description/proof of that for group, normal subgroup, and quotient group, classification map is group homomorphism whose kernel normal subgroup is
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of quotient group of group by normal subgroup.
- The reader knows a definition of kernel of group homomorphism.
- The reader admits the proposition that for any group, the multiplication map with any fixed element from left or right is a bijection.
Target Context
- The reader will have a description and a proof of the proposition that for any group, any normal subgroup, and the quotient group by the subgroup, the classification map is a group homomorphism whose kernel the normal subgroup is.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the normal subgroups of } G'\}\)
\(G' / G\): \(= \text{ the quotient group }\)
\(f\): \(G' \to G' / G, g' \mapsto [g']\)
//
Statements:
\(f \in \{\text{ the group homomorphisms }\}\)
\(\land\)
\(G = \text{ the kernel of } f\)
//
2: Note
By the proposition that for any group homomorphism, the kernel of the homomorphism is a normal subgroup of the domain, the kernel of any group homomorphism is a normal subgroup of the domain; by this proposition, any normal subgroup of any group is the kernel of a group homomorphism.
So, the normal subgroups are the kernels of the group homomorphisms.
3: Proof
Whole Strategy: Step 1: see that \(f\) is a group homomorphism; Step 2: see that \(G\) is the kernel of \(f\).
Step 1:
Let us see that \(f\) is a group homomorphism.
\(f (1) = [1]\), which is the identity in \(G' / G\).
For each \(g'_1, g'_2 \in G'\), \(f (g'_1 g'_2) = [g'_1 g'_2] = [g'_1] [g'_2] = f (g'_1) f (g'_2)\).
For each \(g' \in G'\), \(f (g'^{-1}) = [g'^{-1}] = [g']^{-1} = f (g')^{-1}\).
So, \(f\) is a group homomorphism.
Step 2:
Let us see that \(G\) is the kernel of \(f\).
\(f (g') = [g'] = [1]\) means that \(g' G = 1 G\), which means that \(g' g = 1\) for a \(g \in G\), which means that \(g' = g^{-1} \in G\).
On the other hand, for each \(g \in G\), \(f (g) = [g] = [1]\), because \(g G = 1 G\), by the proposition that for any group, the multiplication map with any fixed element from left or right is a bijection.
