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| In [[mathematics]], a [[group (mathematics)|group]] ''G'' is said to be '''complete''' if every [[automorphism]] of ''G'' is [[inner automorphism|inner]], and the group is a centerless group; that is, it has a trivial [[outer automorphism group]] and trivial [[Center (group theory)|center]].
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| Equivalently, a group is complete if the conjugation map <math>G \to \mbox{Aut}(G)</math> (sending an element ''g'' to conjugation by ''g'') is an isomorphism: one-to-one implies centerless, as no inner automorphisms are the identity, while onto corresponds to no outer automorphisms.
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| == Examples ==
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| As an example, all the [[symmetric group]]s ''S''<sub>''n''</sub> 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 [[Automorphisms of the symmetric and alternating groups#exceptional outer automorphism|outer automorphism]].
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| The automorphism group of a simple group ''G'' is an [[almost simple group]];
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| for a nonabelian [[simple group]] ''G'', the automorphism group of ''G'' is complete.
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| == Properties ==
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| A complete group is always [[isomorphism|isomorphic]] to its [[automorphism group]] (via sending an element to conjugation by that element), although the reverse need not hold: for example, the [[dihedral group]] of eight elements is isomorphic to its automorphism group, but it is not complete. For a discussion, see {{harv|Robinson|1996|loc=section 13.5}}.
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| == Extensions of complete groups ==
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| Assume that a group ''G'' is a group extension given as a [[short exact sequence]] of groups
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| :<math> 1 \rightarrow N \rightarrow G \rightarrow G' \rightarrow 1 </math>
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| with [[Kernel (group theory)|kernel]] ''N'' and quotient '' G' ''. If the kernel ''N'' is a complete group then the extension splits: G is [[isomorphic]] to the [[direct product of groups|direct product]] N × G'. A proof using homomorphisms and exact sequences can be given in a natural way: The action of ''G'' (by [[Conjugation (group theory)|conjugation]]) on the normal subgroup ''N'' gives rise to a [[group homomorphism]] <math> \phi: G \rightarrow \mathrm{Aut}(N) \cong N </math>. 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<sub>''G''</sub>(''N'') of ''N'' in ''G'', and so ''G'' is at least a [[semidirect product]] C<sub>''G''</sub>(''N'') ⋊ ''N'', but the action of ''N'' on C<sub>''G''</sub>(''N'') is trivial, and so the product is direct. This proof is somewhat interesting since the original exact sequence is reversed during the proof.
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| 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<sub>''G''</sub>(''N'') × ''N'' is a direct product. The proof follows directly from the definition: ''N'' is centerless giving C<sub>''G''</sub>(''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''<sup>−1</sup> is the identity on ''N'' and so ''gn''<sup>−1</sup> is in C<sub>''G''</sub>(''N'') and every element ''g'' of ''G'' is a product (''gn''<sup>−1</sup>)''n'' in C<sub>''G''</sub>(''N'')''N''.
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| == References ==
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| *{{Citation | last1=Robinson | first1=Derek John Scott | title=A course in the theory of groups | publisher=[[Springer-Verlag]] | location=Berlin, New York | isbn=978-0-387-94461-6 | year=1996}}
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| * {{Citation | last1=Rotman | first1=Joseph J. | title=An introduction to the theory of groups | publisher=[[Springer-Verlag]] | location=Berlin, New York | isbn=978-0-387-94285-8 | year=1994}} (chapter 7, in particular theorems 7.15 and 7.17).
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| == External links ==
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| * [[Joel David Hamkins]]: [http://arxiv.org/abs/math/9808094v1 How tall is the automorphism tower of a group?]
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| [[Category:Properties of groups]]
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