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
[edit]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
[edit]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).
Extensions of complete groups
[edit]Assume that a group G is a group extension given as a short exact sequence of groups
- 1 ⟶ N ⟶ G ⟶ G′ ⟶ 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, 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 CG(N) of N in G, and so G is at least a semidirect product, CG(N) ⋊ N, but the action of N on CG(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 = CG(N) × N is a direct product. The proof follows directly from the definition: N is centerless giving CG(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 CG(N) and every element, g, of G is a product (gn−1)n in CG(N)N.
References
[edit]- Robinson, Derek John Scott (1996), A course in the theory of groups, Berlin, New York: Springer-Verlag, ISBN 978-0-387-94461-6
- Rotman, Joseph J. (1994), An introduction to the theory of groups, Berlin, New York: Springer-Verlag, ISBN 978-0-387-94285-8 (chapter 7, in particular theorems 7.15 and 7.17).