Subgroup is Normal iff Left Cosets are Right Cosets

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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.


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.


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:

$g^{-1} \circ h \in N$


\(\displaystyle N \circ g\) \(=\) \(\displaystyle \paren {g \circ g^{-1} } \circ \paren {h \circ N}\) $\quad$ Definition of Inverse Element and Coset by Identity $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g \circ \paren {\paren { g^{-1} \circ h } \circ N}\) $\quad$ Subset Product within Semigroup is Associative: Corollary $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g \circ N\) $\quad$ $g^{-1} \circ h \in N$ and Left Coset Equals Subgroup iff Element in Subgroup $\quad$

Since this holds for all $g \in G$, $N$ is normal in $G$ (by definition 1).


Also see