Extensible implies subgroup-conjugating

The result stated here is superseded by the following result, which is both stronger and simpler: extensible equals inner. In other words, the latter result has weaker and easier-to-verify hypotheses, and/or stronger and easier-to-use conclusions.
The main purpose of including this result is that it has a considerably easier proof, and/or was historically proved before the stronger result.

Statement

Every extensible automorphism of a group must be a subgroup-conjugating automorphism: it must send every subgroup to a conjugate subgroup.

NOTE: This is superseded by the result that the extensible automorphisms are precisely the inner automorphism. Inner automorphisms are obviously subgroup-conjugating, which makes the result trivial in light of that fact.

Related facts

Corollaries

• Extensible implies normal (superseded by extensible equals inner)
• Extensible automorphism-invariant equals normal (superseded by extensible equals inner)

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.
3. Equivalence of definitions of subgroup-conjugating automorphism

Proof

Using the "permutation-extensible" language by fact (3)

Fact (3) says that subgroup-conjugating automorphisms are the same as permutation-extensible automorphisms, i.e., automorphisms that can be extended to inner automorphisms for every embedding in a symmetric group. For our concrete proof, we use the "permutation-extensible" formulation.

Proof in the permutation-extensible language

Given: A group $G$, an extensible automorphism $\sigma$ of $G$. A set $S$ with an embedding $G\to \mathrm{S}\mathrm{y}\mathrm{m}(S)$.

To prove: $\sigma$ extends to an inner automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(S)$.

Proof: We consider the following cases:

• $S$ is infinite: By assumption, $\sigma$ extends to an automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(S)$. By fact (2), this automorphism must be inner. Hence, $\sigma$ extends to an inner automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(S)$.
• $S$ is finite, and its cardinality is different from $2$ or $6$: By assumption, $\sigma$ extends to an automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(S)$. By fact (1), this automorphism must be inner. Hence, $\sigma$ extends to an inner automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(S)$.
• $S$ is finite with cardinality $2$: By assumption, $\sigma$ extends to an automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(S)$. 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 $\mathrm{S}\mathrm{y}\mathrm{m}(S\setminus \{s\})$, which is the symmetric group on a set of size five. Since $\sigma$ is extensible, it extends to an automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(S\setminus \{s\})$, and by fact (1), this automorphism must be inner. This inner automorphism can further be extended to an inner automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(S)$, by using the same permutation.
  • There is no element of $S$ fixed by all elements of $G$: Let $T=S\bigsqcup \{x_0\}$ with $G$ acting on $x_0$ trivially. Thus, $G$ acts on $T$, with $G\le \mathrm{S}\mathrm{y}\mathrm{m}(S)\le \mathrm{S}\mathrm{y}\mathrm{m}(T)$. $T$ is a set of size seven. Since $\sigma$ is extensible, it extends to an automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(T)$, and by fact (1), this automorphism must be inner. Suppose $h\in \mathrm{S}\mathrm{y}\mathrm{m}(T)$ is a permutation giving this inner automorphism. Then, since $G$ fixes $x_0$, $hG{h}^{-1}$ fixes $h{x}_0$. Since $hG{h}^{-1}=\sigma (G)=G$, we get that $G$ fixes $h{x}_0$. Since no element of $S$ is fixed by the whole of $G$, $h{x}_0=x_0$. Thus, the permutation $h$ restricts to a permutation on the subset $S$, and this inner automorphism gives the required inner automorphism extending $\sigma$ to $\mathrm{S}\mathrm{y}\mathrm{m}(S)$.