Coset Product of Normal Subgroup is Consistent with Subset Product Definition
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Theorem
Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Let $a \circ N$ and $b \circ N$ be the left cosets of $a$ and $b$ by $N$.
Then the coset product:
- $\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is consistent with the definition of the coset product as the subset product of $a \circ N$ and $b \circ N$:
- $\paren {a \circ N} \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
Proof
Consider the set:
- $\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
As $e \in N$, have:
- $\paren {a \circ b} \circ N = \paren {a \circ e} \circ \paren {b \circ N} \subseteq \paren {a \circ N} \circ \paren {b \circ N}$
by Group Axiom $\text G 1$: Associativity of $\circ$.
Hence $\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$.
Now let $x \in a \circ N$ and $y \in b \circ N$.
Then by the definition of subset product:
- $\exists n_1 \in N: x = a \circ n_1$
- $\exists n_2 \in N: y = b \circ n_2$
It follows that:
\(\ds x \circ y\) | \(=\) | \(\ds \paren {a \circ n_1} \circ \paren {b \circ n_2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \circ n_1} \circ \paren {n_3 \circ b}\) | for some $n_3 \in N$: Definition of Normal Subgroup | |||||||||||
\(\ds \) | \(=\) | \(\ds a \circ \paren {\paren {n_1 \circ n_3} \circ b}\) | Group Axiom $\text G 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds a \circ \paren {n_4 \circ b}\) | for some $n_4 \in N$: Definition of Subgroup | |||||||||||
\(\ds \) | \(=\) | \(\ds a \circ \paren {b \circ n_5}\) | for some $n_5 \in N$: Definition of Normal Subgroup | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \circ b} \circ n_5\) | Group Axiom $\text G 1$: Associativity | |||||||||||
\(\ds \) | \(\in\) | \(\ds \paren {a \circ b} \circ N\) | Definition of Subset Product |
So the definition by subset product:
- $\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
leads to the definition of coset product as:
- $\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
$\blacksquare$