description/proof of that for group and its subgroup, subgroup is normal subgroup if its conjugate subgroup by each element of group is contained in it
Topics
About: group
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of normal subgroup.
- The reader knows a definition of conjugate subgroup of subgroup by element.
Target Context
- The reader will have a description and a proof of the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
//
Statements:
\(\forall p \in G' \setminus G (p G p^{-1} \subseteq G)\)
\(\implies\)
\(G \in \{\text{ the normal subgroups of } G'\}\)
//
2: Natural Language Description
For any group, \(G'\), and any subgroup, \(G\), of \(G'\), if for each \(p \in G' \setminus G\), \(p G p^{-1} \subseteq G\), then, \(G\) is a normal subgroup of \(G'\).
3: Proof
Let us suppose that for each \(p \in G' \setminus G\), \(p G p^{-1} \subseteq G\).
Obviously, for each \(p \in G\), \(p G p^{-1} \subseteq G\). So, for each \(p \in G'\), \(p G p^{-1} \subseteq G\).
For each \(p \in G'\), \(G \subseteq p G p^{-1}\)? For any \(p' \in G\), let us take \(p '' := p^{-1} p' p \in G\), because \(p^{-1}\) can be taken in lieu of \(p\). \(p' = p p'' p^{-1} \in p G p^{-1}\). So, \(G \subseteq p G p^{-1}\). So, \(p G p^{-1} = G\), which means that \(G\) is a normal subgroup of \(G'\).
