Subgroup is Normal iff Left Cosets are Right Cosets
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Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) if and only if:
- Every right coset of $N$ in $G$ is a left coset
or equivalently:
- The right coset space of $N$ in $G$ equals its left coset space.
Proof
Necessary Condition
Let $N$ be a normal subgroup of $G$ by Definition 1.
Then the equality of the coset spaces follows directly from definition of normal subgroup and coset.
$\Box$
Sufficient Condition
Suppose that every right coset of $N$ in $G$ is a left coset of $N$ in $G$.
Let $g \in G$.
Since every right coset of $N$ in $G$ is a left coset, there exists an $h \in G$ such that $N \circ g = h \circ N$.
By Element of Group is in its own Coset:
- $g \in N \circ g = h \circ N$
From Element in Left Coset iff Product with Inverse in Subgroup:
- $g^{-1} \circ h \in N$
Then:
\(\ds N \circ g\) | \(=\) | \(\ds \paren {g \circ g^{-1} } \circ \paren {h \circ N}\) | Definition of Inverse Element and Coset by Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds g \circ \paren {\paren { g^{-1} \circ h } \circ N}\) | Subset Product within Semigroup is Associative: Corollary | |||||||||||
\(\ds \) | \(=\) | \(\ds g \circ N\) | $g^{-1} \circ h \in N$ and Left Coset Equals Subgroup iff Element in Subgroup |
Since this holds for all $g \in G$, $N$ is normal in $G$ (by definition 1).
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \alpha$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Proposition $7.4 \ \text{(e)}$