Coset Product is Well-Defined/Proof 1

From ProofWiki
Jump to navigation Jump to search

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
\(\ds \leadsto \ \ \) \(\ds a^{-1} \circ a'\) \(\in\) \(\ds N\) 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
\(\ds \leadsto \ \ \) \(\ds \paren{b^{-1} \circ a^{-1} \circ a' } \circ b'\) \(=\) \(\ds \paren{n \circ b^{-1} } \circ b'\) Group Axiom $\text G 0$: Closure
\(\ds \leadsto \ \ \) \(\ds \paren{b^{-1} \circ a^{-1} } \circ \paren {a' \circ b'}\) \(=\) \(\ds \paren{n \circ b^{-1} } \circ b'\) Group Axiom $\text G 1$: Associativity
\(\ds \leadsto \ \ \) \(\ds \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}\) \(=\) \(\ds \paren{n \circ b^{-1} } \circ b'\) Inverse of Group Product
\(\ds \leadsto \ \ \) \(\ds \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}\) \(=\) \(\ds n \circ \paren{ b^{-1} \circ b' }\) Group Axiom $\text G 1$: Associativity
\(\ds \leadsto \ \ \) \(\ds \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}\) \(\in\) \(\ds N\) 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.

$\blacksquare$