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This article gives a proof/explanation of the equivalence of multiple definitions for the term subgroup-conjugating automorphism
View a complete list of pages giving proofs of equivalence of definitions

Contents 1 The definitions that we have to prove as equivalent 1.1 Subgroup-conjugating automorphism 1.2 Permutation-extensible automorphism 1.3 Permutation-pushforwardable automorphism 2 Related facts 2.1 Applications 3 Proof 3.1 Permutation-extensible implies subgroup-conjugating 3.2 Subgroup-conjugating implies permutation-pushforwardable 3.3 Puermutation-pushforwardable implies permutation-extensible

The definitions that we have to prove as equivalent

Subgroup-conjugating automorphism

An automorphism σ of a group G is termed a subgroup-conjugating automorphism if, for any subgroup H of G, H and σ(H) are conjugate subgroups.

Permutation-extensible automorphism

An automorphism σ of a group G is termed a permutation-extensible automorphism if, for any injective homomorphism , where is the symmetric group on a set, there is a permutation α of S such that , where c_α denotes conjugation by α.

Permutation-pushforwardable automorphism

An automorphism σ of a group G is termed a permutation-extensible automorphism if, for any homomorphism (not necessarily injective) , where is the symmetric group on a set, there is a permutation α of S such that , where c_α denotes conjugation by α.

Related facts

Applications

• Extensible implies subgroup-conjugating

Proof

Permutation-extensible implies subgroup-conjugating

In this proof, we use the notation c_g for conjugation by g, which is the map .

Given: A group G, a subgroup H, a permutation-extensible automorphism σ of G.

To prove: σ(H) is a conjugate subgroup to H.

Proof: The case that H is the trivial subgroup is obvious, so we give the proof for H nontrivial.

Let S = G / H and T = G, with G acting on S and T both by left multiplication. Let U be the disjoint union of S and T. G acts faithfully on U, so we have an embedding:

.

By the condition, there exists such that σ extends to conjugation by α in . Consider the element . Clearly, the isotropy subgroup of the element is the subgroup c_α(H) in G, which equals σ(H).

Now, observe that αH cannot be in T, because then its isotropy group would be trivial, and σ(H) cannot be trivial if H is nontrivial. Thus, , so there exists such that αH = gH. Thus, the isotropy subgroup of α(H) is the conjugate subgroup c_g(H), and thus σ(H) = c_g(H).

Subgroup-conjugating implies permutation-pushforwardable

Given: A group G, an automorphism σ of G such that σ sends every subgroup to a conjugate subgroup. A homomorphism for some set S.

To prove: There exists a permutation α of S such that .

Proof: Let be the orbit of some point in S under the induced action of G. Let , and let H be the isotropy subgroup of x. Let x' be any point in whose isotropy subgroup is σ(H). Such a point exists because H and σ(H) are conjugate subgroups. Now define:

.

This is well-defined and gives a permutation of the orbit . If we define α in this way for each orbit, we get a permutation of S and it satisfies the condition .

Puermutation-pushforwardable implies permutation-extensible

This implication is obvious.