Finite solvable-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., finite solvable-extensible automorphism) must also satisfy the second automorphism property (i.e., inner automorphism)
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Suppose is a finite solvable group and is a finite solvable-extensible automorphism: in other words, extends to an automorphism of for any finite solvable group containing . Then, is an inner automorphism of .
- Finite-quotient-pullbackable implies inner
- Hall-semidirectly extensible implies inner
- Finite-extensible implies inner
- 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 .
- Prime power order implies nilpotent, Nilpotent implies solvable
- Solvability is extension-closed
Given: A finite group , a finite solvable-extensible automorphism of .
To prove: is inner.
Proof: Let be a prime not dividing the order of . Consider the group constructed by fact (1). Note first that is a -group, hence by fact (2), is solvable. Thus, both and are solvable, and hence, by fact (3), itself is a finite solvable group.
Since is finite solvable-extensible, 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.