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”.