Coset Product is Well-Defined
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 $a, a', b, b' \in G$ 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.
$\blacksquare$
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:
- $\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$
Let $n \in N$ be arbitrary.
Then:
\(\ds \paren {a \circ b} \circ n\) | \(\in\) | \(\ds \paren {a \circ b} \circ N\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a \circ e \circ b} \circ n\) | \(\in\) | \(\ds \paren {a \circ b} \circ N\) | Group Axiom $\text G 2$: Existence of Identity Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a \circ e} \circ \paren {b \circ n}\) | \(\in\) | \(\ds \paren {a \circ b} \circ N\) | Group Axiom $\text G 1$: Associativity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a \circ e} \circ \paren {b \circ n}\) | \(\in\) | \(\ds \paren {a \circ N} \circ \paren {b \circ N}\) | Definition of Subset Product | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ \paren {b \circ n}\) | \(\in\) | \(\ds \paren {a \circ N} \circ \paren {b \circ N}\) | Definition of Identity Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ \paren {b \circ N}\) | \(\subseteq\) | \(\ds \paren {a \circ N} \circ \paren {b \circ N}\) | by hypothesis | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a \circ b} \circ N\) | \(\subseteq\) | \(\ds \paren {a \circ N} \circ \paren {b \circ N}\) | Subset Product within Semigroup is Associative: Corollary |
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 $a, a', b, b' \in G$ 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 |
$\blacksquare$
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 $a, a', b, b' \in G$ 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 |
$\blacksquare$
Also see
- Coset Product of Normal Subgroup is Consistent with Subset Product Definition
- Congruence Modulo Normal Subgroup is Congruence Relation
Sources
- 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.2$: Homomorphisms