Semidirect product with self-normalizing subgroup of automorphism group of coprime order implies every automorphism is inner

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Suppose N is a finite group, and H is a nontrivial Self-normalizing subgroup (?) of the automorphism group \operatorname{Aut}(N), i.e., the normalizer of H in \operatorname{Aut}(N) is H. Further, suppose the orders of N and H are relatively prime. Consider the external semidirect product:

G = N \rtimes H.

Then, G is a Group in which every automorphism is inner (?). Note that G need not be a Complete group (?), though it is a complete group in many practical situations. One example where G is not complete is when N is a cyclic group of order two.

Related facts

Facts used

  1. Equivalence of definitions of normal Hall subgroup
  2. Schur-Zassenhaus theorem: We use here the conjugacy part of the theorem: any two permutable complements to a normal Hall subgroup are conjugate.
  3. Automorphism group action lemma: Suppose G is a group, and N,H \le G are subgroups such that H \le N_G(N). Suppose \sigma is an automorphism of G such that the restriction of \sigma to N gives an automorphism \alpha of N, and such that \sigma also restricts to an automorphism of H, say \sigma'. Consider the map:

\rho: H \to \operatorname{Aut}(N)

that sends an element h \in H to the automorphism of N induced by conjugation by h (note that this is an automorphism since H \le N_G(N)). Then, we have:

\rho \circ \sigma' = c_\alpha \circ \rho

where c_\alpha denotes conjugation by \alpha in the group \operatorname{Aut}(N).


First part

Given: A finite group N, a self-normalizing subgroup H of \operatorname{Aut}(N), such that the order of N and H are relatively prime. G is the semidirect product of N and H.

To prove: Every automorphism of G is inner.


  1. N is a normal Hall subgroup of G: Since the order of N and H are relatively prime, we see that the index [G:N], which equals the order of H, is relatively prime to the order of N. Thus, N is a Hall subgroup of G. It is also clearly normal, by the definition of semidirect product.
  2. N is characteristic in G: This follows from step (1) and fact (1).
  3. If \sigma is an automorphism of G, then \sigma(N) = N: This follows from step (2).
  4. If \sigma is an automorphism of G, then there exists a g \in G such that if c_g denotes conjugation by G, c_g(H) = \sigma(H). In particular, if \tau = c_{g^{-1}} \circ \sigma, then \tau(H) = H: By the previous step, \sigma(N) = N, so \sigma must send H to a permutable complement of N. By fact (2) (Schur-Zassenhaus theorem) any two such permutable complements are conjugate by some g \in G, yielding the required result.
  5. If \tau is an automorphism of G such that \tau(H) = H, then there exists h \in H such that \tau = c_h: Note that since N is characteristic in G, \tau(N) = N as well. We see that the conditions are satisfied for the automorphism group action lemma (fact (3)), with \rho being the inclusion of H in \operatorname{Aut}(N). So, if \alpha is the restriction of \tau to N, we obtain that the action of \tau on H is given by conjugation by \alpha in \operatorname{Aut}(N). Thus, we have \alpha \in \operatorname{Aut}(N) such that \alpha conjugates H to itself. By the assumption that H is self-normalizing in \operatorname{Aut}(N), \alpha must be equal to an element inside H. Call this element h. Then, \tau agrees with conjugation by h on H as well as on N. Since G = NH, we get \tau = c_h on G.
  6. If \sigma is an automorphism of G, \sigma is inner: By step (4), \sigma = c_g \circ\tau for g \in G and \tau an automorphism that preserves H, and by step (5), \tau = c_h for some h \in H. Thus, \sigma = c_g \circ c_h = c_{gh}, so \sigma is an inner automorphism of G.