ON CERTAIN PRODUCTS OF PERMUTABLE SUBGROUPS

Abstract

In this paper, we study the structure of finite groups $G=AB$ which are a weakly mutually $sn$ -permutable product of the subgroups A and B, that is, A permutes with every subnormal subgroup of B containing $A \cap B$ and B permutes with every subnormal subgroup of A containing $A \cap B$ . We obtain generalisations of known results on mutually $sn$ -permutable products.

1 Introduction

All groups considered here will be finite.

Mutually permutable products, that is, products $G=AB$ such that A permutes with every subgroup of B and B permutes with every subgroup of A, have been extensively studied by many authors [3]. In recent years, some other permutability connections between the factors have also been considered. In particular, the rich normal structure of a mutually permutable product of two nilpotent groups [3, Ch. 5] has motivated interest in the study of mutually $sn$ -permutable products.

Definition 1.1. We say that a group $G = AB$ is the mutually $sn$ -permutable product of the subgroups A and B if A permutes with every subnormal subgroup of B and B permutes with every subnormal subgroup of A.

Carocca [5] showed that a mutually $sn$ -permutable product of two soluble groups is also soluble. In [1], the authors analyse the structure of mutually $sn$ -permutable products and prove the following extension of a classical result of Asaad and Shaalan [2].

Theorem 1.2 [1, Theorem B]

Let $G = AB$ be the mutually $sn$ -permutable product of the subgroups A and B, where A is supersoluble and B is nilpotent. If B permutes with each Sylow subgroup of A, then the group G is supersoluble.

Following [8], we say that a subgroup H of a group G is $\mathbb {P}$ -subnormal in G whenever either $H=G$ or there exists a chain of subgroups $H=H_{0} \leq H_{1} \leq \cdots \leq H_{n-1} \leq H_{n}=G$ such that $| H_{i} : H_{i-1} |$ is a prime for every $i=1, \dots , n$ . It turns out that supersoluble groups are exactly those groups in which every subgroup is $\mathbb {P}$ -subnormal. Having in mind this result and the influence of the embedding of Sylow subgroups on the structure of a group, the following extension of the class of supersoluble groups introduced in [8] seems to be natural.

Definition 1.3. A group G is called widely supersoluble, w-supersoluble for short, if every Sylow subgroup of G is $\mathbb {P}$ -subnormal in G.

The class of all finite $ w $ -supersoluble groups, denoted by $ w\mathcal {U} $ , is a saturated formation of soluble groups containing $ \mathcal {U} $ , the class of all supersoluble groups, which is locally defined by a formation function f, such that for every prime p, $f(p)$ is composed of all soluble groups G whose Sylow subgroups are abelian of exponent dividing $p-1$ [8, Theorems 2.3 and 2.7]. Not every group in $w\mathcal {U}$ is supersoluble [8, Example 1]. However, every group in $w\mathcal {U}$ has an ordered Sylow tower of supersoluble type [8, Proposition 2.8].

In [4], mutually $sn$ -permutable products in which the factors are w-supersoluble are analysed. The following extension of Theorem 1.2 holds.

Theorem 1.4 [4, Theorem 4]

Let $G = AB$ be the mutually $sn$ -permutable product of the subgroups A and B, where A is w-supersoluble and B is nilpotent. If B permutes with each Sylow subgroup of A, then the group G is w-supersoluble.

Assume that $G = AB$ is the mutually $sn$ -permutable product of the subgroups A and B. Then, by [3, Proposition 4.1.16 and Corollary 4.1.17], $A \cap B$ is subnormal in G and permutes with every subnormal subgroup of A and B. Assume now that $G = AB$ and $A \cap B$ satisfy this condition. Then G is the mutually $sn$ -permutable product of A and B if and only if A permutes with every subnormal subgroup V of B such that $A\cap B\leqslant V $ and B permutes with every subnormal subgroup U of A such that $A\cap B\leqslant U$ . This motivates the following definition.

Definition 1.5. Let A and B be two subgroups of a group G such that $G = AB$ . We say that G is the weakly mutually $sn$ -permutable product of A and B if A permutes with every subnormal subgroup V of B such that $A\cap B\leqslant V $ and B permutes with every subnormal subgroup U of A such that $A\cap B\leqslant U$ .

Obviously, mutually $sn$ -permutable products are weakly mutually $sn$ -permutable, but the converse is not true in general, as the following example shows.

Example 1.6. Let $G = \Sigma _{4}$ be the symmetric group of degree $4$ . Consider a maximal subgroup A of G which is isomorphic to $\Sigma _{3}$ , and $B = A_4$ , the alternating group of degree $4$ . Then $G = AB$ is the weakly mutually $sn$ -permutable product of the subgroups A and B. However, the product is not mutually $sn$ -permutable because A does not permute with a subnormal subgroup of order $2$ of B.

The first goal of this paper to prove weakly mutually $sn$ -permutable versions of the aforesaid theorems. We show that Theorem 1.4 holds for weakly mutually $sn$ -permutable products.

Theorem A. Let $G = AB$ be the weakly mutually $sn$ -permutable product of the subgroups A and B, where A is w-supersoluble and B is nilpotent. If B permutes with each Sylow subgroup of A, then the group G is w-supersoluble.

The next corollary follows from the proof of Theorem A and generalises Theorem 1.2.

Corollary B. Let $G = AB$ be the weakly mutually $sn$ -permutable product of the subgroups A and B, where A is supersoluble and B is nilpotent. If B permutes with each Sylow subgroup of A, then the group G is supersoluble.

The second part of the paper is concerned with weakly mutually $sn$ -permutable products with nilpotent derived subgroups. Our starting point is the following extension of a classical result of Asaad and Shaalan [2].

Theorem 1.7 [1, Theorem C]

Let $G=AB$ be the mutually $sn$ -permutable product of the supersoluble subgroups A and B. If the derived subgroup $G'$ of G is nilpotent, then G is supersoluble.

A natural question is whether this result is true for weakly mutually $sn$ -permutable products under the same conditions. The following example answers this question negatively.

Example 1.8. Let $G=\langle a,b,c :a^5=b^5=c^6=1,ab=ba, a^c=a^2b^3,b^c=a^{-1}b^{-1} \rangle \simeq [C_{5} \times C_{5}]C_{6}$ . Then $G=AB$ is the weakly mutually $sn$ -permutable product of $A=\langle c \rangle $ and $B=[\langle a \rangle \times \langle b \rangle ]\langle c^3 \rangle $ . Note that B is a normal subgroup of G; therefore, it permutes with every subgroup of A. Moreover, $A \cap B=\langle c^3 \rangle $ and the unique subnormal subgroup of B containing $A \cap B$ is the whole of B. It is not difficult to see that B is supersoluble. Therefore, A and B are supersoluble and $G'$ is nilpotent. Moreover, A is nilpotent and B is a normal subgroup of G. Thus, in particular, it permutes with every Sylow subgroup of A.

However, an additional assumption allows us to get supersolubility.

Theorem C. Let $G=AB$ be the weakly mutually sn-permutable product of the supersoluble subgroups A and B. If B permutes with each Sylow subgroup of A, A permutes with every Sylow subgroup of B and the derived subgroup $G'$ of G is nilpotent, then G is supersoluble.

By [7, Theorem 2.6], a group G is w-supersoluble if and only if every metanilpotent subgroup of G is supersoluble. In particular, if we have a group with $G'$ nilpotent, every w-supersoluble subgroup is supersoluble. Therefore, the following result is clear.

Corollary D. Let $G=AB$ be the weakly mutually sn-permutable product of the w-supersoluble subgroups A and B. If B permutes with each Sylow subgroup of A, A permutes with every Sylow subgroup of B and the derived subgroup $G'$ of G is nilpotent, then G is w-supersoluble.

2 Preliminary results

In this section we will prove some results needed for the proofs of our main results. We start by showing that factor groups of weakly mutually $sn$ -permutable products are also weakly mutually $sn$ -permutable products.

Lemma 2.1. Let $G = AB$ be the weakly mutually $sn$ -permutable product of A and B, and let $ N $ be a normal subgroup of $ G $ . Then $ G/N=(AN/N)(BN/N) $ is the weakly mutually $sn$ -permutable product of $AN/N$ and $BN/N$ .

Proof. Let us consider $ G/N=(AN/N)(BN/N) $ . Suppose that $ HN/N $ is a subnormal subgroup of $ AN/N $ such that $ AN/N\cap BN/N\leqslant HN/N $ . Note that $U=HN \cap A$ is a subnormal subgroup of $ A $ such that $UN=HN$ and $A \cap B \leq U$ . Since $ U $ permutes with $ B $ , it follows that $ HN=UN $ permutes with $ BN $ .

Interchanging $ A $ and $ B $ and arguing in the same manner proves the result.

Lemma 2.2. Let $ G=AB $ be the weakly mutually $sn$ -permutable product of A and B.

1. (a) If $ H $ is a subnormal subgroup of $ A $ such that $ A\cap B \leqslant H $ , then $ HB $ is a weakly mutually $sn$ -permutable product of H and B.

2. (b) If $ A\cap B=1 $ , then $ G=AB $ is a totally $sn$ -permutable product of A and B.

Proof. Since every subnormal subgroup of $ H $ is a subnormal subgroup of $ A $ , it follows that $ B $ permutes with every subnormal subgroup $ L $ of $ H $ such that $ A\cap B\leqslant L $ . On the other hand, let $ M $ be a subnormal subgroup of $ B $ such that $ A\cap B\leqslant M $ . Then we have $HM=H(A\cap B)M=(A\cap HB)M=AM\cap HB=MA\cap BH= M(A\cap BH)= M(A\cap B) H=MH $ . Hence $ HB $ is a weakly mutually $sn$ -permutable product of H and B.

For (b), every subnormal subgroup of $ A $ permutes with $ B $ by (a) and every subnormal subgroup of $ B $ permutes with $ A $ . So $ G=AB $ is the mutually $sn$ -permutable product of A and B. Hence $ G=AB $ is the totally $sn$ -permutable product of A and B since $ A\cap B=1 $ .

Observe that Lemma 2.2 implies that if $ G=AB $ is the weakly mutually $ sn $ -permutable product of $ A $ and $ B $ , $ H $ is a subnormal subgroup of $ A $ such that $ A\cap B\leqslant H $ and $ K $ is a subnormal subgroup of $ B $ such that $ A\cap B\leqslant K $ , then $ HK $ is a weakly mutually $ sn $ -permutable product of $ H $ and $ K $ . In the next result we analyse the behaviour of minimal normal subgroups of weakly mutually $sn$ -permutable products containing the intersection of the factors.

Lemma 2.3. Let $ G=AB $ be the weakly mutually $sn$ -permutable product of A and B. If $ N $ is a minimal normal subgroup of $ G $ such that $ A\cap B\leqslant N $ , then either $ A\cap N=B\cap N=1 $ or $ N=(N\cap A)(N\cap B) $ .

Proof. Observe that $ A\cap N $ is a normal subgroup of $ A $ such that $ A\cap B\leqslant A\cap N$ and consequently $ H=(A\cap N)B $ is a subgroup of G. Note that $ N\cap H = N\cap (A\cap N)B = (A\cap N)(B\cap N)$ . Since $ N\cap H $ is a normal subgroup of $ H $ , it follows that $ B $ normalises $ N\cap H=(A\cap N)(B\cap N) $ .

By the same argument, $ K=A(B\cap N) $ is a subgroup of $ G $ such that $ K\cap N= A(B\cap N)\cap N=(A\cap N)(B\cap N) $ . Moreover, $ A $ normalises $ K\cap N=(A\cap N)(B\cap N) $ . It follows that $ (A\cap N)(B\cap N) $ is a normal subgroup of $ G $ . By the minimality of $ N $ , $ A\cap N=B\cap N=1 $ or $ N=(N\cap A)(N\cap B) $ as required.

Lemma 2.4. Let $G = AB$ be the weakly mutually $sn$ -permutable product of the subgroups A and B, where B is nilpotent. If B permutes with each Sylow subgroup of A, then $ A\cap B $ is a nilpotent subnormal subgroup of $ G $ .

Proof. It is clear that $ A\cap B $ is nilpotent. The Sylow subgroups of $ B $ are normal in $ B $ , so $ A\cap B $ permutes with every Sylow subgroup of $ B $ . Let $A_{q}$ be a Sylow subgroup of A, with q a prime dividing $| A |$ . Since $ B $ permutes with every Sylow subgroup of $ A $ , it follows that $BA_{q}$ is a subgroup of G. Hence $BA_{q} \cap A=A_{q}(A \cap B)$ . Therefore $ A\cap B $ permutes with every Sylow subgroup of $ A $ . Applying [3, Theorem 1.2.14(3)], $ A\cap B $ is a subnormal subgroup of both $ A $ and $ B $ . By [3, Theorem 1.1.7], $ A\cap B $ is a subnormal subgroup of $ G $ .

Lemma 2.5. Let $G = AB$ be the weakly mutually $sn$ -permutable product of the subgroups A and B, where A is soluble and B is nilpotent. If B permutes with each Sylow subgroup of A, then the group G is soluble.

Proof. Suppose that the theorem is false, and let $ G $ be a minimal counterexample. If $ N $ is a minimal normal subgroup of $ G $ , then $ G/N=(AN/N)(BN/N) $ is the weakly mutually $sn$ -permutable product of the subgroups $AN/N$ and $BN/N$ by Lemma 2.1. Since $BN/N$ permutes with each Sylow subgroup of $AN/N$ , it follows that $ G/N $ is soluble by the minimality of G. Let $ N_{1} $ and $ N_{2} $ be two minimal subgroups of $ G $ . Then $ G\cong G/(N_{1}\cap N_{2}) $ is soluble, a contradiction. Hence $ G $ has a unique minimal normal subgroup $ N $ of $ G $ and we may assume that $ N $ is nonabelian. This means that $ \textbf {F}(G)=1 $ .

On the other hand, $ A\cap B \leqslant \textbf {F}(G) $ using Lemma 2.4. Therefore $ A\cap B=1 $ and then $ G=AB $ is the totally $sn$ -permutable product of $ A $ and $ B $ . The result then follows by applying [5, Theorem 6].

Lemma 2.6 [1, Lemma 3]

Let $ G $ be a primitive group and let $ N $ be its unique minimal normal subgroup. Assume that $ G/N $ is supersoluble. If $ N $ is a $ p $ -group, where $ p $ is the largest prime dividing $ |G| $ , then $ N=\mathbf {F}(G)=\mathbf {O}_{p}(G) $ is a Sylow $ p $ -subgroup of G.

3 Main results

We are ready to prove our main results.

Proof of Theorem A

Suppose the theorem is not true and let $ G $ be a minimal counterexample. We shall prove our theorem in five steps.

(a) $ G $ is a primitive soluble group with a unique minimal normal subgroup $ N $ and $ N=\textbf {C}_{G}(N)=\textbf {F}(G)=\textbf {O}_{p}(G) $ for a prime $ p $ . Let $ N $ be a minimal normal subgroup of $ G $ . By Lemma 2.1, $ G/N=(AN/N)(BN/N) $ is a weakly mutually $ sn $ -permutable product of $ AN/N $ and $ BN/N $ and it is clear that $ BN/N $ permutes with every Sylow subgroup of $ AN/N $ . Moreover, $ AN/N $ is $ w $ -supersoluble and $ BN/N $ is nilpotent. By the minimality of $ G $ , it follows that $ G/N $ is $ w $ -supersoluble. Note that $ w\mathcal {U} $ , the class of finite $ w $ -supersoluble groups, is a saturated formation of soluble groups by [8, Theorems 2.3 and 2.7]. This implies that $ G $ is a primitive soluble group and so $ G $ has a unique minimal normal subgroup $ N $ with $ N=\textbf {C}_{G}(N)=\textbf {F}(G)=\textbf {O}_{p}(G) $ for some prime $ p $ as required.

(b) The subgroup $ BN $ is $ w $ -supersoluble and $ 1\not = A\cap B\leqslant N $ . If $ A\cap B=1 $ , then the result follows by Lemma 2.2 and Theorem 1.4. Applying Lemma 2.4, it follows that $ A\cap B $ is a nilpotent subnormal subgroup of $ G $ . Therefore $ 1\not = A\cap B \leqslant \textbf {F}(G)=N $ and so $ N= (N\cap A)(N\cap B) $ by Lemma 2.3. Hence $ NB=(N\cap A)(N\cap B)B=(N\cap A)B $ is a weakly mutually $ sn $ -permutable product of $ N\cap A $ and $ B $ . Also note that $ B $ permutes with every Sylow subgroup of $ N\cap A $ (there is only one Sylow subgroup of $ N\cap A $ , which is $ N\cap A $ ). If $ NB < G $ , then $ NB $ is $ w $ -supersoluble by the choice of $ G $ . So we may assume that $ G=NB $ . In this case, consider a subgroup $ N_{1} \leqslant A\cap B\leqslant N $ . Note that $ N_{1} $ is normal in $ N $ since $ N $ is abelian. Hence $ N=N_{1}^{G}=N_{1}^{NB}=N_{1}^{B}\leqslant B $ and $ G=B $ , a contradiction. Hence the result follows.

(c) $ N $ is the Sylow $ p\!$ -subgroup of G and $ p $ is the largest prime dividing $ |G| $ . Let $ q $ be the largest prime dividing $ |G| $ and suppose that $ q\not = p $ . Suppose first that $ q $ divides $ |BN| $ . Since $ BN $ has a Sylow tower of supersoluble type, $ BN $ has a unique Sylow $ q $ -subgroup, say $ (BN)_{q} $ . This means that $ (BN)_{q} $ centralises $ N $ . Thus $ (BN)_{q}=1 $ , since $ \mathbf {C}_{G}(N)=N $ , a contradiction.

We may assume that $ q $ divides $ |A| $ but does not divide $ |BN| $ . Since $ A $ has a Sylow tower of supersoluble type, $ A $ has a unique Sylow $ q $ -subgroup, $ A_{q} $ say. This means that $ A_{q} $ is normalised by $ N\cap A $ . Consider $ A_{q}(N\cap B)=A_{q}(A\cap B)(N\cap B) $ , a weakly mutually permutable product of $ A_{q}(A\cap B) $ and $ N\cap B $ by Lemma 2.2. Also $ N\cap B $ permutes with each Sylow subgroup of $ A_{q}(A\cap B) $ . Suppose that $ A_{q}(N\cap B) < G $ . Then $ A_{q}(N\cap B) $ is $ w $ -supersoluble by the choice of $ G $ . It follows that $ A_{q}(N\cap B) $ has a unique Sylow $ q $ -subgroup since it has a Sylow tower of supersoluble type. In other words, $ A_{q} $ is normalised by $ N\cap B $ . Hence $ A_{q} $ is normalised by $ (N\cap A)(N\cap B)=N $ . This means that $ A_{q} $ centralises $ N $ , a contradiction. We may assume that $ A_{q}(N\cap B)=G $ . Then $ N \cap B=B $ and so $ B $ is an elementary abelian $ p $ -group. Moreover, $ A=A_{q}(A\cap B) $ . Since $ A\cap B $ is a Sylow $ p $ -subgroup of $ A $ which is subnormal in $ A $ , it is normal in $ A $ . Hence $ A\cap B $ is normal in $ G $ because $ A\cap B $ is normal in the abelian group $ B $ . By the minimality of $ N $ , it follows that $ N=A\cap B $ , that is, $ G=A_{q}(N\cap B)=A_{q}(A \cap B)=A $ , a contradiction. Therefore $ p $ is the largest prime dividing $ |G| $ .

We now prove that $ N $ is the Sylow $ p $ -subgroup of $ G $ . Since $ G $ is a primitive soluble group, $ G=NM $ , where $ M $ is a maximal subgroup of $ G $ and $ N\cap M=1 $ . Then $ M\cong G/N $ is $ w $ -supersoluble. By [6, Theorem A.15.6], $ \textbf {O}_{p}(M)=1 $ . If $ p $ divides $ |M| $ , then since $ M $ has a Sylow tower of supersoluble type, $ \textbf {O}_{p}(M)\not =1 $ , a contradiction. Hence $ p $ does not divide $ |M| $ and therefore $ N $ is the unique Sylow $ p $ -subgroup of $ G $ .

(d) $ N $ is contained in $ A $ and $ N $ is not contained in $ B $ . Suppose that $ B $ is a $ p $ -group. Then $ G=AN $ . Let $ N_{1}\leqslant A\cap B $ . Since $ B $ is abelian, $ N \leq N_{1}^{G}=N_{1}^{AN}=N_{1}^{A}\leqslant A $ and so $ G=AN=A $ , a contradiction. So we may assume that $ B $ is not a $ p $ -group. If $ N $ is contained in $ B $ , then since $ B $ is nilpotent and $ N =\textbf {C}_{G}(N)$ , it follows that $ B $ is a $ p $ -group, a contradiction. Therefore $ N $ is not contained in $ B $ . Hence $ B $ has a nontrivial Hall $ p' $ -subgroup, $ B_{p'} $ , which is normal in $ B $ . Consequently, $ AB_{p'}=A(A\cap B)B_{p'} $ is a subgroup of G. Then $ 1\not = B_{p'}^{G}\leqslant AB_{p'} $ and so $ N \leqslant AB_{p'}$ . Hence $ N\leqslant A $ as required.

(e) Final contradiction. Let $ A_{p'} $ be a Hall $ p' $ -subgroup of $ A $ . If $ A_{p'}=1 $ , then $ G=BN $ is $ w $ -supersoluble by (b), a contradiction. Hence $ A_{p'}\not = 1 $ . Since $ B $ permutes with every Sylow subgroup of $ A $ , it follows that $ A_{p'}B $ is a subgroup of $ G $ . But $ N $ is not contained in $ B $ , so $ A_{p'}B $ is a proper subgroup of $ G $ . Since $ NA_{p'}B=G $ , it follows that $ N\cap A_{p'}B=N\cap B $ is normal in $ G $ . The minimality of $ N $ implies that $ N=N\cap B $ or $ N\cap B=1 $ . If $ N=N\cap B $ , we get a contradiction with (d). Therefore $ N\cap B=1 $ , and then $ A\cap B\leqslant N\cap B=1 $ , contradicting (b).

Proof of Theorem C

Assume the result is not true and let G be a minimal counterexample. It is clear that $G' \neq 1$ , A and B are proper subgroups of G, and G is a primitive soluble group. Hence there exists a unique minimal normal subgroup N of G, such that $N=F(G)=C_{G}(N)$ . Moreover, $G'=N$ . We may assume that $A' \neq 1$ and $B' \neq 1$ , otherwise A or B is nilpotent and the result follows from Corollary B. If $A \cap B=1$ , then G is the mutually $sn$ -permutable product of A and B. By [1, Theorem C], the group is supersoluble, a contradiction. Thus we may assume $A \cap B \neq 1$ . Since A permutes with every Sylow subgroup of B and B permutes with every Sylow subgroup of A, it follows that $A \cap B$ permutes with every Sylow subgroup of A and every Sylow subgroup of B. Hence $A \cap B$ is subnormal in A and it is a subnormal subgroup of B. Let $N_{1}$ denote a minimal normal subgroup of A such that $N_{1} \leq A'$ . Since A is supersoluble, it is clear that $| N_{1} | =p$ . Note that $N_{1}(A \cap B)$ is a subnormal subgroup of A. Therefore $BN_{1}(A \cap B)=BN_{1}$ is a subgroup of G. Now $1 \neq N_{1}^G=N_{1}^B \leq BN_{1}$ . Hence $N \leq BN_{1}$ and then $N=N_{1}(N \cap B)$ . Consequently, either $N_{1} \leq N \cap B$ or $N_{1} \leq N \cap B$ . Denote $T=BN$ . If $N_{1} \leq N \cap B$ , then $T=B$ is a supersoluble normal subgroup of G. Assume $N_{1} \cap (N \cap B)=1$ . Then $N \cap B$ is a maximal subgroup of N and so T is the weakly mutually $sn$ -permutable product of B and N. Consequently, T satisfies the hypotheses of the theorem. If T is a proper subgroup of G, then $T=BN$ is supersoluble. Assume that $G = BN$ . Then B is a maximal subgroup of G such that $B \cap N = 1$ , $B' \leq N \cap B=1$ and B is nilpotent. By Corollary B, G is supersoluble, contrary to assumption. Hence either B is a normal subgroup of G or $BN$ is a supersoluble normal subgroup of G.

Arguing in an analogous manner with A shows that if $AN$ is a proper subgroup of G, then it is supersoluble. Consequently if $BN$ and $AN$ are both proper subgroups of G, then G is the product of two supersoluble normal subgroups with $G'$ nilpotent. Then G is supersoluble, a contradiction. Therefore we may assume that $G=BN$ or $G=AN$ . Suppose without loss of generality that $G=BN$ . Then $N \cap B$ is a normal subgroup of G. If $N \cap B=N$ , then $G=B$ , a contradiction. Hence $N \cap B=1$ . Now $B' \leq N \cap B=1$ and B is nilpotent, the final contradiction.