717: For Group, Multiplication Map with Fixed Element from Left or Right Is Bijection

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description/proof of that for group, multiplication map with fixed element from left or right is bijection

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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: Proof

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Starting Context

• The reader knows a definition of group.

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Target Context

• The reader will have a description and a proof of the proposition that for any group, the multiplication map with any fixed element from left or right is a bijection.

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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:
$G$: $\in \{\text{ the groups }\}$
$p$: $\in G$
$f_p$: $:G\to G,p^{\prime }\mapsto p{p}^{\prime }$
$f_p^{\prime }$: $:G\to G,p^{\prime }\mapsto p^{\prime }p$
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Statements:
$f_p\in \{\text{ the bijections }\}$
$\wedge$
$f_p^{\prime }\in \{\text{ the bijections }\}$
//

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2: Natural Language Description

For any group, $G$, and any element, $p\in G$, the multiplication map for $G$ with $p$ from left, $f_p:G\to G,p^{\prime }\mapsto p{p}^{\prime }$, or from right, $f_p^{\prime }:G\to G,p^{\prime }\mapsto p^{\prime }p$, is a bijection.

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3: Proof

Whole Strategy: Step 1: see that $f_p$ is injective by supposing some 2 distinct elements having the same image and finding a contradiction; Step 2: see that $f_p$ is surjective by finding a domain element that maps to an arbitrary codomain element; Step 3: see that $f_p^{\prime }$ is injective by supposing some 2 distinct elements having the same image and finding a contradiction; Step 4: see that $f_p^{\prime }$ is surjective by finding a domain element that maps to an arbitrary codomain element.

Step 1:

Let us see that $f_p$ is injective.

Let us suppose that there were some elements, $p_1,p_2\in G$, such that $p_1\ne p_2$ and $f_p(p_1)=f_p(p_2)$.

$f_p(p_1)=p{p}_1=p{p}_2=f_p(p_2)$. $p_1=p^{-1}p{p}_1=p^{-1}p{p}_2=p_2$, a contradiction.

So, for each $p_1,p_2\in G$ such that $p_1\ne p_2$, $f_p(p_1)\ne f_p(p_2)$.

Step 2:

Let us see that $f_p$ is surjective.

Let $p^{\prime }\in G$ be any. $f_p(p^{-1}{p}^{\prime })=p{p}^{-1}{p}^{\prime }=p^{\prime }$.

Step 3:

Let us see that $f_p^{\prime }$ is injective.

Let us suppose that there were some elements, $p_1,p_2\in G$, such that $p_1\ne p_2$ and $f_p^{\prime }(p_1)=f_p^{\prime }(p_2)$.

$f_p^{\prime }(p_1)=p_{1}p=p_{2}p=f_p(p_2)$. $p_1=p_{1}p{p}^{-1}=p_{2}p{p}^{-1}=p_2$, a contradiction.

So, for each $p_1,p_2\in G$ such that $p_1\ne p_2$, $f_p^{\prime }(p_1)\ne f_p^{\prime }(p_2)$.

Step 4:

Let us see that $f_p^{\prime }$ is surjective.

Let $p^{\prime }\in G$ be any. $f_p^{\prime }(p^{\prime }{p}^{-1})=p^{\prime }{p}^{-1}p=p^{\prime }$.

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References

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