# Hall-semidirectly extensible 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., Hall-semidirectly extensible automorphism) must also satisfy the second automorphism property (i.e., inner automorphism)

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## Contents

## Statement

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

## Related facts

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

## Facts used

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

## Proof

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

**To prove**: is inner.

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

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