description/proof of that for group, normal subgroup is kernel of canonical group homomorphism onto quotient group of group by normal subgroup
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of normal subgroup of group.
- 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 is the kernel of the canonical group homomorphism onto the quotient group of the group 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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the normal subgroups of } G'\}\)
\(G' / G\): \(= \text{ the quotient group }\)
\(f\): \(G' \to G' / G, g' \mapsto [g']\)
//
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 subgroup 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.
