with kernel, , and quotient, . If the kernel, , is a complete group then the extension splits: is isomorphic to the direct product, . A proof using homomorphisms and exact sequences can be given in a natural way: The action of (by conjugation) on the normal subgroup, gives rise to a group homomorphism, . Since and has trivial center the homomorphism is surjective and has an obvious section given by the inclusion of in . The kernel of is the centralizer of in , and so is at least a semidirect product, , but the action of on is trivial, and so the product is direct. This proof is somewhat interesting since the original exact sequence is reversed during the proof.