Subgroup is Normal iff Contains Conjugate Elements

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:


 * $\forall g \in G: \left({n \in N \iff g \circ n \circ g^{-1} \in N}\right)$
 * $\forall g \in G: \left({n \in N \iff g^{-1} \circ n \circ g \in N}\right)$

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


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

Necessary Condition
Suppose that $g \circ N = N \circ g$, by definition 1 of normality in $G$.

Let $n \in N$.

Then:

Sufficient Condition
Suppose that:


 * $\forall g \in G: \left({n \in N \iff g \circ n \circ g^{-1} \in N}\right)$

Let $g \circ n \circ g^{-1} \in N$.

Similarly, $n \circ g \in N \circ g \implies N \circ G = g \circ N$.

As $g$ is arbitrary, then so is $g^{-1}$.

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

By definition of set equality:
 * $g \circ N = N \circ g$

Also see

 * Equivalence of Definitions of Normal Subgroup