Finite solvable-extensible 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 solvable-extensible automorphism) must also satisfy the second automorphism property (i.e., inner automorphism)
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Statement

Suppose H is a finite solvable group and \sigma is a finite solvable-extensible automorphism: in other words, \sigma extends to an automorphism of G for any finite solvable group G containing H. Then, \sigma is an inner automorphism of H.

Related facts

Weaker facts

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 G = F(G) \rtimes H.
  2. Prime power order implies nilpotent, Nilpotent implies solvable
  3. Solvability is extension-closed

Proof

Given: A finite group H, a finite solvable-extensible automorphism \sigma of H.

To prove: \sigma is inner.

Proof: Let p be a prime not dividing the order of H. Consider the group G constructed by fact (1). Note first that F(G) is a p-group, hence by fact (2), is solvable. Thus, both F(G) and G/F(G) \cong H are solvable, and hence, by fact (3), G itself is a finite solvable group.

Since \sigma is finite solvable-extensible, \sigma extends to an automorphism \sigma' of G. Further, since G is complete, there exists g \in G such that \sigma' is conjugation by g.

Let \rho:G \to H 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 \rho(g) \in H. However, since the restriction of \rho to the subgroup H is the identity map, we conclude that conjugation by g has the same effect on H as conjugation by \rho(g). In particular, \sigma equals conjugation by \rho(g), and hence is inner.