On p-nilpotency and minimal subgroups of finite groups

SCIENCE IN CHINA (Series A)

Vol. 46 No. 2

March 2003

On p-nilpotency and minimal subgroups of finite groups GUO Xiuyun (

)

1

& K. P. Shum2

1. Department of Mathematics, Shanxi University, Taiyuan 030006, China; 2. Department of Mathematics, The Chinese University of Hong Kong, Shatin, N. T., Hong Kong, China (SAR) Correspondence should be addressed to Guo Xiuyun (email:[email protected]) and K. P. Shum (email: [email protected]) Received August 31, 2001; revised March 29, 2002 Abstract We call a subgroup H of a finite group G c-supplemented in G if there exists a subgroup K of G such that G = HK and H ∩ K core(H). In this paper it is proved that a finite group G is p-nilpotent if G is S4 -free and every minimal subgroup of P ∩ GN is c-supplemented in NG (P ), and when p = 2 P is quaternion-free, where p is the smallest prime number dividing the order of G, P a Sylow p-subgroup of G. As some applications of this result, some known results are generalized. Keywords:

1

p -nilpotent groups, minimal subgroups, formation.

Introduction

All groups considered in this paper are finite groups. Our notation and terminology are standard. The reader may refer to ref. [1]. The relationship between the properties of minimal subgroups of a group G and the structure of G has been extensively studied by a number of authors (for example, see refs. [2—5]). A well known result is the following Itô’s lemma: Itˆ o’s lemma[6]. Let p be a prime dividing the order of a group G. If each element of G of order p lies in Z(G) and in addition, each element of G of order 4 still lies in Z(G) when p = 2, then G is p-nilpotent. Buckley[4] showed in 1970 that a group G of odd order is supersolvable if all minimal subgroups of G are normal in G. Since then, there are a number of papers dealing with the generalizations of this result. In this paper, we focus on the p-nilpotency of groups, which relates with the wellknown Burnside theorem for p-nilpotence. Burnside theorem states that if p is a prime dividing the order of a group G and P is a Sylow p-subgroup of G such that P is in the center of its normalizer then G is p-nilpotent. As inspired by the Burnside’s theorem and Itô’s lemma, one might wonder whether a group G is p-nilpotent if every element of G with order p lies in the center of NG (P ) and every element of G with order 4 is also in the center of NG (P ) when p = 2, where P is a Sylow p-subgroup of G. In this aspect, Ballester-Bolinches and Guo[3] gave an answer to this question. They proved the following result. Theorem A (ref. [3] Theorems 1 and 2).

Let p be a prime number dividing the order of

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a group G and P a Sylow p-subgroup of G. If every element of P ∩ G with order p lies in the center of NG (P ) and when p = 2 either every element of P ∩ G with order 4 lies in the center of NG (P ) or P is quaternion-free and NG (P ) is 2-nilpotent, then G is p-nilpotent, where G is the commutator subgroup of G. On the other hand, Guo and Shum[5] have recently discussed the p-nilpotency of a group G by considering the complemented minimal subgroups of G. The following question naturally arises: Is the group G p-nilpotent if every minimal subgroup of P ∩ G with order p is either normal or complemented in NG (P )? To tackle the above question, we first recall that a subgroup H of a group G is c-supplemented in G if there exists a subgroup K of G such that G = HK and H ∩ K core(H)[7] . By this concept, we can see that a minimal subgroup A of G which is either normal or complemented in G is equivalent to A being c-supplemented in G. Also since GN G , we may ask whether the group G is p-nilpotent or not if every subgroup of P ∩ GN with order p is c-supplemented in NG (P ), where GN is the nilpotent residual of G. By the nilpotent residual GN , we mean the intersection of all normal subgroups X of G such that G/X is nilpotent. The best result we can obtain is the following: Main theorem.

Let G be a group such that G is S4 -free. Also let p be the smallest prime

number dividing the order of G and P a Sylow p-subgroup of G. If every minimal subgroup of P ∩ GN is c-supplemented in NG (P ) and when p = 2 P is quaternion-free, then G is p-nilpotent, where GN is the nilpotent residual of G. It is clear that if G is a group with S4 -free and p = 2, then both the results of BallesterBolinches and Guo (see Theorem 2 of ref. [3]) and the result of Guo and Shum (see Theorem 2 of ref. [5]) become the special case of our main theorem. It is also clear that if p is the smallest prime number dividing the order of a group G and p is odd, then our main theorem is a generalization of Theorem 1 of ref. [3] and Theorem 2 of ref. [5]. As some applications of this result, we may generalize a result of Thompson and a result of Buckley[4]. It can be easily seen that the hypothesis that G is S4 -free and P is quaternion-free in our main theorem cannot be removed. For example, if we let G = S4 , the symmetric group of degree 4, and P a Sylow 2-subgroup of G. Then P is a dihedral group of order 8 and NG (P ) = P . It is easy to see that every minimal subgroup of P ∩ GN is c-supplemented in NG (P ) and G is quaternion-free, but G is not 2-nilpotent. If we let A = a, b|a4 = 1, b2 = a2 , b−1 ab = a−1 be a quaternion group, then A has an automorphism α of order 3. Let G = α< A. Then, it is clear to see that every element of G with order 2 lies in the center of G and therefore it is c-supplemented in G and, of course, G is S4 -free, but G itself is not 2-nilpotent. We also observe that the assumption that p is the smallest prime number dividing the order of G cannot be removed in our main theorem. In fact, if we let G = A5 , the alternating group of degree 5, then it is clear that G is S4 -free and every subgroup of P = P ∩ G = P ∩ GN with order 5 is c-supplemented in NG (P ), where P is a Sylow 5-subgroup of G. Obviously, in the above case, G itself is not 5-nilpotent.

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Preliminary results Recall that if F is a formation, then F -residual GF of a group G is the smallest normal

subgroup of G such that G/GF is in F . Therefore the nilpotent residual GN of a group G is the smallest normal subgroup of G such that G/GN is nilpotent. We first recall the following basic lemma. Lemma 2.1[7].

Let G be a group and H a subgroup of G.

(i) Let K be a subgroup of G such that H is contained in K. If H is c-supplemented in G, then H is c-supplemented in K. (ii) Let N be a normal subgroup of G such that N is contained in H. Then H is csupplemented in G if and only if H/N is c-supplemented in G/N . (iii) Let π be a set of primes. Let N be a normal π -subgroup of G and H a π-subgroup of G. If H is c-supplemented in G then HN/N is c-supplemented in G/N . If, furthermore, N normalizes H, then the converse also holds. (iv) Let L be a subgroup of G and H Φ(L). If H is c-supplemented in G then H is normal in G and H Φ(G). The following lemma is almost obvious, but it will be useful in the sequel. Lemma 2.2. Let P be a p-subgroup of a group G and N a normal p -subgroup of G for some prime p. If A is a minimal subgroup of P and A is c-supplemented in NG (P ), then AN/N is c-supplemented in NG (P )N/N . Proof.

If A is normal in NG (P ), then it is clear that AN/N is normal in NG (P )N/N .

If A is not normal in NG (P ), then, by the hypotheses, there exists a subgroup K of NG (P ) such that NG (P ) = AK and A ∩ K = 1. It is clear that NG (P )N/N = (AN/N )(KN/N ). If (AN/N ) ∩ (KN/N ) = 1, then A KN and therefore NG (P )N/N = KN/N . By comparing the orders of two sides of the above equation, we can immediately find a contradiction. Hence, (AN/N ) ∩ (KN/N ) = 1. Consequently AN/N is c-supplemented in NG (P )N/N . The following lemma relates with p-nilpotent groups. It is clear that it is a generalization of Itô’s lemma when p is the smallest prime number dividing the order of a group. Lemma 2.3.

Let p be the smallest prime dividing the order of a group G and P a Sylow

p-subgroup of G. If every minimal subgroup of P ∩ GN is c-supplemented in G, and when p = 2 either every cyclic subgroup of P ∩GN with order 4 is c-supplemented in G or P is quaternion-free, then G is p-nilpotent, where GN is the nilpotent residual of G. Proof.

Assume that the lemma is false and let G be a counterexample of minimal order.

Then G is not p-nilpotent. Noticing that all its Sylow p-subgroups are conjugate in G, we see that the hypotheses of our lemma is subgroup-closure by Lemma 2.1. Consequently, G is a minimal non-p-nilpotent group (that is, every proper subgroup of a group is p-nilpotent but itself is not p-nilpotent). Now, by a result of Itˆ o (see Theorem 10.3.3 of ref. [1]), G must be a minimal non-nilpotent group. By a result of Schmidt (see Theorem 9.1.9 and Exercises 9.1.11 of ref. [1]), we know that G is of order pα q β , where q is a prime which is different from p, P is normal in G and any Sylow q-subgroup Q of G is cyclic. Moreover, P = G = GN and P is of exponent p

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when p is odd and of exponent at most 4 when p = 2. Let A be a minimal subgroup of P . Then by our hypotheses, there exists a subgroup K of G such that G = AK and A ∩ K core(A). If A is not normal in G then K is a maximal subgroup of G with index p. Since p is the smallest prime dividing the order of G, we see that K is a normal subgroup of G and therefore the Sylow q-subgroup of K is normal in G. This leads to the nilpotence of G, a contradiction. Hence, every minimal subgroup of P must be normal in G and therefore every minimal subgroup of P must be in the center of G. If p is odd, then G is p-nilpotent by Itˆ o’s lemma, a contradiction. If p = 2 and every cyclic subgroup of P ∩ GN with order 4 is c-supplemented in G, then we claim that every cyclic subgroup of P of order 4 is also normal in G. In fact, if we let B = b be a subgroup of P with order 4, then, by our hypotheses, there exists a subgroup K of G such that G = BK and B ∩ K core(B). If [G : K] = 4, then Kb2 is a subgroup of G with index 2 and therefore Kb2 is normal in G. This implies that the Sylow q-subgroup of Kb2 is normal in G and therefore G is nilpotent, this is a contradiction. If [G : K] = 2, then K itself is a normal subgroup of G with index 2. We still get a contradiction. Hence, B must be 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 therefore G = P × Q, a contradiction. If P = B, then it is clear that G is p-nilpotent, another contradiction. If p = 2 and P is quaternion-free, then by applying Theorem 2.8 of ref. [8], we have Ω1 (P ) P ∩ GN ∩ Z(G) = 1, a contradiction. By all these contradictions, we show that the lemma is true. Lemma 2.4[2]. Let F be a saturated formation. Assume that G is a group such that G does not belong to F and there exists a maximal subgroup M of G such that M ∈ F and G = M F (G), where F (G) is the Fitting subgroup. Then GF /(GF ) is a chief factor of G, GF is a p-group for some prime p, GF has exponent p if p > 2 and exponent at most 4 if p = 2. Moreover, GF is either an elementary abelian group or (GF ) = Z(GF ) = Φ(GF ) is an elementary abelian group.

3

Main theorem We now establish our main theorem for p-nilpotent groups. The proof is rather lengthy and

we will divide our proof by establishing several claims. Theorem 3.1. Let G be a group such that G is S4 -free. Also let p be the smallest prime number dividing the order of G and P a Sylow p-subgroup of G. If every minimal subgroup of P ∩ GN is c-supplemented in NG (P ) and when p = 2 P is quaternion-free, then G is p-nilpotent, where GN is the nilpotent residual of G. Proof.

Assume that the theorem is false and let G be a counterexample of minimal order.

Then (i) Op (G) = 1. If Op (G) = 1, then we may pick a minimal normal subgroup N of G such that N is contained in Op (G). Now consider the quotient group G/N . Then P N/N is a Sylow p-subgroup of G/N . Since (G/N )N = GN N/N and P ∩ GN ∈ Sylp (GN ), we see that (G/N )N ∩ (P N/N ) = (GN ∩ P )N/N . By Lemma 2.2, every subgroup of (G/N )N ∩ (P N/N ) = (GN ∩ P )N/N with order

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p is c-supplemented in NG/N (P N/N ) = NG (P )N/N . The minimality of G implies that G/N is p-nilpotent and hence G is p-nilpotent, a contradiction. Thus our claim (i) is established. (ii) For every subgroup M of G satisfying NG (P ) M < G, M must be p-nilpotent. In particular, NG (P ) is p-nilpotent. If NG (P ) = G, then, by Lemma 2.3, G is p-nilpotent. Hence NG (P ) < G. It is clear that M satisfies the hypotheses of the theorem for every subgroup M of G with NG (P ) M < G. By the choice of G, we see that claim (ii) holds. (iii) Op (G) = 1. Because G is not p-nilpotent, by Frobenius’s theorem (see Theorem 10.3.2 of ref. [1]), there exists a subgroup H of P such that NG (H) is not p-nilpotent. By using our claim (ii), we may choose a subgroup H of P such that NG (H) is not p-nilpotent but NG (K) is p-nilpotent for every subgroup K of P with H < K P . Now we show that NG (H) = G. Suppose on the contrary that NG (H) < G, then, we have H < P ∗ P for some P ∗ ∈ Sylp (NG (H)). Since P ∗ ∩(NG (H))N P ∩GN , by Lemma 2.1, we see that every minimal subgroup of P ∗ ∩(NG (H))N is c-supplemented in P ∗ . By the choice of H, we know that NG (P ∗ ) is p-nilpotent and therefore NNG (H) (P ∗ ) is p-nilpotent. It follows that every minimal subgroup of P ∗ ∩ (NG (H))N is csupplemented in NNG (H) (P ∗ ). Now, by the minimality of G, we immediately see that NG (H) is p-nilpotent, a contradiction. This proves claim (iii). (iv) G/Op (G) is p-nilpotent and CG (Op (G)) Op (G). From the proof of claim (iii), we also know that NG (K) is p-nilpotent for every subgroup K of P with Op (G) < K P . Hence, by Frobenius theorem (see Theorem 10.3.2 of ref. [1]), we see that G/Op (G) is p-nilpotent and therefore G is p-solvable with the following upper p p-series 1 < Op (G) < Opp (G) < Opp p (G) = G. Consequently, we have CG (Op (G)) Op (G). Thus, claim (iv) is established. (v) G = P Q, where Q is an elementary abelian Sylow q-subgroup of G for a prime q = p. Moreover, P is a maximal subgroup of G and QOp (G)/Op (G) is a minimal normal subgroup of G/Op (G). In fact, for any q ∈ π(G) with q = p, since G is p-solvable, there exists a Sylow q-subgroup Q of G such that G1 = P Q is a subgroup of G (see Theorem 6.3.5 of ref. [9]). Clearly, G1 N ∩P GN ∩P and NG1 (P ) NG (P ) and therefore G1 satisfies the hypotheses of our theorem. If G1 < G, then G1 is p-nilpotent. This leads to Q CG (Op (G)) Op (G), a contradiction. Thus, G = P Q is solvable. Now we let T /Op (G) be a minimal normal subgroup of G/Op (G) which is contained in Opp (G)/Op (G). Then T = Op (G)(T ∩Q). If T ∩Q is a proper subgroup of Q, then P T is a proper subgroup of G and therefore P T is p-nilpotent. It follows that T ∩ Q CG (Op (G)) Op (G), a contradiction. Thus T = Opp (G) and hence QOp (G)/Op (G) is an elementary abelian q-group complementing P/Op (G). This implies that P is a maximal subgroup of G and our claim (v) holds. (vi) [P : Op (G)] = p and therefore GN Opp (G). If [P : Op (G)] > p, then there exists a subgroup G1 of G such that [G : G1 ] = p and Opp (G) G1 . Let P1 ∈ Sylp (G1 ) with P1 P . Then Op (G) < P1 < P . Clearly P NG (P1 ). Thus, by the

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maximality of P , we have NG (P1 ) = P and therefore NG1 (P1 ) = P1 . It is now easy to see that G1 satisfies the hypotheses of the theorem. Hence, G1 is p-nilpotent and Q CG (Op (G)) Op (G), a contradiction. So claim (vi) holds. (vii) P ∩ GN is normal in G and G = (P ∩ GN )L, where L = a Q 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 (by conjugate) on Q is irreducible. By claim (vi), we have GN Opp (G). It follows from (i) that 1 = P ∩ GN Op (G) and therefore P ∩ GN = Op (G) ∩ GN is normal in G. Now by the definition of GN , we know that G/P ∩ GN is p-nilpotent. Let D/P ∩ GN be a normal p-complement of G/P ∩ GN . By Schur-Zassenhaus theorem we may assume that D = (P ∩ GN )Q. Let P1 /P ∩ GN be a maximal subgroup of P/P ∩ GN . Then P NG (P1 ). The maximality of P implies that NG (P1 ) = P or G. If NG (P1 ) = P , then NH (P1 ) = P1 , where H = P1 D = P1 Q. It is clear that P1 ∩ H N P ∩ GN and therefore H satisfies the hypotheses of our theorem. By the minimality of G, H is p-nilpotent. It follows that D = (P ∩ GN )Q = (P ∩ GN ) × Q and therefore Q is a normal subgroup of G, a contradiction. Hence P1 is normal in G. Since P is not normal in G, we know that Op (G) = P1 and P/P ∩ GN is a cyclic group. On the other hand, by the Frattini argument we have G = (P ∩ GN )NG (Q). Thus, we may assume that G = (P ∩ GN )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), we see that ap ∈ Z(L). Also since P is a maximal subgroup of G, we know that (P ∩ GN )Q/P ∩ GN is a minimal normal subgroup of G/P ∩ GN and therefore the action of a (by conjugate) on Q is irreducible. Then claim (vii) is proved. (viii) If P ∩ GN is an elementary abelian p-group, then the theorem holds. In fact, if P ∩ GN Z(P ), then, by using the Frattini argument, we have G = NG (P ∩ GN ) = CG (P ∩ GN )NG (P ). Since NG (P ) = P and P CG (P ∩ GN ), we obtain that G = NG (P ∩ GN ) = CG (P ∩ GN ). Hence Q(P ∩ GN ) = Q × (P ∩ GN ) and so Q is normal in G, a contradiction. Now assume that P ∩ GN Z(P ). Let A1 be a minimal subgroup of P ∩ GN and A1 Z(P ). Then, by our hypotheses, there exists a subgroup K1 of P such that P = A1 K1 and A1 ∩ K1 = 1. In general, we may find minimal subgroups A1 , A2 , ..., As of P ∩ GN and also subgroups K1 , K2 , ..., Ks of P such that P = Ai Ki , Ai ∩ Ki = 1 for i = 1, 2, ..., s and 1 = P ∩ GN ∩ K1 ∩ ... ∩ Ks Z(P ). Furthermore, we may assume that Ai K1 ∩ ... ∩ Ki−1 (i = 2, 3, ..., s) and therefore we can have K1 ∩ ... ∩ Ki−1 = Ai (K1 ∩ ... ∩ Ki )

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for i = 2, 3, ..., s. Since Ki is maximal in P and therefore P ∩ GN ∩ Ki is normal in P , it is easy to see that (P ∩ GN ∩ Ki )a is a complement of Ai in P . Hence we may replace Ki by (P ∩ GN ∩ Ki )a and therefore we may further assume that a Ki for i = 1, 2, ..., s. Since P = (P ∩ GN )a, we see that K1 ∩ ... ∩ Ks = (P ∩ GN ∩ K1 ∩ ... ∩ Ks )a. It follows that K1 ∩ ... ∩ Ks /P ∩ GN ∩ K1 ∩ ... ∩ Ks a/P ∩ GN ∩ K1 ∩ ... ∩ Ks ∩ a is cyclic. Because P ∩ GN ∩ K1 ∩ ... ∩ Ks Z(P ), we know that K1 ∩ ... ∩ Ks is abelian. If p is an odd prime, then, by using Theorem 6.5.2 of ref. [9], we have K1 ∩ ... ∩ Ks Op (G) and therefore P = (P ∩ GN )(K1 ∩ ... ∩ Ks ) Op (G), a contradiction. The remaining case is p = 2. Case 1. |a| > 2. Since K1 ∩...∩Ks is an abelian normal subgroup of P and a ∈ K1 ∩...∩Ks , we obtain Φ(K1 ∩ ... ∩ Ks ) = a2 and therefore Ω1 (a2 ) Z(P ). On the other hand, by (vii) Ω1 (a2 ) Z(L) and so Ω1 (a2 ) Z(G). By Theorem 2.8 in ref. [8], we see that Ω1 (a2 ) ∩ (P ∩ GN ) = 1. Now, we consider the quotient group G/Ω1 (a2 ). Because P ∩ GN is an elementary abelian group and every element of P ∩ GN is c-supplemented in NG (P ) = P , it is easy to see that every minimal subgroup of GN Ω1 (a2 )/Ω1 (a2 )∩P/Ω1 (a2 ) = (GN ∩P )Ω1 (a2 )/Ω1 (a2 ) is c-supplemented in NG/Ω1 (a2 ) (P/Ω1 (a2 )) = P/Ω1 (a2 ). The minimality of G implies that G/Ω1 (a2 ) is 2-nilpotent and consequently Q is normal in G. This contradicts to the fact that G is not 2-nilpotent. Case 2.

|a| = 2. Then, since P = (P ∩ GN )a and P is not normal in G, we see that

a ∩ (P ∩ GN ) = 1. On the other hand, since a acts on Q, we see that a is an automorphism of Q with order two. Noticing that the action of a (by conjugate) on Q is irreducible, we know that Q is a cyclic group of order q and ba = b−1 , where Q = b. In this case, P ∩ GN is a minimal normal subgroup of G. In fact, let N be a minimal normal subgroup of G with N P ∩ GN and G1 = N L. Then it is easy to see that QN/N is a minimal normal subgroup of G1 /N and therefore the Sylow p-subgroup N a/N of G1 /N is a maximal subgroup of G1 /N . Moreover, NG1 (N a) = N a. Noticing that (N a)∩(G1 )N N and applying Lemma 2.1 (i), we know that every minimal subgroup of (N a)∩(G1 )N is c-supplemented in NG1 (N a) = N a. So G1 satisfies the hypotheses of the theorem. If G1 < G, then the minimality of G implies that G1 is p-nilpotent and therefore N Q = N × Q. Consequently, 1 = N ∩ Z(P ) P ∩ Z(G) ∩ GN , in contradiction to Theorem 2.8 of ref. [8]. Hence P ∩GN is a minimal normal subgroup of G. Since P ∩GN = Op (G) and Op (G) ∩ NG (Q) is normal in NG (Q), we see that Op (G) ∩ NG (Q) = 1 and therefore the action of b on P ∩ GN is fixed-point-free. Now we may assume that P ∩ GN = {1, c1 , c2 , ..., cq } and b = (c1 , c2 , ..., cq ) is a permutation on the set {c1 , c2 , ..., cq }. If we assume that c1 ∈ Z(P ), then, −1

by using ba = b−1 and considering (c1 )a −1 i

−i

ba

−1

= (c1 )b , we see that (c2 )a = cq . If we use (bi )a = b−i

and consider (c1 )a b a = (c1 )b , we can have (ci+1 )a = cq−i+1 for i = 1, 2, ..., (q + 1)/2. It follows that a = (c2 , cq )(c3 , cq−1 )...(c(q+1)/2 , c(q+3)/2 ) and therefore Z(P ) = {1, c1 }. Since P ∩ GN , K1 , ..., Ks are maximal subgroups of P and 1 = P ∩ GN ∩ K1 ∩ ... ∩ Ks Z(P ), we have Φ(P ) Z(P ). Noticing that P is not abelian, we see that Φ(P ) = P = Z(P ). Thus, P is an extra-special 2-group. By using Theorem 5.3.8 of ref. [1], we see that |P | = 22n+1 for some positive number n and therefore |P ∩ GN | = 22n . Since 22n − 1 = (2n + 1)(2n − 1) and q = 22n − 1 is a prime number, we see that n = 1 and |P | = 23 and therefore q = 3. Now it is easy to see that

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G S4 , a contradiction. Thus (viii) holds. (ix) The final step. By (viii) we may assume that Φ(P ∩GN ) = 1. For any minimal subgroup A of Φ(P ∩GN ), by the hypotheses, we know that A is normal in P and so Ω1 (Φ(P ∩ GN )) Z(P ). By the Frattini argument we have G = NG (Ω1 (Φ(P ∩ GN ))) = CG (Ω1 (Φ(P ∩ GN )))NG (P ). Noticing that NG (P ) = P and P CG (Ω1 (Φ(P ∩ GN ))), we have Ω1 (Φ(P ∩ GN )) Z(G). However, this result contradicts to Theorem 2.8 of ref. [8] while p = 2. So the remaining case is p = 2. Let the exponent of Φ(P ∩ GN ) be pe . By Theorem 1 of ref. [4], the following series 1 Ω1 (Φ(P ∩ GN )) Ω2 (Φ(P ∩ GN )) ... Ωe (Φ(P ∩ GN )) = Φ(P ∩ GN ) k

is a central series of Φ(P ∩ GN ) and Ωk (Φ(P ∩ GN )) = {x|xp = 1} for k = 1, 2, ..., e. Let x ∈ Ω2 (Φ(P ∩ GN )), g ∈ P . Since Ω2 (Φ(P ∩ GN )) is normal in G, there exists an element y ∈ Ω2 (Φ(P ∩ GN )) such that xg = xy. Then xp = (xp )g = (xg )p = (xy)p . But since (Ω2 (Φ(P ∩ GN ))) Ω1 (Φ(P ∩ GN )) Z(G) and Ω1 (Φ(P ∩ GN )) has exponent p, we see that (xy)p = xp y p . So we obtain xp = xp y p whence y ∈ Ω1 (Φ(P ∩ GN )). Hence every minimal subgroup of Ω2 (Φ(P ∩ GN ))/Ω1 (Φ(P ∩ GN )) is normal in P/Ω1 (Φ(P ∩ GN )). By using the above arguments again, we may have Ω2 (Φ(P ∩ GN ))/Ω1 (Φ(P ∩ GN )) Z(G/Ω1 (Φ(P ∩ GN ))). By induction we may have the following Ω1 (Φ(P ∩ GN )) Z(G), Ω2 (Φ(P ∩ GN ))/Ω1 (Φ(P ∩ GN )) Z(G/Ω1 (Φ(P ∩ GN ))), ............. Ωe (Φ(P ∩ GN ))/Ωe−1 (Φ(P ∩ GN )) Z(G/Ωe−1 (Φ(P ∩ GN ))), and therefore Q stabilizes the chain of subgroups 1 Ω1 (Φ(P ∩ GN )) Ω2 (Φ(P ∩ GN )) ... Ωe (Φ(P ∩ GN )) = Φ(P ∩ GN ). Now we consider the quotient group P ∩ GN /Φ(P ∩ GN ). For any minimal subgroup X of P ∩ GN /Φ(P ∩ GN ), there exists a subgroup X = x of P ∩ GN such that X = XΦ(P ∩ GN )/Φ(P ∩ GN ). If |x| = p, then, by the hypotheses, X is c-supplemented in P and therefore X is c-supplemented in P/Φ(P ∩ GN ). If |x| > p, then xp ∈ Φ(P ∩ GN ) and (xΩe−1 (Φ(P ∩ GN ))/Ωe−1 (Φ(P ∩ GN )))p Z(G/Ωe−1 (Φ(P ∩ GN )). For the sake of convenience, we set N = Ωe−1 (Φ(P ∩ GN )). For any g ∈ P , there exists an element y ∈ P ∩ GN such that xg = xy. Then (xN/N )p = ((xN/N )p )g = ((xN/N )g )p = ((xN/N )(yN/N ))p . But since p is odd and (P ∩GN ) Ωe (Φ(P ∩ GN )) = Φ(P ∩ GN ), we see that ((xN/N )(yN/N ))p = (xN/N )p (yN/N )p . So we obtain (xN/N )p = (xN/N )p (yN/N )p . It follows that y p ∈ N and therefore y ∈ Ωe (Φ(P ∩ GN )). Hence X = XΦ(P ∩GN )/Φ(P ∩GN ) is normal in P/Φ(P ∩GN ). Now we have shown that G/Φ(P ∩GN )

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satisfies the hypotheses of our theorem. By induction G/Φ(P ∩ GN ) is p-nilpotent. Hence we see that Q stabilizes the chain of subgroups 1 Ω1 (Φ(P ∩ GN )) ... Ωe (Φ(P ∩ GN )) = Φ(P ∩ GN ) < P ∩ GN . It follows from Theorem 5.3.2 of ref. [9] that [P ∩GN , Q] = 1 and therefore Q is a normal subgroup of G, which is the final contradiction. The proof of the theorem is complete.

4

Applications In this section, we give some applications of Theorem 3.1. In the literature of solvable groups,

there is an important result of Thompson which asserts that a finite group G is solvable if G has a nilpotent maximal subgroup of odd order (see Theorem 10.4.2 of ref. [1]). Later on, Deskins and Janko proved that Thompson’s result still holds if the Sylow 2-subgroup of nilpotent maximal subgroup is allowed to have class at most two (see IV, Satz 7.4 of ref. [6]). By using the well-known odd-order theorem of Feit and Thompson[10] , we can give another generalization of Thompson’s result on solvable groups: Theorem 4.1. Let G be a group such that G is S4 -free. Also let M be a nilpotent maximal subgroup of G and P a Sylow 2-subgroup of M . If every minimal subgroup of P ∩ GN is c-supplemented in P and P is quaternion-free, then G is solvable. Proof. Assume that the theorem is not true and G is a counterexample of minimal order. Let M2 be the 2 -Hall subgroup of M . By Theorem 1 of ref. [11], M2 is normal in G. If M2 = 1, we may let N be a minimal normal subgroup of G such that N M2 . Then M/N is a nilpotent maximal subgroup of G/N and P N/N satisfies the hypotheses of theorem by Lemma 2.1. Now, by the minimality of G, it implies that G/N is solvable and therefore G is solvable since N is nilpotent, a contradiction. Hence M2 = 1 and P is a maximal subgroup of G. It follows that P is a Sylow 2-subgroup of G and P = NG (P ). Now, by applying Theorem 3.1, we see that G is 2-nilpotent. By using the well-known odd-order theorem of Feit and Thompson, we see that G is solvable. This contradiction illustrates that Theorem 4.1 holds. As we have mentioned above that Buckley showed that a group G of odd-order is supersolvable if each minimal subgroup of G is normal in G. Now we generalize this result. We not only replace the normal assumption of minimal subgroups by the c-supplement assumption of minimal subgroups, but also replace all minimal subgroups of G by some of minimal subgroups of G. In fact, our result is more general. Theorem 4.2. Let F be a saturated formation containing the class of supersolvable groups U and G a group such that G is S4 -free. Also let N be a normal subgroup of G such that G/N is in F . If for every prime p dividing the order of N and for every Sylow p-subgroup P of N , every minimal subgroup of P ∩ GN is c-supplemented in NG (P ) and when p = 2 P is quaternion-free, then G is in F , where GN is the nilpotent residual of G. Proof. Assume that the theorem is false and we may let G be a counterexample of minimal order. By using Lemma 2.1 and Theorem 3.1, we know that N is a Sylow tower group. Thus, if p is the largest prime dividing the order of N and P is a Sylow p-subgroup of N , then P must be normal in G and therefore every minimal subgroup of P ∩ GN is c-supplemented in NG (P ) = G.

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Now, we consider the quotient group G/P . Since (G/P )/(N/P ) G/N and (G/P )N = GN P/P , we know that G/P satisfies the hypotheses of our theorem by Lemma 2.2. Then, the minimality of G implies that G/P is in F . Observing that GF GN and G is not in F, we see that 1 = GF is contained in P ∩ GN . Therefore GF is a p-group, where GF is the F -residual of G. By Theorem 3.5 of ref. [12], there exists a maximal subgroup M of G such that G = M F (G), where F (G) = Soc(G mod Φ(G)) and G/coreG (M ) is not in F . Then G = M GF and therefore G = M F (G) since GF is a p-group, where F (G) is the Fitting subgroup of G. It is now clear that M satisfies the hypotheses of our theorem for its normal subgroup M ∩ P . Hence, by the minimality of G, it leads to M must be in F . Now, by Lemma 2.4, GF has exponent p when p = 2 and exponent at most 4 when p = 2. If GF is an elementary abelian group, then GF is a minimal normal subgroup of G. For any minimal subgroup A of GF , we know that A is c-supplemented in NG (P ) = G by our hypotheses. If A is not normal in G then there exists a subgroup K of G such that G = AK and A ∩ K = 1. It is clear that K ∩ GF is normal in G. The minimality of GF implies that K ∩ GF = 1 and A is normal in G, a contradiction. Hence A is normal in G and GF = A is cyclic. Now suppose that GF is not an elementary abelian group. Then (GF ) = Z(GF ) = Φ(GF ) is an elementary abelian group by Lemma 2.4. Noticing that Φ(GF ) Φ(G), we know that every minimal subgroup of (GF ) is not complemented in G. It now follows from our hypotheses that every minimal subgroup of (GF ) must be normal in G. For any minimal subgroup A of GF /(GF ) , there exists a subgroup A of GF such that A = A(GF ) /(GF ) . Assume that A is of prime order. If A is not normal in G, then, by our hypotheses, there exists a subgroup K of G such that G = AK and A ∩ K = 1. Noticing that (GF ) = Φ(GF ) Φ(G), we see that K/(GF ) is a complement of A. The minimality of GF /(GF ) implies that A = GF /(GF ) is normal in G/(GF ) , and therefore GF /(GF ) is a cyclic group of prime order. Hence we may assume that p = 2 and every generated element of GF is of order 4. It follows immediately that Ω1 (GF ) = (GF ) = Φ(GF ) and therefore every minimal subgroup of Ω1 (GF ) is normal in G. Hence Ω1 (GF ) Z(G). By Lemma 2.15 of ref. [8], every p -element of G acts trivially on GF . Since GF /(GF ) is a chief factor of G, we see that GF /(GF ) is a cyclic group of prime order. We have now shown that for all cases, GF /(GF ) is always a cyclic group of prime order. Noticing that GF /(GF ) is G-isomorphic to Soc(G/coreG (M )), it follows that G/coreG (M ) is supersolvable, a contradiction. Thus, our proof is completed. Corollary 4.1. Let G be a group such that G is S4 -free. If for every prime p dividing the order of G and for every Sylow p-subgroup P of G, every minimal subgroup of P ∩ GN is c-supplemented in NG (P ) and when p = 2 P is quaternion-free, then G is supersolvable, where GN is the nilpotent residual of G. Acknowledgements This work was supported by a research grant of Shanxi Province for the first author and partially supported by a fund of UGC(HK) for the second author (Grant No. 2160126, 1999/2000).

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