Equivalence of definitions of subgroup-conjugating automorphism

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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

The definitions that we have to prove as equivalent

Subgroup-conjugating automorphism

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

Permutation-extensible automorphism

An automorphism \sigma of a group G is termed a permutation-extensible automorphism if, for any injective homomorphism i: H \to \operatorname{Sym}(S), where \operatorname{Sym}(S) is the symmetric group on a set, there is a permutation \alpha of S such that i \circ \sigma = c_\alpha \circ i, where c_\alpha denotes conjugation by \alpha.

Permutation-pushforwardable automorphism

An automorphism \sigma of a group G is termed a permutation-extensible automorphism if, for any homomorphism (not necessarily injective) \rho: H \to \operatorname{Sym}(S), where \operatorname{Sym}(S) is the symmetric group on a set, there is a permutation \alpha of S such that \rho \circ \sigma = c_\alpha \circ \rho, where c_\alpha denotes conjugation by \alpha.

Related facts

Applications

Proof

Permutation-extensible implies subgroup-conjugating

In this proof, we use the notation c_g for conjugation by g, which is the map x \mapsto gxg^{-1}.

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

To prove: \sigma(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:

G \to \operatorname{Sym}(U).

By the condition, there exists \alpha \in \operatorname{Sym}(U) such that \sigma extends to conjugation by \alpha in \operatorname{Sym}(U). Consider the element H \in U. Clearly, the isotropy subgroup of the element \alpha H \in U is the subgroup c_\alpha(H) in G, which equals \sigma(H).

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

Subgroup-conjugating implies permutation-pushforwardable

Given: A group G, an automorphism \sigma of G such that \sigma sends every subgroup to a conjugate subgroup. A homomorphism \rho:G \to \operatorname{Sym}(S) for some set S.

To prove: There exists a permutation \alpha of S such that \rho \circ \sigma = c_\alpha \circ \rho.

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

\alpha(g \cdot x) = \sigma(g) \cdot x'.

This is well-defined and gives a permutation of the orbit \mathcal{O}. If we define \alpha in this way for each orbit, we get a permutation of S and it satisfies the condition \rho \circ \sigma = c_\alpha \circ \rho.

Puermutation-pushforwardable implies permutation-extensible

This implication is obvious.