Nilpotent Hall subgroups of same order are conjugate
Contents
History
The result was proved by Wielandt in 1954.
Statement
Verbal statement
In a finite group, any two nilpotent Hall subgroups of the same order are conjugate subgroups.
Statement with symbols
Suppose is a finite group and is a set of primes. Let and be Hall -subgroups of . Then, and are conjugate subgroups inside : in other words, there exists such that .
Facts used
- Equivalence of definitions of finite nilpotent group: Specifically, the fact that in a finite nilpotent group, all the Sylow subgroups are normal
- Normality is upper join-closed: If and is normal in both and , then is normal in
- Sylow implies order-conjugate: Specifically, the fact that any two -Sylow subgroups of a finite group are conjugate
Proof
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Given: A finite group , a set </math> of prime divisors of . Two nilpotent Hall subgroups of
To prove: and are conjugate.
Proof: We prove the result by a double induction: induction on the order of , and, for a given , induction on the size of .
Outer induction on order: Let's assume the result is true for all groups of order smaller than the order of .
Inner induction on size of set of primes: Note that the result is true if has size one, by Fact (3). The hard part is the inductive step. We will assume that the result holds for the particular group for all smaller sized sets of primes.
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | where are -Sylow subgroups of and respectively. Further, are normal in respectively. | Fact (1) | are nilpotent. | Follows directly from Fact (1). | |
2 | are both -Sylow subgroups of . | are -Hall in | Step (1) | ||
3 | Consider the subgroups and . These are both nilpotent -Hall subgroups of . | Steps (1), (2) | Step-direct | ||
4 | and are conjugate in . | inductive assumption on the size of the set of primes | Steps (2), (3) | direct from the step and the inductive assumption. | |
5 | We can use the conjugating element of Step (4) to conjugate to a nilpotent subgroup of that looks like where is a -Sylow subgroup of . | is nilpotent | Steps (1), (4) | ||
6 | Suppose is a proper subgroup of . Then, and are conjugate in , hence also in . | inductive assumption on the order of the group, is nilpotent | Step (5) | Direct from the inductive assumption | |
7 | Suppose . Then, is normal in . | Fact (2) | is normal in and in on account of being a Hall subgroup of a nilpotent group. Hence, it is normal in . | ||
8 | Suppose . Construct as in Step (7). Let . Then, and are -Sylow subgroups of . | Steps (5), (7) | direct from the construction | ||
9 | Suppose . Continuing notation from Step (8), and are conjugate in by the image of some element in . Then must conjugate to . | Fact (3) | Steps (7), (8) | direct from Step (8). | |
10 | and are conjugate. | Steps (6), (9) | Step (6) shows that and are conjugate if . Step (9) shows that and are conjugate if . Thus, and are conjugate in every case. | ||
11 | and are conjugate. | Steps (5), (10) | Step (5) says that and are conjugate. Step (10) says that and are conjugate. Since conjugacy is an equivalence relation, we obtain that and are conjugate. |
References
Textbook references
- Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 236, Exercise 2 (Chapter 6)