FINITE GROUPS WITH HEREDITARILY G-PERMUTABLE SCHMIDT SUBGROUPS

Abstract

A subgroup A of a group G is said to be hereditarily G-permutable with a subgroup B of G, if $AB^x = B^xA$ for some element $x \in \langle A, B \rangle $. A subgroup A of a group G is said to be hereditarily G-permutable in G if A is hereditarily G-permutable with every subgroup of G. In this paper, we investigate the structure of a finite group G with all its Schmidt subgroups hereditarily G-permutable.

1 Introduction

All groups considered in the paper are finite.

Recall that a group G is said to be a minimal nonnilpotent group or Schmidt group if G is not nilpotent and every proper subgroup of G is nilpotent. It is clear that every nonnilpotent group contains Schmidt subgroups, and their embedding has a strong structural impact (see, for example, [2, 3, 10]).

However, the following extensions of permutability turn out to be important in the structural study of groups and were introduced by Guo et al. in [6].

Definition 1.1. Let A and B be subgroups of a group G.

1. (1) A is said to be G-permutable with B if there exists some $g \in G$ such that $AB^g = B^gA$ .

2. (2) A is said to be hereditarily G-permutable with B (or G-h-permutable with B, for short) if there exists some $g \in \langle A,B\rangle $ such that $AB^g = B^gA$ .

3. (3) A is said to be G-permutable in G if A is G-permutable with all subgroups of G.

4. (4) A is said to be hereditarily G-permutable (or G-h-permutable, for short) in G if A is hereditarily G-permutable with all subgroups of G.

It is clear that permutability implies G-permutability but the converse does not hold in general as the Sylow $2$ -subgroups of the symmetric group of degree $3$ show.

Our main goal here is to complete the structural study of groups in which every Schmidt subgroup of a group G is G-h-permutable. This study was started in [2] where we prove the following important fact.

Theorem 1.2 [2, Theorem B].

If every Schmidt subgroup of a group G is G-h-permutable in G, then G is soluble.

Observe that the alternating group of degree $4$ is a nonsupersoluble Schmidt group.

Let $p_1> p_2 > \cdots > p_r$ be the primes dividing $|G|$ and let $P_i$ be a Sylow $p_i$ -subgroup of G, for each $i = 1, 2, \ldots , r$ . Then we say that G is a Sylow tower group of supersoluble type if all subgroups $P_1, P_1P_2, \ldots , P_1P_2\cdots P_{r-1}$ are normal in G. The class of all Sylow tower groups of supersoluble type is denoted by $\mathfrak {D}$ .

Recall that if $\mathfrak {F}$ is a nonempty class of groups and $\pi $ is a set of primes, then $\mathfrak {F}_\pi $ is the class of all $\pi $ -groups in $\mathfrak {F}$ . In particular, if p is a prime, then $\mathfrak {N}_p$ is the class of all p-groups and $\mathfrak {D}_{\pi (p-1)}$ is the class of all Sylow tower groups G of supersoluble type such that every prime dividing $|G|$ also divides $p-1$ .

If G is a group, then $\operatorname {\mathrm {Soc}}(G)$ is the product of all minimal normal subgroups of G and $\Phi (G)$ is the Frattini subgroup of G, that is, the intersection of all maximal subgroups of G.

Our main goal here is to describe completely the groups G with trivial Frattini subgroup which have their Schmidt subgroups G-h-permutable.

Theorem 1.3. Let G be a group with $\Phi (G) = 1$ . Assume that $\mathfrak {F} = LF (F)$ is the saturated formation locally defined by the canonical local definition F such that $F(p) = \mathfrak {N}_p \mathfrak {D}_{\pi (p-1)}$ for every prime p. If every Schmidt subgroup of G is G-h-permutable in G, then the following statements hold:

1. (1) $G = [\operatorname {\mathrm {Soc}}(G)]M$ is the semidirect product of $\operatorname {\mathrm {Soc}}(G)$ with an $\mathfrak {F}$ -group M;

2. (2) if $\Phi (M) = 1$ , then M is supersoluble.

We shall adhere to the notation and terminology of [1, 4].

2 Definitions and preliminary results

Our first lemma collects some basic properties of G-h-permutable subgroups which are very useful in induction arguments. Its proof is straightforward.

Lemma 2.1. Let A, B and K be subgroups of G with K normal in G. Then, the following statements hold.

1. (1) If A is G-h-permutable with B, then $AK/K$ is $G/K$ -h-permutable with $BK/K$ in $G/K$ .

2. (2) If $K \subseteq A$ , then $A/K$ is $G/K$ -h-permutable with $BK/K$ in $G/K$ if and only if A is G-h-permutable with B in G.

3. (3) If A is G-h-permutable in G, then $AK/K$ is $G/K$ -h-permutable in $G/K$ .

4. (4) If $A \subseteq B$ and A is G-h-permutable in G, then A is B-h-permutable in B.

The following result describes the structure of Schmidt groups.

Lemma 2.2 [5, 8].

Let S be a Schmidt group. Then S satisfies the following properties:

1. (1) the order of S is divisible by exactly two prime numbers p and q;

2. (2) S is a semidirect product $S = [P]\langle a\rangle $ , where P is a normal Sylow p-subgroup of S and $\langle a\rangle $ is a nonnormal Sylow q-subgroup of S and $\langle a^q\rangle \in \operatorname {\mathrm {Z}}(S)$ ;

3. (3) P is the nilpotent residual of S, that is, the smallest normal subgroup of S with nilpotent quotient;

4. (4) $P/ \Phi (P)$ is a noncentral chief factor of S and $\Phi (P) = P{'} \subseteq \operatorname {\mathrm {Z}}(S)$ ;

5. (5) $\Phi (S) = \operatorname {\mathrm {Z}}(S) = P{'} \times \langle a^q \rangle $ ;

6. (6) $\Phi (P)$ is the centraliser $C_P(a)$ of a in P;

7. (7) if $\operatorname {\mathrm {Z}}(S) = 1$ , then $|S| = p^mq$ , where m is the order of p modulo q.

In what follows, $\operatorname {\mathrm {Sch}}(G)$ denotes the set of all Schmidt subgroups of a group G. Following [3], a Schmidt group with a normal Sylow p-subgroup will be called an $S_{\langle p,q \rangle }$ -group.

The proof of Theorem 1.3 follows after a series of lemmas. They give us an interesting picture of the groups with supersoluble Schmidt subgroups.

Lemma 2.3. Let $\mathfrak {F} = \{H \mid \operatorname {\mathrm {Sch}}(H) \subseteq \mathfrak {U} \}$ , where $\mathfrak {U}$ is the class of all supersoluble groups. Then, $\mathfrak {F}$ satisfies the following properties:

1. (1) if $G \in \mathfrak {F}$ , then G is a Sylow tower group of supersoluble type; in particular, G is a soluble group;

2. (2) $\mathfrak {F}$ is a subgroup-closed saturated Fitting formation;

3. (3) $\mathfrak {U} \subseteq \mathfrak {F}$ ;

4. (4) $\mathfrak {F} = LF(F)$ , where F is the canonical local definition such that $F(p) = \mathfrak {N}_p \mathfrak {D}_{\pi (p-1)}$ for every prime p.

Proof. Statements (1), (2) and (3) follow from [7, Lemma 4 and Theorem 2].

Let $\mathfrak {H} = LF (F)$ be a local formation defined by the formation function F with $F(p) = \mathfrak {N}_p \mathfrak {D}_{\pi (p-1)}$ for every prime p. Assume that $\mathfrak {F} \nsubseteq \mathfrak {H}$ . Let G be a group in $\mathfrak {F} \setminus \mathfrak {H}$ of minimal order. Since $\mathfrak {F}$ is a saturated formation, it follows that G is a primitive soluble group. Let $N = \operatorname {\mathrm {Soc}}(G)$ be the unique minimal normal subgroup of G. Then $G/N \in \mathfrak {H}$ . Since G is a Sylow tower group of supersoluble type and $\operatorname {\mathrm {C}}_G(N) = N$ , we see that N is a Sylow p-subgroup of G, where p is the largest prime dividing $|G|$ .

Let $q \in \pi (G)$ with $q \neq p$ and let Q be a Sylow q-subgroup of G. Since $N = \operatorname {\mathrm {C}}_G(N)$ , it follows that $PQ$ is not nilpotent. Hence, G has an $S_{\langle p,q\rangle }$ -subgroup S, which is supersoluble p-closed because $G \in \mathfrak {F}$ . Then, by statements (4) and (5) of Lemma 2.2, $|S/\operatorname {\mathrm {Z}}(S)| = pq$ and therefore, by statement (7) of Lemma 2.2, q divides $p-1$ . Since G is a Sylow tower group of supersoluble type, it follows that

$$ \begin{align*} G/N = G/\operatorname{\mathrm{C}}_G(N) \in \mathfrak{D}_{\pi(p-1)},\end{align*} $$

and thus $G \in \mathfrak {H}$ , which is a contradiction. Hence, $\mathfrak {F} \subseteq \mathfrak {H}$ .

Assume that $\mathfrak {F} \neq \mathfrak {H}$ , and let G be a group in $\mathfrak {H} \setminus \mathfrak {F}$ of minimal order. Since $\mathfrak {H}$ is a saturated formation and $F(p)$ is a formation of soluble groups for all primes p, it follows that G is a primitive soluble group. Let N be a unique minimal normal subgroup of G. The choice of G yields $G \in \mathfrak {H}$ and $G/N \in \mathfrak {F}$ . Since G is soluble, N is a p-group for some prime p, and from $G \in \mathfrak {H}$ , it follows that

$$ \begin{align*}G/N = G/\operatorname{\mathrm{C}}_G(N) \in \mathfrak{N}_p \mathfrak{D}_{\pi(p-1)}.\end{align*} $$

We conclude that $G/N \in \mathfrak {D}_{\pi (p-1)}$ because $\operatorname {\mathrm {O}}_p(G/N) = 1$ by [4, Lemma A.13.6].

Let S be an $S_{\langle r,q\rangle }$ -subgroup of G. If $r \neq p$ , then S is contained in some Hall $p'$ -subgroup H of G. Since $H \cong G/N \in \mathfrak {F}$ , we see that $S \in \mathfrak {U}$ . If $r = p$ , then from $G/N \in \mathfrak {D}_{\pi (p-1)}$ , it follows that q divides $p-1$ . Thus, by Lemma 2.2, $S \in \mathfrak {U}$ . Consequently, every Schmidt subgroup of G is supersoluble, which is a contradiction. Hence, $\mathfrak {F} = \mathfrak {H}$ .

The following examples show that groups in Lemma 2.3 may not be supersoluble.

Example 2.4. Let

$$ \begin{align*}Q = \langle a,b \mid a^4 = b^4 = 1,\, a^2 = b^2,\, b^{-1}ab = a^{-1} \rangle \end{align*} $$

be the quaternion group of order $8$ . Then G has a faithful and irreducible module A over the field of $5$ elements of dimension $2$ . Let $G = [A]Q$ be the corresponding semidirect product. Then G is not supersoluble and $C = [A]\langle a\rangle $ and $D = [A]\langle b\rangle $ are supersoluble and normal subgroups of $G = CD$ . By Lemma 2.3, $G \in \mathfrak {F} = \{H\, |\, \operatorname {\mathrm {Sch}}(H) \subseteq \mathfrak {U}\}$ .

Example 2.5. Assume that M is a nonabelian group of order $21$ . Then M has a faithful and irreducible module N over $\text {GF}(43)$ , the field of $43$ elements (see, for example, [4, Corollary B.11.8]). Consider the semidirect product $G = [N]M$ . It is obvious that G is not supersoluble. By Lemma 2.3, $G \in \mathfrak {F} = \{H\,|\, \operatorname {\mathrm {Sch}}(H) \subseteq \mathfrak {U}\}$ .

The following result is of interest although it is not needed for the proof of Theorem 1.3.

Proposition 2.6. Let $\mathfrak {F} = \{H \mid \operatorname {\mathrm {Sch}}(H) \subseteq \mathfrak {U}\}$ . Then, for every $n\in \mathbb {N}$ , there exists a group $G \in \mathfrak {F}$ of nilpotent length n.

Proof. Let $n \geq 2$ and let $p_1, p_2, \ldots , p_n$ be primes such that $p_1 < p_2 < \cdots < p_n$ and $p_i$ divides $p_j-1$ for all $i < j$ , where $i = 1,2, \ldots , n-1$ , $j = 2, \ldots , n$ . By Dirichlet’s theorem, there exists an infinite set of primes of the form

$$ \begin{align*} p_1p_2\cdots p_{n_0} + 1, \end{align*} $$

where $n_0 \in \mathbb {N}$ . Assume that $p_{n+1}$ is one of them. It is obvious that $p_i$ divides $p_{n+1} - 1$ for any $i = 1,2, \ldots , n$ .

Assume that $G_1$ is a cyclic group of order $p_1$ . Assume that $i \geq 2$ and $G_{i -1}$ is in $\mathfrak F$ and of nilpotent length $i-1$ . By [4, Corollary B.11.8], $G_{i -1}$ has a faithful and irreducible module $V_{p_i}$ over the field of $p_i$ elements. Let $G_i = [V_{p_i}]G_{i-1}$ be the corresponding semidirect product. Then $\operatorname {\mathrm {F}}(G_i) = V_{p_i}$ and hence the nilpotent length of $G_i$ is equal to i. Furthermore, by Lemma 2.3, $G_i \in \mathfrak {F}$ . In particular, $G_n$ is an $\mathfrak {F}$ -group of nilpotent length n.

The following subgroup embedding property was introduced by Vasil’ev, Vasil’eva and Tyutyanov in [9].

Definition 2.7. A subgroup H of a group G is said to be $\mathbb {P}$ -subnormal in G if there exists a chain of subgroups

$$ \begin{align*} H = H_0 \subseteq H_1 \subseteq \cdots \subseteq H_{n-1} \subseteq H_n = G \end{align*} $$

such that for every $i = 1, 2, \ldots , n$ , either $|H_i : H_{i-1}| \in \mathbb {P}$ or $H_{i-1}$ is normal in $H_i$ .

Note that $\mathbb {P}$ -subnormality coincides with K- $\mathfrak {U}$ -subnormality (see [1, Ch. 6]) in the soluble universe.

Lemma 2.8. Let A be a G-h-permutable subgroup of a soluble group G. Then, A is $\mathbb {P}$ -subnormal in G. In particular, the supersoluble residual $A^{\mathfrak {U}}$ of A is subnormal in G.

Proof. Let G be a group of smallest order for which the lemma is not true, and let L be a minimal normal subgroup of G. Since G is soluble, $|L| = p^n$ for some prime $p \in \pi (G)$ and $n \geq 1$ . Suppose that $G = AL$ . Then A is a maximal subgroup of G and $A \cap L = 1$ . Let $L_1$ be a subgroup of prime order of L. Then, $AL_1^x = L_1^xA$ for some $x \in G$ . Consequently, $AL_1^x$ is a subgroup of G. Since A is maximal in G and $A \neq AL_1^x$ , we see that $AL_1^x = G$ . Because

$$ \begin{align*}|G:A| = |AL_1^x|/|A| =|L_1^x|/|A \cap L_1^x| = |L_1^x|,\end{align*} $$

we conclude that $|G:A| = p$ and then A is $\mathbb {P}$ -subnormal in G, which is a contradiction. Hence, $G \neq AL$ . Since $|AL| < |G|$ , by Lemma 2.1, it follows that A is $\mathbb {P}$ -subnormal in $AL$ . By Lemma 2.1, $AL/L$ is $(G/L)$ -h-permutable in $G/L$ , and from $|G/L| < |G|$ , it follows that $AL/L$ is $\mathbb {P}$ -subnormal in $G/L$ . In particular, $AL$ is $\mathbb {P}$ -subnormal in G by [1, Lemma 6.1.6]. However, then A is a $\mathbb {P}$ -subnormal subgroup of G by [1, Lemma 6.1.7], which is a contradiction. Consequently, A is $\mathbb {P}$ -subnormal in G. Applying [1, Lemma 6.1.9], we conclude that $A^{\mathfrak {U}}$ is subnormal in G.

Example 2.9. Let G be a group isomorphic to the alternating group of degree $6$ . Since G does not have maximal subgroups of prime index, the identity subgroup $1$ of G is G-h-permutable but not $\mathbb {P}$ -subnormal in G. Thus, the solubility of the group G in Lemma 2.8 is essential.

Lemma 2.10. Let $G \in \mathfrak {F} = \{H\,|\, \operatorname {\mathrm {Sch}}(H) \subseteq \mathfrak {U} \}$ . If $\Phi (G) = 1$ and every Schmidt subgroup of G is G-h-permutable in G, then G is supersoluble.

Proof. We argue by induction on $|G|$ . Let N be a minimal normal subgroup of G. Since G is soluble by Lemma 2.3, it follows that N is p-elementary abelian for some prime p. Since $\Phi (G) = 1$ , it follows that $G = NM$ for some maximal subgroup M of G and $N \cap M = 1$ .

Suppose that $NM_{p'}$ is p-nilpotent. Then $NM_{p'} \subseteq \operatorname {\mathrm {C}}_G(N)$ . Then $G/\operatorname {\mathrm {C}}_G(N)$ is a p-group. Since $\operatorname {\mathrm {O}}_p(G/\operatorname {\mathrm {C}}_G(N)) = 1$ by [4, Lemma A.13.6], we have $N \subseteq \operatorname {\mathrm {Z}}(G)$ . Then $G = N \times M$ . Now, M belongs to $\mathfrak {F}$ and $\Phi (M) \subseteq \Phi (G) = 1$ by [4, Theorem A.9.2]. By induction, M is supersoluble. Hence, G is supersoluble.

Assume that $NM_{p'}$ is not p-nilpotent. Consequently, $NM_{p'}$ contains a minimal non-p-nilpotent group X. By [1, Corollary 6.4.5], X is an $S_{\langle p,q \rangle }$ -subgroup $X = [P]Q$ and $P \subseteq N$ . We can assume without loss of generality that $Q \subseteq M_{p'}$ . Since the subgroup $[P]Q$ is G-h-permutable, we may assume that $([P]Q)M = PM$ is a subgroup of G. Consequently, $P = N$ and $NQ$ is an $S_{\langle p,q\rangle }$ -subgroup G. By hypothesis, $NQ$ is supersoluble. Hence, in view of Lemma 2.2, $|N/\Phi (N)| = p$ by Lemma 2.2. Since $\Phi (N) = 1$ , it follows that $|N| = p$ .

Consequently, we may assume that every minimal normal subgroup of G is cyclic. Then, by [4, Theorem A.10.6], $\operatorname {\mathrm {F}}(G)$ is a direct product of normal subgroups of G of prime order and so $G/\operatorname {\mathrm {C}}_G(\operatorname {\mathrm {F}}(G))$ is abelian. Since $\operatorname {\mathrm {C}}_G(\operatorname {\mathrm {F}}(G)) \subseteq \operatorname {\mathrm {F}}(G)$ by [4, Theorem A.10.6], it follows that $G/\operatorname {\mathrm {F}}(G)$ is abelian. In particular, G is supersoluble.

3 Proof of Theorem 1.3

Since G is soluble and $\Phi (G) = 1$ , we conclude that $\operatorname {\mathrm {F}}(G) = \operatorname {\mathrm {Soc}}(G)$ and $G = [\operatorname {\mathrm {Soc}}(G)]M$ for some subgroup M of G, that is, $\operatorname {\mathrm {Soc}}(G) \cap M = 1$ by [4, Theorem A.10.6].

Let S be a Schmidt subgroup of M. Suppose that S is an $S_{\langle p,q \rangle }$ -subgroup. Then, by hypothesis, S is G-h-permutable in G. Consequently, by Lemma 2.8, S is $\mathbb {P}$ -subnormal in G and $S^{\mathfrak {U}}$ is subnormal in G. In view of Lemma 2.2, we see that either $S^{\mathfrak {U}} \neq 1$ is a p-subgroup of S or $S^{\mathfrak {U}} = 1$ . Assume that $S^{\mathfrak {U}} \neq 1$ . Then,

$$ \begin{align*} S^{\mathfrak{U}} \subseteq F(G) \cap M = 1, \end{align*} $$

which is a contradiction. Therefore, every Schmidt subgroup of M is supersoluble.

By Lemma 2.3, it follows that $M \in \mathfrak {F} = LF (F)$ , where F is the formation function given by $F(r) = \mathfrak {N}_r \mathfrak {D}_{\pi (r-1)}$ for any prime r.

By Lemma 2.10, M is supersoluble provided that $\Phi (M) = 1$ .