# Complete group

In mathematics, a group *G* is said to be **complete** if every automorphism of *G* 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, *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.

The automorphism group of a simple group is an almost simple group; for a non-abelian simple group *G*, the automorphism group of *G* is complete.

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 (Robinson 1996, section 13.5).

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, 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*.