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 \pi is a set of primes. Let H and K be Hall \pi-subgroups of G. Then, H and K are conjugate subgroups inside G: in other words, there exists g \in G such that gHg^{-1} = 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 N \le H, L \le G 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 \pi = \{ p_1,p_2 , \dots, p_r\}</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 \pi.

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 \pi 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 = P_1P_2 \dots P_r, K = Q_1Q_2 \dots Q_r where P_i,Q_i are p-Sylow subgroups of H and K respectively. Further, P_i, Q_i are normal in H,K respectively. Fact (1) H,K are nilpotent. Follows directly from Fact (1).
2 P_i,Q_i are both p_i-Sylow subgroups of G. H,K are \pi-Hall in G Step (1)
3 Consider the subgroups P_1P_2\dots P_{r-1} and Q_1Q_2 \dots Q_{r-1}. These are both nilpotent \pi \setminus \{ p_r \}-Hall subgroups of G. Steps (1), (2) Step-direct
4 P_1P_2 \dots P_{r-1} and Q_1Q_2 \dots Q_{r-1} 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 = P_1P_2 \dots P_{r-1}J where J is a p_r-Sylow subgroup of G. K is nilpotent Steps (1), (4)
6 Suppose T = \langle H, L \rangle 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 \langle H, L \rangle = G. Then, N := P_1P_2 \dots P_{r-1} 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 \langle H, L \rangle = G.
8 Suppose \langle H, L \rangle = G. Construct N as in Step (7). Let \overline{G} = G/N, \overline{H} = H/N, \overline{L} = L/N. Then, \overline{H} and \overline{L} are p_r-Sylow subgroups of \overline{G}. Steps (5), (7) direct from the construction
9 Suppose \langle H,L \rangle = G. Continuing notation from Step (8), \overline{H} and \overline{L} are conjugate in \overline{G} by the image \overline{g} of some element g \in G in \overline{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 \langle H, L \rangle \ne G. Step (9) shows that H and L are conjugate if \langle H,L \rangle = 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