Subgroup is Subset of Conjugate iff Normal

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

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


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

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

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

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

Then:

Then by $(1)$ above:
 * $\paren {\forall g \in G: N \subseteq g \circ N \circ g^{-1} } \implies \paren {\forall g \in G: N \subseteq g^{-1} \circ N \circ g}$

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

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

Then:

Similarly:

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

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

Hence the result.

Also see

 * Equivalence of Definitions of Normal Subgroup
 * Subgroup is Superset of Conjugate iff Normal