Coset Product is Well-Defined/Proof 2

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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

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.

$\blacksquare$


Sources

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