# Coset Product is Well-Defined

## 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 1

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.

$\blacksquare$

## Proof 2

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.

$\blacksquare$

## Proof 3

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

$\blacksquare$

## Proof 4

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

$\blacksquare$

## Proof 5

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

$\blacksquare$