# Coset Product is Well-Defined/Proof 3

## 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 $a, a', b, b' \in G: N \circ a = N \circ a', N \circ b = N \circ b'$.

We need to show that $N \circ a \circ b = N \circ a' \circ b'$.

So:

 $\ds N \circ a \circ b$ $=$ $\ds N \circ a' \circ b$ as $N \circ a = N \circ a'$ $\ds$ $=$ $\ds a' \circ N \circ b$ $N \circ a' = a' \circ N$ as $N$ is normal $\ds$ $=$ $\ds a' \circ N \circ b'$ as $N \circ b = N \circ b'$ $\ds$ $=$ $\ds N \circ a' \circ b'$ $N \circ a' = a' \circ N$ as $N$ is normal

$\blacksquare$