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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)
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Statement

Suppose H is a finite group and σ is a finite-quotient-pullbackable automorphism of G: in other words, for any surjective homomorphism , there is an automorphism σ' of G such that .

Then, σ is an inner automorphism of G.

Facts used

1. Every finite group is the Fitting quotient of a p-dominated group for any prime p not dividing its order: Suppose H is a finite group and p is a prime not dividing the order of H. Then, there exists a finite complete group G such that the Fitting subgroup F(G) is a p-group, and G is a semidirect product of F(G) and H. In particular, .

Proof

Given: A finite group H, an automorphism σ of H such that for any surjective homomorphism there is an automorphism σ' of G such that .

To prove: σ is inner.

Proof: Let p be a prime not dividing the order of H. Using fact (1), there is a finite group G such that the Fitting subgroup F(G) is a p-group, and G / F(G) is isomorphic to H. Let be the quotient map.

By assumption, there exists an automorphism σ' of G such that . Since G is complete, σ' is inner, so there exists such that σ' is conjugation by g. It is easy to see then that σ equals conjugation by the element ρ(g), and thus, σ is inner.