Coset Product is Well-Defined/Proof 4

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.

By Left Congruence Class Modulo Subgroup is Left Coset, it can be shown that the left congruence modulo $N$ is an equivalence relation.

Let $\RR^l_N$ denote the equivalence relation on $G$

$\RR^l_N := \set {\tuple {x, y} \in G \times G: x^{-1} y \in N}$

Let $\struct {G / N, \circ_N}$ be the quotient group of $G$ by $N$, where $\circ_N$ denotes the operation induced on $G / N$ by $\circ$.

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By Left Congruence Class Modulo Subgroup is Left Coset again, the equivalence class $\eqclass g {\RR^l_N} \in \struct {G / N, \circ_N}$ of an element $g \in G$ is the left coset $g N$.

Let $a, a', b, b' \in G$, such that:

$\eqclass a {\RR^l_N} = \eqclass {a'} {\RR^l_N}$

and:

$\eqclass b {\RR^l_N} = \eqclass {b'} {\RR^l_N}$

It follows that if:

$a, a', b, b' \in G$

then:

$ a \circ N = a' \circ N$

and:

$ b \circ N = b' \circ N$.

We need to show that:

$\eqclass {a \circ b} {\RR^l_N} = \eqclass {a' \circ b'} {\RR^l_N}$.

That is:

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

We have:

Hence:

$\eqclass {a \circ b} {\RR^l_N} = \eqclass {a' \circ b'} {\RR^l_N}$.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 50.1$ Quotient groups