description/proof of that normal subgroup of group is normal subgroup of subgroup of group multiplied by normal 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
- The reader knows a definition of normal subgroup of group.
- The reader admits the proposition that for any group, any subgroup of the group multiplied by any normal subgroup of the group is a subgroup of the group.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any group, any normal subgroup of the group is a normal subgroup of any subgroup of the group multiplied by the normal 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:
\(G\): \(\in \{\text{ the groups }\}\)
\(G_1\): \(\in \{\text{ the subgroups of } G\}\)
\(G_2\): \(\in \{\text{ the normal subgroups of } G\}\)
//
\(G_2 \in \{\text{ the normal subgroups of } G_1 G_2\}\)
\(\land\)
\(G_2 \in \{\text{ the normal subgroups of } G_2 G_2\}\)
//
2: Natural Language Description
For any group, \(G\), any subgroup, \(G_1\), of \(G\), and any normal subgroup, \(G_2\), of \(G\), \(G_2\) is a normal subgroup of \(G_1 G_2\) and \(G_2\) is a normal subgroup of \(G_2 G_1\).
3: Proof
\(G_1 G_2 \subseteq G\) and \(G_2 G_1 \subseteq G\) are indeed some subgroups of \(G\), by the proposition that for any group, any subgroup of the group multiplied by any normal subgroup of the group is a subgroup of the group.
1st let us think of being a normal subgroup of \(G_1 G_2\).
\(G_2 \subseteq G_1 G_2\) is obvious. So, \(G_2\) is a subgroup of \(G_1 G_2\).
For any \(p \in G_1 G_2\), \(p G_2 p^{-1} \subseteq G_2\), because \(p \in G\). So, \(G_2\) is a normal subgroup of \(G_1 G_2\), 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.
Let us think of being a normal subgroup of \(G_2 G_1\).
\(G_2 \subseteq G_2 G_1\) is obvious. So, \(G_2\) is a subgroup of \(G_2 G_1\).
For any \(p \in G_2 G_1\), \(p G_2 p^{-1} \subseteq G_2\), because \(p \in G\). So, \(G_2\) is a normal subgroup of \(G_2 G_1\), 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.
