Finite-quotient-pullbackable implies inner
This article gives the statement and possibly, proof, of an implication relation between two automorphism properties. That is, it states that every automorphism satisfying the first automorphism property (i.e., finite-quotient-pullbackable automorphism) must also satisfy the second automorphism property (i.e., inner automorphism)
View all automorphism property implications | View all automorphism property non-implications
Get more facts about finite-quotient-pullbackable automorphism|Get more facts about inner automorphism
Then, is an inner automorphism of .
- Every finite group is the Fitting quotient of a p-dominated group for any prime p not dividing its order: Suppose is a finite group and is a prime not dividing the order of . Then, there exists a finite complete group such that the Fitting subgroup is a -group, and is a semidirect product of and . In particular, .
Given: A finite group , an automorphism of such that for any surjective homomorphism there is an automorphism of such that .
To prove: is inner.
Proof: Let be a prime not dividing the order of . Using fact (1), there is a finite group such that the Fitting subgroup is a -group, and is isomorphic to . Let be the quotient map.
By assumption, there exists an automorphism of such that . Since is complete, is inner, so there exists such that is conjugation by . It is easy to see then that equals conjugation by the element , and thus, is inner.