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.

Examples
As an example, all the symmetric groups, $Sn$, are complete except when $n ∈ {2, 6}$. For the case $n = 2$, the group has a non-trivial center, while for the case $n = 6$, there is an outer automorphism.

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.

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

with kernel, $1 ⟶ N ⟶ G ⟶ G′ ⟶ 1$, and quotient, $N$. If the kernel, $G′$, is a complete group then the extension splits: $N$ is isomorphic to the direct product, $G$. A proof using homomorphisms and exact sequences can be given in a natural way: The action of $N × G′$ (by conjugation) on the normal subgroup, $G$, gives rise to a group homomorphism, $N$. Since $φ : G → Aut(N) ≅ N$ and $Out(N) = 1$ has trivial center the homomorphism $N$ is surjective and has an obvious section given by the inclusion of $φ$ in $N$. The kernel of $G$ is the centralizer $φ$ of $CG(N)$ in $N$, and so $G$ is at least a semidirect product, $G$, but the action of $C_{G}(N) ⋊ N$ on $N$ is trivial, and so the product is direct.

This can be restated in terms of elements and internal conditions: If $C_{G}(N)$ is a normal, complete subgroup of a group $N$, then $G$ is a direct product. The proof follows directly from the definition: $G = CG(N) × N$ is centerless giving $N$ is trivial. If $C_{G}(N) &cap; N$ is an element of $g$ then it induces an automorphism of $G$ by conjugation, but $N$ and this conjugation must be equal to conjugation by some element $N = Aut(N)$ of $n$. Then conjugation by $N$ is the identity on $gn−1$ and so $N$ is in $gn^{−1}$ and every element, $C_{G}(N)$, of $g$ is a product $G$ in $(gn^{−1})n$.