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

By Cosets are Equal iff Product with Inverse in Subgroup:

$\paren {a \circ b}^{-1} \circ \paren {a' \circ b'} \in N \implies \paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$

and the proof is complete.

$\blacksquare$