \(g' \in G' \setminus G\).
So, there is no such \(g\).
That means that \(G g' \subseteq G' \setminus G\).
