Let $N$ 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 $\phi:G \to H$ such that the kernel of $\phi$ 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 $gN + hN = (g+h)N$. This group is well-defined (proof).

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

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

• Certainly $N \subseteq \{ g : gN = N \}$ (because $nN = N$ for all $n$, since $N$ is closed as a subgroup of $G$);
• We have $\{ g : gN = N \} \subseteq N$ because if $gN = N$ then in particular $g e \in N$ (where $e$ is the group identity) so $g \in N$.

"Is a kernel" implies "normal"

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

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

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