# Coset Product is Well-Defined/Proof 1

## 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 \ \$ $\ds a^{-1} \circ a'$ $\in$ $\ds N$ Cosets are Equal iff Product with Inverse in Subgroup $\ds \leadsto \ \$ $\ds b^{-1} \circ a^{-1} \circ a'$ $\in$ $\ds b^{-1} \circ N$ Definition of Subset Product $\ds \leadsto \ \$ $\ds b^{-1} \circ a^{-1} \circ a'$ $\in$ $\ds N \circ b^{-1}$ $N$ is a normal subgroup $\ds \leadsto \ \$ $\ds \exists n \in N: \,$ $\ds b^{-1} \circ a^{-1} \circ a'$ $=$ $\ds n \circ b^{-1}$ Definition of Subset Product $\ds \leadsto \ \$ $\ds \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}$ $=$ $\ds n \circ b^{-1} \circ b'$ Group Properties $\ds \leadsto \ \$ $\ds \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}$ $\in$ $\ds N$ Definition of Subset Product
$\paren {a \circ b}^{-1} \circ \paren {a' \circ b'} \in N \implies \paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$

and the proof is complete.

$\blacksquare$