Subgroup is normal if and only if it is the kernel of a homomorphism

by Patrick Stevens Jun 17 2016 updated Jun 17 2016

The "correct way" to think about normal subgroups is as kernels of homomorphisms.

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$~$.


"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$~$:

"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.