Coset Product is Well-Defined/Proof 5

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

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

and:

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

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:

\(\ds a \circ N\)

\(=\)

\(\ds a' \circ N\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds a\)

\(\in\)

\(\ds a' \circ N\)

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 $n_3 \in N$

\(\ds \leadsto \ \ \)

\(\ds a' \circ b'\)

\(\in\)

\(\ds \paren{ a \circ b } \circ N\)

as $n_3 \circ n_2 \in N$

\(\ds \leadsto \ \ \)

\(\ds \paren{ a' \circ b' } \circ N\)

\(=\)

\(\ds \paren{ a \circ b }\circ N\)

Definition of Left Coset

$\blacksquare$

Sources

• 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.10$