# Coset Product of Normal Subgroup is Consistent with Subset Product Definition

## Theorem

Let $\struct {G, \circ}$ be a group.

Let $N$ be a normal subgroup of $G$.

Let $a, b \in G$.

Let $a \circ N$ and $b \circ N$ be the left cosets of $a$ and $b$ by $N$.

Then the coset product:

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

is consistent with the definition of the coset product as the subset product of $a \circ N$ and $b \circ N$:

$\paren {a \circ N} \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$

## Proof

Consider the set:

$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$

As $e \in N$, have:

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

by Group Axiom $\text G 1$: Associativity of $\circ$.

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

Now let $x \in a \circ N$ and $y \in b \circ N$.

Then by the definition of subset product:

$\exists n_1 \in N: x = a \circ n_1$
$\exists n_2 \in N: y = b \circ n_2$

It follows that:

 $\ds x \circ y$ $=$ $\ds \paren {a \circ n_1} \circ \paren {b \circ n_2}$ $\ds$ $=$ $\ds \paren {a \circ n_1} \circ \paren {n_3 \circ b}$ for some $n_3 \in N$: Definition of Normal Subgroup $\ds$ $=$ $\ds a \circ \paren {\paren {n_1 \circ n_3} \circ b}$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds a \circ \paren {n_4 \circ b}$ for some $n_4 \in N$: Definition of Subgroup $\ds$ $=$ $\ds a \circ \paren {b \circ n_5}$ for some $n_5 \in N$: Definition of Normal Subgroup $\ds$ $=$ $\ds \paren {a \circ b} \circ n_5$ Group Axiom $\text G 1$: Associativity $\ds$ $\in$ $\ds \paren {a \circ b} \circ N$ Definition of Subset Product

So the definition by subset product:

$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$

leads to the definition of coset product as:

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

$\blacksquare$