340: For Group, Symmetric Subset, Element of Group, and Subset, Element Multiplied by Symmetric Subset from Right or Left and Symmetric Subset Multiplied by Subset from Right or Left Are Disjoint if Element Multiplied by Symmetric Subset from Left and Right and Subset Are Disjoint

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A description/proof of that for group, symmetric subset, element of group, and subset, element multiplied by symmetric subset from right or left and symmetric subset multiplied by subset from right or left are disjoint if element multiplied by symmetric subset from left and right and subset are disjoint

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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: Description
• 2: Proof

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Starting Context

• The reader knows a definition of multiplication of subsets of group.
• The reader knows a definition of symmetric subset of group.

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Target Context

• The reader will have a description and a proof of the proposition that for any group, any symmetric subset of the group, any element of the group, and any subset of the group, the element multiplied by the symmetric subset from right or left and the symmetric subset multiplied by the subset from right or left, respectively, are disjoint if the element multiplied by the symmetric subset from left and right and the subset are disjoint.

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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: Description

For any group, $G$, any symmetric subset, $S_1\subseteq G$, any element, $p\in G$, and any subset, $S_2\subseteq G$, $p{S}_1\cap S_1{S}_2=\mathrm{\varnothing }$ and $S_{1}p\cap S_2{S}_1=\mathrm{\varnothing }$ if $(S_{1}p{S}_1)\cap S_2=\mathrm{\varnothing }$.

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2: Proof

Let us suppose that $(S_{1}p{S}_1)\cap S_2=\mathrm{\varnothing }$ hereafter.

Let us suppose that $p{S}_1\cap S_1{S}_2\ne \mathrm{\varnothing }$. $p{s}_{11}=s_{12}{s}_2$ where $s_{11},s_{12}\in S_1$ and $s_2\in S_2$. ${s_{12}}^{-1}p{s}_{11}=s_2$, which means that $(S_{1}p{S}_1)\cap S_2\ne \mathrm{\varnothing }$, a contradiction.

Let us suppose that $S_{1}p\cap S_2{S}_1\ne \mathrm{\varnothing }$. $s_{11}p=s_2{s}_{12}$ where $s_{11},s_{12}\in S_1$ and $s_2\in S_2$. $s_{11}p{s_{12}}^{-1}=s_2$, which means that $(S_{1}p{S}_1)\cap S_2\ne \mathrm{\varnothing }$, a contradiction.

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References

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