Normal restriction in finite groups by Hung Phi Tong-Viet | Papers by Hung Phi

NORMAL RESTRICTION IN FINITE GROUPS HUNG P. TONG VIET Abstract. A subgroup H is called an N R−subgroup (Normal Restriction) if, whenever K H, then K G ∩ H = K, where K G is the normal closure of K in G. In this paper, we will prove some sufficient conditions for the solvability of finite groups which possess many N R-subgroups. We also prove a criterion for the existence of a normal p-complement in finite groups. 1. Introduction Let G be a finite group. A subgroup H of G is called a CR-subgroup (Character Restriction) of G if every complex irreducible character of H is a restriction of some irreducible character of G. It is well known that if H is a CR-subgroup of G and K H, then K G ∩H = K. This leads to the following definitions: a triple (G, H, K) is said to be special in G if K H ≤ G and H ∩ K G = K, where K G is the normal closure of K in G. A subgroup H is called an N R−subgroup (Normal Restriction) if, whenever K H, then (G, H, K) is special in G. From definition, we see that if H ≤ G and H is simple then H is an N R-subgroup of G. Li Shirong [12] called a subgroup K, an N E-subgroup if (G, NG (K), K) is special in G. He showed that if every minimal subgroup of G is an N E-subgroup then G is solvable. Such a group is called a P E-group and the structure of a minimal non-PEgroup was investigated. Yangming Li[10] showed that if every minimal subgroup of prime order or of order 4 is an N E-subgroup of G, then G is supersolvable. The author also classified non-abelian simple groups whose second maximal subgroups are P E-groups. In Tong-Viet[15], it is shown that if every maximal subgroup of G is an N Rsubgroup then G is solvable. This gives a positive answer to a question posed in Berkovich[2]. We see that in the symmetric group S4 only self-normalizing nonnilpotent maximal subgroups are N R-subgroups but this group is still solvable. Our first result is a generalization of Theorem 1.1 in Tong-Viet[15]. Theorem 1.1. If every non-nilpotent maximal subgroup of G is either normal or N R in G, then G is solvable. A maximal subgroup of G is said to be 1-maximal. For n ≥ 2, a subgroup H is called n-maximal if it is maximal in some (n − 1)-maximal subgroup of G. In view of Theorem 1.1, it is natural to ask whether or not a group G is solvable if every non-nilpotent 2-maximal subgroup of G is either subnormal or N R in G. The answer to this question is ‘No’. The minimal counter-example is the alternating group A5 . It is easy to see that every 2-maximal subgroup of A5 is nilpotent. In the Date: October 28, 2009. 1991 Mathematics Subject Classification. Primary 20D10; Secondary 20D05. Key words and phrases. solvable, p-nilpotent. 1 2 HUNG P. TONG VIET next theorem, we will show that in fact A5 is the unique non-abelian composition factor of groups satisfying the assumption of the above question. Denote by S(G) the solvable radical of a group G, that is, a maximal normal solvable subgroup of G. Theorem 1.2. If every non-nilpotent 2-maximal subgroup of G is either subnormal or N R in G, then G/S(G) is trivial or isomorphic to A5 . As a consequence, we obtain another sufficient condition for the solvability of finite groups as follows: Corollary 1.3. If every 2-maximal subgroup of G is either subnormal or N R in G then G is solvable. We cannot extend Corollary 1.3 further as all 3-maximal subgroups of A5 are N R in A5 . A subgroup H is said to have a normal complement in G if there exists a normal subgroup L of G such that G = HL and H ∩ L = 1. In Isaacs[8], it is shown that if N = NG (P ) is a CR-subgroup, where P ∈ Sylp (G), then N has a normal complement in G. This result has been generalized in Berkovich[2]. In that paper, the author replaces the property CR by N R and still obtains the same conclusion. However we can see that the argument may apply to a broader class of subgroups, say p-subgroups instead of p-Sylow subgroups. Proposition 1.4. Let P be a p-subgroup of G and N = NG (P ). If the triples (G, N, P ) and (G, N, Φ(P )) are special in G then N has a normal complement T in G, with (p, |T |) = 1. Applying Lemma 2.1(e), we get the following corollary: Corollary 1.5. Let P be a p-subgroup of G and N = NG (P ). Assume that for any T ∈ {P, Φ(P )}, there exists a subgroup L of G such that G = N L and N ∩ L = T then N has a normal complement in G. Recall that a group G is said to be p-nilpotent if there exists a normal p -subgroup N of G such that G = SN where S ∈ Sylp (G) for some prime p and we call N a normal p-complement in G. As an application of Proposition 1.4, we prove a new criterion for the existence of a normal p-complement in finite groups. Theorem 1.6. Let P be a p-subgroup of G and N = NG (P ). Assume that the triples (G, N, P ) and (G, N, Φ(P )) are special in G. If N is p-nilpotent then G is p-nilpotent. Observe that the existence of a supersolvable maximal N R-subgroup does not imply the supersolvability of a group. The minimal counter-example is A4 with a supersolvable maximal N R-subgroup A3 . Theorem 1.7. Let H be an N R-subgroup of prime index in G. If H is supersolvable then G is supersolvable. All groups are finite. We adapt the notations in Conway et al.[4] for finite simple groups. All other notations for finite groups are standard. NORMAL RESTRICTION IN FINITE GROUPS 3 2. Preliminaries Lemma 2.1. Let K H ≤ T ≤ G. Assume the triple (G, H, K) is special in G. (a) The triple (T, H, K) is special in T. (b) If L/K G HK G /K G and triple (G, H, L ∩ H) is special in G then the triple (G/K G , HK G /K G , L/K G ) is special in G/K G . In particular, if K G then the triple (G/K, H/K, L/K) is special in G/K. (c) If H is an N R-subgroup of G then HK G /K G is an N R-subgroup of G/K G . (d) If K G and every non-nilpotent maximal subgroup of G is either normal or N R in G, then every non-nilpotent maximal subgroup of G/K is either normal or N R in G/K. (e) Let K H ≤ G. If there is a subgroup L such that G = HL and H ∩ L = K then the triple (G, H, K) is special. Proof. (a) − (c) are in Lemma 4 in Berkovich[2]. As nilpotence and self-normalizing are preserved under taking quotient group, this together with (c) yield (d). Finally (e) is Lemma 9 in Berkovich[2]. Lemma 2.2. (Tate[14]). Let H is p-nilpotent. G and P ∈ Sylp (G). If H ∩ P ≤ Φ(P ) then H Theorem 2.3. (Zsigmondy[16]). Let q and n be integers with q ≥ 2 and n ≥ 3. If (q, n) = (2, 6) then there is a prime r such that r | q n − 1 but r does not divide q i − 1 for i < n. We call such a prime r a primitive prime divisor of q n − 1 and denote by qn . Lemma 2.4. Let t ≥ 0 be an integer. (a) 2t − 1, and 2t−1 − 1 are both prime powers if and only if t = 3. (b) 2t + 1, and 2t+1 + 1 are both prime powers if and only if 0 ≤ t ≤ 3. Proof. (a) Observe first that t and t − 1 are both primes if and only if t = 3, in which case 2t − 1 = 7 and 2t−1 − 1 = 3 are both primes. If t = 6 then 26 − 1 = 9.7 is not a prime power. Assume that t > 3 and t = 6. Then either t or t−1 is not prime. Assume that n > 3, n = 6 is not a prime. We will show that 2n − 1 is not a prime power. By way of contradiction, assume that 2n − 1 = pm for some m. By Theorem 2.3, p is a primitive prime divisor of 2n − 1. Let s < n be a non-trivial prime divisor of n and write n = sa. Then pm = 2n − 1 = 2sa − 1 is divisible by 2s − 1. It follows that p is a divisor of 2s − 1 with s < n, contradicting to the definition of primitive prime divisor. Thus if either t or t − 1 is not prime then either 2t − 1 or 2t−1 − 1 is not a prime power. (b) If 0 ≤ t ≤ 3, then we can check that both 2t + 1 and 2t+1 + 1 are prime powers. Assume that t ≥ 4, and 2t + 1, 2t+1 + 1 are both prime powers. We see that t or t + 1 must be odd. Now let n be an odd integer with n ≥ 4. We will show that 2n + 1 is not a prime power. By way of contradiction, assume that 2n + 1 = pm for some prime p. Since n is odd, we have 3|2n + 1 so that p = 3. Thus 2n + 1 = 3m or equivalently 3m − 1 = 2n . As n ≥ 4, we see that m ≥ 3. By Zsigmondy’s Theorem, 3m − 1 has a primitive prime divisor which is not a divisor of 3 − 1 = 2, a contradiction. Thus 2t + 1 and 2t+1 + 1 cannot be both prime powers when t ≥ 4. Lemma 2.5. (Thompson [6, Bemerkung II.7.5]). If G is a minimal simple group then G is isomorphic to one of the following groups: 4 HUNG P. TONG VIET (1) (2) (3) (4) (5) L2 (p), p > 3 is a prime, and p2 − 1 ≡ 0 (mod 5); L2 (3r ), r is an odd prime; L2 (2r ), r is a prime; Sz(2r ), r is an odd prime; L3 (3). Lemma 2.6. (i) (Corollary 2.2, King[9]). Assume that G ∼ L2 (q) is a minimal = simple group. If M is a maximal subgroup of G then M is one of the following groups: (a) Dq−1 for q ≥ 13 odd and D2(q−1) for q even; (b) Dq+1 for q = 7, 9 odd and D2(q+1) for q even; (c) a Frobenius group of order q(q − 1)/2 for q odd and q(q − 1) for q even; (d) S4 when q ≡ ±1 (mod 8) with q prime or q = p2 and p ≡ ±3 (mod 8); (e) A4 when q ≡ ±3 (mod 8) with q prime; (ii) (Theorem 9, Suzuki[13]). Assume that G ∼ Sz(q), q = 2r , r an odd prime. The = maximal subgroup of G are as follows: (a) a Frobenius group of order q 2 (q − 1); (b) a dihedral subgroup of order 2(q − 1); (c) a Frobenius group of order 4(q ± t + 1), with t2 = 2q. (iii) (Conway et al.[4], page 13). Assume that G ∼ L3 (3). The maximal subgroup = of G are as follows: (a) a group of order 32 : 2S4 (b) a group of order 13 : 3 (c) S4 . Let p be a prime. Recall that Op (G) is the smallest normal subgroup of G such that G/Op (G) is a p-group, or equivalently, Op (G) is a subgroup of G generated by all p -elements in G. Also F (G) is the Fitting subgroup of G, that is the largest nilpotent normal subgroup of G. Theorem 2.7. (Satz Baumann[1]). Let G be a non-solvable group possessing a nilpotent maximal subgroup. Then O2 (G/F (G)) is a direct product of simple groups with dihedral 2-Sylow subgroups. The simple groups that can appear are L2 (q) with q = 9 or q a prime of the form 2m ± 1 > 5. Remark 2.8. We have L2 (9) ∼ A6 , Out(A6 ) ∼ Z2 × Z2 and S6 ∼ A6 .21 . The 2= = = Sylow subgroups of A6 and S6 are not maximal but the 2-Sylow subgroups of other extensions of A6 are maximal. (see Conway et al.[4] page 4). For any p-group P, we denote by A(P ) the set of abelian subgroups of P of maximal order. The Thompson subgroup J(P ) is a subgroup of P generated by all members of A(P ). It is well known that J(P ) is characteristic in P, Z(J(P )) is characteristic in J(P ) and hence Z(J(P )) is characteristic in P. The following theorem gives a sufficient condition for the existence of a normal p-complement in finite groups. Theorem 2.9. (Glauberman-Thompson, Theorem 8.3.1, Gorenstein[5]). Let P ∈ Sylp (G) with p odd. If NG (Z(J(P ))) has a normal p-complement then G has a normal p-complement. Recall that a group G is called minimal non-nilpotent if G is non-nilpotent but every maximal subgroup of G is nilpotent. The structure of minimal non-nilpotent groups are given by the following theorem: NORMAL RESTRICTION IN FINITE GROUPS 5 Theorem 2.10. (O.J. Schmidt, 9.1.9, Robinson[11]). Assume that G is a minimal non-nilpotent group. Then (a) G is solvable; (b) |G| = pm q n where p, q are distinct primes. Morever, there is a unique p-Sylow subgroup P and a cyclic q-Sylow subgroup Q. Hence G = P Q and P G. Lemma 2.11. (a) If S is a non-abelian simple group and S A ≤ Aut(S), then there exists a subgroup K ≤ S such that A = SNA (K) and every proper-over group of K in S is local in S. Moreover when S is a finite group of Lie type and S = 2 F4 (2) , we can choose K to be a Borel subgroup of S. (b) If every maximal subgroup of G is an N R-subgroup then G is solvable. Proof. (a) follows from Theorem 1.2[15] and its proof. (b) is Theorem 1.1[15]. The following results are obvious. Lemma 2.12. Let A be a normal subgroup of prime order of G. Then G is supersolvable if and only if G/A is supersolvable. Lemma 2.13. Let H be a normal subgroup of G. Assume that H is p-nilpotent with a normal p-complement K. Then K G. Lemma 2.14. Let t ≥ 0 be an integer. (a) D2t is nilpotent if and only if t is a power of 2. (b) D2t is minimal non-nilpotent if and only if t is an odd prime. 3. N R-subgroups and solvability The following result is a generalization of exercise 9.1.10 in Robinson[11]. Lemma 3.1. If every self-normalizing maximal subgroup of G is nilpotent then G is solvable. Proof. As nilpotent groups are solvable, we can assume that G is not nilpotent. If every maximal subgroup of G is self-normalizing then every maximal subgroup of G is nilpotent by hypothesis, hence G is a minimal non-nilpotent group, and so by Theorem 2.10, G is solvable. Thus G contains a maximal subgroup which is not self-normalizing, hence G is not simple. If A is any non-trivial normal subgroup of G then G/A satisfies the hypothesis of the lemma, so that by induction, G/A is solvable. Therefore G has a unique minimal normal subgroup N and G/N is solvable. Assume N is not solvable, hence N is a direct product of some nonabelian simple groups. Let P ∈ Sylp (N ), with p odd. By Frattini’s argument, we have G = NG (P )N. Clearly NG (P ) < G, hence there exists a maximal subgroup M of G such that NG (P ) ≤ M < G. We have G = NG (P )N = M N and hence M is not normal in G. Therefore M is nilpotent. Let Q ∈ Sylp (M ). As M is nilpotent, we have Q M so that M = NG (Q). We will show that Q ∈ Sylp (G). By way of contradiction, assume that Q is not a p-Sylow subgroup of G. Let S be a pSylow subgroup of G containing Q. We have Q < NS (Q), and hence Q < NS (Q) ≤ NG (Q) = M, a contradiction as Q ∈ Sylp (M ). As Q M, the Thompson subgroup J(Q) is characteristic in Q and Z(J(Q)) is characteristic in J(Q), and so Z(J(Q)) is characteristic in Q. It follows that Z(J(Q)) M so that M = NG (Z(J(Q))). Since M is nilpotent, it is p-nilpotent and hence by Theorem 2.9, G has a normal p-complement H. By the uniqueness of N, we have N ≤ H. This is a contradiction 6 HUNG P. TONG VIET as p divides the order of N but not that of H. Thus N is solvable and so G is solvable. Proof of Theorem 1.1. Let G be a minimal counter-example to Theorem 1.1. Assume first that G is non-abelian simple. Then every maximal subgroup of G is either nilpotent or N R. By Lemma 2.11(b), G contains a nilpotent maximal subgroup H. By Theorem 2.7, we have G ∼ L2 (q) for q a prime of the form 2m ±1 > = 5 prime. By Lemma 2.6(i), G has a maximal subgroup B of order q(q−1)/2 which is a Frobenius group with Frobenius kernel U of order q and a Frobenius complement T with |T | = (q − 1)/2 > 1. Since B is not nilpotent, from hypothesis, B is an N R-subgroup of G. As G is simple and U is non-trivial, we have G = U G , hence U G ∩ B = G ∩ B = B > U, contradicting to the fact that B is an N R-subgroup of G. Therefore G is not non-abelian simple. Since the hypothesis of the Theorem inherits to proper quotient of G, G has a unique minimal normal subgroup N which is not solvable. Then N = S x1 × S x2 × · · · × S xt , where S is a non-abelian simple group, and x1 , x2 , · · · , xt ∈ G. Let K be the subgroup of S obtained from Lemma 2.11(a), and let R = K1 × K2 × · · · × Kt , where Ki = K xi , i = 1, · · · , t. Then R is a non-trivial proper subgroup of N. Since N is a unique minimal normal subgroup of G, NG (R) < G. We will show that G = NG (R)N. For any g ∈ G, since N g = N, there exists a permutation π ∈ St such that S xi g = S xiπ . Let gi = xi gx−1 . Then gi ∈ NG (S). We have iπ Rg = K x1 g × K x2 g × · · · × K xt g = K g1 x1π × K g2 x2π × · · · × K gt xtπ = = K g1π−1 x1 × K g2π−1 x2 × · · · × K gtπ−1 xt = K h1 x1 × K h2 x2 × · · · × K ht xt = s s s = K x1 s1 × K x2 s2 × · · · × K xt st = K1 1 × K2 2 × · · · × Kt t , where K giπ−1 = K hi with hi ∈ S by Lemma 2.11(a), and si = hxi ∈ S xi . Let i s s s = s1 .s2 . . . st ∈ N. Since [S xi , S xj ] = 1 if i = j ∈ {1, 2, . . . , t}, Ki = Ki i . Thus g s R = R , where s ∈ N. Therefore G = NG (R)N. Let M be a maximal subgroup of G containing NG (R). We have G = M N. Clearly M is not normal in G, otherwise N ≤ M and hence G = M, a contradiction. Assume that M is nilpotent. As F (G) = 1, by Theorem 2.7, O2 (G) G is a product of simple groups, it follows that N ≤ O2 (G) and hence S ∼ L2 (q) for q = 9 = or q a prime of the form q = 2m ± 1 > 5. We will show that π(M ) ∩ π(N ) = {2}. Assume p ∈ π(M ) ∩ π(N ) and p is odd. As in proof of Lemma 3.1, the p-Sylow subgroup P of M is also a p-Sylow subgroup of G. Apply Glauberman-Thompson Theorem, G has a normal p-complement. This leads to a contradiction as in the proof of the previous lemma. Thus M ∩ N is a 2-group. Since R ≤ M ∩ N, R must be a 2-group and hence K is a 2-group. However by the choice of K in Lemma 2.11(a), |K| = q(q − 1)/2 which is not a 2-group as q is odd. Thus M is not nilpotent. Therefore we can assume that M is not nilpotent nor normal in G. Then M is an N R-subgroup of G. The argument below is exactly the same as in the last part of the proof of Theorem 1.1 in [15]. Let Q = M ∩ N. We have G = M N, and Q = M ∩N M. As G/N = M N/N M/Q, M/Q is solvable. If Q is solvable then M is solvable. By Proposition 7 in [2], G = M, a contradiction. Thus Q is non-solvable. NORMAL RESTRICTION IN FINITE GROUPS 7 Let L be any non-trivial normal subgroup of M. Since M is an N R-subgroup of G, we have L = LG ∩ M. It follows from the fact that N is the unique minimal normal subgroup of G, N ≤ LG . We have Q = N ∩ M ≤ LG ∩ M = L. We conclude that Q is a minimal normal subgroup of M. Since Q is a minimal normal subgroup of M and Q is non-solvable, Q = W1 ×W2 ×· · ·×Wk , where Wi W for all 1 ≤ i ≤ k and W is a non-abelian simple group. Suppose that there exists j ∈ {1, 2, . . . , t} such that S xj ≤ Q. As S xj is normal in N, (S xj )G = (S xj )N M = (S xj )M ≤ M. However as (S xj )G = N, G = M N = M, a contradiction. Therefore S xj ∩ Q < S xj for any j ∈ {1, 2, . . . , t}. Since Kj ≤ S xj N, we have Kj ≤ S xj ∩ Q Q. As Q is a direct product of non-abelian simple groups and S xj ∩ Q is a non-trivial normal subgroup of Q, there exists a non-empty set J ⊆ {1, 2, . . . , t} such that S xj ∩ Q = i∈J Wi . Hence Kj ≤ i∈J Wi < S xj , and so K ≤ i∈J Wi x−1 j < S, where Wi x−1 Wi j i∈J x−1 j are non- abelian simple for any i ∈ J. However by Lemma 2.11(a), is local in S. This final contradiction completes the proof. Proof of Theorem 1.2. Let M be any maximal subgroup of G. From hypothesis, every non-nilpotent maximal subgroup H of M is either subnormal or N R in G. By Lemma 2.1(a) and the maximality of H in M, H is either normal or N R in M, so that M satisfies the hypothesis of Theorem 1.1. Thus M is solvable. It follows that every maximal subgroup of G is solvable. If N is any non-trivial proper normal subgroup of G, then N is solvable and hence N ≤ S(G). By Lemma 2.1(d), G/N satisfies the hypothesis, so that (G/N )/S(G/N ) is trivial or isomorphic to A5 . Since S(G/N ) = S(G)/N, we have (G/N )/S(G/N ) ∼ (G/N )/(S(G)/N ) ∼ G/S(G). = = Thus G/S(G) is either trivial or isomorphic to A5 , and we are done. Therefore we can assume that G is simple and so G is a minimal simple group. By Theorem 2.10(a), G contains a non-nilpotent maximal subgroup. Now let M be any nonnilpotent maximal subgroup of G. Let H be any maximal subgroup of M, as H is not subnormal in G, H is either nilpotent or N R in G. We will show that if H is N R in G then H is of prime order. Assume that H is N R in G. Let A be any nontrivial normal subgroup of H. As H is an N R-subgroup of G, the triple (G, H, A) is special in G so that AG ∩ H = A. As G is simple, we have AG = G, hence H = A, so that H is a simple group. Since H is solvable, H must be cyclic of prime order. Therefore M is a minimal non-nilpotent group. By Theorem 2.10(b), |M | = pm q n where p, q are distinct primes and m, n are a positive integer, the p-Sylow subgroup P of M is normal in M while the q-Sylow subgroup Q of M is cyclic. Now by Lemma 2.5, we consider the following cases: (a) Case G ∼ L2 (q), q > 3 prime and q 2 − 1 ≡ 0 (mod 5) or G ∼ L2 (q), q = 3r , r = = odd prime. If q = 5 then G = L2 (5) ∼ A5 and we are done. If q = 7 then G = L2 (7) = and since 7 ≡ −1(mod 8), S4 is a maximal subgroup of G. However S4 is neither nilpotent nor minimal non-nilpotent as it contains a non-nilpotent subgroup S3 . Therefore we can assume that q ≥ 13. By Lemma 2.6(i), the Frobenius F of order q(q−1)/2 and dihedral groups Dq−1 , Dq+1 are maximal subgroup of G. Since q ≥ 13 and 1 < (q − 1)/2 < q, F is not nilpotent, and so (q − 1)/2 = sa , s prime. We have q + 1 = 2(sa + 1). If Dq+1 is nilpotent then q + 1 = 2(sa + 1) = 2t for some integer t ≥ 0, by Lemma 2.14(a). Hence q = 2t − 1 and (q − 1)/2 = 2t−1 − 1 are both prime powers. Since q ≥ 13, we have t ≥ 4 and so 2t − 1 and 2t−1 − 1 cannot be prime powers at the same time by Lemma 2.4(a). Thus Dq+1 is not nilpotent, and hence Dq+1 is minimal non-nilpotent. Thus (q + 1)/2 = sa + 1 must be an odd prime by 8 HUNG P. TONG VIET Lemma 2.14(b). It follows that s is an even prime and so s = 2. Hence q = 2a+1 + 1 and 2a + 1 are both prime powers. By Lemma 2.4(b), we have a = 0, 1, 2, 3 and so q = 3, 5, 9, 17 respectively. As q ≥ 13, we have q = 17. However (17 + 1)/2 = 32 is not an odd prime. Thus these cases cannot happen unless q = 5. (b) Case G ∼ L2 (2r ), r prime. As L2 (4) ∼ L2 (5) ∼ A5 , we can assume that r is = = = an odd prime. By Lemma 2.6(i), G has a maximal subgroup isomorphic to D2(q+1) . Clearly D2(q+1) is not nilpotent and hence it must be minimal non-nilpotent. By Lemma 2.14(b), q + 1 = 2r + 1 is prime. As r is odd, 2r + 1 is divisible by 3. Thus 2r + 1 = 3 so that r = 1, a contradiction. (c) Case G ∼ Sz(q), q = 2r , r odd prime. By Lemma 2.6(ii), and Theorem = 3.10[7], G has a maximal subgroup M = NG (A), where A is cyclic of order q + t + 1 with 2t2 = q, |M : A| = 4 and M is a Frobenius group with Frobenius kernel A. Moreover CG (u) = A for any 1 = u ∈ A. As q + t + 1 is odd and CG (u) = A for any 1 = u ∈ A, we see that M is not nilpotent. As |M : A| = 4, M contains a subgroup of index 2 which is not nilpotent. Thus M is not a minimal non-nilpotent. (d) Case G ∼ L3 (3). By Lemma 2.6(iii), G has a maximal subgroup M which is = isomorphic to S4 . As in case (a), S4 is neither nilpotent nor minimal non-nilpotent. The proof is now completed. Proof of Corollary 1.3 As G satisfies the hypothesis of Theorem 1.2, G is solvable or G/S(G) ∼ A5 . If the first possibility holds then we are done. Thus we = assume that G/S(G) ∼ A5 . As the hypothesis of the corollary inherits to quotient = group and since A5 is simple, it follows that every 2-maximal subgroup of A5 is an N R-subgroup. As V4 A4 ≤ A5 and V4 is maximal in A4 , A4 is maximal in A5 , we deduce that V4 is a 2-maximal subgroup of A5 . Let K be any subgroup of order 2 in V4 . Then K V4 ≤ A5 . It follows that K A5 ∩ V4 = K. However as A5 is simple, and 1 = K A5 A5 , we have K A5 = A5 , and hence K A5 ∩ V4 = V4 > K. This contradicts to the fact that H is an N R-subgroup of A5 . Therefore G is solvable. The proof is now completed. 4. Normal Complement Proof of Proposition 1.4 Let K = P G G. As (G, N, P ) is special in G, K ∩ N = P. Also NK (P ) = K ∩ N = P, P is a self-normalizing p-subgroup of K so that P is a self-normalizing p-Sylow subgroup of K. By Frattini’s argument, we have G = N K. Let L = Φ(P )G . We have L G and since the triple (G, N, Φ(P )) is special in G, Φ(P ) ≤ P ∩ L ≤ N ∩ L = Φ(P ), hence P ∩ L = Φ(P ). As L ≤ K and P L/L ∼ P/(P ∩L) ∼ P/Φ(P ), P L/L is an elementary abelian p-Sylow subgroup of = = K/L. We will show that P L/L is self-normalizing in K/L. In fact, assume Lg ∈ K/L normalizes P L/L where g ∈ K. Then we have P g ≤ P L ≤ K. Since P is a p-Sylow subgroup of K, by Sylow’s Theorem, P g = P u for some u ∈ P L. Indeed, we can take u ∈ L. Thus g ∈ NK (P )L = P L. This proves our claim. Now by Burnside Normal p-Complement Theorem, (Theorem 7.4.3 Gorenstein[5]), P L/L has a normal pcomplement H/L in K/L. It follows that K = P LH = P H and P L ∩ H = L. As K/L G/L and H/L is a normal p-complement in K/L, by Lemma 2.13, H/L G/L and hence H G. Observe that P ∩ H ≤ P ∩ (P L ∩ H) = P ∩ L ≤ N ∩ L = Φ(P ) NORMAL RESTRICTION IN FINITE GROUPS 9 as the triple (G, N, Φ(P )) is special in G. By Lemma 2.2, H has a normal pcomplement T in H. Thus K = P H = P T, P ∩ T = 1 and so G = NK = NPT = NT and N ∩ T = (N ∩ P T ) ∩ T = (N ∩ K) ∩ T = P ∩ T = 1. Applying Lemma 2.13 again for H G, we have T G. Thus T is a normal complement for N in G. The proof is now completed. Proof of Theorem 1.6. By Proposition 1.4, G = N L for some normal p subgroup L of G. Assume that N is p-nilpotent. Then N = S.K where P ≤ S ∈ Sylp (N ) and K is a normal p -subgroup of N. We have G = N L = SKL. Clearly KL G, KL is a p -subgroup and so S ∈ Sylp (G). Hence G is p-nilpotent. Proof of Theorem 1.7. Assume that H is supersolvable. Let A be a minimal normal subgroup of H. Then A is a cyclic subgroup of prime order p. Assume first that A G. Then H/A is an N R-subgroup of G/A by Lemma 2.1(c). Moreover H/A is supersolvable and has prime order in G/A. 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