On the Finite Group Which Is a Product of Two Subnormal Supersolvable Subgroups

Abstract

Let G be a finite group that is a product of two subnormal ( normal) supersolvable subgroups. The following are interesting topics in the study of the structure of G: obtaining the conditions in addition to guarantee that G is supersolvable and giving the detailed structure of G when G is non-supersolvable. In this paper, we obtain a characteristic property of G being non-supersolvable and two new sufficient conditions for G being a supersolvable group.

1. Introduction

All groups in this paper are assumed to be finite. Let G be a group. We denote $\left|G\right|$ as the order of G and $\pi \left(G\right)$ the set of prime divisors dividing $\left|G\right|$. We denote $e\left(G\right)$ as the exponent of G. For some fixed prime $p\in \pi \left(G\right)$, we denote the Sylow p-subgroup of G by $G_p$. Further, we will use the following notations: $C_n$ is the cyclic group of order n, $Q_8$ is the quaternion group of order 8, $D_8$ is the dihedral group of order 8, and $\Phi \left(G\right)$, $G^{\prime }$, $Z\left(G\right)$, $F\left(G\right)$ are the Frattini subgroup, the derived subgroup, the center and the Fitting subgroup of G, respectively. The semidirect product of a normal subgroup A and a subgroup B is denoted by $A⋊B$.

Following Tang et al. in [1], we denote by ${\mathfrak{B}}_1$ the class of groups which can be factorized as the product of two normal supersolvable subgroups. Additionally, we denote by ${\mathfrak{B}}_2$ the class of groups which can be factorized as the product of two subnormal supersolvable subgroups. Other notations and terminology of this paper are standard (refer [2]).

It is well-known that the group being the product of two normal solvable subgroups is solvable, while the group being the product of two normal nilpotent subgroups is nilpotent. However, the group which is the product of two normal supersolvable subgroups may not be supersolvable; that is, the groups in ${\mathfrak{B}}_1$ are not always supersolvable. In [3], Huppert constructed an example. Hence a natural and interesting problem arises:

Problem 1.

Assume that G is a product of two normal supersolvable subgroups, i.e., $G\in {\mathfrak{B}}_1$. Under which conditions in addition can we guarantee that G is supersolvable ?

Many scholars have studied Problem 1 and nice works about it can be found in the literature. For example, let $G\in {\mathfrak{B}}_1$. Baer obtained that G is supersolvable if and only if $G^{\prime }$ is nilpotent ([4]); Friesen obtained that G is supersolvable if G is the product of two normal supersolvable subgroups of the coprime index ([5]). Vasil’ev and Vasil’eva obtained that G is a supersolvable group if G possesses a nilpotent normal subgroup such that every Sylow subgroup of the corresponding factor group is abelian ([6]). Recently, Monakhov and Chirik identified that many results still hold if one weakens the normality to subnormality, i.e., $G\in {\mathfrak{B}}_2$ ([7]). Meanwhile, Guo raised the following problem from another point of view:

Problem 2.

([8] Chapter II, Problem 6.34) Assume that $G\in {\mathfrak{B}}_{\mathfrak{1}}$(or, $G\in {\mathfrak{B}}_2$) and G is non-supersolvable. Then, what is the structure of G?

Recently, many results about Problem 2 have appeared. For example, Guo and Kondratl’ev described the structure of the minimal non-supersolvable groups in ${\mathfrak{B}}_1$ ([9]). In [1], Tang Ye and Guo described the structure of non-supersolvable groups in ${\mathfrak{B}}_1$ whose proper ${\mathfrak{B}}_1$-subgroups and proper ${\mathfrak{B}}_1$-factor groups are supersolvable.

In this paper, we continue to study these problems. First, we give an answer to Problem 2:

Theorem 1.

Let $G=AB\in {\mathfrak{B}}_2$, where A and B are subnormal supersolvable subgroups of G. Assume that G is not supersolvable. Then there exists a chain $N⊴S⊲⊲G$ such that $S/N\cong X$, where $X=WH$, W is a minimal normal subgroup of X, W a p-group, p is an odd prime and the subgroup H is isomorphic to one of the following:

(i)

$Q_8$;

(ii)

$M_p(n,1)$;

(iii)

$M_p(1,1,1)$.

Here the subgroups $M_p(n,1)$ and $M_p(1,1,1)$ are defined below (Lemma 1).

Then, we give two new results for Problem 1:

Theorem 2.

Let $G=AB\in {\mathfrak{B}}_2$, where A and B are subnormal supersolvable subgroups of G. If $G/G^{\prime }$ is cyclic, then G is supersolvable.

Theorem 3.

Let $G=AB\in {\mathfrak{B}}_2$, where A and B are subnormal supersolvable subgroups of G. If $\left(\right|A/A^{\prime }|,|B/B^{\prime }\left|\right)=1$, then G is supersolvable.

Remark 1.

Thanks to an anonymous reviewer, we learnt that Theorem 3 has already been proved in reference [10], which just appeared in the literature. We repeat it here for the sake of the completeness of this paper. Furthermore, inspired by reference [10] and applying the same argument of Theorem 2, we can see that a slightly general form of Theorem 2 holds: Let $G=\langle A,B\rangle$ be a group generated by two subnormal supersolvable subgroups A and B of G. If $G/G^{\prime }$ is cyclic, then G is supersolvable.

Finally, we show that our results are generalizations of the main results in [11].

2. Preliminaries

Lemma 1

([12] Theorem 2.3.7). Let G be a minimal non-abelian p-group. Then, G is isomorphic to one of the following types:

(1)

$Q_8=\langle a,b|a^4=1,a^2=b^2,a^b=a^{-1}\rangle$ (quaternion group);

(2)

$M_p(n,m)=\langle a,b|a^{p^n}=b^{p^m}=1,a^b=a^{1+p^{n-1}}\rangle$ (metacyclic group);

(3)

$M_p(n,m,1)=\langle a,b|a^{p^n}=b^{p^m}=c^p=1,[a,b]=c,[a,c]=[b,c]=1$ (non-metacyclic group).

Corollary 1.

Let G be a minimal non-abelian p-group and $Z\left(G\right)$ be a cyclic group. Then, G is isomorphic to one of the following types:

(1)

$Q_8=\langle a,b|a^4=1,a^2=b^2,a^b=a^{-1}\rangle$ (quaternion group);

(2)

$M_p(n,1)$ (metacyclic group);

(3)

$M_p(1,1,1)$ (non-metacyclic group).

In particular, $M_2(2,1)\cong M_2(1,1,1)\cong D_8$ (dihedral group of order 8).

Proof. 

Applying Lemma 1,

(1)

Let $G=Q_8$. Clearly, $Z\left(Q_8\right)\cong C_2$ is cyclic;

(2)

Let $G=M_p(n,m)$. Then $Z\left(G\right)=\langle a^p,b^p\rangle$. If $m=1$, then $Z\left(G\right)=\langle a^p\rangle$ is cyclic. Further, if $m>1$, then $Z\left(G\right)$ is not cyclic;

(3)

Let $G=M_p(n,m,1)$. Then $Z\left(G\right)=\langle a^p,b^p,c\rangle$. If $m=n=1$, then $Z\left(G\right)=\langle c\rangle$ is a cyclic group. Clearly, if $m>1$ or $n>1$, we have that $Z\left(G\right)$ is not cyclic. □

Lemma 2

([7] Lemma 10).Let $G=AB\in {\mathfrak{B}}_2$. If A, $B⊲⊲G$ and A is supersolvable and B is nilpotent, then G is supersolvable.

The following lemma implies that the properties of a group G in ${\mathfrak{B}}_i$ ($i=1,2$) are determined by the structure of $G/F\left(G\right)$. From this point of view, one can deeply understand the structure of the groups in ${\mathfrak{B}}_i$, $i=1,2$.

Lemma 3.

Let G be a group. Then $G\in {\mathfrak{B}}_1$ ($G\in {\mathfrak{B}}_2$, respectively) if and only if $G/F\left(G\right)=A/F\left(G\right)·B/F\left(G\right)$ is the product of two normal (subnormal, respectively) subgroups of G, where $A,B$ are supersolvable groups. In particular, $G/F\left(G\right)$ is nilpotent.

Proof. 

It follows form [7] Lemma 9 and Lemma 2. □

Lemma 4

([7] Theorem 3). Let G be a group and $G\in {\mathfrak{B}}_2$. If $G^{\prime }$ is nilpotent, then G is supersolvable.

Lemma 5

([2] III, Theorem 4.5). Let G be a group. If $\Phi \left(G\right)=1$, then $F\left(G\right)$ is the direct product of minimal normal subgroups of G.

Lemma 6

([13] Chapter 1, Lemma 1.3). Let V be a vector space with dimension $n\ge 1$ over the field $GF\left(p\right)$. Assume that G is abelian and $e\left(G\right)\mid p-1$ and G acts irreducibly on V. Then, $n=1$ and G is cyclic.

Lemma 7

([14] Chapter B, Corollary 9.4). Let G be a group. Assume that K is a field and V is a simple $KG$-module. If V is faithful for $Z\left(G\right)$, then $Z\left(G\right)$ is cyclic.

3. Proofs

Proof of Theorem 1.

Suppose that $A_1$ is proper and subnormal and contains A. Then, $A_1=A(A_1\cap B)$. It is easy to see that $A_1$ satisfies the conditions of the theorem. By Induction, we have that $A_1$ is supersolvable. Similarly, suppose that $B_1$ is proper and subnormal and contains B. Then, $B_1$ is supersolvable. Now we have that $G=A_1{B}_1$ and $A_1$ and $B_1$ are supersolvable. Hence we can assume that G is a group of minimal order verifying the following: 1. G is a non-supersolvable group; 2. $G=AB$, where A and B are two normal supersolvable subgroups of G.

It is easy to see that G is solvable. By Lemma 2, we may assume that $F\left(G\right)\le A\cap B$.

If $\Phi \left(G\right)\ne 1$, then $G/\Phi \left(G\right)=(A/\Phi (G\left)\right)·(B/\Phi (G\left)\right)$; that is, $G/\Phi \left(G\right)$ satisfies our conditions. By our choice of G, $G/\Phi \left(G\right)$ is supersolvable. It follows from [13] Chapter 1, Corollary 3.2 that G is supersolvable, which is a contradiction. Therefore, $\Phi \left(G\right)=1$. Let N be a minimal normal subgroup of G. Then, $N\le F\left(G\right)$. Hence, $G/N=A/N·B/N$ is the product of two normal supersolvable subgroups $A/N$ and $B/N$. By the choice of G, $G/N$ is supersolvable. Assume that G has two distinct minimal normal subgroups, say $M,N$. Then $G/M$ and $G/N$ are supersolvable. Therefore, $G=G/(M\cap N)\lesssim G/M\times G/N$ is supersolvable, which is a contradiction. Hence, G has the unique minimal normal subgroup, say, W. We have that W is an elementary abelian p-group, where $p\in \pi \left(G\right)$. Note that $\Phi \left(G\right)=1$. Then, $G=WH$, where H is a complement to W in G ([2] Hilfssatz III.4.4) and $W=F\left(G\right)=C_G\left(F\left(G\right)\right)=F\left(A\right)=F\left(B\right)$. By Lemma 3, H is nilpotent.

Denote $A/F\left(G\right)=U$, $B/F\left(G\right)=V$, then $H=UV$. By Lemma 3, $U,V$ are abelian and normal in H. Further, since $A,B⊲G$ and $F\left(G\right)=C_G\left(F\left(G\right)\right)$, we have that $\Phi \left(A\right)=\Phi \left(B\right)=1$ and $C_A\left(F\left(A\right)\right)=A\cap C_G\left(F\left(A\right)\right)=A\cap C_G\left(F\left(G\right)\right)=A\cap F\left(A\right)=F\left(A\right)$. By Lemma 5, we may assume that $F\left(A\right)=N_1\times N_2\times \cdots \times N_r$, where $N_i$ is a minimal normal subgroup of A, $i=1,2,\cdots ,r$. Note that A is supersolvable. Then, $N_i$ is a cyclic subgroup of order p, $i\in \{1,2,\cdots ,r\}$. Therefore,

$$U=A/F\left(A\right)=A/C_A\left(F\left(A\right)\right)\lesssim A/C_A\left(N_1\right)\times A/C_A\left(N_2\right)\times \cdots \times C_A\left(N_r\right).$$

This implies that $e\left(U\right)\mid p-1$. Similarly, $e\left(V\right)\mid p-1$. If H is abelian, then $e\left(H\right)\mid p-1$. By Lemma 6, $W\cong C_p$. Therefore, G is supersolvable, which is a contradiction. Now we have that H is non-abelian. This implies that there exists a Sylow q-subgroup $H_q=U_q{V}_q$ of H such that $H_q$ is not abelian, where $p\ne q$. By the choice of G, $H=H_q$. Further, Lemma 7 implies that $Z\left(H\right)$ is cyclic.

On the other hand, clearly, $Z\left(H\right)U$, $Z\left(H\right)V$ are abelian. Therefore, by our choice of G, we may assume that $Z\left(H\right)\le U\cap V$. Further, note that $U\cap V\le Z\left(H\right)$; then, $Z\left(H\right)=U\cap V$. It is easy to see that $C_V\left(U\right)\le Z\left(H\right)$. Then, $N_V\left(U\right)>C_V\left(U\right)$ since H is a p-group. This implies that there exists a $v\in N_V\left(U\right)\setminus C_V\left(U\right)$ such that $v^q\in Z\left(H\right)$. Therefore,

$$|WU\langle v,Z(H)\rangle :WU|=|\langle v,Z(H)\rangle :Z(H\left)\right(\langle v\rangle \cap WU\left)\right|=|\langle v,Z(H)\rangle :Z(H\left)\right|=q.$$

(1)

Clearly, $WU\langle v,Z\left(H\right)\rangle$ satisfies our conditions. If $U\langle v,Z\left(H\right)\rangle <H$, then the choice of G implies that $WU\langle v,Z\left(H\right)\rangle$ is supersolvable. Hence, $U\langle v,Z\left(H\right)\rangle$ is abelian ([7] Theorem 3), which is contrary to the choice of v. Now we have that $V=\langle v,Z\left(H\right)\rangle$. Similarly, there exists a $u\in N_U\left(V\right)\setminus C_U\left(V\right)$ such that $U=\langle u,Z\left(H\right)\rangle$. By $\left(1\right)$, we have that $U,V$ are maximal subgroups of H. Moreover, $H^{\prime }=U^{\prime }{V}^{\prime }[U,V]=[U,V]\le U\cap V=Z\left(H\right)$. Then, the nilpotency class of H equals to 2 and $H^{\prime }=\langle [u,v]\rangle$. Note that ${[u,v]}^q=[u^q,v]=[u,v^q]=1$ ([2] III, 1.3). Then, $|H^{\prime }|=q$, and it follows that H is a minimal non-abelian group ([12] Theorem 2.3.6). By Corollary 1, we have the desired conclusion. □

Proof of Theorem 2.

By Lemma 3, $G/F\left(G\right)$ is nilpotent. Then, ${(G/F\left(G\right))}^{\prime }\le \Phi (G/F\left(G\right))$ ([2] III Satz 3.11). Note that $(G/F\left(G\right))/{(G/F\left(G\right))}^{\prime }$ is isomorphic to a factor group of $G/G^{\prime }$. It follows from $G/G^{\prime }$ being a cyclic group that $(G/F\left(G\right))/{(G/F\left(G\right))}^{\prime }$ is cyclic. Therefore, $G/F\left(G\right)$ is cyclic. We have either $G/F\left(G\right)=AF\left(G\right)/F\left(G\right)$ or $G/F\left(G\right)=BF\left(G\right)/F\left(G\right)$, which implies that either $G=AF\left(G\right)$ or $G=BF\left(G\right)$. Now, by Lemma 2, G is supersolvable. □

Proof of Theorem 3.

Since $A\cap F\left(G\right)=F\left(A\right)$, then $AF\left(G\right)/F\left(G\right)\cong A/(A\cap F(G\left)\right)=A/F\left(A\right)$. By Lemma 2, $AF\left(G\right)$ is a supersolvable group. Note that A is a subnormal supersolvable subgroup of G. Then, $A^{\prime }\le F\left(A\right)$. It follows that $\left|AF\right(G)/F(G\left)\right|=|A/F(A\left)\right|$ divides $|A/A^{\prime }|$. Similarly, $\left|BF\right(G)/F(G\left)\right|=|B/F(B\left)\right|$ divides $|B/B^{\prime }|$, and $BF\left(G\right)$ is a supersolvable group. Therefore,

$$\left(\right|AF\left(G\right)/F\left(G\right)|,|BF\left(G\right)/F\left(G\right)\left|\right)=1.$$

Moreover, by Lemma 4, $AF\left(G\right)/F\left(G\right)$, $BF\left(G\right)/F\left(G\right)$ are subnormal abelian subgroups of $G/F\left(G\right)$. Therefore, $G/F\left(G\right)$ is abelian; that is, $G^{\prime }$ is nilpotent. Now by Lemma 4, we have that G is supersolvable. □

4. Application

In [3], huppert constructed a non-supersolvable group in ${\mathfrak{B}}_1$. In [11], Zhang introduced the following four examples for the propose of studying Problem 2:

Example 1.

$X^1:8p^2,4|(p-1)$,

$a^4=1,a^2=b^2,ba=a^{-1}b,c_1^p=c_2^p=1,[c_1,c_2]=1$,

$a^{-1}{c}_{1}a=c_2,a^{-1}{c}_{2}a=c_1^{-1}$, $b^{-1}{c}_{1}b=c_1^s,b^{-1}{c}_{2}b=c_2^{-s}$,

where $s^4\equiv 1\left(\mathrm{mod}p\right)$.

Example 2.

$X^2:p^q{q}^{\alpha +\beta },\beta \ge 2,q^{max(\alpha ,\beta )}\mid (p-1)$,

$a^{q^{\alpha }}=b^{q^{\beta }}=1,ab=b^{1+q^{\beta -1}}a,c_k^p=1,[c_i,c_j]=1,1\le k,i,j\le q$,

$a^{-1}{c}_{i}a=c_{i+1},1\le i\le q-1;a^{-1}{c}_{q}a=c_1^t$, where $t^{q^{\alpha -1}}\equiv 1\left(\mathrm{mod}p\right)$,

$b^{-1}{c}_{i}b=c_i^{u^{1+i{q}^{\beta -1}}},1\le i\le q$, where $u^{q^{\beta }}\equiv 1\left(\mathrm{mod}p\right)$.

Example 3.

$X^3:p^q{q}^{\alpha +\beta +1},q^{max(\alpha ,\beta )}|(p-1)$, $min(\alpha ,\beta )=1$,

$a^{q^{\alpha }}=b^{q^{\beta }}=c^q=1,ba=abc,ca=ac,cb=bc,c_k^p=1,[c_i,c_j]=1,1\le k,i,j\le q$,

$a^{-1}{c}_{i}a=c_{i+1},1\le i\le q-1;a^{-1}{c}_{q}a=c_1^t$, where $t^{q^{\alpha -1}}\equiv 1\left(\mathrm{mod}p\right)$,

$b^{-1}{c}_{i}b=c_i^{v{u}^{q-i+1}}$, $1\le i\le q,v^{q^{\beta }}\equiv 1\left(\mathrm{mod}p\right)$, where $u^q\equiv 1\left(\mathrm{mod}p\right)$.

Example 4.

$X^4:p^2·2^3,4\nmid (p-1)$,

$a^4=d^2=1$, $d^{-1}ad=a^{-1}$, $c_1^p=c_2^p=1$, $[c_i,c_j]=1$,

$a^{-1}{c}_{1}a=c_2$, $a^{-1}{c}_{2}a=c_1^{-1}$, $d^{-1}{c}_{1}d=c_2$, $d^{-1}{c}_{2}d=c_1$.

Remark 2.

Note that $X^1$, $X^2$, $X^3$ are minimal non-supersolvable groups, but $X^4$ is not. For example, let $p=2$. Then $X^4=(C_3\times C_3)⋊D_8$. The subgroup $(C_3\times C_3)⋊C_4$ is not supersolvable.

We find that Zhang in [11] obtained a result for Problem 2:

Theorem 4.

([11] Theorem 2) Let $G\in {\mathfrak{B}}_2$. Assume that G is non-supersolvable. Then, G has a subgroup chain $N⊴S⊲⊲G$ such that $S/N\cong X^i$, where $i\in \{1,2,3,4\}$.

Now we analyse the structure of $X^i$, $i=1,2,3,4$.

Lemma 8.

Assume that $i\in \{1,2,3,4\}$. Then, $F\left(X^i\right)$ is the unique minimal normal subgroup of $X^i$, and $X^i/F\left(X^i\right)$ is a nilpotent minimal non-abelian group.

Proof. 

(1) Let $i=1$. We have $X^1=(\langle c_1\rangle \times \langle c_2\rangle )⋊Q_8$, $F\left(X^1\right)=\langle c_1\rangle \times \langle c_2\rangle$. Then $X^1/F\left(X^1\right)\cong Q_8$.

(2) Let $i=2$. We have $\langle a,b\rangle \cong M_q(\alpha ,\beta )$, $C=\langle c_1,c_2,\dots ,c_q\rangle$, $C⋊\langle a\rangle$ is a $p^q{q}^{\alpha }$-group, $F\left(X^2\right)=\langle c_1,c_2,\dots ,c_q\rangle$. Then $X^2/F\left(X^2\right)\cong M_q(\alpha ,\beta )$.

(3) Let $i=3$. We have $\langle a,b\rangle \cong M_q(\alpha ,\beta ,1)$, $F\left(X^3\right)=\langle c_1,c_2,\dots ,c_q\rangle$. Then $X^3/F\left(X^3\right)\cong M_q(\alpha ,\beta ,1)$.

(4) Let $i=4$. We have $X^4=\langle c_1,c_2\rangle ⋊D_8$, $F\left(X^4\right)=\langle c_1,c_2\rangle$. Then $X^4/F\left(X^4\right)\cong D_8$.

The result follows from the defining relations of generators and Lemma 1. □

By applying Lemma 8, we can assert that our Theorem 1 is a deepening of Theorem 4.

Corollary. 2.

([11] Theorem 2) Let $G=AB\in {\mathfrak{B}}_2$, where A and B are subnormal supersolvable subgroups of G. Assume that G is not supersolvable. Then, there exists a chain $N⊴S⊲⊲G$ such that $S/N\cong X$, where $X=WH$, W is a minimal normal subgroup of X, W a p-group, H is a minimal non-abelian q-group, $p\ne q$, and p is an odd prime.

Since ${\mathfrak{B}}_1\subseteq {\mathfrak{B}}_2$, we have the following by our Theorem 2.

Corollary. 3.

([11] Theorem 4) Let $G=AB\in {\mathfrak{B}}_1$, where A and B are normal supersolvable subgroups of G. If $G/G^{\prime }$ is cyclic, then G is supersolvable.