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

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