How to show that all normals subgroups are exactly the kernel of morphisms definite on G?

Question

I want to show that all normals subgroups are exactly the kernel of morphisms definite on G. I don't know how to do. Moreover i want to have an explanation on what is a morphism definite on G. Is a morphism in G to another group G' is a morphism definite on G ? Or is that mean that morphism has to be in G itself or in a subgroup of G?

Thanks

Answer 1

Let $G$ be a group and $N$ be a normal subgroup. Then you can consider the factor group $G/N$ with the projection $$\pi:G\to G/N:x\in G\mapsto xN$$ Then $\pi$ is a group homomorphism of $G$ onto the factor group $G/N$ with kernel $N$, that's $\mathrm{Ker}(\pi)=N$. This proves that each normal subgroup $N$ of $G$ is the kernel of some group homomorphism of $G$ into another group.