337: For Coset Map with Respect to Subgroup, Preimage of Image of Subset Is Subgroup Multiplied by Subset

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A description/proof of that for coset map with respect to subgroup, preimage of image of subset is subgroup multiplied by subset

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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 left or right coset of subgroup by element of group.
• The reader admits the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets.

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Target Context

• The reader will have a description and a proof of the proposition that for the left or right coset map with respect to any subgroup, the preimage of the image of any subset is the subgroup multiplied by the subset from left or right, respectively.

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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, $G_1$, any subgroup, $G_2\subseteq G_1$, the left or right coset map, $\pi :G_1\to G_1/G_2$, and any subset, $S\subseteq G_1$, ${\pi }^{-1}\pi (S)=S{G}_2$ or ${\pi }^{-1}\pi (S)=G_{2}S$.

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2: Proof

Let us prove it for the left coset map. For any element, $p\in {\pi }^{-1}\pi (S)$, $\pi (p)\in \pi (S)$. There is an element, $p^{\prime }\in S$, such that $\pi (p)=\pi (p^{\prime })$, which means that $p{G}_2=p^{\prime }{G}_2$. So, $p\in p^{\prime }{G}_2$, so, $p\in S{G}_2$. For any element, $p\in S{G}_2$, $p\in p^{\prime }{G}_2$ for an element, $p^{\prime }\in S$. By the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets, $p{G}_2=p^{\prime }{G}_2$. $\pi (p)\in \pi (S)$. $p\in {\pi }^{-1}\pi (S)$.

Let us prove it for the right coset map. For any element, $p\in {\pi }^{-1}\pi (S)$, $\pi (p)\in \pi (S)$. There is an element, $p^{\prime }\in S$, such that $\pi (p)=\pi (p^{\prime })$, which means that $G_{2}p=G_2{p}^{\prime }$. So, $p\in G_2{p}^{\prime }$, so, $p\in G_{2}S$. For any element, $p\in G_{2}S$, $p\in G_2{p}^{\prime }$ for an element, $p^{\prime }\in S$. By the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets, $G_{2}p=G_2{p}^{\prime }$. $\pi (p)\in \pi (S)$. $p\in {\pi }^{-1}\pi (S)$.

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References

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