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 $\varphi :G\to H$ such that the kernel of $\varphi$ 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 $\varphi :G\to G/N$ 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 $ge\in N$ (where $e$ is the group identity) so $g\in N$.

"Is a kernel" implies "normal"

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

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

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