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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., Hall-semidirectly extensible automorphism) must also satisfy the second automorphism property (i.e., inner automorphism)
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Contents 1 Statement 2 Related facts 3 Facts used 4 Proof

Statement

Suppose H is a finite group and σ is a Hall-semidirectly extensible automorphism: in other words, H extends to an automorphism of G for any finite group G containing H as a Hall retract. Then, σ is an inner automorphism of H.

Related facts

• Finite-quotient-pullbackable implies inner
• Finite-extensible implies inner
• Finite solvable-extensible implies inner

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 p-dominated group G with H as Fitting quotient: in other words, there exists a finite complete group G such that the Fitting subgroup F(G) is a p-group, and H is a subgroup of G such that .

Proof

Given: A finite group H, a Hall-semidirectly extensible automorphism σ of H.

To prove: σ is inner.

Proof: Let p be a prime not dividing the order of H. Consider the group G constructed by fact (1). Since σ is Hall-semidirectly extensible and H is a Hall retract (it is a complement to the normal Hall subgroup F(G)), σ extends to an automorphism σ' of G. Further, since G is complete, there exists such that σ' is conjugation by g.

Let be the retraction with kernel F(G). Note that conjugation by g preserves F(G), hence it induces a conjugation map on H as a quotient, namely, conjugation by the element . However, since the restriction of ρ to the subgroup H is the identity map, we conclude that conjugation by g has the same effect on H as conjugation by ρ(g). In particular, σ equals conjugation by ρ(g), and hence is inner.