# Coset Product is Well-Defined/Proof 5

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

To show that the coset product is well-defined, we need to demonstrate that:

$\paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$

So:

 $\ds a \circ N$ $=$ $\ds a' \circ N$ $\ds \leadsto \ \$ $\ds a$ $\in$ $\ds a' \circ N$ Definition of Left Coset $\ds \leadsto \ \$ $\ds a$ $=$ $\ds a' \circ n_1$ for some $n_1 \in N$ Similarly, $b' = b \circ n_2$ for some $n_2 \in N$. $\ds \leadsto \ \$ $\ds a' \circ b'$ $=$ $\ds a \circ n_1 \circ b \circ n_2$ But $N \circ b = b \circ N$, as $N$ is normal, and so: $\ds \leadsto \ \$ $\ds a' \circ b'$ $=$ $\ds a \circ b \circ n_3 \circ n_2$ as $n_1 \circ b = b \circ n_3$ for some $n_3 \in N$ $\ds \leadsto \ \$ $\ds a' \circ b'$ $\in$ $\ds a \circ b \circ N$ as $n_3 \circ n_2 \in N$ $\ds \leadsto \ \$ $\ds a' \circ b' \circ N$ $=$ $\ds a \circ b \circ N$ Definition of Left Coset

$\blacksquare$