Coset Product is Well-Defined/Proof 4

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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.


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

Let $a, b \in G$.

We have:

\(\ds \paren {a \circ N} \circ \paren {b \circ N}\) \(=\) \(\ds a \circ N \circ b \circ N\) Subset Product within Semigroup is Associative
\(\ds \) \(=\) \(\ds a \circ b \circ N \circ N\) Definition of Normal Subgroup
\(\ds \) \(=\) \(\ds \paren {a \circ b} \circ N\) Product of Subgroup with Itself