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.