Let be a subgroup of group $G$ . Then $N$ is normal in $G$ if and only if there is a group $H$ and a group homomorphism $ϕ : G \to H$ such that the kernel of $ϕ$ is $N$ .

Proof

"Normal" implies "is a kernel"

Let $N$ be normal, so it is closed under conjugation. Then we may define the quotient_group $G / N$ , whose elements are the left cosets of $N$ in $G$ , and where the operation is that $g N + h N = (g + h) N$ . This group is well-defined (proof).

Now there is a homomorphism $ϕ : G \to G / N$ given by $g \mapsto g N$ . This is indeed a homomorphism, by definition of the group operation $g N + h N = (g + h) N$ .

The kernel of this homomorphism is precisely ${g : g N = N}$ ; that is simply $N$ :

Certainly $N \subseteq {g : g N = N}$ (because $n N = N$ for all $n$ , since $N$ is closed as a subgroup of $G$ );

We have ${g : g N = N} \subseteq N$ because if $g N = N$ then in particular $g e \in N$ (where $e$ is the group identity) so $g \in N$ .

"Is a kernel" implies "normal"

Let $ϕ : G \to H$ have kernel $N$ , so $ϕ (n) = e$ if and only if $n \in N$ . We claim that $N$ is closed under conjugation by members of $G$ .

Indeed, $ϕ (h n h^{- 1}) = ϕ (h) ϕ (n) ϕ (h^{- 1}) = ϕ (h) ϕ (h^{- 1})$ since $ϕ (n) = e$ . But that is $ϕ (h h^{- 1}) = ϕ (e)$ , so $h n h^{- 1} \in N$ .

That is, if $n \in N$ then $h n h^{- 1} \in N$ , so $N$ is normal.