Finite Groups with $\mathbb{P}$-Subnormal Schmidt Subgroups

Abstract

A subgroup $H$ of a group $G$ is called $\mathbb{P}$-subnormal in $G$ whenever either $H=G$ or there is a chain of subgroups $H=H_{0}\subset H_{1}\subset\mathinner{\ldotp\ldotp\ldotp}\subset H_{n}=G$ such that $|H_{i}:H_{i-1}|$ is a prime for every $i=1,2,\mathinner{\ldotp\ldotp\ldotp},n$. We study the structure of a finite group $G$ all of whose Schmidt subgroups are $\mathbb{P}$-subnormal. The obtained results complement the answer to Problem 18.30 in the Kourovka Notebook..

INTRODUCTION

We consider only finite groups.

A subgroup $H$ of a group $G$ is called $\mathbb{P}$-subnormal if it either coincides with the group $G$ or is connected with $G$ by a chain of subgroups all of whose indices are primes. The notion of $\mathbb{P}$-subnormal subgroup was proposed in [1] in connection with the development of the famous Huppert theorem that a group $G$ is supersoluble if and only if any of its proper subgroups can be connected with $G$ by a chain of subgroups with prime indices.

Groups with a system $\Sigma$ of given $\mathbb{P}$-subnormal subgroups were studied in many papers. In particular, groups in which every Sylow subgroup is $\mathbb{P}$-subnormal were described in [2]. The supersolubility of a group in the cases where $\Sigma$ is the set of normalizers of all Sylow subgroups of $G$ and $\Sigma$ is the set of all Hall subgroups of $G$ was proved in [3]. Classes of groups with $\mathbb{P}$-subnormal primary subgroups and $\mathbb{P}$-subnormal primary cyclic subgroups were considered in [4]. The structure of groups representable as a product of $\mathbb{P}$-subnormal subgroups was studied in [5].

A special place in the study of groups with a given system of $\mathbb{P}$-subnormal subgroups is occupied by the case when $\Sigma=\mathrm{Sch}(G)$ is the set of all Schmidt subgroups of $G$. Recall that a Schmidt group is a nonnilpotent group all of whose proper subgroups are nilpotent. A simple check shows that every nonnilpotent group contains at least one Schmidt subgroup (i.e., a subgroup that is a Schmidt group). The study of groups with a given system of $\mathbb{P}$-subnormal subgroups was motivated by Problem 18.30 from the Kourovka Notebook[6]:

Is a finite group soluble if all its Schmidt subgroups are $\mathbb{P}$-subnormal?

Tyutyanov used the classification of finite simple groups to obtain a positive answer to this question in [7]. In connection with this result, it is natural to formulate a more general problem:

Investigate the normal structure of a group all of whose Schmidt subgroups are $\mathbb{P}$-subnormal.

Particular aspects of this problem were addressed in [8], where the metanilpotency of a group with $\mathbb{P}$-subnormal generalized Schmidt subgroups was established. A generalized Schmidt group was understood as any $B$-group, i.e., a group whose quotient group by the Frattini subgroup is a Schmidt group (the notion of $B$-group was proposed by Berkovich in [9]). It is clear that any Schmidt group is a $B$-group. At the same time, a dihedral group of order $18$ is a $B$-group and not a Schmidt group. As follows from the structure of Schmidt groups, $G$ is a $B$-group if and only if $G/\Phi(G)$ is a biprimary Miller–Moreno group, i.e., a nonnilpotent group all of whose proper subgroups are abelian.

According to the results of [10], groups with $\mathbb{P}$-subnormal Schmidt subgroups are much more complex than groups with $\mathbb{P}$-subnormal $B$-subgroups.

For a group $G$, let $\pi(G)=\{p_{1},p_{2},\mathinner{\ldotp\ldotp\ldotp},p_{r}\}$ with $p_{1}>p_{2}>\mathinner{\ldotp\ldotp\ldotp}>p_{r}$, and let $P_{i}$ be a Sylow $p_{i}$-subgroup of $G$ for $i=1,2,\mathinner{\ldotp\ldotp\ldotp},r$. We will say that a group $G$ has a Sylow tower of supersoluble type (or $G$is Ore dispersive) if the subgroups $P_{1}$, $P_{1}P_{2}$, $\mathinner{\ldotp\ldotp\ldotp}$, $P_{1}P_{2}\mathinner{\ldotp\ldotp\ldotp}P_{r-1}$ are normal in $G$. In what follows, we will denote by $\mathfrak{D}$ the class of all groups $G$ having a Sylow tower of supersoluble type. Further, for a given prime $p$, we denote by $\mathfrak{D}_{\pi(p-1)}$ the class of all Ore dispersive groups $G$ such that $\pi(G)\subseteq\pi(p-1)$, where $\pi(p-1)$ is the set of all prime divisors of $p-1$.

Our main goal is to prove the following theorem.

Theorem 1

Let $\mathfrak{F}=\{H\,|\,\mathrm{Sch}(H)\subseteq\mathfrak{U}\},$ where $\mathfrak{U}$ is the class of all supersoluble groups. Then the following statements hold:

• (1) $\mathfrak{F}$ is a local formation with canonical local definition $F$ such that $F(p)=\mathfrak{N}_{p}\mathfrak{D}_{\pi(p-1)}$ for each prime $p$;

• (2) if each Schmidt subgroup of $G$ is $\mathbb{P}$-subnormal$,$ then $G/F(G)\in\mathfrak{F}$;

• (3) if each Schmidt subgroup of $G$ is $\mathbb{P}$-subnormal and $H$ is its nonsupersoluble Schmidt subgroup with normal Sylow $p$-subgroup $P,$ then $P\subseteq G^{\mathfrak{F}}\cap F(G)$.

Obviously, every subnormal subgroup of a soluble group is $\mathbb{P}$-subnormal. The groups in which every Schmidt subgroup is subnormal were described in [11].

1 DEFINITIONS AND PRELIMINARY RESULTS

In this paper we use the definitions and notation adopted in [12].

Fix the following notation:

• – $\mathfrak{U}$ is the class of all supersoluble groups;

• – $\mathfrak{N}$ is the class of all nilpotent groups;

• – if $\mathfrak{F}$ is a nonempty class and $\pi$ is a set of primes, then $\mathfrak{F}_{\pi}$ is the class of all $\pi$-groups from $\mathfrak{F}$;

• – if $\mathfrak{F}$ is a formation, then $G^{\mathfrak{F}}$ is the intersection of all normal subgroups $N$ of a group $G$ for which $G/N\in\mathfrak{F}$ (the subgroup $G^{\mathfrak{F}}$ is called the $\mathfrak{F}$-residual of $G$);

• – $\mathbb{P}$ is the set of all primes;

• – if $n$ is a positive integer, then $\pi(n)$ is the set of all primes dividing $n$ (in particular, $\pi(G)=\pi(|G|)$);

• – $\mathrm{Sch}(G)$ is the set of all Schmidt subgroups of a group $G$;

• – if $A$ and $B$ are subgroups of a group $G$, then $[A]B$ is their semidirect product with the normal subgroup $A$.

The basic structure of Schmidt groups, which is described in the following lemma, was established in [13, 14].

Lemma 1

Let $S$ be a Schmidt group. Then the following statements hold:

• (1) $\pi(S)=\{p,q\}$;

• (2) $S=[P]\langle a\rangle,$ where $P$ is a normal Sylow $p$-subgroup of $S$ and $\langle a\rangle$ is its Sylow $q$-subgroup such that $\langle a^{q}\rangle\subseteq Z(S)$;

• (3) $P$ is the $\mathfrak{N}$-residual of $S$;

• (4) $P/\Phi(P)$ is a minimal normal subgroup of $S/\Phi(P)$ and$,$ in addition$,$$\Phi(P)=P^{{}^{\prime}}\subseteq Z(S)$;

• (5) $\Phi(S)=Z(S)=P^{{}^{\prime}}\times\langle a^{q}\rangle$;

• (6) if $Z(S)=1,$ then $|S|=p^{m}q,$ where $m$ is the exponent of $p$ modulo $q$.

Following [15], we call a Schmidt $\{p,q\}$-group with a normal Sylow $p$-subgroup and a nonnormal cyclic Sylow $q$-subgroup an $S_{<p,q>}$-group. In addition, we call a Schmidt group $S=[P]\langle a\rangle$ with a normal Sylow $p$-subgroup $P$ for which $|P/\Phi(P)|=p^{m}$, where $m$ is the exponent of $p$ modulo $q$, and a nonnormal cyclic Sylow $q$-subgroup $\langle a\rangle$ an $S_{<p,q,m>}$-group. Note that an $S_{<p,q,m>}$-group $S$ is supersoluble if and only if $m=1$.

A subgroup $H$ of a group $G$ is called $\mathbb{P}$-subnormal in $G$ if either $H=G$ or there exists a chain of subgroups

$$H=H_{0}\subset H_{1}\subset\mathinner{\ldotp\ldotp\ldotp}\subset H_{n-1} \subset H_{n}=G$$

such that $|H_{i}:H_{i-1}|\in\mathbb{P}$ for any $i=1,2,\mathinner{\ldotp\ldotp\ldotp},n$. If $H$ is a $\mathbb{P}$-subnormal subgroup of $G$, then we write $H$ $\mathbb{P}$-sn $G$ according to [1].

In the following lemma we give the main properties of $\mathbb{P}$-subnormal subgroups.

Lemma 2

Suppose that $H,$$K,$ and $N$ are subgroups of $G,$ and $N$ is normal in $G$. Then:

• (1) if $H$ $\mathbb{P}$-$\mathrm{sn}$ $G,$ then $H\cap N$ $\mathbb{P}$-$\mathrm{sn}$ $N$ and $HN/N$ $\mathbb{P}$-$\mathrm{sn}$ $G/N$;

• (2) if $N\subseteq H$ and $H/N$ $\mathbb{P}$-$\mathrm{sn}$ $G/N,$ then $H$ $\mathbb{P}$-$\mathrm{sn}$ $G$;

• (3) if $H$ $\mathbb{P}$-$\mathrm{sn}$ $K$ and $K$ $\mathbb{P}$-$\mathrm{sn}$ $G,$ then $H$$\mathbb{P}$-$\mathrm{sn}$ $G$;

• (4) if $G^{\mathfrak{U}}\subseteq H,$ then $H$ $\mathbb{P}$-$\mathrm{sn}$ $G$;

• (5) if $H$ $\mathbb{P}$-$\mathrm{sn}$ $G$ and $H\subseteq K,$ then $H$ $\mathbb{P}$-$\mathrm{sn}$ $K$;

• (6) if the group $G$ is soluble and $H$ $\mathbb{P}$-$\mathrm{sn}$ $G,$ then the subgroup $H^{\mathfrak{U}}$ is subnormal in $G$;

• (7) if $H$ is a Schmidt subgroup of a soluble group $G$ and $H$ $\mathbb{P}$-$\mathrm{sn}$ $G,$ then either the subgroup $H$ is supersoluble or $H^{\mathfrak{U}}\subseteq F(G)$.

Proof

Statements (1)–(4) are proved in Lemma 3.1 from [2], and statement (5) is proved in Lemma 3.4 from [2].

Let us prove statement (6). If $H=G$, then it is obvious. Therefore, we can assume that $H\neq G$ and there exists a subgroup chain

$$H=H_{0}\subset H_{1}\subset\mathinner{\ldotp\ldotp\ldotp}\subset H_{n-1} \subset H_{n}=G$$

such that $|H_{i}:H_{i-1}|\in\mathbb{P}$ for any $i=1,2,\mathinner{\ldotp\ldotp\ldotp},n$. Since the group $G$ is soluble, we have $H_{i}^{\mathfrak{U}}\subseteq H_{i-1}$ for all $i=1,2,\mathinner{\ldotp\ldotp\ldotp},n$ (see, e.g., Lemma 3.3 from [2]). Since the formation $\mathfrak{U}$ is hereditary, we have $H_{i-1}H_{i}^{\mathfrak{U}}/H_{i}^{\mathfrak{U}}\in\mathfrak{U}$. Therefore, since

$$H_{i-1}H_{i}^{\mathfrak{U}}/H_{i}^{\mathfrak{U}}\cong H_{i-1}/H_{i-1}\cap H_{i }^{\mathfrak{U}},$$

we have $H_{i-1}^{\mathfrak{U}}\subseteq H_{i}^{\mathfrak{U}}$. Consequently, $H_{i-1}^{\mathfrak{U}}\subseteq H_{i}^{\mathfrak{U}}\subseteq H_{i}$ and $H_{i-1}^{\mathfrak{U}}\unlhd H_{i}^{\mathfrak{U}}$. Thus, the subgroup $H^{\mathfrak{U}}$ is subnormal in $G$.

Let us prove statement (7). Let $H$ be a $\mathbb{P}$-subnormal Schmidt subgroup of a soluble group $G$. Then, by statement (6), the supersoluble residual $H^{\mathfrak{U}}$ of the subgroup $H$ is subnormal in $G$. By Lemma 1, $\pi(H)=\{p,q\}$ and $H=[P]\langle a\rangle$, where $P$ is a normal Sylow $p$-subgroup of $H$ and $\langle a\rangle$ is its Sylow $q$-subgroup. In addition, $P$ is the $\mathfrak{N}$-residual of $H$ and $P/\Phi(P)$ is a minimal normal subgroup of $H/\Phi(P)$. Then either $H^{\mathfrak{U}}=P$ or $H^{\mathfrak{U}}\subseteq\Phi(P)$. If $H^{\mathfrak{U}}=P$, then it follows from statement (6) that the subgroup $P$ is subnormal in $G$. Then, obviously, $P=H^{\mathfrak{N}}\subseteq F(G)$. Let $H^{\mathfrak{U}}\subseteq\Phi(P)$. Since the subgroup $P$ is normal in $H$, we have $H^{\mathfrak{U}}\subseteq\Phi(H)$. Now it follows from the facts that the formation $\mathfrak{U}$ of all supersoluble groups is saturated (see Example IV.3.4.(f) in [12]) and $H/\Phi(H)\in\mathfrak{U}$ that $H$ is a supersoluble subgroup.

The lemma is proved.    $\square$

Remark 1

The requirement of solubility of the group $G$ in statements (6) and (7) of Lemma 2 is essential and cannot be discarded in the general case. For example, in the group $PSL_{2}(7)$, the subgroup $H\cong S_{4}$ is $\mathbb{P}$-subnormal, but its $\mathfrak{U}$-residual, obviously, is not a subnormal subgroup. In the alternating group $A_{5}$, the subgroup $A_{4}$ is a $\mathbb{P}$-subnormal Schmidt subgroup, but its supersoluble residual is not contained in the Fitting subgroup of $A_{5}$.

Recall that a formation is a class of groups closed under taking homomorphic images and finite subdirect products. A formation $\mathfrak{F}$ is called

–  saturated if, for any group $G$, the membership $G/\Phi(G)\in\mathfrak{F}$ always implies $G\in\mathfrak{F}$;

–  hereditary if it is closed under taking subgroups.

A class $\mathfrak{F}$ is called a Fitting class if it satisfies the following requirements:

(1)  $\mathfrak{F}$ is a normal hereditary class;

(2)  if $G=AB$, where $A\unlhd G$, $B\unlhd G$, $A\in\mathfrak{F}$, and $B\in\mathfrak{F}$, then $G\in\mathfrak{F}$.

A Fitting formation is a formation that is a Fitting class.

A minimal supplement to a normal subgroup $N$ of a group $G$ is a subgroup $L$ of $G$ such that $LN=G$, but $L_{1}N\neq G$ for any proper subgroup $L_{1}$ of $L$.

Lemma 3

If $K$ and $D$ are subgroups of a group $G,$ the subgroup $D$ is normal in $G,$ and $K/D$ is an $S_{<p,q>}$-subgroup$,$ then a minimal supplement $L$ to the subgroup $D$ in $K$ has the following properties:

• (1) $L$ is a $p$-closed $\{p,q\}$-subgroup;

• (2) all proper normal subgroups of $L$ are nilpotent;

• (3) the subgroup $L$ contains an $S_{<p,q>}$-subgroup $[P]Q$ such that $Q$ is not contained in $D$ and $L=([P]Q)^{L}=Q^{L}$;

• (4) if an $S_{<p,q>}$-subgroup $[P]Q$ is $\mathbb{P}$-subnormal in $G,$ then one of the following holds:

  • (i) $K/D=([P]Q)D/D$; in particular$,$ the subgroup $K/D$ is $\mathbb{P}$-subnormal in $G/D$;

  • (ii) the subgroup $K/D$ is supersoluble$,$ has a Sylow $p$-subgroup of order $p,$$q$ divides $p-1,$ and a Sylow $q$-subgroup of $K/D$ is $\mathbb{P}$-subnormal in $G/D$;

• (5) $K/D$ is an $S_{<p,q,m>}$-group if and only if $[P]Q$ is an $S_{<p,q,m>}$-group.

Proof

Statements (1)–(3) are proved in Lemma 2 from [17].

Let us prove statement (4). Assume that an $S_{<p,q>}$-subgroup $[P]Q$ is $\mathbb{P}$-subnormal in $G$. By Lemma 11.1 from [16], $LD=K$ and $L\cap D\subseteq\Phi(L)$. The Frattini subgroup consists of nongenerating elements; therefore, if $([P]Q)(L\cap D)=L$, then $[P]Q=L$. In this case, by statement (1) of Lemma 2,

$$([P]Q)D/D=LD/D=K/D$$

is a $\mathbb{P}$-subnormal subgroup of $G/D$.

Now, let $[P]Q$ be a proper subgroup of $L$. Then it follows from $L\cap D\subseteq\Phi(L)$ that $([P]Q)(L\cap D)$ is a proper subgroup of $L$. If the subgroup $([P]Q)(L\cap D)$ is contained in some subnormal subgroup of $L$, then by statement (2) it is nilpotent, which is impossible, since $([P]Q)(L\cap D)$ contains a Schmidt subgroup $[P]Q$. Consequently, $([P]Q)(L\cap D)$ is not contained in any normal maximal subgroup of $L$. From

$$K/D=LD/D\cong L/L\cap D$$

we conclude that $L/L\cap D$ is a Schmidt group. Now, according to statement (2) of Lemma 1, $Q(L\cap D)/(L\cap D)$ is a Sylow $q$-subgroup of $L/L\cap D$. Further, $([P]Q)(L\cap D)/L\cap D$ is a proper subgroup of $L/L\cap D$; hence, by statement (5) of Lemma 1,

$$([P]Q)(L\cap D)/L\cap D\subseteq\Phi(L/L\cap D)(Q(L\cap D)/L\cap D).$$

By statement (4) of Lemma 1, $\Phi(L/L\cap D)(Q(L\cap D)/L\cap D)$ is a maximal subgroup of $L/L\cap D$. Since the subgroup $[P]Q$ is $\mathbb{P}$-subnormal in $G$ by the hypothesis, it follows in view of Lemma 2 that the subgroup $\Phi(L/L\cap D)(Q(L\cap D)/L\cap D)$ is $\mathbb{P}$-subnormal in $L/L\cap D$. Hence, the index of the maximal subgroup

$$\Phi(L/L\cap D)(Q(L\cap D)/L\cap D)$$

in the group $L/L\cap D$ is a prime $p$, and therefore the group $(L/L\cap D)/\Phi(L/L\cap D)$ is supersoluble. However, since the formation $\mathfrak{U}$ is saturated, it follows that the group $L/L\cap D$ is supersoluble. This and the isomorphism $K/D\cong L/L\cap D$ imply that $K/D$ is a supersoluble group. Now, by Lemma 1 from [18], the group $K/D$ has a Sylow $p$-subgroup of order $p$ and $q$ divides $p-1$. Since $|K/D|=pq^{n}$, $([P]Q)(L\cap D)/L\cap D$ is a Sylow $q$-subgroup of $L/L\cap D$. Using Lemma 2, we conclude that a Sylow $q$-subgroup of $K/D$ is $\mathbb{P}$-subnormal in $G/D$.

Let us prove statement (5). Assume that $K/D$ is an $S_{<p,q,m>}$-group. Then, by statement (6) of Lemma 1,

$$|(K/D)/Z(K/D)|=p^{m}q,$$

where $m$ is the exponent of $p$ modulo $q$. By statement (4) of Lemma 3, the subgroup $[P]Q$ is an $S_{<p,q,k>}$-group for some natural $k\geq 1$ that is the exponent of $p$ modulo $q$. This and the definition of the exponent imply that $m=k$. Arguing in the reverse order, we can show that if $[P]Q$ is an $S_{\langle p,q,m\rangle}$-group, then $K/D$ is also an $S_{<p,q,m>}$-group.

The lemma is proved.    $\square$

Recall the definition of a local formation. A function

$$f:\mathbb{P}\to\{\mbox{formations of finite groups}\}$$

is called a formation function.

For a formation function $f$, a chief factor $A/B$ of a group $G$ is called $f$-central ($f$-excentral) if

$$G/C_{G}(A/B)\cong\mbox{Aut}_{G}(A/B)\in f(p)$$

for all primes $p\in\pi(A/B)$ ($G/C_{G}(A/B)$ does not belong to $f(p)$ for at least one prime $p\in\pi(A/B)$, respectively). A class of groups $\mathfrak{F}=LF(f)$ is called a local formation if it consists of all the groups $G$ such that either $G=1$ or $G\neq 1$ and any chief factor $A/B$ of the group $G$ is $f$-central. In this case, the local formation $\mathfrak{F}$ is said to be defined by means of the formation function$f$, and $f$ is said to be a local definition of the formation $\mathfrak{F}$.

Assume that $f$ is a formation function and $\mathfrak{F}=LF(f)$. Then $f$ is called

(a)  internal  if $f(p)\subseteq\mathfrak{F}$ for all $p\in\mathbb{P}$;

(b)  complete if $f(p)=\mathfrak{N}_{p}f(p)$ for all $p\in\mathbb{P}$;

(c)  canonical  if it is complete and internal.

As shown in Theorem IV.3.7 from [12], for any local formation $\mathfrak{F}$ there exists a unique canonical formation function $F$ such that $\mathfrak{F}=LF(F)$. This function is called the canonical local definition of the formation $\mathfrak{F}$.

Note that, according to the Gaschütz–Lubeseder–Schmid theorem ([12], Theorem IV.4.6), a formation $\mathfrak{F}$ is saturated if and only if it is local. Hence, in particular, for any saturated formation $\mathfrak{F}$ there exists a canonical local definition $F$ such that $\mathfrak{F}=LF(F)$.

Lemma 4

Let $\mathfrak{F}=\{G\,|\,\mathrm{Sch}(G)\subseteq\mathfrak{U}\}$. Then the following statements hold:

• (1) $G\in\mathfrak{F}$ if and only if the group $G$ is soluble and$,$ for any primes $p,q\in\pi(G)$ and a Hall $\{p,q\}$-subgroup$,$ either this subgroup is nilpotent or it is $p$-closed and $q$ divides $p-1$;

• (2) if $G\in\mathfrak{F},$ then the group $G$ has a Sylow tower of supersoluble type;

• (3) the class $\mathfrak{F}$ is a hereditary saturated Fitting formation;

• (4) $\mathfrak{U}\subseteq\mathfrak{F}$;

• (5) $\mathfrak{F}$ is a local formation with a canonical local definition $F$ such that $F(p)=\mathfrak{N}_{p}\mathfrak{D}_{\pi(p-1)}$.

Proof

Statement (1) follows from Lemma 5 and Theorem 1 in [18], and statements (2)–(5) follow from Lemma 2.3 in [10].

The lemma is proved.    $\square$

Remark 2

It follows from Proposition 2.6 of paper [10] that if a group $G$ belongs to the class $\mathfrak{F}=\{H\,|\,\mathrm{Sch}(H)\in\mathfrak{U}\}$, then $G$ can have any nilpotent length greater than $1$. In particular, there exist groups $G\in\mathfrak{F}$ that are not supersoluble. Let, for example, $M$ be a nonabelian group of order $21$. Then there exists a faithful irreducible $M$-module $N$ over a field of $43$ elements (see, e.g., Corollary B.11.8 in [12]). Obviously, the group $G=[N]M$ is not supersoluble, but it belongs to the class $\{H\,|\,\mathrm{Sch}(H)\in\mathfrak{U}\}$ by Lemma 4.

2 PROOF OF THEOREM 1

(1)  By statement (5) of Lemma 4, the local formation $\mathfrak{F}=\{G\,|\,\mathrm{Sch}(G)\subseteq\mathfrak{U}\}$ has a canonical local definition $F$ such that $F(p)=\mathfrak{N}_{p}\mathfrak{D}_{\pi(p-1)}$.

(2)  Let each Schmidt subgroup of $G$ be $\mathbb{P}$-subnormal. Then the group $G$ is soluble by the main result of [7].

Let $D$ be a minimal supplement to $F(G)$ in the group $G$. In this case, in particular, $DF(G)=G$ and $D\cap F(G)\subseteq\Phi(D)$. Let $K/D\cap F(G)$ be an arbitrary Schmidt subgroup of $D/D\cap F(G)$. Without loss of generality, we can assume that $K/D\cap F(G)$ is an $S_{<p,q>}$-subgroup for some primes $p$ and $q$. By statement (4) of Lemma 3, a minimal supplement $L$ to the subgroup $D\cap F(G)$ in $K$ contains an $S_{<p,q>}$-subgroup $[P]Q$ such that $Q$ is not contained in $D\cap F(G)$.

The subgroup $[P]Q$ is $\mathbb{P}$-subnormal in $G$ by the hypothesis. Then, by statement (3) of Lemma 3, we have one of the following statements:

(i)  $K/D\cap F(G)=([P]Q)(D\cap F(G))/D\cap F(G)$; in particular, the subgroup $K/D\cap F(G)$ is $\mathbb{P}$-subnormal in $D/D\cap F(G)$;

(ii)  the subgroup $K/D\cap F(G)$ is supersoluble, has a Sylow $p$-subgroup of order $p$, $q$ divides $p-1$, and a Sylow $q$-subgroup of $K/D\cap F(G)$ is $\mathbb{P}$-subnormal in $D/D\cap F(G)$.

Consider case (i). Assume that the group $K/D\cap F(G)$ is not supersoluble. Then it follows from

$$K/D\cap F(G)=([P]Q)(D\cap F(G))/D\cap F(G)$$

and Lemma 1 that the subgroup $[P]Q$ is not supersoluble, and hence $P=([P]Q)^{\mathfrak{N}}=([P]Q)^{\mathfrak{U}}$ by statement (3) of Lemma 1. By statement (7) of Lemma 2, we have $P\subseteq D\cap F(G)$. Then, however, $K/D\cap F(G)$ is a $q$-group, which is impossible.

Thus, all Schmidt subgroups of $D/D\cap F(G)$ are supersoluble; i.e., $D/D\cap F(G)\in\mathfrak{F}$. Therefore, by the isomorphism

$$G/F(G)=DF(G)/F(G)\cong D/D\cap F(G),$$

we have $G/F(G)\in\mathfrak{F}$.

(3)  Let $H$ be a nonsupersoluble $\{p,q\}$-Schmidt subgroup of a group $G$ with a normal Sylow $p$-subgroup $P$. Then $H$ is an $S_{<p,q,m>}$-group for some natural $m>1$. Assume that $P$ is not contained in $F(G)$. By statement (3) of Lemma 1, $P$ is the $\mathfrak{N}$-residual of the subgroup $H$. Therefore, a Sylow $q$-subgroup $Q$ of $H$ is not contained in $F(G)$. Now, applying statement (5) of Lemma 1, we find that $HF(G)/F(G)$ is an $S_{<p,q>}$-subgroup of $G/F(G)$. Since $G/F(G)\in\mathfrak{F}$, it follows that $HF(G)/F(G)$ is an $S_{<p,q,1>}$-group, which is impossible according to statement (5) of Lemma 3. Hence, $P$ is contained in $F(G)$. It also follows from $G/G^{\mathfrak{F}}\in\mathfrak{F}$ that $H\subseteq G^{\mathfrak{F}}$.

The theorem is proved.

Corollary 1

Let $\mathfrak{F}=\{H\,|\,\mathrm{Sch}(H)\in\mathfrak{U}\}$. If $\Phi(G)=1$ and each Schmidt subgroup of $G$ is $\mathbb{P}$-subnormal in $G,$ then the group $G$ can be presented in the form $G=[F(G)]M,$ where $M\in\mathfrak{F}$.