A generalization of Burnside’s p-nilpotency criterion

Theorem 1 ([8, Theorem 7.2.1])

Let p be a prime number, let G be a finite group, and let P be a Sylow p-subgroup of G. Suppose that NG⁢(P) is p-nilpotent. Then G is p-nilpotent if P is abelian.

Theorem 2 ([4])

Let p be a prime number, let G be a finite group, and let P be a Sylow p-subgroup of G. Suppose that NG⁢(P) is p-nilpotent. Then G is p-nilpotent if the nilpotency class of P is less than p, i.e., P≤Zp-1⁢(P), where Zp-1⁢(P) denotes the (p-1)-th term of the ascending central series of P.

Theorem 2 can be deduced from Yoshida’s transfer theorem (see [7, Theorem 10.1]).

Theorem 4 ([2, Theorem D])

Let p be an odd prime number, let G be a finite group, and let P be a Sylow p-subgroup of G. Suppose that NG⁢(P) is p-nilpotent. Then G is p-nilpotent if Ω⁢(P)≤Zp-1⁢(P).

Theorem 5 ([1, Theorem 1 and Theorem 2])

Let p be a prime number, let G be a finite group, and let P be a Sylow p-subgroup of G. Assume that NG⁢(P) is p-nilpotent. Then G is p-nilpotent if either of the following conditions holds:

1. Ω⁢(P∩G′)≤Z⁢(P).

2. When p=2, Ω1⁢(P∩G′)≤Z⁢(P) and P is quaternion-free.

where G′ is the commutator subgroup of G.

Let p be a prime number, let G be a finite group, and let P be a Sylow p-subgroup of G. Assume that NG⁢(P) is p-nilpotent. Then G is p-nilpotent if either of the following holds:

1. G satisfied the (p-1)-th Engel condition for (P,Ω⁢(P∩Op⁢(G))).

2. p=2, Ω1⁢(P)≤Z⁢(P∩O2⁢(G)) and P is quaternion-free.

Suppose that p is a prime. Let G be a group and P a Sylow p-subgroup of G. Assume that NG⁢(P) is p-nilpotent. Then G is p-nilpotent if either of the following holds:

1. Ω⁢(P∩Op⁢(G))≤Zp-1⁢(P).

2. p=2, Ω1⁢(P∩Op⁢(G))≤Z⁢(P) and P is quaternion-free.

Lemma 1 ([9, Lemma 2.3])

Let P be a 2-group and A an automorphism group of P of odd order. If A acts trivially on Ω1⁢(P) and P is quaternion-free, then A=1.

Lemma 2 ([5, Theorem B])

Let H be a p-soluble linear group over a field of characteristic p satisfying Op⁢(H)=1. If g is an element of order pm in H, then the minimal polynomial of g is (x-1)r, where r=pm, unless there is an integer m0, not greater than m, such that pm0-1 is a power of a prime q for which cq⁢(H)>1 (where cq⁢(H) denotes the class of a Sylow q-subgroup of H; that is, the length of the upper or lower central series of H). In the latter case, if m0 is the smallest such integer, one has

Proof of the Main Theorem.

Assume that either condition (1) or (2) holds. Suppose that the assertion is false, and let G be a counterexample with minimal order.

Step 1: Op′⁢(G)=1. Clearly, the hypotheses hold for G/Op′⁢(G). Hence the minimal choice of G implies that Op′⁢(G)=1.

Step 2: For any subgroup S of G with P≤S⊊G, S is p-nilpotent. Since we have NS⁢(P)≤NG⁢(P), NS⁢(P) is p-nilpotent. Note that Op⁢(S)≤Op⁢(G), the hypotheses hold for S as well, and thus S is p-nilpotent by the minimal choice of G.

Step 3: We have P∩Op⁢(G)⊴G and thus Op⁢(G) is p-soluble. We first prove that P∩Op⁢(G)⊊P and thus Op⁢(G)⊊G. Assume that P∩Op⁢(G)=P.

Suppose that p=2. If condition (1) holds, then G satisfies the 1-st Engel condition for (P,Ω⁢(P)). This means that Ω⁢(P)≤Z⁢(P). Hence G is p-nilpotent by Theorem 5 (1), a contradiction. If condition (2) holds, applying Theorem 5 (2) directly, we have G is p-nilpotent, a contradiction. Hence p is odd. Write

Let H be a subgroup of G such that H/Op⁢(G)=NG¯⁢(Z⁢(J⁢(P¯))), where J⁢(P¯) is the Thompson subgroup of P¯. Since Op⁢(G¯)=1, H⊊G. Noticing that P≤H, we have that H is p-nilpotent by Step 2. Thus H/Op⁢(G)=NG¯⁢(Z⁢(J⁢(P¯))) is p-nilpotent as well. It then follows from the Glauberman–Thompson theorem (see [3, Theorem 8.3.1]) that G¯ is p-nilpotent. If G¯ is a p′-group, then P=Op⁢(G) and G=NG⁢(P) is p-nilpotent by hypothesis, a contradiction. Hence G¯ is not a p′-group. Let K/Op⁢(G) be the normal p-complement of G¯. Then we have that G/K≅(G/Op′⁢(G))/(K/Op′⁢(G)) is a p-group, and thus Op⁢(G)≤K. Moreover, K⊊G as G¯ is not a p′-group. Therefore P≰K. But P≤Op⁢(G)≤K, a contradiction. Hence P∩Op⁢(G)⊊P, as desired.

Suppose that NG⁢(P∩Op⁢(G))⊊G. Noticing that P≤NG⁢(P∩Op⁢(G)), we have that NG⁢(P∩Op⁢(G)) is p-nilpotent by Step 2. Then NOp⁢(G)⁢(P∩Op⁢(G)) is p-nilpotent. It follows that the hypotheses of the theorem hold for Op⁢(G). Since Op⁢(G)⊊G, Op⁢(G) is p-nilpotent by the minimal choice of G. This implies that G is p-nilpotent, a contradiction. Therefore NG⁢(P∩Op⁢(G))=G. It follows that P∩Op⁢(G) is a normal Sylow p-subgroup of Op⁢(G) and thus Op⁢(G) is p-soluble.

Step 4: The group G is p-soluble and CG⁢(Op⁢(G))≤Op⁢(G)≠1. By Step 3, G is p-soluble. Hence CG⁢(Op⁢(G))≤Op⁢(G) by Step 1 and [3, Theorem 6.3.2].

Step 5: NG⁢(P)=P is a maximal subgroup of G. Let T be the normal p-complement of NG⁢(P). Since T≤CG⁢(Op⁢(G))≤Op⁢(G) by Step 4, T=1. Hence NG⁢(P)=P.

Suppose that M is a maximal subgroup of G containing NG⁢(P). Then M is p-nilpotent by Step 2. Hence [Op′⁢(M),Op⁢(G)]=1 and thus Op′⁢(M)=1. Therefore M=P is a maximal subgroup of G.

Step 6: G=P⁢Q, where Q is a Sylow q-subgroup of G for some q≠p. Moreover, P=(P∩Op⁢(G))⁢〈a〉 for some a∈NG⁢(Q) and Q has no proper subgroups that are 〈a〉-invariant. Moreover, Op⁢(G)⁢Q/Op⁢(G) is a chief factor of G and Q is elementary abelian. Clearly, G is not a p-group. Let q be a prime such that q∈π⁢(G) and q≠p. Since G is p-soluble, there exists a Sylow q-subgroup Q of G such that PQ is a subgroup of G (see [3, Theorem 6.3.5]). By Step 5, we have G=P⁢Q.

Now let P1/(P∩Op⁢(G)) be a maximal subgroup of P/(P∩Op⁢(G)). Then P≤NG⁢(P1). The maximality of P implies that NG⁢(P1)=P or NG⁢(P1)=G. Let H=P1⁢Op⁢(G). By construction, H≠G. Suppose that NG⁢(P1)=P. Then NH⁢(P1)=P1 is p-nilpotent. It follows that H satisfies the hypotheses of the Main Theorem. By the minimality of G, we have that Op⁢(G)≤H is p-nilpotent and thus G is p-nilpotent, a contradiction. Therefore, NG⁢(P1)=G and P1 is normal in G. As G is not p-nilpotent, this yields Op⁢(G)=P1. In particular, P/P∩Op⁢(G) has a unique maximal subgroup and hence is cyclic. On the other hand, by the Frattini argument, we have that

As Op⁢(G)=P∩Op⁢(G)⁢Q, this yields

Clearly, P=P∩[(P∩Op⁢(G))⁢NG⁢(Q)]=(P∩Op⁢(G))⁢(P∩NG⁢(Q)). Since P/(P∩Op⁢(G)) is a cyclic group, we can find a cyclic group 〈a〉≤P∩NG⁢(Q) such that P=(P∩Op⁢(G))⁢〈a〉.

If X is a proper subgroup of Q that is 〈a〉-invariant, then PX will be a subgroup strictly between P and G, which is contrary to Step 5.

Let T/Op⁢(G) be a chief factor of G. Then T/Op⁢(G) is an elementary abelian q-group and there exists a Sylow q-subgroup Q1 of T such that T=Q1⁢Op⁢(G). It is clear that P⁢T=P⁢Q1. By Step 5, we have P⁢T=P⁢Q=G. Hence Q≅Q1 is elementary abelian.

Step 7: Let P2=Ω⁢(P∩Op⁢(G))⁢〈a〉 when condition (1) of the Main Theorem holds, and let P2=Ω1⁢(P∩Op⁢(G))⁢〈a〉 when condition (2) of the Main Theorem holds. Then P=P2. Suppose that P>P2. Since P∩Op⁢(G)⊴G, we have Ω⁢(P∩Op⁢(G))⊴G and Ω1⁢(P∩Op⁢(G))⊴G. Clearly, 〈a〉⁢Q is a subgroup of G. It follows that G1=P1⁢Q is a subgroup of G. Since P>P2, G1⊊G.

Assume that K is a subgroup of G1 containing P2. Then

Since K∩Q is a subgroup of Q that is 〈a〉-invariant, it follows that K∩Q=1 or Q. Hence K=P2 or G1. This means that P2 is maximal in G1.

Hence NG1⁢(P2)=P2 or G1. Suppose that NG1⁢(P2)=G1. Then we have Q≤NG⁢(P2). It follows that

and thus NG⁢(P)=G is p-nilpotent, a contradiction.

Suppose that NG1⁢(P2)=P2. Then NG1⁢(P2) is p-nilpotent. It follows that G1 satisfies the hypotheses of the theorem and thus G1 is p-nilpotent by the minimal choice of G. If condition (1) of the Main Theorem holds, then Q centralizes Ω⁢(P∩Op⁢(G)). It follows that Q centralizes P∩Op⁢(G) by [3, Theorem 5.3.2] and thus Op⁢(G) is p-nilpotent, a contradiction. If condition (2) of the Main Theorem holds, then Q centralizes Ω1⁢(P∩Op⁢(G)) and P∩Op⁢(G) is quaternion-free. In this case we also have that Q centralizes P∩Op⁢(G) by Lemma 2 and Op⁢(G) is p-nilpotent, a contradiction.

Therefore we have P=P2.

Step 8: The final contradiction. We want to apply Lemma 2 to obtain the final contradiction. First of all we shall construct the groups in Lemma 2. Put G¯=G/Φ⁢(Op⁢(G)).

Step 8.1: Op′⁢(G¯)=1. Suppose that Op′⁢(G¯)≠1. Let K be a normal subgroup of G such that Op′⁢(G¯)=K/Φ⁢(Op⁢(G)). We can write K=Kp′⁢Φ⁢(Op⁢(G)), where Kp′ is a Hall p′-subgroup of K. Since Φ⁢(Op⁢(G)) is a normal Sylow p-subgroup of K, it follows that any two Hall p′-subgroups of K are conjugated in K by the Schur–Zassenhaus Theorem. Then, by Frattini’s argument, we have

As Φ⁢(Op⁢(G))≤Φ⁢(G), this yields that G=NG⁢(Kp′) and thus Kp′⊴G. But then 1≠Kp′≤Op′⁢(G), contrary to Step 1.

Step 8.2: CG¯⁢(Op⁢(G¯))=Op⁢(G¯). By Step 8.1 and [3, Theorem 6.3.2].

Step 8.3: Put H=G¯/Op⁢(G¯). Then H is a p-soluble linear group over a field of characteristic p satisfying Op⁢(H)=1. It is easy to see that

acts faithfully on Op⁢(G¯), where Op⁢(G¯)=Op⁢(G)/Φ⁢(Op⁢(G)) is an elementary abelian p-group. So H is a p-soluble linear group over a field of characteristic p. Clearly, Op⁢(H)=Op⁢(G¯/Op⁢(G¯))=1. Hence Step 8.3 holds.

Step 8.4: Let g denote the image of a in H=G¯/Op⁢(G¯). Then g≠1. Suppose g=1. Then a⁢Φ⁢(Op⁢(G))∈Op⁢(G¯)=Op⁢(G)/Φ⁢(Op⁢(G)) and thus a∈Op⁢(G). From Step 3 one concludes that P∩Op⁢(G)≤Op⁢(G). But P=(P∩Op⁢(G))⁢〈a〉 by Step 7. So P=Op⁢(G) and G=NG⁢(P) is p-nilpotent by the hypothesis, a contradiction.

Step 8.5: There exists an element y∈Op⁢(G¯) such that Ep-1⁢(g,y)≠1. Since Q is abelian by Step 6, Lemma 1 shows that the minimal polynomial of g is (x-1)r, where r is the order of g. From Step 8.4, one deduces that r≥p. This means that (g-1)p-1≠0 in H. Hence Step 8.5 holds.

Step 8.6: Completing Step 8. For any u∈P, by Step 7, we can write u=s⁢t, where t∈〈a〉 and s∈Ω⁢(P∩Op⁢(G)) when condition (1) of the Main Theorem holds, or s∈Ω1⁢(P∩Op⁢(G)) when condition (2) of the Main Theorem holds. Hence, by hypotheses, Ep-1⁢(a,u)=Ep-1⁢(a,s)=1. The above equation implies that, for any z∈Op⁢(G¯), we have Ep-1⁢(g,z)=1. This is contrary to Step 8.5.

These completes the proof of Main Theorem. ∎