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 | lo= \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 | lo= \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: {{begin-eqn}} {{eqn | lo= \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}}