# 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

$\blacksquare$