Equivalently, a group is complete if the conjugation map, G → Aut(G) (sending an element g to conjugation by g), is an isomorphism: injectivity implies that only conjugation by the identity element is the identity automorphism, meaning the group is centerless, while surjectivity implies it has no outer automorphisms.
If the kernel, N, is a complete group then the extension splits: G is isomorphic to the direct product, N × G′.
A proof using homomorphisms and exact sequences can be given in a natural way: The action of G (by conjugation) on the normal subgroup, N, gives rise to a group homomorphism, φ : G → Aut(N) ≅ N. Since Out(N) = 1 and N has trivial center the homomorphism φ is surjective and has an obvious section given by the inclusion of N in G. The kernel of φ is the centralizer CG(N) of N in G, and so G is at least a semidirect product, CG(N) ⋊ N, but the action of N on CG(N) is trivial, and so the product is direct.
This can be restated in terms of elements and internal conditions: If N is a normal, complete subgroup of a group G, then G = CG(N) × N is a direct product.
The proof follows directly from the definition: N is centerless giving CG(N) ∩ N is trivial.