Extensible implies permutation-extensible

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

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
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 $$\operatorname{Sym}(S)$$.

Applications

 * Extensible implies subgroup-conjugating
 * Extensible implies normal
 * Extensible automorphism-invariant equals normal

Facts used

 * 1) uses::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) uses::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 $$\operatorname{Sym}(S)$$.