Coset Product is Well-Defined

Theorem

Let $\struct {G, \circ}$ be a group.

Let $N$ be a normal subgroup of $G$.

Let $a, b \in G$.

Then the coset product:

$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$

is well-defined.

Proof 1

Let $N \lhd G$ where $G$ is a group.

Let $a, a', b, b' \in G$ such that:

$a \circ N = a' \circ N$

and:

$b \circ N = b' \circ N$

To show that the coset product is well-defined, we need to demonstrate that:

$\paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$

So:

$\ds a \circ N$

$=$

$\ds a' \circ N$

by hypothesis

$\ds \leadsto \ \ $

$\ds a^{-1} \circ a'$

$\in$

$\ds N$

Cosets are Equal iff Product with Inverse in Subgroup

$\ds \leadsto \ \ $

$\ds b^{-1} \circ a^{-1} \circ a'$

$\in$

$\ds b^{-1} \circ N$

Definition of Subset Product

$\ds \leadsto \ \ $

$\ds b^{-1} \circ a^{-1} \circ a'$

$\in$

$\ds N \circ b^{-1}$

$N$ is a normal subgroup

$\ds \leadsto \ \ $

$\ds \exists n \in N: \, $

$\ds b^{-1} \circ a^{-1} \circ a'$

$=$

$\ds n \circ b^{-1}$

Definition of Subset Product

$\ds \leadsto \ \ $

$\ds \paren{b^{-1} \circ a^{-1} \circ a' } \circ b'$

$=$

$\ds \paren{n \circ b^{-1} } \circ b'$

Group Axiom $\text G 0$: Closure

$\ds \leadsto \ \ $

$\ds \paren{b^{-1} \circ a^{-1} } \circ \paren {a' \circ b'}$

$=$

$\ds \paren{n \circ b^{-1} } \circ b'$

Group Axiom $\text G 1$: Associativity

$\ds \leadsto \ \ $

$\ds \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}$

$=$

$\ds \paren{n \circ b^{-1} } \circ b'$

Inverse of Group Product

$\ds \leadsto \ \ $

$\ds \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}$

$=$

$\ds n \circ \paren{ b^{-1} \circ b' }$

Group Axiom $\text G 1$: Associativity

$\ds \leadsto \ \ $

$\ds \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}$

$\in$

$\ds N$

Definition of Subset Product

By Cosets are Equal iff Product with Inverse in Subgroup:

$\paren {a \circ b}^{-1} \circ \paren {a' \circ b'} \in N \implies \paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$

and the proof is complete.

$\blacksquare$

Proof 2

Let $N \lhd G$ where $G$ is a group.

Consider $\paren {a \circ N} \circ \paren {b \circ N}$ as a subset product:

$\paren {a \circ N} \circ \paren {b \circ N} = \set {a \circ n_1 \circ b \circ n_2: n_1, n_2 \in N}$

This is justified by Coset Product of Normal Subgroup is Consistent with Subset Product Definition.

Since $N$ is normal, each conjugate $b^{-1} \circ N \circ b$ is contained in $N$.

So for each $n_1 \in N$ there is some $n_3 \in N$ such that $b^{-1} \circ n_1 \circ b = n_3$.

So, if $a \circ n_1 \circ b \circ n_2 \in \paren {a \circ N} \circ \paren {b \circ N}$, it follows that:

$\ds a \circ n_1 \circ b \circ n_2$

$=$

$\ds a \circ b \circ b^{-1} \circ n_1 \circ b \circ n_2$

$\ds $

$=$

$\ds a \circ b \circ n_3 \circ n_2$

$\ds $

$\in$

$\ds \paren {a \circ b} \circ N$

Definition of Subset Product

$\ds $

$\in$

$\ds N \circ b^{-1}$

Definition of Normal Subgroup

That is:

$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$

Let $n \in N$ be arbitrary.

Then:

$\ds \paren {a \circ b} \circ n$

$\in$

$\ds \paren {a \circ b} \circ N$

$\ds \leadsto \ \ $

$\ds \paren {a \circ e \circ b} \circ n$

$\in$

$\ds \paren {a \circ b} \circ N$

Group Axiom $\text G 2$: Existence of Identity Element

$\ds \leadsto \ \ $

$\ds \paren {a \circ e} \circ \paren {b \circ n}$

$\in$

$\ds \paren {a \circ b} \circ N$

Group Axiom $\text G 1$: Associativity

$\ds \leadsto \ \ $

$\ds \paren {a \circ e} \circ \paren {b \circ n}$

$\in$

$\ds \paren {a \circ N} \circ \paren {b \circ N}$

Definition of Subset Product

$\ds \leadsto \ \ $

$\ds a \circ \paren {b \circ n}$

$\in$

$\ds \paren {a \circ N} \circ \paren {b \circ N}$

Definition of Identity Element

$\ds \leadsto \ \ $

$\ds a \circ \paren {b \circ N}$

$\subseteq$

$\ds \paren {a \circ N} \circ \paren {b \circ N}$

by hypothesis

$\ds \leadsto \ \ $

$\ds \paren {a \circ b} \circ N$

$\subseteq$

$\ds \paren {a \circ N} \circ \paren {b \circ N}$

Subset Product within Semigroup is Associative: Corollary

So:

$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$

and

$\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$

The result follows by definition of set equality.

$\blacksquare$

Proof 3

Let $N \lhd G$ where $G$ is a group.

Let $a, a', b, b' \in G$ such that:

$N \circ a = N \circ a'$

and:

$N \circ b = N \circ b'$

We need to show that:

$N \circ \paren {a \circ b} = N \circ \paren {a' \circ b'}$

So:

$\ds N \circ \paren {a \circ b}$

$=$

$\ds \paren {N \circ a} \circ b$

Subset Product within Semigroup is Associative: Corollary

$\ds $

$=$

$\ds \paren {N \circ a'} \circ b$

by hypothesis

$\ds $

$=$

$\ds \paren {a' \circ N} \circ b$

Definition of Normal Subgroup

$\ds $

$=$

$\ds a' \circ \paren {N \circ b}$

Subset Product within Semigroup is Associative: Corollary

$\ds $

$=$

$\ds a' \circ \paren {N \circ b'}$

by hypothesis

$\ds $

$=$

$\ds \paren {a' \circ N} \circ b'$

Subset Product within Semigroup is Associative: Corollary

$\ds $

$=$

$\ds \paren {N \circ a'} \circ b'$

Definition of Normal Subgroup

$\ds $

$=$

$\ds N \circ \paren {a' \circ b'}$

Subset Product within Semigroup is Associative: Corollary

$\blacksquare$

Proof 4

Let $N \lhd G$ where $G$ is a group.

By Left Congruence Class Modulo Subgroup is Left Coset, it can be shown that the left congruence modulo $N$ is an equivalence relation.

Let $\RR^l_N$ denote the equivalence relation on $G$

$\RR^l_N := \set {\tuple {x, y} \in G \times G: x^{-1} y \in N}$

Let $\struct {G / N, \circ_N}$ be the quotient group of $G$ by $N$, where $\circ_N$ denotes the operation induced on $G / N$ by $\circ$.

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By Left Congruence Class Modulo Subgroup is Left Coset again, the equivalence class $\eqclass g {\RR^l_N} \in \struct {G / N, \circ_N}$ of an element $g \in G$ is the left coset $g N$.

Let $a, a', b, b' \in G$, such that:

$\eqclass a {\RR^l_N} = \eqclass {a'} {\RR^l_N}$

and:

$\eqclass b {\RR^l_N} = \eqclass {b'} {\RR^l_N}$

It follows that if:

$a, a', b, b' \in G$

then:

$ a \circ N = a' \circ N$

and:

$ b \circ N = b' \circ N$.

We need to show that:

$\eqclass {a \circ b} {\RR^l_N} = \eqclass {a' \circ b'} {\RR^l_N}$.

That is:

$ \paren { a \circ b } \circ N = \paren {a' \circ b'} \circ N$

We have:

$\ds \paren {a \circ b} \circ N$

$=$

$\ds a \circ b \circ \paren{N \circ N}$

Product of Subgroup with Itself

$\ds $

$=$

$\ds a \circ \paren{ b \circ N} \circ N$

Subset Product within Semigroup is Associative

$\ds $

$=$

$\ds a \circ \paren{ N \circ b} \circ N$

Definition of Normal Subgroup

$\ds $

$=$

$\ds \paren {a \circ N} \circ \paren {b \circ N}$

Subset Product within Semigroup is Associative

$\ds $

$=$

$\ds \paren {a' \circ N} \circ \paren {b' \circ N}$

By Hypothesis

$\ds $

$=$

$\ds a' \circ \paren{ N \circ b'} \circ N$

Subset Product within Semigroup is Associative

$\ds $

$=$

$\ds a' \circ \paren{ b' \circ N} \circ N$

Definition of Normal Subgroup

$\ds $

$=$

$\ds \paren{a' \circ b'} \circ \paren{N \circ N}$

Subset Product within Semigroup is Associative

$\ds $

$=$

$\ds \paren {a' \circ b'} \circ N$

Product of Subgroup with Itself

Hence:

$\eqclass {a \circ b} {\RR^l_N} = \eqclass {a' \circ b'} {\RR^l_N}$.

$\blacksquare$

Proof 5

Let $N \lhd G$ where $G$ is a group.

Let $a, a', b, b' \in G$ such that:

$a \circ N = a' \circ N$

and:

$b \circ N = b' \circ N$

To show that the coset product is well-defined, we need to demonstrate that:

$\paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$

So:

$\ds a \circ N$

$=$

$\ds a' \circ N$

by hypothesis

$\ds \leadsto \ \ $

$\ds a$

$\in$

$\ds a' \circ N$

Definition of Left Coset

$\ds \leadsto \ \ $

$\ds a$

$=$

$\ds a' \circ n_1$

for some $n_1 \in N$

Similarly, $b' = b \circ n_2$ for some $n_2 \in N$.

$\ds \leadsto \ \ $

$\ds a' \circ b'$

$=$

$\ds a \circ n_1 \circ b \circ n_2$

But $N \circ b = b \circ N$, as $N$ is normal, and so:

$\ds \leadsto \ \ $

$\ds a' \circ b'$

$=$

$\ds a \circ b \circ n_3 \circ n_2$

as $n_1 \circ b = b \circ n_3$ for some $n_3 \in N$

$\ds \leadsto \ \ $

$\ds a' \circ b'$

$\in$

$\ds \paren{ a \circ b } \circ N$

as $n_3 \circ n_2 \in N$

$\ds \leadsto \ \ $

$\ds \paren{ a' \circ b' } \circ N$

$=$

$\ds \paren{ a \circ b }\circ N$

Definition of Left Coset

$\blacksquare$

Also see

Sources

1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.2$: Homomorphisms
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): normal subgroup
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): normal subgroup