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On the Schur-Zassenhaus theorem for groups of finite Morley rank
Published online by Cambridge University Press: 12 March 2014
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The Schur-Zassenhaus Theorem is one of the fundamental theorems of finite group theory. Here is its statement:
Fact 1.1 (Schur-Zassenhaus Theorem). Let G be a finite group and let N be a normal subgroup of G. Assume that the order ∣N∣ is relatively prime to the index [G:N]. Then N has a complement in G and any two complements of N are conjugate in G.
The proof can be found in most standard books in group theory, e.g., in [S, Chapter 2, Theorem 8.10]. The original statement stipulated one of N or G/N to be solvable. Since then, the Feit-Thompson theorem [FT] has been proved and it forces either N or G/N to be solvable. (The analogous Feit-Thompson theorem for groups of finite Morley rank is a long standing open problem).
The literal translation of the Schur-Zassenhaus theorem to the finite Morley rank context would state that in a group G of finite Morley rank a normal π-Hall subgroup (if it exists at all) has a complement and all the complements are conjugate to each other. (Recall that a group H is called a π-group, where π is a set of prime numbers, if elements of H have finite orders whose prime divisors are from π. Maximal π-subgroups of a group G are called π-Hall subgroups. They exist by Zorn's lemma. Since a normal π-subgroup of G is in all the π-Hall subgroups, if a group has a normal π-Hall subgroup then this subgroup is unique.)
The second assertion of the Schur-Zassenhaus theorem about the conjugacy of complements is false in general. As a counterexample, consider the multiplicative group ℂ* of the complex number field ℂ and consider the p-Sylow for any prime p, or even the torsion part of ℂ*. Let H be this subgroup. H has a complement, but this complement is found by Zorn's Lemma (consider a maximal subgroup that intersects H trivially) and the use of Zorn's Lemma is essential. In fact, by Zorn's Lemma, any subgroup that has a trivial intersection with H can be extended to a complement of H. Since ℂ* is abelian, these complements cannot be conjugated to each other.
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- Copyright © Association for Symbolic Logic 1992
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