description/proof of that for group, normal subgroup, and element of group, left coset of subgroup by element is right coset of subgroup by element
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of normal subgroup of group.
- The reader knows a definition of left or right coset of subgroup by element of group.
- The reader admits the proposition that for any group, any finite product of subsets of the group is associative.
Target Context
- The reader will have a description and a proof of the proposition that for any group, any normal subgroup, and any element of the group, the left coset of the subgroup by the element is the right coset of the subgroup by the element.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the normal subgroups of } G'\}\)
\(g'\): \(\in G'\)
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Statements:
\(g' G = G g'\)
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2: Proof
Whole Strategy: Step 1: see that \(g' G {g'}^{- 1} = G\) implies that \(g' G = G g'\).
Step 1:
\(g' G {g'}^{- 1} = G\), by the definition of normal subgroup.
So, \(g' G {g'}^{- 1} g' = G g'\).
But the left hand side is \(g' G ({g'}^{- 1} g')\), by the proposition that for any group, any finite product of subsets of the group is associative, \(= g' G 1 = (g' G) 1 = g' G\).
So, \(g' G = G g'\).
