942: The 1st Isomorphism Theorem for Groups

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description/proof of the 1st isomorphism theorem for groups

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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 %structure kind name% homomorphism.
• The reader knows a definition of %category name% isomorphism.
• The reader knows a definition of quotient group of group by normal subgroup.
• The reader admits the proposition that for any group homomorphism, the kernel of the homomorphism is a normal subgroup of the domain.
• The reader admits the proposition that for any group homomorphism, the range of the homomorphism is a subgroup of the codomain.
• The reader admits the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that for any group homomorphism, the quotient group of the domain by the kernel of the homomorphism is 'groups - homomorphisms' isomorphic to the range of the homomorphism: the 1st isomorphism theorem for groups.

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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:
$G_1$: $\in \{\text{ the groups }\}$
$G_2$: $\in \{\text{ the groups }\}$
$f$: $:G_1\to G_2$, $\in \{\text{ the group homomorphisms }\}$
$Ker(f)$: $=\text{ the kernel of }f$
$Ran(f)$: $=\text{ the range of }f$
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Statements:
$Ker(f)\in \{\text{ the normal subgroups of }{G}_1\}$
$\wedge$
$Ran(f)\in \{\text{ the subgroups of }{G}_2\}$
$\wedge$
$G_1/Ker(f){\cong }_{groups}Ran(f)$ by $f^{\prime }:G_1/Ker(f)\to Ran(f),[g]\mapsto f(g)$, where ${\cong }_{groups}$ denotes being 'groups - homomorphisms' isomorphic
//

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2: Proof

Whole Strategy: Step 1: see that $Ker(f)$ is a normal subgroup of $G_1$; Step 2: see that $Ran(f)$ is a subgroup of $G_2$; Step 3: see that $f^{\prime }$ is well-defined; Step 4: see that $f^{\prime }$ is bijective; Step 5: see that $f^{\prime }$ is a group homomorphism; Step 6: conclude the proposition.

Step 1:

$Ker(f)$ is a normal subgroup of $G_1$, by the proposition that for any group homomorphism, the kernel of the homomorphism is a normal subgroup of the domain.

So, $G_1/Ker(f)$ is well-defined.

Step 2:

$Ran(f)$ is a subgroup of $G_2$, by the proposition that for any group homomorphism, the range of the homomorphism is a subgroup of the codomain.

Step 3:

$f^{\prime }$ is well-defined, because for any $g^{\prime }\in G_1$ such that $[g^{\prime }]=[g]$, $g^{\prime }=g{g}^{″}$ where $g^{″}\in Ker(f)$, $f(g^{\prime })=f(g)f(g^{″})=f(g)1=f(g)$.

Step 4:

$f^{\prime }$ is injective, because for any $[g_1]\ne [g_2]$, $f(g_1{g_2}^{-1})=f(g_1){f(g_2)}^{-1}$, but $g_1{g_2}^{-1}\notin Ker(f)$, so, $f(g_1){f(g_2)}^{-1}\ne 1$, $f(g_1)\ne f(g_2)$.

$f^{\prime }$ is surjective, obviously.

So, $f^{\prime }$ is bijective.

Step 5:

$f^{\prime }$ is a group homomorphism, because $f^{\prime }([g_1][g_2])=f^{\prime }([g_1{g}_2])=f(g_1{g}_2)=f(g_1)f(g_2)=f^{\prime }([g_1])f^{\prime }([g_2])$; $f^{\prime }([1])=f(1)=1$; $f^{\prime }([g{]}^{-1})=f^{\prime }([g^{-1}])=f(g^{-1})={f(g)}^{-1}={f^{\prime }([g])}^{-1}$.

Step 6:

So, $f^{\prime }$ is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.

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References

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