338: With Respect to Subgroup, Coset by Element of Group Equals Coset iff Element Is Member of Latter Coset

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A description/proof of that with respect to subgroup, coset by element of group equals coset iff element is member of latter coset

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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.

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Target Context

• The reader will have a description and a proof of 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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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, $G_2\subseteq G_1$, and any element, $p^{\prime }\in G_1$, the left or right coset, $p^{\prime }{G}_2=p{G}_2$ or $G_2{p}^{\prime }=G_{2}p$, if and only if $p^{\prime }\in p{G}_2$ or $p^{\prime }\in G_{2}p$, respectively.

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2: Proof

Let us suppose that $p^{\prime }\in p{G}_2$. $p^{\prime }=p{q}^{\prime }$ for a $q^{\prime }\in G_2$. For any $q\in G_2$, $p^{\prime }q=p{q}^{\prime }q\in p{G}_2$, so, $p^{\prime }{G}_2\subseteq p{G}_2$. As $p=p^{\prime }{q}^{\prime -1}$, $p\in p^{\prime }{G}_2$, so, likewise, $p{G}_2\subseteq p^{\prime }{G}_2$. So, $p{G}_2=p^{\prime }{G}_2$.

Let us suppose that $p^{\prime }\in G_{2}p$. $p^{\prime }=q^{\prime }p$ for a $q^{\prime }\in G_2$. For any $q\in G_2$, $q{p}^{\prime }=q{q}^{\prime }p\in G_{2}p$, so, $G_2{p}^{\prime }\subseteq G_{2}p$. As $p=q^{\prime -1}{p}^{\prime }$, $p\in G_2{p}^{\prime }$, so, likewise, $G_{2}p\subseteq G_2{p}^{\prime }$. So, $G_{2}p=G_2{p}^{\prime }$.

Let us suppose that $p^{\prime }{G}_2=p{G}_2$. $p^{\prime }\in p{G}_2$, because $e\in G_2$.

Let us suppose that $G_2{p}^{\prime }=G_{2}p$. $p^{\prime }\in G_{2}p$, because $e\in G_2$.

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References

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