Coset Product is Well-Defined/Proof 3

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

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

and:

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

We need to show that:

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


So:

\(\ds N \circ \paren {a \circ b}\) \(=\) \(\ds \paren {N \circ a} \circ b\) Subset Product within Semigroup is Associative: Corollary
\(\ds \) \(=\) \(\ds \paren {N \circ a'} \circ b\) by hypothesis
\(\ds \) \(=\) \(\ds \paren {a' \circ N} \circ b\) Definition of Normal Subgroup
\(\ds \) \(=\) \(\ds a' \circ \paren {N \circ b}\) Subset Product within Semigroup is Associative: Corollary
\(\ds \) \(=\) \(\ds a' \circ \paren {N \circ b'}\) by hypothesis
\(\ds \) \(=\) \(\ds \paren {a' \circ N} \circ b'\) Subset Product within Semigroup is Associative: Corollary
\(\ds \) \(=\) \(\ds \paren {N \circ a'} \circ b'\) Definition of Normal Subgroup
\(\ds \) \(=\) \(\ds N \circ \paren {a' \circ b'}\) Subset Product within Semigroup is Associative: Corollary

$\blacksquare$


Sources

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