description/proof of that for group, \(2\) elements, and subgroup, cosets of subgroup by elements are same iff product of inverse of element and element is contained in 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.
- The reader admits the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets.
Target Context
- The reader will have a description and a proof of the proposition that for any group, any \(2\) elements, and any subgroup, the left or right cosets of the subgroup by the elements are same if and only if the product of the inverse of an element and the other element or the product of an element and the inverse of the other element is contained in 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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(g'_1\): \(\in G'\)
\(g'_2\): \(\in G'\)
//
Statements:
(
\(g'_1 G = g'_2 G\)
\(\iff\)
\({g'_1}^{-1} g'_2 \in G\)
)
\(\land\)
(
\(G g'_1 = G g'_2\)
\(\iff\)
\(g'_1 {g'_2}^{-1} \in G\)
)
//
2: Note
\({g'_1}^{-1} g'_2 \in G\) can be replaced with \({g'_2}^{-1} g'_1 \in G\), because the order of \(g'_1, g'_2\) does not matter.
Likewise, \(g'_1 {g'_2}^{-1} \in G\) can be replaced with \(g'_2 {g'_1}^{-1} \in G\).
But \({g'_1}^{-1} g'_2 \in G\) cannot be replaced with \(g'_1 {g'_2}^{-1} \in G\) and \(g'_1 {g'_2}^{-1} \in G\) cannot be replaced with \({g'_1}^{-1} g'_2 \in G\).
3: Proof
Whole Strategy: Step 1: suppose that \(g'_1 G = g'_2 G\); Step 2: see that \({g'_1}^{-1} g'_2 \in G\); Step 3: suppose that \({g'_1}^{-1} g'_2 \in G\); Step 4: see that \(g'_1 G = g'_2 G\); Step 5: suppose that \(G g'_1 = G g'_2\); Step 6: see that \(g'_1 {g'_2}^{-1} \in G\); Step 7: suppose that \(g'_1 {g'_2}^{-1} \in G\); Step 8: see that \(G g'_1 = G g'_2\).
Step 1:
Let us suppose that \(g'_1 G = g'_2 G\).
Step 2:
\(g'_2 1 = g'_1 g\) for a \(g \in G\).
So, \({g'_1}^{-1} g'_2 = g \in G\).
Step 3:
Let us suppose that \({g'_1}^{-1} g'_2 \in G\).
Step 4:
\({g'_1}^{-1} g'_2 = g\) where \(g \in G\).
So, \(g'_2 = g'_1 g \in g'_1 G\).
\(g'_2 G = g'_1 G\), by the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets.
Step 5:
Let us suppose that \(G g'_1 = G g'_2\).
Step 6:
\(1 g'_1 = g g'_2\) for a \(g \in G\).
So, \(g'_1 {g'_2}^{-1} = g \in G\).
Step 7:
Let us suppose that \(g'_1 {g'_2}^{-1} \in G\).
Step 8:
\(g'_1 {g'_2}^{-1} = g\) where \(g \in G\).
So, \(g'_1 = g g'_2 \in G g'_2\).
\(G g'_1 = G g'_2\), by the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets.
