description/proof of that kernel of group homomorphism is normal subgroup of domain
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of kernel of group homomorphism.
- 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 homomorphism, the kernel of the homomorphism is a normal subgroup of the domain.
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:
\(G_1\): \(\in \{\text{ the groups }\}\)
\(G_2\): \(\in \{\text{ the groups }\}\)
\(f\): \(: G_1 \to G_2\), \(\in \{\text{ the group homomorphisms }\}\)
\(Ker (f)\): \(= \text{ the kernel of } f\)
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Statements:
\(Ker (f) \in \{\text{ the normal subgroup of } G_1\}\)
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2: Proof
Whole Strategy: Step 1: see that \(Ker (f)\) is a subgroup of \(G_1\); Step 2: see that for any \(g_1 \in G_1\), \(g_1 Ker (f) {g_1}^{-1} = Ker (f)\).
Step 1:
For any elements, \(g_1, g_2 \in Ker (f)\), \(g_1 g_2 \in Ker (f)\), because \(f (g_1 g_2) = f (g_1) f (g_2) = 1 1 = 1\).
\(1 \in Ker (f)\), because \(f (1) = 1\).
For any element, \(g \in Ker (f)\), \(g^{-1} \in Ker (f)\), because \(f (g^{-1}) = {f (g)}^{-1} = 1^{-1} = 1\).
\(Ker (f)\) satisfies the associative law, because the elements are elements of \(G_1\) and the operation is inherited from that of \(G_1\).
So, \(Ker (f)\) is a subgroup of \(G_1\).
Step 2:
For any \(g_1 \in G_1\), \(g_1 Ker (f) {g_1}^{-1} = Ker (f)\)?
For any \(g_2 \in g_1 Ker (f) {g_1}^{-1}\), \(g_2 = g_1 g_3 {g_1}^{-1}\) for a \(g_3 \in Ker (f)\). \(f (g_2) = f (g_1) f (g_3) f ({g_1}^{-1}) = f (g_1) 1 f ({g_1}^{-1}) = f (g_1) {f (g_1)}^{-1} = 1\). So, \(g_2 \in Ker (f)\). For any \(g_2 \in Ker (f)\), let us define \(g_3 := {g_1}^{-1} g_2 g_1 \in G_1\). \(g_2 = g_1 g_3 {g_1}^{-1}\). \(f (g_3) = f ({g_1}^{-1}) f (g_2) f (g_1) = {f (g_1)}^{-1} 1 f (g_1) = 1\). So, \(g_3 \in Ker (f)\), and \(g_2 \in g_1 Ker (f) {g_1}^{-1}\).
So, yes, \(g_1 Ker (f) {g_1}^{-1} = Ker (f)\).
So, \(Ker (f)\) is a normal subgroup of \(G_1\).
