Complete group
In mathematics, a group is said to be complete if every automorphism of is inner, and it is centerless; that is, it has a trivial outer automorphism group and trivial center. Equivalently, a group is complete if the conjugation map, (sending an element to conjugation by ), 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. Examples. As an example, all the symmetric groups, , are complete except when }. For the case , the group has a non-trivial center, while  for the case , there is an outer automorphism. The automorphism group of a simple group is an almost simple group; for a non-abelian simple group , the automorphism group of is complete. Properties. A complete group is always isomorphic to its automorphism group (via sending an element to conjugation by that element), although the converse need not hold: for example, the dihedral group of 8 elements is isomorphic to its automorphism group, but it is not complete. For a discussion, see . Extensions of complete groups. Assume that a group is a group extension given as a short exact sequence of  groups 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 can be restated in terms of elements and internal conditions: If is a normal, complete subgroup of a group , then is a direct product. The proof follows directly from the definition: is centerless giving is trivial. If is an element of then it induces an automorphism of by conjugation, but and this conjugation must be equal to conjugation by some element of . Then conjugation by is the identity on and so is in and every element, , of is a product in .