Subgroup is Superset of Conjugate iff Normal

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

Proof
By definition, a subgroup is normal in $G$ iff:


 * $\forall g \in G: g \circ N = N \circ g$

First note that:
 * $(1): \quad \left({\forall g \in G: g \circ N \circ g^{-1} \subseteq N}\right) \iff \left({\forall g \in G: g^{-1} \circ N \circ g \subseteq N}\right)$

which is show by, for example, setting $h := g^{-1}$ and substituting.

Necessary Condition
Suppose that $N$ is normal in $G$.

Then:

Similarly:

Sufficient Condition
Let $N$ be a subgroup of $G$ such that:
 * $\forall g \in G: g \circ N \circ g^{-1} \subseteq N$

and so from $(1)$ above:
 * $\forall g \in G: g^{-1} \circ N \circ g \subseteq N$

Then:

Similarly:

Thus we have:
 * $N \circ g \subseteq g \circ N$
 * $g \circ N \subseteq N \circ g$

By Equality of Sets, it follows that:
 * $g \circ N = N \circ g$

hence the result.

Also see

 * Normal Subgroup Equivalent Definitions

Also as definition
Some sources use this property as the definition of a normal subgroup.