Extensible implies subgroup-conjugating: Difference between revisions

Latest revision as of 03:03, 5 August 2026

The result stated here is superseded by the following result, which is both stronger and simpler: extensible equals inner.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NOTE: This is superseded by the result that the extensible automorphisms are precisely the inner automorphism.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Extensible implies normal (superseded by extensible equals inner)
Extensible automorphism-invariant equals normal (superseded by extensible equals inner)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

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.
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.
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 $σ$ of $G$ . A set $S$ with an embedding $G \to S y m (S)$ .

To prove: $σ$ extends to an inner automorphism of $S y m (S)$ .

Proof: We consider the following cases:

$S$ is infinite: By assumption, $σ$ extends to an automorphism of $S y m (S)$ . By fact (2), this automorphism must be inner. Hence, $σ$ extends to an inner automorphism of $S y m (S)$ .
$S$ is finite, and its cardinality is different from $2$ or $6$ : By assumption, $σ$ extends to an automorphism of $S y m (S)$ . By fact (1), this automorphism must be inner. Hence, $σ$ extends to an inner automorphism of $S y m (S)$ .
$S$ is finite with cardinality $2$ : By assumption, $σ$ extends to an automorphism of $S y m (S)$ . 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 $S y m (S ∖ {s})$ , which is the symmetric group on a set of size five. Since $σ$ is extensible, it extends to an automorphism of $S y m (S ∖ {s})$ , and by fact (1), this automorphism must be inner. This inner automorphism can further be extended to an inner automorphism of $S y m (S)$ , by using the same permutation.
- There is no element of $S$ fixed by all elements of $G$ : Let $T = S ⊔ {x_{0}}$ with $G$ acting on $x_{0}$ trivially. Thus, $G$ acts on $T$ , with $G \leq S y m (S) \leq S y m (T)$ . $T$ is a set of size seven. Since $σ$ is extensible, it extends to an automorphism of $S y m (T)$ , and by fact (1), this automorphism must be inner. Suppose $h \in S y m (T)$ is a permutation giving this inner automorphism. Then, since $G$ fixes $x_{0}$ , $h G h^{- 1}$ fixes $h x_{0}$ . Since $h G h^{- 1} = σ (G) = G$ , 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 $σ$ to $S y m (S)$ .

@@ Line 23: / Line 23: @@
 ==Proof==
-===Using the "permutation-extensible" language by Fact (3)===
+===Using the "permutation-extensible" language by fact (3)===
-By Fact (3), 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.
+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===
@@ Line 36: / Line 36: @@
 * <math>S</math> is infinite: By assumption, <math>\sigma</math> extends to an automorphism of <math>\operatorname{Sym}(S)</math>. By fact (2), this automorphism must be inner. Hence, <math>\sigma</math> extends to an inner automorphism of <math>\operatorname{Sym}(S)</math>.
-* <math>S</math> is finite, and its cardinality is different from <math>2</math> or <math>6</math>: By assumption, <math>\sigma</math> extends to an automorphism of <math>\operatorname{Sym}(S)</math>. By fact (2), this automorphism must be inner. Hence, <math>\sigma</math> extends to an inner automorphism of <math>\operatorname{Sym}(S)</math>.
+* <math>S</math> is finite, and its cardinality is different from <math>2</math> or <math>6</math>: By assumption, <math>\sigma</math> extends to an automorphism of <math>\operatorname{Sym}(S)</math>. By fact (1), this automorphism must be inner. Hence, <math>\sigma</math> extends to an inner automorphism of <math>\operatorname{Sym}(S)</math>.
 * <math>S</math> is finite with cardinality <math>2</math>: By assumption, <math>\sigma</math> extends to an automorphism of <math>\operatorname{Sym}(S)</math>. 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.
 * <math>S</math> is finite with cardinality <math>6</math>: We consider two cases.
 ** There is an element <math>s \in S</math> such that every element of <math>G</math> fixes <math>s</math>: In this case, <math>G</math> is a subgroup of the subgroup <math>\operatorname{Sym}(S \setminus \{ s \})</math>, which is the symmetric group on a set of size five. Since <math>\sigma</math> is extensible, it extends to an automorphism of <math>\operatorname{Sym}(S \setminus \{ s \})</math>, and by fact (1), this automorphism must be inner. This inner automorphism can further be extended to an inner automorphism of <math>\operatorname{Sym}(S)</math>, by using the same permutation.
 ** There is no element of <math>S</math> fixed by all elements of <math>G</math>: Let <math>T = S \sqcup \{ x_0 \}</math> with <math>G</math> acting on <math>x_0</math> trivially. Thus, <math>G</math> acts on <math>T</math>, with <math>G \le \operatorname{Sym}(S) \le \operatorname{Sym}(T)</math>. <math>T</math> is a set of size seven. Since <math>\sigma</math> is extensible, it extends to an automorphism of <math>\operatorname{Sym}(T)</math>, and by fact (1), this automorphism must be inner. Suppose <math>h \in \operatorname{Sym}(T)</math> is a permutation giving this inner automorphism. Then, since <math>G</math> fixes <math>x_0</math>, <math>hGh^{-1}</math> fixes <math>hx_0</math>. Since <math>hGh^{-1} = \sigma(G) = G</math>, we get that <math>G</math> fixes <math>hx_0</math>. Since no element of <math>S</math> is fixed by the whole of <math>G</math>, <math>hx_0 = x_0</math>. Thus, the permutation <math>h</math> restricts to a permutation on the subset <math>S</math>, and this inner automorphism gives the required inner automorphism extending <math>\sigma</math> to <math>\operatorname{Sym}(S)</math>.