694: Map Between Groups That Maps Product of 2 Elements to Product of Images of Elements Is Group Homomorphism

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description/proof of that map between groups that maps product of 2 elements to product of images of elements is group homomorphism

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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: Natural Language Description
• 3: Note 1
• 4: Proof
• 5: Note 2

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Starting Context

• The reader knows a definition of group.
• The reader knows a definition of %structure kind name% homomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.

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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:
$G_1$: $\in \{\text{ the groups }\}$
$G_2$: $\in \{\text{ the groups }\}$
$f$: $:G_1\to G_2$
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Statements:
$\mathrm{\forall }{p}_1,p_2\in G_1(f(p_1{p}_2)=f(p_1)f(p_2))$
$⟹$
$f\in \{\text{ the group homomorphisms }\}$
//

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2: Natural Language Description

For any groups, $G_1,G_2$, and any map, $f:G_1\to G_2$, if for each $p_1,p_2\in G_1$, $f(p_1{p}_2)=f(p_1)f(p_2)$, $f$ is a group homomorphism.

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3: Note 1

By a frequently seen definition, $\mathrm{\forall }{p}_1,p_2\in G_1(f(p_1{p}_2)=f(p_1)f(p_2))$ is all that is required, but we do not take that stance: refer to Note for the definition of %structure kind name% homomorphism.

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4: Proof

Whole Strategy: Step 1: see that $f$ maps the identity to the identity; Step 2: see that $f$ maps the inverse of each element to the inverse of the image of the element.

Step 1:

Let us see that $f$ maps the identity to the identity.

$f(1)=f(11)=f(1)f(1)$, so, $1=f(1){f(1)}^{-1}=f(1)f(1){f(1)}^{-1}=f(1)$.

Step 2:

Let us see that $f$ maps the inverse of each element to the inverse of the image of the element.

For each $p\in G_1$, $1=f(1)=f(p{p}^{-1})=f(p)f(p^{-1})$ and $1=f(1)=f(p^{-1}p)=f(p^{-1})f(p)$, which means that $f(p^{-1})={f(p)}^{-1}$.

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5: Note 2

We cannot prove that $f$ maps the identity to the identity like this: "for each $p\in G_1$, $f(p)=f(1p)=f(p1)=f(1)f(p)=f(p)f(1)$", because $f(p)$ s do not necessarily cover the whole $G_2$. In fact, that is the reason why even the frequently seen definition of ring homomorphism (refer to Note for the definition of %structure kind name% homomorphism) needs to have the requirement of mapping the multiplicative identity to the multiplicative identity in the definition.

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References

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