Complete group

Assume that a group G is a group extension given as a short exact sequence of groups

1 ⟶ N ⟶ G ⟶ G′ ⟶ 1

with kernel, N, and quotient, G′. 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 C_G(N) of N in G, and so G is at least a semidirect product, C_G(N) ⋊ N, but the action of N on C_G(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 = C_G(N) × N is a direct product. The proof follows directly from the definition: N is centerless giving C_G(N) ∩ N is trivial. If g is an element of G then it induces an automorphism of N by conjugation, but N = Aut(N) and this conjugation must be equal to conjugation by some element n of N. Then conjugation by gn⁻¹ is the identity on N and so gn⁻¹ is in C_G(N) and every element, g, of G is a product (gn⁻¹)n in C_G(N)N.

• Robinson, Derek John Scott (1996), A course in the theory of groups, Berlin, New York: Springer-Verlag, ISBN 978-0-387-94461-6
• Rotman, Joseph J. (1994), An introduction to the theory of groups, Berlin, New York: Springer-Verlag, ISBN 978-0-387-94285-8 (chapter 7, in particular theorems 7.15 and 7.17).

• Joel David Hamkins: How tall is the automorphism tower of a group?