339: With Respect to Normal Subgroup, Set of Cosets Forms Group

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A description/proof of that with respect to normal subgroup, set of cosets forms group

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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 with respect to any normal subgroup, the set of the cosets forms a group with the canonical multiplication and inversion.

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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 normal subgroup, $G_2\subseteq G_1$, and the left or right coset map, $\pi :G_1\to G_1/G_2$, the set of the cosets, $G_1/G_2$, forms a group with the canonical multiplication, $⟨p_1{G}_2,p_2{G}_2⟩\mapsto p_1{p}_2{G}_2$ or $⟨G_2{p}_1,G_2{p}_2⟩\mapsto G_2{p}_1{p}_2$ and the canonical inverse, $p{G}_2\mapsto p^{-1}{G}_2$ or $G_{2}p\mapsto G_2{p}^{-1}$, respectively.

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2: Proof

Let us prove it for the left coset map. Let us prove that the multiplication is well-defined, which is about when $p_{11}{G}_2=p_{12}{G}_2$ and $p_{21}{G}_2=p_{22}{G}_2$, $p_{11}{p}_{21}{G}_2=p_{12}{p}_{22}{G}_2$. 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, it is about $p_{11}{p}_{21}\in p_{12}{p}_{22}{G}_2=p_{12}{p}_{21}{G}_2$. While $p_{11}=p_{12}{q}_1$ for a $q_1\in G_2$, $p_{11}{p}_{21}=p_{12}{q}_1{p}_{21}\in p_{12}{p}_{21}{G}_2$? Is there a $q\in G_2$ such that $p_{12}{q}_1{p}_{21}=p_{12}{p}_{21}q$? $q_1{p}_{21}=p_{21}q$? ${p_{21}}^{-1}{q}_1{p}_{21}=q$? Yes, because $G_2$ is a normal subgroup.

The multiplication is associative, because $(p_1{G}_2{p}_2{G}_2)p_3{G}_3=p_1{p}_2{G}_2{p}_3{G}_2=p_1{p}_2{p}_3{G}_2=p_1{G}_2{p}_2{p}_3{G}_2=p_1{G}_2(p_2{G}_2{p}_3{G}_2)$.

Let us prove that the inverse is well-defined, which is about when $p_1{G}_2=p_2{G}_2$, ${p_1}^{-1}{G}_2={p_2}^{-1}{G}_2$. 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, it is about ${p_1}^{-1}\in {p_2}^{-1}{G}_2$. Is there a $q\in G_2$, ${p_1}^{-1}={p_2}^{-1}q$? While $p_1=p_2{q}_1$ for a $q_1\in G_2$, ${p_1}^{-1}={q_1}^{-1}{p_2}^{-1}$ and ${p_1}^{-1}={q_1}^{-1}{p_2}^{-1}={p_2}^{-1}q$? $p_2{q_1}^{-1}{p_2}^{-1}=q$? Yes, because $G_2$ is a normal subgroup.

$G_2$ is the identity element, because $G_2{p}_1{G}_2=e{p}_1{G}_2=p_1{G}_2=p_{1}e{G}_2=p_1{G}_2{G}_2$.

${p_1}^{-1}{G}_2$ is the inverse of $p_1{G}_2$, because ${p_1}^{-1}{G}_2{p}_1{G}_2={p_1}^{-1}{p}_1{G}_2=e{G}_2=G_2=e{G}_2=p_1{p_1}^{-1}{G}_2=p_1{G}_2{p_1}^{-1}{G}_2$.

Let us prove it for the right coset map.

Let us prove that the multiplication is well-defined, which is about when $G_2{p}_{11}=G_2{p}_{12}$ and $G_2{p}_{21}=G_2{p}_{22}$, $G_2{p}_{11}{p}_{21}=G_2{p}_{12}{p}_{22}$. 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, it is about $p_{11}{p}_{21}\in G_2{p}_{12}{p}_{22}=G_2{p}_{11}{p}_{22}$. While $p_{21}=q_1{p}_{22}$ for a $q_1\in G_2$, $p_{11}{p}_{21}=p_{11}{q}_1{p}_{22}\in G_2{p}_{11}{p}_{22}$? Is there a $q\in G_2$ such that $p_{11}{q}_1{p}_{22}=q{p}_{11}{p}_{22}$? $p_{11}{q}_1=q{p}_{11}$? $p_{11}{q}_1{p_{11}}^{-1}=q$? Yes, because $G_2$ is a normal subgroup.

The multiplication is associative, because $(G_2{p}_1{G}_2{p}_2)G_3{p}_3=G_2{p}_1{p}_2{G}_2{p}_3=G_2{p}_1{p}_2{p}_3=G_2{p}_1{G}_2{p}_2{p}_3=G_2{p}_1(G_2{p}_2{G}_2{p}_3)$.

Let us prove that the inverse is well-defined, which is about when $G_2{p}_1=G_2{p}_2$, $G_2{p_1}^{-1}=G_2{p_2}^{-1}$. 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, it is about ${p_1}^{-1}\in G_2{p_2}^{-1}$. Is there a $q\in G_2$, ${p_1}^{-1}=q{p_2}^{-1}$? While $p_1=q_1{p}_2$ for a $q_1\in G_2$, ${p_1}^{-1}={p_2}^{-1}{q_1}^{-1}$ and ${p_1}^{-1}={p_2}^{-1}{q_1}^{-1}=q{p_2}^{-1}$? ${p_2}^{-1}{q_1}^{-1}{p}_2=q$? Yes, because $G_2$ is a normal subgroup.

$G_2$ is the identity element, because $G_2{G}_2{p}_1=G_{2}e{p}_1=G_2{p}_1=G_2{p}_{1}e=G_2{p}_1{G}_2$.

$G_2{p_1}^{-1}$ is the inverse of $G_2{p}_1$, because $G_2{p_1}^{-1}{G}_2{p}_1=G_2{p_1}^{-1}{p}_1=G_{2}e=G_2=G_{2}e=G_2{p}_1{p_1}^{-1}=G_2{p}_1{G}_2{p_1}^{-1}$.

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References

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