941: Kernel of Group Homomorphism Is Normal Subgroup of Domain

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description/proof of that kernel of group homomorphism is normal subgroup of domain

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Topics

About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of kernel of group homomorphism.
• The reader knows a definition of normal subgroup of group.

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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.

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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.

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Main Body

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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 $f(g_1{g}_2)=f(g_1)f(g_2)=11=1$.

$1\in Ker(f)$, 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 $g_3\in Ker(f)$. $f(g_2)=f(g_1)f(g_3)f({g_1}^{-1})=f(g_1)1f({g_1}^{-1})=f(g_1){f(g_1)}^{-1}=1$. So, $g_2\in Ker(f)$. For any $g_2\in Ker(f)$, let us define $g_3:={g_1}^{-1}{g}_2{g}_1\in G_1$. $g_2=g_1{g}_3{g_1}^{-1}$. $f(g_3)=f({g_1}^{-1})f(g_2)f(g_1)={f(g_1)}^{-1}1f(g_1)=1$. 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 $G_1$.

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References

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