Kernel of homomorphism is normal subgroup

Question

When trying to prove that a normal subgroup is a kernel, you map x to Nx(the coset). f(x) = Nx
Why exactly do we do that? Why when you take an element in the normal subgroup, and map it to Nx, the solution is e? Could you show it to me?

I mean, let n be in N which is a normal subgroup. why f(n) = e?

Answer 1

A normal subgroup $N$ of a group $G$ is, in particular, the kernel of the homomorphism $\phi:G \rightarrow G/N$. This map is defined by $x \mapsto xN$ (or $x \mapsto Nx$; it doesn't matter since $xN = Nx$ whenever $N$ is normal), and multiplication in $G/N$ is defined by $(xN)(yN) = xyN$. You can check that this is indeed a homomorphism, and we have $x \in \ker(\phi) \iff x \in N$. To prove that last fact:

If $x \in N$, then $\phi(x) = xN = N = eN$, and $eN$ is the identity element in $G/N$. Therefore, $x \in \ker(\phi)$.

On the other hand, if $x \in \ker(\phi)$, then $xN = eN \implies (x \cdot e) \in N \implies x \in N$.