Theorem: Subgroup is Subset of Conjugate iff Normal

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

Proof:
By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
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:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = g \circ N
      | r = N \circ g
      | c = {{Defof|Normal Subgroup}}
}}
{{eqn | ll= \leadsto
      | l = g^{-1} \circ \paren {g \circ N}
      | r = g^{-1} \circ \paren {N \circ g}
      | c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
      | l = \paren {g^{-1} \circ g} \circ N
      | r = g^{-1} \circ N \circ g
      | c = Subset Product within Semigroup is Associative: Corollary
}}
{{eqn | ll= \leadsto
      | l = e \circ N
      | r = g^{-1} \circ N \circ g
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = N
      | r = g^{-1} \circ N \circ g
      | c = Coset by Identity
}}
{{eqn | ll= \leadsto
      | l = N
      | o = \subseteq
      | r = g^{-1} \circ N \circ g
      | c = {{Defof|Set Equality|index = 2}}
}}
{{end-eqn}}
{{qed|lemma}}
Then by $(1)$ above:
:$\paren {\forall g \in G: N \subseteq g \circ N \circ g^{-1} } \implies \paren {\forall g \in G: N \subseteq g^{-1} \circ N \circ g}$
{{qed|lemma}}
=== 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:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = N
      | o = \subseteq
      | r = g \circ N \circ g^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = N \circ g
      | o = \subseteq
      | r = \paren {g \circ N \circ g^{-1} } \circ g
      | c = Subset Relation is Compatible with Subset Product: Corollary 2
}}
{{eqn | ll= \leadsto
      | l = N \circ g
      | o = \subseteq
      | r = \paren {g \circ N} \circ \paren {g^{-1} \circ g}
      | c = Subset Product within Semigroup is Associative: Corollary
}}
{{eqn | ll= \leadsto
      | l = N \circ g
      | o = \subseteq
      | r = \paren {g \circ N} \circ e
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = N \circ g
      | o = \subseteq
      | r = g \circ N
      | c = Coset by Identity
}}
{{end-eqn}}
Similarly:
{{improve|Do not repeat this verification again. This part was done at the beginning of the proof.}}
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = N
      | o = \subseteq
      | r = g^{-1} \circ N \circ g
      | c = 
}}
{{eqn | ll= \leadsto
      | l = g \circ N
      | o = \subseteq
      | r = g \circ \paren {g^{-1} \circ N \circ g}
      | c = Subset Relation is Compatible with Subset Product: Corollary 2
}}
{{eqn | ll= \leadsto
      | l = g \circ N
      | o = \subseteq
      | r = \paren {g \circ g^{-1} } \circ \paren {N \circ g}
      | c = Subset Product within Semigroup is Associative: Corollary
}}
{{eqn | ll= \leadsto
      | l = g \circ N
      | o = \subseteq
      | r = e \circ \paren {N \circ g}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = g \circ N
      | o = \subseteq
      | r = N \circ g
      | c = Coset by Identity
}}
{{end-eqn}}
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.
{{qed}}