Hall-semidirectly 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., Hall-semidirectly extensible automorphism) must also satisfy the second automorphism property (i.e., inner automorphism)
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Suppose H is a finite group and \sigma 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, \sigma is an inner automorphism of H.

Related 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.


Given: A finite group H, a Hall-semidirectly 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). Since \sigma is Hall-semidirectly extensible and H is a Hall retract (it is a complement to the normal Hall subgroup F(G)), \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.