description/proof of that subgroup of group multiplied by normal subgroup of group is subgroup of group
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.
Target Context
- The reader will have a description and a proof of 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.
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:
\(G_1 G_2 \in \{\text{ the subgroups of } G\}\)
\(\land\)
\(G_2 G_1 \in \{\text{ the subgroups of } G\}\)
//
2: Natural Language Description
For any group, \(G\), any subgroup, \(G_1\), of \(G\), and any normal subgroup, \(G_2\), of \(G\), \(G_1 G_2\) is a subgroup of \(G\) and \(G_2 G_1\) is a subgroup of \(G\).
3: Proof
1st let us think of \(G_1 G_2\).
For the identity element, \(1 \in G\), \(1 \in G_1\) and \(1 \in G_2\), so, \(1 \in G_1 G_2\).
For any \(p_1, p'_1 \in G_1\) and any \(p_2, p'_2 \in G_2\), \(p_1 p_2 p'_1 p'_2 = p_1 p'_1 p''_2 {p'_1}^{-1} p'_1 p'_2\), where \(p''_2 \in G_2\), because as \(p'_1 G_2 {p'_1}^{-1} = G_2\), there is such a \(p''_2\); \(= p_1 p'_1 p''_2 p'_2 \in G_1 G_2\).
For any \(p_1 \in G_1\) and any \(p_2 \in G_2\), \((p_1 p_2)^{-1} = {p_2}^{-1} {p_1}^{-1} = {p_1}^{-1} p'_2 p_1 {p_1}^{-1}\), where \(p'_2 \in G_2\), because as \({p_1}^{-1} G_2 p_1 = G_2\), there is such a \(p'_2\); \(= {p_1}^{-1} p'_2 \in G_1 G_2\).
The associativity of multiplications holds, because it holds in the ambient \(G\).
Let us think of \(G_2 G_1\).
For the identity element, \(1 \in G\), \(1 \in G_1\) and \(1 \in G_2\), so, \(1 \in G_2 G_1\).
For any \(p_1, p'_1 \in G_1\) and any \(p_2, p'_2 \in G_2\), \(p_2 p_1 p'_2 p'_1 = p_2 p_1 {p_1}^{-1} p''_2 p_1 p'_1\), where \(p''_2 \in G_2\), because as \({p_1}^{-1} G_2 p_1 = G_2\), there is such a \(p''_2\); \(= p_2 p''_2 p_1 p'_1 \in G_2 G_1\).
For any \(p_1 \in G_1\) and any \(p_2 \in G_2\), \((p_2 p_1)^{-1} = {p_1}^{-1} {p_2}^{-1} = {p_1}^{-1} p_1 p'_2 {p_1}^{-1}\), where \(p'_2 \in G_2\), because as \(p_1 G_2 {p_1}^{-1} = G_2\), there is such a \(p'_2\); \(= p'_2 {p_1}^{-1} \in G_2 G_1\).
The associativity of multiplications holds, because it holds in the ambient \(G\).
