981: Group Homomorphism Is Injective iff Kernel Is 1 Subgroup

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description/proof of that group homomorphism is injective iff kernel is 1 subgroup

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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 group.
• The reader knows a definition of injection.

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Target Context

• The reader will have a description and a proof of the proposition that any group homomorphism is injective if and only if the kernel is the 1 subgroup.

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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:
$G_1$: $\in \{\text{ the groups }\}$
$G_2$: $\in \{\text{ the groups }\}$
$f$: $:G_1\to G_2$, $\in \{\text{ the group homomorphisms }\}$
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Statements:
$Ker(f)=\{1\}$
$⟺$
$f\in \{\text{ the injections }\}$
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2: Proof

Whole Strategy: Step 1: suppose that $Ker(f)=\{1\}$, take any $g,g^{\prime }\in G_1$ such that $g\ne g^{\prime }$, suppose that $f(g)=f(g^{\prime })$, and find a contradiction; Step 2: suppose that $f$ is injective, and see that $Ker(f)=\{1\}$.

Step 1:

Let us suppose that $Ker(f)=\{1\}$.

Let $g,g^{\prime }\in G_1$ be any such that $g\ne g^{\prime }$.

Let us suppose that $f(g)=f(g^{\prime })$ and find a contradiction.

$1=f(g)f(g^{\prime }{)}^{-1}=f(g)f(g^{\prime -1})=f(g{g}^{\prime -1})$, which implies that $g{g}^{\prime -1}=1$, which implies that $g=g^{\prime }$, a contradiction.

Step 2:

Let us suppose that $f$ is injective.

$f(1)=1$.

Let us suppose that there is a $g\in G_1$ such that $f(g)=1$. As $f(1)=1$ and $f$ is injective, there is no other $g$ than $1$, which means that $Ker(f)=\{1\}$.

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References

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