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

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.

Parents:

• Normal subgroup

Normal subgroups are subgroups which are in some sense “the same from all points of view”.