# 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”.