Coset Product is Well-Defined

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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 1

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.

$\blacksquare$


Proof 2

Let $N \lhd G$ where $G$ is a group.


Consider $\paren {a \circ N} \circ \paren {b \circ N}$ as a subset product:

$\paren {a \circ N} \circ \paren {b \circ N} = \set {a \circ n_1 \circ b \circ n_2: n_1, n_2 \in N}$

This is justified by Coset Product of Normal Subgroup is Consistent with Subset Product Definition.


Since $N$ is normal, each conjugate $b^{-1} \circ N \circ b$ is contained in $N$.

So for each $n_1 \in N$ there is some $n_3 \in N$ such that $b^{-1} \circ n_1 \circ b = n_3$.

So, if $a \circ n_1 \circ b \circ n_2 \in \paren {a \circ N} \circ \paren {b \circ N}$, it follows that:

\(\ds a \circ n_1 \circ b \circ n_2\) \(=\) \(\ds a \circ b \circ b^{-1} \circ n_1 \circ b \circ n_2\)
\(\ds \) \(=\) \(\ds a \circ b \circ n_3 \circ n_2\)
\(\ds \) \(\in\) \(\ds \paren {a \circ b} \circ N\) Definition of Subset Product
\(\ds \) \(\in\) \(\ds N \circ b^{-1}\) Definition of Normal Subgroup

That is:

$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$


Then:

\(\ds a \circ b \circ n\) \(\in\) \(\ds \paren {a \circ b} \circ N\)
\(\ds \leadsto \ \ \) \(\ds a \circ e \circ b \circ n\) \(\in\) \(\ds \paren {a \circ N} \circ \paren {b \circ N}\)
\(\ds \leadsto \ \ \) \(\ds \paren {a \circ b} \circ N\) \(\subseteq\) \(\ds \paren {a \circ N} \circ \paren {b \circ N}\)


So:

$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$

and

$\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$

The result follows by definition of set equality.

$\blacksquare$


Proof 3

Let $N \lhd G$ where $G$ is a group.


Let $a, a', b, b' \in G: N \circ a = N \circ a', N \circ b = N \circ b'$.

We need to show that $N \circ a \circ b = N \circ a' \circ b'$.


So:

\(\ds N \circ a \circ b\) \(=\) \(\ds N \circ a' \circ b\) as $N \circ a = N \circ a'$
\(\ds \) \(=\) \(\ds a' \circ N \circ b\) $N \circ a' = a' \circ N$ as $N$ is normal
\(\ds \) \(=\) \(\ds a' \circ N \circ b'\) as $N \circ b = N \circ b'$
\(\ds \) \(=\) \(\ds N \circ a' \circ b'\) $N \circ a' = a' \circ N$ as $N$ is normal

$\blacksquare$


Proof 4

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

$\blacksquare$


Proof 5

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:

\(\ds a \circ N\) \(=\) \(\ds a' \circ N\)
\(\ds \leadsto \ \ \) \(\ds a\) \(\in\) \(\ds a' \circ N\) 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 $n_3 \in N$
\(\ds \leadsto \ \ \) \(\ds a' \circ b'\) \(\in\) \(\ds a \circ b \circ N\) as $n_3 \circ n_2 \in N$
\(\ds \leadsto \ \ \) \(\ds a' \circ b' \circ N\) \(=\) \(\ds a \circ b \circ N\) Definition of Left Coset

$\blacksquare$


Also see


Sources

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