Finite-quotient-pullbackable implies inner

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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 \sigma is a finite-quotient-pullbackable automorphism of G: in other words, for any surjective homomorphism \rho:G \to H, there is an automorphism \sigma' of G such that \rho \circ \sigma' = \sigma \circ \rho.

Then, \sigma 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, G/F(G) \cong H.

Proof

Given: A finite group H, an automorphism \sigma of H such that for any surjective homomorphism \rho:G \to H there is an automorphism \sigma' of G such that \rho \circ \sigma' = \sigma \circ \rho.

To prove: \sigma 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 \rho:G \to H be the quotient map.

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