description/proof of that intersection of subgroup of group and normal subgroup of group is normal subgroup of subgroup
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
Target Context
- The reader will have a description and a proof of the proposition that for any group, the intersection of any subgroup of the group and any normal subgroup of the group is a normal subgroup 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:
\(G\): \(\in \{\text{ the groups }\}\)
\(G_1\): \(\in \{\text{ the subgroups of } G\}\)
\(G_2\): \(\in \{\text{ the normal subgroups of } G\}\)
//
Statements:
\(G_1 \cap G_2 \in \{\text{ the normal subgroups of } G_1\}\)
//
2: Natural Language Description
For any group, \(G\), any subgroup, \(G_1 \subseteq G\), and any normal subgroup, \(G_2 \subseteq G\), \(G_1 \cap G_2 \subseteq G_1\) is a normal subgroup of \(G_1\).
3: Proof
Whole Strategy: Step 1: see that for each \(p_1 \in G_1\), \(p_1 (G_1 \cap G_2) {p_1}^{-1} \subseteq G_1 \cap G_2\) and conclude the proposition.
Step 1:
By 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, it suffices to show that for each \(p_1 \in G_1\), \(p_1 (G_1 \cap G_2) {p_1}^{-1} \subseteq G_1 \cap G_2\).
\(p_1 (G_1 \cap G_2) {p_1}^{-1} \subseteq p_1 G_2 {p_1}^{-1} = G_2\), because \(p_1 \in G\). \(p_1 (G_1 \cap G_2) {p_1}^{-1} \subseteq p_1 G_1 {p_1}^{-1} \subseteq G_1\), because \(p_1 \in G_1\). So, \(p_1 (G_1 \cap G_2) {p_1}^{-1} \subseteq G_1 \cap G_2\).
