Coset Product is Well-Defined/Proof 4
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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 $a, b \in G$.
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 |
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 50.1$ Quotient groups