Hall retract implies order-conjugate
This article describes a fact or result that is not basic but it still well-established and standard. The fact may involve terms that are themselves non-basic
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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Hall retract) must also satisfy the second subgroup property (i.e., order-conjugate subgroup)
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This article states and (possibly) proves a fact that involves two finite groups of relatively prime order, requiring the additional datum that at least one of them is solvable. Due to the Feit-Thompson theorem, we know that for two finite groups of relatively prime orders, one of them is solvable. Hence, the additional datum of solvability can be dropped. However, the proof of the Feit-Thompson theorem is considered heavy machinery.
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Contents
Name
This result is often stated as the conjugacy part of the Schur-Zassenhaus theorem.
Statement
Suppose is a finite group and is a Hall retract of . In other words, is a Hall subgroup of that is also a retract: there exists a normal complement to in . Note that is thus a normal Hall subgroup. Assume, further, that either or is a solvable group.
Then, if there is any subgroup of of the same order as , and are conjugate subgroups.
Note that the assumption that either or is a solvable group is superfluous because, as a corollary of the odd-order theorem, given two groups of coprime order, one of them is solvable.
Proof
We reduce the problem to basic cases by induction on the order.
Given: A finite group . A normal -Hall subgroup of , and two Hall subgroups of .
To prove: is conjugate to .
Proof:
- Reduction to the case that is minimal normal in : Suppose is a nontrivial normal subgroup of properly contained in . Let be the quotient map. is normal -Hall in , and are -Hall subgroups. By the inductive hypothesis, and are conjugate in . Suppose is such that conjugates to . Then, conjugates to some -Hall subgroup inside . Thus, it suffices to show that any -Hall subgroup of is conjugate to . This again follows by the inductive hypothesis on the order, since is normal -Hall in .
- Reduction to the case that is trivial; in other words, has no nontrivial normal -subgroups: Suppose is a nontrivial normal -subgroup of . Then, by the product formula, and are -subgroups, and since are Hall, this forces and . Consider the quotient map . Under this quotient map, is a normal -Hall subgroup, and are -Hall. The inductive hypothesis yields that and are conjugate in . Since and both contain the kernel of , this yields that and are conjugate in .
- Resolution of the case where is solvable: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
- Resolution of the case where is solvable: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
References
- Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 221, Theorem 2.1(ii), Chapter 6 (Solvable and pi-solvable groups), Section 6.2 (The Schur-Zassenhaus theorem), ^{More info}