Subgroup equals Conjugate iff Normal/Proof 2

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $N$ be a subgroup of $G$.

Then $N$ is normal in $G$ :


 * $\forall g \in G: g \circ N \circ g^{-1} = N$
 * $\forall g \in G: g^{-1} \circ N \circ g = N$

Proof
From Subgroup Superset of Conjugate iff Normal, $N$ is normal in $G$ :
 * $\forall g \in G: N \supseteq g \circ N \circ g^{-1}$
 * $\forall g \in G: N \supseteq g^{-1} \circ N \circ g$

From Subgroup Subset of Conjugate iff Normal, $N$ is normal in $G$ :
 * $\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
 * $\forall g \in G: N \subseteq g^{-1} \circ N \circ g$

The result follows by definition of set equality.