Complete group

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In mathematics, a group G is said to be complete if every automorphism of G is inner, and the group is a centerless group; that is, it has a trivial center. Equivalently, if the conjugation map $G \to \mbox{Aut}(G)$ (sending an element g to conjugation by g) is an isomorphism: 1-to-1 corresponds to centerless, onto corresponds to no outer automorphisms.

As an example, all the symmetric groups S_n are complete except when n = 2 or 6. For the case n = 2 the group has a nontrivial center, while for the case n = 6 there is an outer automorphism.

For a nonabelian simple group G, the automorphism group of G is complete, i.e.

Inn(Aut(G)) = Aut(Aut(G)).

The automorphism group of a simple group is termed an almost simple group.

[edit] Extensions of complete groups

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

$1 \rightarrow N \rightarrow G \rightarrow G' \rightarrow 1$

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 $\phi: G \rightarrow Aut(N) \cong N$ . 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 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 = 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.

[edit] References

Joseph J. Rotman, An introduction to the theory of groups, Graduate texts in Mathematics, vol. 148, Springer Verlag (chapter 7, in particular theorems 7.15 and 7.17).