Coset Product is Well-Defined/Proof 4

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.

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

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 50.1$ Quotient groups