Extensible implies permutation-extensible

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., extensible automorphism) must also satisfy the second subgroup 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 ${\sigma }^{\prime }$ of $H$ such that the restriction of ${\sigma }^{\prime }$ 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 $\mathrm{S}\mathrm{y}\mathrm{m}(S)$, there exists an element $h\in \mathrm{S}\mathrm{y}\mathrm{m}(S)$ such that if ${\sigma }^{\prime }=c_h$ is conjugation by $h$, the restriction of ${\sigma }^{\prime }$ to $G$ is $\sigma$. In other words, $\sigma$ extends to an inner automorphism of $\mathrm{S}\mathrm{y}\mathrm{m}(S)$.

Related facts

Applications

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

Facts used

1. Symmetric groups 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 $G\to \mathrm{S}\mathrm{y}\mathrm{m}(S)$.

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 (2), 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 $\mathrm{S}\mathrm{y}\mathrm{m}(S)$.