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?

  • $\begingroup$ If $n$ is in $N$ what do the elements of $nN$ look like? The reason we map group elements to cosets in order to show normal subgroups are kernels is because of the first isomorphism theorem. $\endgroup$
    – anon
    Jul 22, 2016 at 18:07

1 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 $xN = eN \implies (x \cdot e) \in N \implies x \in N$.

  • $\begingroup$ The last part is what I don't get. Why is it that x∈ker(ϕ)⟺x∈N? $\endgroup$
    – Matam
    Jul 22, 2016 at 18:11
  • $\begingroup$ Added some detail @Matam $\endgroup$
    – Kaj Hansen
    Jul 22, 2016 at 18:15
  • $\begingroup$ @Matam If $x\notin N$ then $\phi(x)=xN\ne N$ because $x\in xN$ and $x\notin xN$. And if $x\in N$ then xN=N$ hence $x\in\ker\phi$ $\endgroup$ Jul 22, 2016 at 18:16
  • $\begingroup$ One last thing, why does xN = eN => (x * e) in N? $\endgroup$
    – Matam
    Jul 22, 2016 at 18:28
  • $\begingroup$ We have $eN = N$, and since $N$ is a subgroup, we also know $e \in N$. So, if we write down the elements of the set $xN = \{xn \ | \ n \in N \}$, then $xe$ must be one of those elements. $\endgroup$
    – Kaj Hansen
    Jul 22, 2016 at 18:33

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