Extensions of Complete Groups
Assume that a group G is a group extension given as a short exact sequence of groups
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 . 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 proof is somewhat interesting since the original exact sequence is reversed during the proof.
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. 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 CG(N) and every element g of G is a product (gn-1)n in CG(N)N.
Read more about this topic: Complete Group
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