Extensible implies permutation-extensible

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This article gives the statement and possibly, proof, of an implication relation between two automorphism properties. That is, it states that every automorphism satisfying the first automorphism property (i.e., extensible automorphism) must also satisfy the second automorphism property (i.e., permutation-extensible automorphism)
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Statement

Any extensible automorphism of a group is a permutation-extensible automorphism.

Definitions used

Extensible automorphism

Further information: Extensible automorphism

An automorphism \sigma of a group G is termed extensible if, for any embedding of G in a bigger group H, there exists an automorphism \sigma' of H such that the restriction of \sigma' to G equals \sigma.

Permutation-extensible automorphism

Further information: Permutation-extensible automorphism

An automorphism \sigma of a group G is termed permutation-extensible if, for any embedding of G in a symmetric group \operatorname{Sym}(S), there exists an element h \in \operatorname{Sym}(S) such that if \sigma' = c_h is conjugation by h, the restriction of \sigma' to G is \sigma. In other words, \sigma extends to an inner automorphism of \operatorname{Sym}(S).

Related facts

Applications

Facts used

  1. Symmetric groups on finite sets are complete: For n a natural number other than 2 or 6, the symmetric group on n elements is a complete group. In particular, every automorphism of it is inner.
  2. Symmetric groups on infinite sets are complete: The symmetric group on any infinite set is a complete group. In particular, every automorphism of it is inner.

Proof

Given: A group G, an extensible automorphism \sigma of G. A set S with an embedding G \to \operatorname{Sym}(S).

To prove: \sigma extends to an inner automorphism of \operatorname{Sym}(S).

Proof: We consider the following cases:

  • S is infinite: By assumption, \sigma extends to an automorphism of \operatorname{Sym}(S). By fact (2), this automorphism must be inner. Hence, \sigma extends to an inner automorphism of \operatorname{Sym}(S).
  • S is finite, and its cardinality is different from 2 or 6: By assumption, \sigma extends to an automorphism of \operatorname{Sym}(S). By fact (2), this automorphism must be inner. Hence, \sigma extends to an inner automorphism of \operatorname{Sym}(S).
  • S is finite with cardinality 2: By assumption, \sigma extends to an automorphism of \operatorname{Sym}(S). But there's only one automorphism of the symmetric group on a two-element set: the identity automorphism. This is clearly inner, so we are done.
  • S is finite with cardinality 6: We consider two cases.
    • There is an element s \in S such that every element of G fixes s: In this case, G is a subgroup of the subgroup \operatorname{Sym}(S \setminus \{ s \}), which is the symmetric group on a set of size five. Since \sigma is extensible, it extends to an automorphism of \operatorname{Sym}(S \setminus \{ s \}), and by fact (1), this automorphism must be inner. This inner automorphism can further be extended to an inner automorphism of \operatorname{Sym}(S), by using the same permutation.
    • There is no element of S fixed by all elements of G: Let T = S \sqcup \{ x_0 \} with G acting on x_0 trivially. Thus, G acts on T, with G \le \operatorname{Sym}(S) \le \operatorname{Sym}(T). T is a set of size seven. Since \sigma is extensible, it extends to an automorphism of \operatorname{Sym}(T), and by fact (1), this automorphism must be inner. Suppose h \in \operatorname{Sym}(T) is a permutation giving this inner automorphism. Then, since G fixes x_0, hGh^{-1} fixes hx_0. Since hGh^{-1} = \sigma(G) = G, we get that G fixes hx_0. Since no element of S is fixed by the whole of G, hx_0 = x_0. Thus, the permutation h restricts to a permutation on the subset S, and this inner automorphism gives the required inner automorphism extending \sigma to \operatorname{Sym}(S).