Equivalence of definitions of subgroup-conjugating automorphism
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
- 1 The definitions that we have to prove as equivalent
- 2 Related facts
- 3 Proof
The definitions that we have to prove as equivalent
An automorphism of a group is termed a subgroup-conjugating automorphism if, for any subgroup of , and are conjugate subgroups.
An automorphism of a group is termed a permutation-extensible automorphism if, for any injective homomorphism , where is the symmetric group on a set, there is a permutation of such that , where denotes conjugation by .
An automorphism of a group 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 of such that , where denotes conjugation by .
Permutation-extensible implies subgroup-conjugating
In this proof, we use the notation for conjugation by , which is the map .
Given: A group , a subgroup , a permutation-extensible automorphism of .
To prove: is a conjugate subgroup to .
Proof: The case that is the trivial subgroup is obvious, so we give the proof for nontrivial.
Let and , with acting on and both by left multiplication. Let be the disjoint union of and . acts faithfully on , so we have an embedding:
By the condition, there exists such that extends to conjugation by in . Consider the element . Clearly, the isotropy subgroup of the element is the subgroup in , which equals .
Now, observe that cannot be in , because then its isotropy group would be trivial, and cannot be trivial if is nontrivial. Thus, , so there exists such that . Thus, the isotropy subgroup of is the conjugate subgroup , and thus .
Subgroup-conjugating implies permutation-pushforwardable
Given: A group , an automorphism of such that sends every subgroup to a conjugate subgroup. A homomorphism for some set .
To prove: There exists a permutation of such that .
Proof: Let be the orbit of some point in under the induced action of . Let , and let be the isotropy subgroup of . Let be any point in whose isotropy subgroup is . Such a point exists because and are conjugate subgroups. Now define:
This is well-defined and gives a permutation of the orbit . If we define in this way for each orbit, we get a permutation of and it satisfies the condition .
Puermutation-pushforwardable implies permutation-extensible
This implication is obvious.