Semidirect product with self-normalizing subgroup of automorphism group of coprime order implies every automorphism is inner
Suppose is a finite group, and is a nontrivial Self-normalizing subgroup (?) of the automorphism group , i.e., the normalizer of in is . Further, suppose the orders of and are relatively prime. Consider the external semidirect product:
Then, is a Group in which every automorphism is inner (?). Note that need not be a Complete group (?), though it is a complete group in many practical situations. One example where is not complete is when is a cyclic group of order two.
- Centerless and characteristic in automorphism group implies automorphism group is complete
- Characteristically simple and non-abelian implies automorphism group is complete
- Holomorph of cyclic group of odd prime order is complete
- Equivalence of definitions of normal Hall subgroup
- Schur-Zassenhaus theorem: We use here the conjugacy part of the theorem: any two permutable complements to a normal Hall subgroup are conjugate.
- Automorphism group action lemma: Suppose is a group, and are subgroups such that . Suppose is an automorphism of such that the restriction of to gives an automorphism of , and such that also restricts to an automorphism of , say . Consider the map:
that sends an element to the automorphism of induced by conjugation by (note that this is an automorphism since ). Then, we have:
where denotes conjugation by in the group .
Given: A finite group , a self-normalizing subgroup of , such that the order of and are relatively prime. is the semidirect product of and .
To prove: Every automorphism of is inner.
- is a normal Hall subgroup of : Since the order of and are relatively prime, we see that the index , which equals the order of , is relatively prime to the order of . Thus, is a Hall subgroup of . It is also clearly normal, by the definition of semidirect product.
- is characteristic in : This follows from step (1) and fact (1).
- If is an automorphism of , then : This follows from step (2).
- If is an automorphism of , then there exists a such that if denotes conjugation by , . In particular, if , then : By the previous step, , so must send to a permutable complement of . By fact (2) (Schur-Zassenhaus theorem) any two such permutable complements are conjugate by some , yielding the required result.
- If is an automorphism of such that , then there exists such that : Note that since is characteristic in , as well. We see that the conditions are satisfied for the automorphism group action lemma (fact (3)), with being the inclusion of in . So, if is the restriction of to , we obtain that the action of on is given by conjugation by in . Thus, we have such that conjugates to itself. By the assumption that is self-normalizing in , must be equal to an element inside . Call this element . Then, agrees with conjugation by on as well as on . Since , we get on .
- If is an automorphism of , is inner: By step (4), for and an automorphism that preserves , and by step (5), for some . Thus, , so is an inner automorphism of .