Research Article
Weakly C*-Normal Subgroups and p-Nilpotency of Finite Groups
College of Science, Sichuan University of Science and Engineering, 643000, Zigong, Sichuan, China
A subgroup is quasi normal in G if for every subgroup K of G such that HK = KH, by Ore (1937) which is a generalization of normality. A subgroup H of G is s-quasi normal if H permutes with all Sylow subgroups of G, by Kegel (1962) and extensive studied by Deskins (1963). A subgroup H is c*-normal in G if there exists a normal subgroup K of G such that G = HK and H∩ K is s-quasi normal embedded in G, by Wei and Wang (2007). Recently, Liu (2009) established some results on the base of weakly c*-normal subgroups of finite groups.
SOME DEFINITION AND PRELIMINARIES
Lemma 1
Suppose that U is s-quasi normally embedded in a group G, H≤G and K◁G. Then:
• | If U≤H, then U is s-quasi normally embedded in H |
• | If UK is s-quasi normally embedded in G, then UK/K is s-quasi normally embedded in G/K |
• | K≤H and H/K is s-quasi normally embedded in G/K, then H is s-quasi normally embedded in G |
Proof
Lemma 1 by Ballester-Bolinches and Pearaza-Anguilera (1998).
Lemma 2
Let G be a group. Then we have,
• | If H is weakly c*-normal in G and H≤weakly c*-normal in K |
• | If N◁G and N≤H. Then H is H is weakly c*-normal in G if and only if H/N is H is weakly c*-normal in G/N |
• | Let be a set of primes. H is a -subgroup of G and N a normal -subgroup of G. If H is weakly c*-normal in G. Then HN/N is weakly c*-normal in G/N |
Proof
Lemma 2.2 by Liu (2009).
Lemma 3
Let G be a group, K an s-quasi normal subgroup of G, P a Sylow p-subgroup of K, where, p is prime number of |G|. If P≤Op(G) or KG = 1, then P is s-quasi normal in G.
Proof
Lemma 2.5 by Wei and Wang (2007).
Lemma 4
Let G be a group and P a s-quasi normal p-subgroups of G, where, P is a prime number of |G|, then Op(G)≤NG(P).
Proof
Lemma 2.2 by Li et al. (2003).
Lemma 5
Let G be a group and p a prime dividing |G| with (|G|, p-1) = 1.Then:
• | If N is normal in G of order p, then N is in Z(G) |
• | If G has a cyclic Sylow p-subgroup, then G is p-nilpotent |
• | If M≤G and |G:M| = p, then M◁G |
Proof
Lemma 2.8 by Wei and Wang (2007).
Lemma 6
Let P be a p-subgroup of a group G and N a normal π-subgroup of G for some prime p. If A is a minimal subgroup of P and A is weakly c*-normal in NG(P), then AN/N is weakly c*-normal in NG(P)N/N.
Proof
If A is normal in G, then AN/N is weakly c*-normal in NG(P)N/N. If A is not normal in G, then by hypotheses, there exists a subgroup K of NG(P) such that NG(P) = AK and A∩ K = 1. Obviously NG(P)N/N = (AN/N)(KN/N). If (AN/N)∩(KN/N) ≠ 1, then K≤AN and therefore NG(P)N/N = KN/N. By comparing the order, there has a contradiction. So, AN/N is weakly c*-normal in NG(P)N/N.
Lemma 7
Let p be the smallest prime dividing |G| and P a Sylow p-subgroup of G. If every minimal subgroup of P is AN/N is weakly c*-normal in G and when p = 2 either every cyclic subgroup of P is weakly c*-normal in G or P is quaternion-free, then G is p-nilpotent.
Proof
Suppose that the result is false and Let G be a minimal counter example. By lemma 2(1), the hypotheses is inherited by subgroups. Therefore, G is minimal non-p-nilpotent group. By Ito (Robinson, 2003, Theorem 10.3.3) G is a minimal non-nilpotent. Then G is of order pαqβ, where, q is a prime, q ≠ p, P is normal in G and any Sylow q-subgroup Q of G is cyclic. Moreover, P is of exponent p if p is odd and exponent at most 4 if p = 2 (Robinson, 2003).
Let A be a minimal subgroup of P. Then by hypotheses, there exists a subgroup K of G such that G = AK and A∩ K is s-quasi normally embedded in G. If A is not normal in G, then K is a maximal subgroup of G of index p. Since, P is the smallest prime divisor of G. This leads that K is normal in G and so the Sylow q-subgroup of K are normal in G, thus G is nilpotent, a contradiction. So, A is normal in G, this leads that A is in the center of G. If either p is odd, or p = 2 and every cyclic subgroup of P is weakly c*-normal in G, then G is p-nilpotent by Itolemma. Then let B = <b> be a subgroup of P of order 4. Then by hypotheses, there exists a subnormal subgroup K such that G = BK and B∩K is s-quasinornally embedded in G. If |G:K| = 4, then K<b2> is a subgroup of index 2 and therefore is normal in G. This implies that the Sylow q-subgroup are normal in G. Then G is nilpotent. A contradiction. If |G:K| = 2, then K is normal in G, we also get a contradiction. Then B is normal in G. If B ≠ P, then, since G is a minimal non-nilpotent group and the exponent of P is at most 4, we have P≤CG(Q) and G = PxQ is nilpotent, another contradiction. The lemma is proved.
MAIN RESULTS
Theorem 1
Let G be a group such that G is S4-free. Also let p be the smallest prime dividing the order of G and P a Sylow p-subgroup of G. If every minimal subgroup of P of order p or 4(when p=2) is weakly c*-normal in NG(P) and when p = 2 P is quaternion-free, then G is p-nilpotent.
Proof
Assume that the theorem is false and let G be a counter example of minimal order. Then:
(1) | OP(G) = 1 |
If OP(G) ≠ 1, then we can chose a minimal normal subgroup N of G such that N≤OP(G). Now consider the quotient group G/N. Obviously PN/N is a Sylow p-subgroup of G/N. By lemma 6, AN/N is weakly c*-normal in NG(P)N/N. The minimality of G implies that G/N is p-nilpotent and hence G is p-nilpotent, a contradiction. Thus OP(G) = 1.
(2) | For every subgroup M of G satisfying P≤M<G, M must be p-nilpotent. In particular, NG(P) is p-nilpotent |
If NG(P) = G, then, by lemma 7, G is p-nilpotent. Hence, NG(P)<G. Since NG(P)∩M≤NM(P)≤NG(P), M satisfying the hypotheses of our theorem. The minimal choice of G implies that M is p-nilpotent.
(3) | G is solvable. Furthermore, P is a maximal subgroup of G and a Hall p-subgroup of G is an elementary abelian q-subgroup Q for some prime q |
Since, G is not p-nilpotent, by Frobenius theorem (Robinson, 2003), theorem 10.3.2), there exists a subgroup H of P such that NG(H) is not p-nilotent. So by (2) we think that NG(H) is not p-nilpotent but NG(K) is p-nilpotent for every subgroup K of P such that K<K≤P. Now we show NG(H) = G. Suppose that NG(H)<G. Then, we H<P*≤P for some Sylow p-subgroup P* of NG(H). Since every minimal subgroups of P* of order p or 4 is weakly c*-normal in NG(P). On the other hand, by the choice of H, NG(P*) is p-nilpotent and so NNG(H)(P*) is p-nilpotent. This implies that NG(H) satisfying the hypotheses of our theorem for its Sylow p-subgroup P* of NG(H). Now, the choice of G implies that NG(H) is p-nilpotent, a contradiction. Hence, Op(G) ≠ 1 and NG(K) is p-nilpotent for every subgroup K of P with Op(G)≤K≤P. Now, by Frobenius theorem (Robinson, 2003), theorem 10.3.2), G/Op(G) is p-nilpotent and hence G is p-nilpotent. By the odd order thorem, G is solvable.
Let T/Op(G) be a chief factor of G. Then T/Op(G) is an elementary abelian q-group for some prime q≠p and there exists a Sylow q-subgroup Q of T such that T = QQp(G). It is clear that PT = PQ. If PT<G, then, by (2), PT is p-nilpotent and so Q≤CG(Op(G)), which contradicts the fact CG(Op(G))≤Op(G) since, G is solvable. Hence, G = PQ and Q is a Hall p-subgroup of G. The minimality of T/Op(G) implies that P/Op(G) is a maximal subgroup of G/Op(G) and therefore P is a maximal subgroup of G.
(4) | G = Op(G)L, where, L is a non-abelian split extension of a normal Sylow q-subgroup Q by a cyclic p-subgroup <a>, ap∈Z(L) and the action of a on Q is irreducible |
Let P1/Op(G) be a normal p-complement of P/Op(G). By Schur-Zassenhaus theorem we have D = Op(G)Q.
Let P1/Op(G) be a maximal subgroup of P/Op(G). Then N(G)P1 = P or G. If N(G)P1 = P, then N(H)P1 = P1, where, H = P1D = PQ. Then H satisfying the hypotheses of the theorem. Since the minimality of G, we have that H is p-nilpotent. Then Op(G)Q = Op(G)xQ and Q is normal in G,a contradiction. Hence, P1◁G. So, Op(G) = P1 and P/Op(G) is a cyclic group. On the other hand, by the Frattini argument, G = Op(G)NG(Q). Since, P is not normal in G, so we assume that G = Op(G)L, where, L = <a>▷<Q is a non-abelian split extension of a normal Sylow q-subgroup Q by a cyclic p-subgroup <a>. Since, |P:Op(G) = p| and Op(G)∩NG(Q)◁NG(Q), ap∈Z(L). Also, since P is a maximal subgroup of G, Op(G)Q/Op(G) is a minimal normal subgroup of G/Op(G) and therefore the action of a (by conjugation) on Q is irreducible.
(5) | If Ω1(Op(G))∩<a> = 1, then [Ω1(Op(G)),Q] = 1 |
Set G1 = Ω1(Op(G))L. Obviously, Ω1(Op(G)) is an elementary abelian and characteristic in Op(G). Since, for any 1≠x∈Ω1(Op(G)), <x>is normal in G, <x><a> = <a><x>. Hence xa∈ Ω1(Op(G))∩(<x><a>). This implies that a induces a power automorphism of p-power order in the elementary abelian p-group Ω1(Op(G)). Thus [Ω1(Op(G), a] = 1. If there exists an element 1 ≠ x∈Ω1(Op(G)) and an element 1 ≠ g∈Q such that xg = x1 ≠ x, then and therefore . It follows that <Ω1(Op(G)), <a>, a-1 gag-1>∈CG1. Since, the action of a on Q is irreducible, QΩ1(Op(G))/Ω1(Op(G)) is a minimal normal subgroup of G1/Op(G) and Ω1Op(G)) <a> is a maximal subgroup in G1. Thus, CG1(x) = Ω1(Op(G)) <a> or G1. But 1≠a-1gag-1∈Q. Hence CG1(x) = G1, in contradiction to xg ≠ x. So [Ω1(Op(G)), Q] = 1.
(6) | The final contradiction |
We consider the following two cases:
Case 1
p>2 or p = 2 and P is quaternion-free. Set G1 = Ω1(Op(G))L. If Ω1(Op(G))∩<a> = 1, then by (5) [Ω1(Op(G)), Q] = 1.
Assume that Ω1(Op(G))∩<a> = <apa>. Then <apa> is a cyclic group with order p and <apa>≤Z(G1) since ap≤Z(L). Consider th quotient G1/<apa>. It is clear that (Ω1(Op(G))/<apa>)∩(<a>/<apa>) = 1and every subgroup K of Ω1(Op(G)) of order p is weakly c*-normal in NG(P) by hypotheses. Then there exists a subnormal subgroup H such that NG(P) = HK and H∩ K is s-quasi normally embedded in G. Let W denote H∩ K. Then W is a Sylow p-subgroup of some s-quasi normal subgroup M of G and so W is normal in M with Q≤M. Since MP = PM, WQ = WxQ. So, [Op(G), Q] = 1 and Q is normal in G, a contradiction.
Case 2
p = 2 and every cyclic subgroup of with order 2 or 4 is weakly c*-normal in NG(P). Let G2 = Ω2(Op(G))L. If Ω1(Op(G))∩ <a> = 1, then by (5) [Ω1(Op(G)), Q] = 1. Now assume that Ω1(Op(G))∩<a> = A = <c> is a cyclic group of order 2. It is clear that A≤Z(Ω1(Op(G))). Let x∈Ω2(Op(G)) with order 4. By hypothesis, <x> is weakly c*-normal in NG(P). There exists a subnormal K of NG(P) such that NG(P) = <x>K and <x>∩K is s-quasi normally embedded in G. Set W = <x>∩K, then there exists a s-quasi normal subgroup H of G such that W is a Sylow p-subgroup of H. If W = H, then W is s-quasi normal in G, so WQ = QW is p-nilpotent by lemma 5 and by lemma 4 Q is normal in G, a contradiction. If H=G, there has nothing to prove. So, we have W<H<G. If G = PH, then G = PH = PQ. Therefore, Q<H and PQ<G, a contradiction. Then PH<G and by (2) PH is p-nilpotent. Let Q* be the normal p-complement of PH, PH = PQ* = Q*P and so WQ* = Q*W = WxQ*. So, W is normal in G, by (1) G/W is p-nilpotent, then G is p-nilpotent, contradiction.
Remark 1
The hypothesis that G is S4-free cant be removed. Let G = S4, P the Sylow p-subgroup. Then P = NG(P) and every minimal subgroup of P is weakly c*-normal in NG(P), But G is not 2-nilpotent.
Remark 2
The hypothesis that P is quaternion-free can not be removed. Let A = <a, b: a4 = 1, b2 = a2, b-1ab = a-1> be a quaternion group, then A has an automorphism of order 3. Let G = <α>▷<A, clear then every element of G with order 2 lies in that center of G and is weakly c*-normal in NG(P), Bu G is not 2-nilpotent.
Corollary 1
Let G be a finite group, p a prime dividing the order of G such that (|G|,p-1) = 1. If there exists a normal subgroup N of G such that G/N is p-nilpotent and every subgroup of prime and order 4 of G is s-quasi normally embedded in G, then G is p-nilpotent.
Proof
Theorem 4.1 by Li et al. (2005).
Let G be a group such that G is S4-free. Also let p be the smallest prime dividing the order of G and P a Sylow p-subgroup of G. If every minimal subgroup of P of order p or 4 (when p = 2) is weakly c*-normal in NG(P) and when p = 2 P is quaternion-free, then G is p-nilpotent.