# Coset Product is Well-Defined/Proof 2

## 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.

Consider $\paren {a \circ N} \circ \paren {b \circ N}$ as a subset product:

$\paren {a \circ N} \circ \paren {b \circ N} = \set {a \circ n_1 \circ b \circ n_2: n_1, n_2 \in N}$

This is justified by Coset Product of Normal Subgroup is Consistent with Subset Product Definition.

Since $N$ is normal, each conjugate $b^{-1} \circ N \circ b$ is contained in $N$.

So for each $n_1 \in N$ there is some $n_3 \in N$ such that $b^{-1} \circ n_1 \circ b = n_3$.

So, if $a \circ n_1 \circ b \circ n_2 \in \paren {a \circ N} \circ \paren {b \circ N}$, it follows that:

 $\ds a \circ n_1 \circ b \circ n_2$ $=$ $\ds a \circ b \circ b^{-1} \circ n_1 \circ b \circ n_2$ $\ds$ $=$ $\ds a \circ b \circ n_3 \circ n_2$ $\ds$ $\in$ $\ds \paren {a \circ b} \circ N$ Definition of Subset Product $\ds$ $\in$ $\ds N \circ b^{-1}$ Definition of Normal Subgroup

That is:

$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$

Then:

 $\ds a \circ b \circ n$ $\in$ $\ds \paren {a \circ b} \circ N$ $\ds \leadsto \ \$ $\ds a \circ e \circ b \circ n$ $\in$ $\ds \paren {a \circ N} \circ \paren {b \circ N}$ $\ds \leadsto \ \$ $\ds \paren {a \circ b} \circ N$ $\subseteq$ $\ds \paren {a \circ N} \circ \paren {b \circ N}$

So:

$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$

and

$\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$

The result follows by definition of set equality.

$\blacksquare$