Nilpotent Hall subgroups of same order are conjugate

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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 G is a finite group and π is a set of primes. Let H and K be Hall π-subgroups of G. Then, H and K are conjugate subgroups inside G: in other words, there exists gG such that gHg1=K.

Facts used

  1. Equivalence of definitions of finite nilpotent group: Specifically, the fact that in a finite nilpotent group, all the Sylow subgroups are normal
  2. Normality is upper join-closed: If NH,LG and N is normal in both H and L, then N is normal in G
  3. Sylow implies order-conjugate: Specifically, the fact that any two p-Sylow subgroups of a finite group are conjugate

Proof

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Given: A finite group G, a set π={p1,p2,,pr}</math> of prime divisors of G. Two nilpotent Hall subgroups H,K of G

To prove: H and K are conjugate.

Proof: We prove the result by a double induction: induction on the order of G, and, for a given G, induction on the size of π.

Outer induction on order: Let's assume the result is true for all groups of order smaller than the order of G.

Inner induction on size of set of primes: Note that the result is true if π has size one, by Fact (3). The hard part is the inductive step. We will assume that the result holds for the particular group G for all smaller sized sets of primes.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 H=P1P2Pr,K=Q1Q2Qr where Pi,Qi are p-Sylow subgroups of H and K respectively. Further, Pi,Qi are normal in H,K respectively. Fact (1) H,K are nilpotent. Follows directly from Fact (1).
2 Pi,Qi are both pi-Sylow subgroups of G. H,K are π-Hall in G Step (1)
3 Consider the subgroups P1P2Pr1 and Q1Q2Qr1. These are both nilpotent π{pr}-Hall subgroups of G. Steps (1), (2) Step-direct
4 P1P2Pr1 and Q1Q2Qr1 are conjugate in G. 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 K to a nilpotent subgroup L of G that looks like L=P1P2Pr1J where J is a pr-Sylow subgroup of G. K is nilpotent Steps (1), (4)
6 Suppose T=H,L is a proper subgroup of G. Then, H and L are conjugate in T, hence also in G. inductive assumption on the order of the group, H is nilpotent Step (5) Direct from the inductive assumption
7 Suppose H,L=G. Then, N:=P1P2Pr1 is normal in G. Fact (2) N is normal in H and in L on account of being a Hall subgroup of a nilpotent group. Hence, it is normal in H,L=G.
8 Suppose H,L=G. Construct N as in Step (7). Let G¯=G/N,H¯=H/N,L¯=L/N. Then, H¯ and L¯ are pr-Sylow subgroups of G¯. Steps (5), (7) direct from the construction
9 Suppose H,L=G. Continuing notation from Step (8), H¯ and L¯ are conjugate in G¯ by the image g¯ of some element gG in G¯=G/N. Then g must conjugate H to L. Fact (3) Steps (7), (8) direct from Step (8).
10 H and L are conjugate. Steps (6), (9) Step (6) shows that H and L are conjugate if H,LG. Step (9) shows that H and L are conjugate if H,L=G. Thus, H and L are conjugate in every case.
11 H and K are conjugate. Steps (5), (10) Step (5) says that K and L are conjugate. Step (10) says that H and L are conjugate. Since conjugacy is an equivalence relation, we obtain that H and L are conjugate.

References

Textbook references