Finite Groups with G-Permutable Schmidt Subgroups

Abstract

Given a non-empty set X of a group G, a subgroup A is said to be X-permutable (respectively, X-h-permutable) with a subgroup B, if $AB^x = B^xA$, for some element $x\in X$ (respectively, $x \in X \cap \langle A,B\rangle$). A subgroup A of a group G is said to be X-permutable (respectively, X-h-permutable) in G if A is X-permutable (respectively, X-h-permutable) with every subgroup of G. In this paper, we study the structure of a finite group G with all its Schmidt subgroups G-permutable (respectively, G-h-permutable).

1 Introduction

All groups considered in the paper are finite.

Recall that a group G is said to be a minimal non-nilpotent group or a Schmidt group if G is a non-nilpotent group whose proper subgroups are nilpotent. It is clear that every non-nilpotent group contains Schmidt subgroups, and so how such subgroups are embedded in the group might provide a deep insight into its structure (see [4, 12, 20]).

Knyagina and Monakhov proved in [12] that a group G is metanilpotent if every Schmidt subgroup of G is subnormal. A full description of groups with subnormal Schmidt subgroups was proposed by Vedernikov in [20], and a complete description of groups having all Schmidt subgroups $\sigma$-subnormal, where $\sigma$ is a partition of the set of all primes, was obtained in [4].

Following [14], we say that a subgroup A of a group G permutes with a subgroup B of G if $AB = BA$. A subgroup A is said to be permutable in G if A permutes with every subgroup of G. It is clear that normal subgroups are permutable and every permutable subgroup is subnormal.

Permutability of subgroups is an important subgroup embedding property that has been studied extensively (see [2, Chapters 1 and 2]). In many situations, the requirement for a permutable subgroup is not to permute with all subgroups, but only with the members of some relevant families of subgroups. In this context, Guo, Shum, and Skiba [9] introduced the following interesting extension of permutability.

Definition 1

Let A and B be subgroups of a group G and X a non-empty subset of G.

• A is said to be X-permutable with B if there exists some $x \in X$, such that $AB^x = B^xA$;

• A is said to be X-h-permutable with B if there exists some $x \in X \cap \langle A,B\rangle$, such that $AB^x = B^xA$.

• A is said to be X-permutable (respectively, X-h-permutable) in G if A is X-permutable (respectively, X-h-permutable) with all subgroups of G.

Note that the 1-permutable subgroups are exactly the permutable subgroups, and if A is X-permutable in a group G, for some $\emptyset \ne X\subseteq G$, then A is G-permutable in G. The Sylow 2-subgroups of the symmetric group G of degree 3 are G-permutable but not permutable in G.

Note that if a subgroup A of a group G is G-h-permutable in G, then A is E-permutable in E for every subgroup E containing A.

Assume that G is the alternating group of degree 5. The only proper non-trivial G-permutable subgroups in G are of order 2. If $A = \langle (1, 2)(3, 4)\rangle$, then A is not ${{\,\textrm{A}\,}}_4$-permutable subgroup in ${{\,\textrm{A}\,}}_4$. Thus, G has not proper G-h-permutable subgroups.

In this paper, the impact of G-permutability and G-h-permutability of the Schmidt subgroups of a group G on its structure of a group is analysed. We prove:

Theorem A

If every Schmidt subgroup of a group G is G-permutable in G, then G is not a non-abelian simple group.

For the subgroup A of G, we write $[A]_G = \{ A^x|x \in G \}$. Define the permutability graph $\Gamma _{{\text {per}}} (G)$ of G as the simple undirected graph whose vertices are the conjugacy classes $[A]_G$ of all subgroups of G, and two vertices $[A]_G$ and $[B]_G$ are joined by an edge if, and only if, a conjugate of A permutes with a conjugate of B.

It is clear that a subgroup A of G is G-permutable in G if, and only if, $[A]_G$ is contained in the center ${{\,\textrm{Z}\,}}(\Gamma _{{\text {per}}}(G))$ of $\Gamma _{{\text {per}}}(G)$, that is, $[A]_G$ is adjacent to every vertex of $\Gamma _{{\text {per}}}(G)$.

The following corollary is just Theorem A in graph theory language.

Corollary 2

If every vertex of the permutability graph $\Gamma _{{\text {per}}}(G)$ of a group G determined by a Schmidt subgroup is central in $\Gamma _{{\text {per}}}(G)$, then G is not a non-abelian simple group.

Theorem A leads to the following natural question: suppose that every Schmidt subgroup of a group G is G-permutable in G, is it true that G is soluble?

The best result we can get in this direction is the following.

Theorem B

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

2 Preliminaries

We shall adhere to the notation and terminology of [2, 3].

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

Lemma 3

Let A, B, K be subgroups of a group G with K normal in G. Then, the following statements hold:

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

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

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

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

If A and B are groups, then we write [A]B or A : B to denote the split extension of A by B.

The following result describes the structure of Schmidt groups.

Lemma 4

([7, 15]) Let S be a Schmidt group. Then, S satisfies the following properties:

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

• (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 non-normal Sylow q-subgroup of S and $\langle a^q\rangle \in {{\,\textrm{Z}\,}}(S)$;

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

• (4) $P/ \Phi (P)$ is a non-central chief factor of S, and $\Phi (P) = P^{'} \subseteq {{\,\textrm{Z}\,}}(S)$;

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

• (6) ${{\,\textrm{C}\,}}_P(a) = \Phi (P)$;

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

Following [12], we denote by ${{\,\textrm{S}\,}}_{\langle p, q\rangle }$ the class of all Schmidt groups with a normal Sylow p-subgroup and a non-normal cyclic Sylow q-subgroup.

Recall that a group G is said to be p-closed if G has a normal Sylow p-subgroup. The following result is a direct consequence of the above lemma.

Lemma 5

([12, Lemma 6]) If G has not p-closed Schmidt subgroups, then G is p-nilpotent.

Lemma 6

([12, Lemma 2]) If K and D are subgroups of a group G, such that D is normal in K and K/D is an ${{\,\textrm{S}\,}}_{\langle p,q\rangle }$-subgroup, then each minimal supplement L to 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) L has an ${{\,\textrm{S}\,}}_{\langle p,q\rangle }$-subgroup [P]Q, such that D does not contain Q and $L = ([P]Q)^L = Q^L$.

The following lemmas turn out to be crucial in the proof of our results.

Lemma 7

([19, Lemma 3]) Let G be a simple group of Lie type that does not belong to the set

$$\begin{aligned} \{{{\,\textrm{PSL}\,}}_6(2), {{\,\textrm{PSp}\,}}_6(2), \Omega _8^{+}(2), {{\,\textrm{PSU}\,}}_4(2)\}. \end{aligned}$$

Then, there is a prime $p \in \pi (G)$, such that p does not divide the order of any proper parabolic subgroup of G.

Lemma 8

([16, Statement 1.6]) Let G be a simple group of Lie type and let U be a unipotent subgroup of G. If L is a maximal subgroup of G and $U\subseteq L$, then L is a parabolic subgroup of G.

Lemma 9

([21, Lemma 1.1]) Let m be an integer greater than 8. Then, there are three primes r, such that $m<r<2m$.

Lemma 10

([13, Corollary 5]) Let $G \in \{{{\,\textrm{A}\,}}_n, {{\,\textrm{S}\,}}_n: n \ge 5\}$ and $G = AB$, where A and B are maximal subgroups of G not containing ${{\,\textrm{A}\,}}_n$. Then, one of the following statements holds:

• (1) $A=({{\,\textrm{S}\,}}_{n-k}\times {{\,\textrm{S}\,}}_k)\cap G$, for some k, such that $1\le k\le 5$ and B is k-homogeneous;

• (2) $n=6$, $A={{\,\textrm{PGL}\,}}_2(5)\cap G$ and $B=({{\,\textrm{S}\,}}_3 \wr {{\,\textrm{S}\,}}_2)\cap G$.

For the proof of the following lemma, we need some information concerning the subgroup structure of ${{\,\textrm{PSL}\,}}_2(q)$. It can be found in [10, Theorem II.8.27].

A minimal non-soluble group G is a non-soluble group such that every proper subgroup of G is soluble. A simple check shows that a group G is a minimal non-soluble group if, and only if, $G/\Phi (G)$ is a minimal simple group, i.e., a non-abelian simple group whose proper subgroups are soluble.

The minimal simple groups were completely classified in [18]. They are

• (1) ${{\,\textrm{PSL}\,}}_2(2^p)$, where p is a prime;

• (2) ${{\,\textrm{PSL}\,}}_2(3^p)$, where $p > 3$ is a prime;

• (3) ${{\,\textrm{PSL}\,}}_2(p)$, where $p > 5$ is a prime and $p^2+1 \equiv 0 \pmod {5}$;

• (4) ${{\,\textrm{PSL}\,}}_3(3)$;

• (5) ${{\,\textrm{Sz}\,}}(2^p)$, where p is an odd prime.

Lemma 11

Assume that G is a minimal simple group. Then, G does not contain non-trivial proper G-h-permutable subgroups.

Proof

Assume that the result is not true, and let F be a non-trivial proper G-h-permutable subgroup of G. We will arrive a contradiction by examining the simple groups of the above list.

(1) G is not ${{\,\textrm{PSL}\,}}_2(2^p)$, where p is a prime.

Assume that $G\cong {{\,\textrm{PSL}\,}}_2(2^p) = {{\,\textrm{PSL}\,}}_2(q)$. Write $q = 2^p$. By [10, Theorem II.8.27], G contains a maximal subgroup L which is isomorphic to the dihedral subgroup ${{\,\textrm{D}\,}}_{2(q-1)}$ and a maximal subgroup S that is isomorphic to the dihedral group ${{\,\textrm{D}\,}}_{2(q + 1)}$. According to [11, $\S$3(2) and $\S$3(4)], $G =ND$ is the unique maximal factorisation of G having $D\cong {{\,\textrm{D}\,}}_{2(q+1)}$ as a factor, where N is a normaliser of a Sylow 2-subgroup of G. Note that $N\cong q:(q-1)$. Furthermore, L is not a factor in any factorisation of G. Since L permutes with some conjugate of F in G, it follows that L contains such conjugate of F. In particular, |F| divides |L|.

On the other hand, F permutes with $S^g$ for some $g \in G$. Consequently, either F is contained in $S^g$ or $G = FS^g$.

Suppose that F is contained in $S^g$. Then, $|F|_{2^{'}}$ divides $q-1$ and $q+1$. Thus, $|F|_{2^{'}} = 1$, and therefore, $|F| = 2$. Note that all involutions of G are conjugate. Therefore, we can assume that the subgroup F is contained in N. It is known that N is a Frobenius group and its cyclic complement E of order $q-1$ is self-centralizing in N. On the other hand, F permutes with a conjugate of E in N as F is N-permutable in N. Without loss of generality, we may assume that FE is a subgroup of G. Then, $FE = F \times E$, and so, ${{\,\textrm{C}\,}}_N(E) \ne E$, which is a contradiction.

Suppose that $G = FS^g$. Since F is contained in a maximal subgroup of G and the maximal factorisation of G involving $S^g$ has a subgroup isomorphic to N as a factor, we may assume that F is contained in $N \cap L$. Note that $|N \cap S^g| = 2$ and |F| is either |N| or $|N: F| = 2$. Since |F| divides $2(q-1)$ and $q(q-1)/2$ divides |F|, it follows that $q = 4$. In this case, $G \cong A_5$ and $N \cong A _4$. In this case, F would be a subgroup of order 6 of N and this is not possible.

(2) G is not ${{\,\textrm{PSL}\,}}_2(3^p)$, where $p > 3$ is a prime.

Assume that $G\cong {{\,\textrm{PSL}\,}}_2(3^p)$. Write $q = 3^p$. By [10, Theorem II.8.27], G contains a maximal subgroup L which is isomorphic to the dihedral subgroup ${{\,\textrm{D}\,}}_{q-1}$ and a maximal subgroup A which is isomorphic to ${{\,\textrm{A}\,}}_4$. By [11, $\S$3(1) and $\S$3(4)], neither L nor A can be factors of a factorisation of G. Since L and A permute with some conjugates of F in G, it follows that both L and S contain a subgroup isomorphic to F. Thus, |F| divides |L| and |A|. But then, $|F|_{2^{'}}$ divides $q-1$ and 12, and therefore, either $|F| = 2$ or $|F| = 4$.

Note that if p is odd, $q-1 = 3^p -1 \equiv 2 \pmod {4}$. Thus, $|F| = 2$. Since all involutions of G are conjugate, we may assume that F is contained in A. Then, F is A-permutable and it follows that A has a subgroup of order 6, which is impossible.

(3) G is not $G\cong {{\,\textrm{PSL}\,}}_2(p)$, where $p > 5$ is a prime and $p^2+1 \equiv 0 \pmod {5}$.

Assume that $G\cong {{\,\textrm{PSL}\,}}_2(p)$. If $p \ge 13$, G has dihedral maximal subgroups of orders $p-1$ and $p+1$. Therefore, we get a contradiction arguing as in Cases 1 and 2.

Assume that $p = 11$. Then, G has a maximal subgroup $D \cong {{\,\textrm{D}\,}}_{12}$ and a maximal subgroup $N \cong 11:5$. Since $D \cap N =1$, and D and N permute with some conjugates of F in G, it follows that $G = CD$ or $G = CN$ for some conjugate C of F in G. Suppose that $G = CD$. Then, N contains a conjugate $C^g$ for some $g \in G$. Since $G = C^gD^g$ and $N \cap D^g= 1$, it follows that $N = C^g$. Let $\langle a \rangle \le G$ be a subgroup of order 2 permuting with N. Then, $a \in N$, a contradiction. If $G = CN$, we can argue analogously to get a contradiction.

It only remains to consider the case $G \cong {{\,\textrm{PSL}\,}}_2(7)$. It is known that G contains three classes of conjugate maximal subgroups whose representatives are $H_1 \cong {{\,\textrm{S}\,}}_4$, $H_2\cong {{\,\textrm{S}\,}}_4$, and $H_3 \cong 7:3$, respectively. It is clear that we may assume that F is contained in one of the subgroups $H_1, H_2, H_3$.

Suppose that $F \subseteq H_3$. If $F = H_3$, then G has a subgroup of G of order 42, because F permutes with a subgroup of G of order 2. This is a contradiction. Analogously, if F is cyclic of order 7, then G has a subgroup of G of order 14, which is also impossible. Thus, F has order 3 and F is then a Sylow 3-subgroup of G. Let C be a cyclic subgroup of G of order 4. Then, for some $h \in G$, $FC^h$ is a subgroup of G of order 12 that must be contained in a maximal subgroup of G isomorphic to ${{\,\textrm{S}\,}}_4$. This is a contradiction.

Suppose that $F \subseteq H_1$. By the above argument, F cannot be cyclic of order 3. Furthermore, F permutes with a Sylow 7-subgroup S of G, and the maximal subgroups of G whose order is divisible by 7 are conjugates of $H_3$ in G. Therefore, $G = FS$ and $F = H_1$. But then, $G = H_1H_2^g$ for some element $g \in G$, which is impossible.

(4) G is not ${{\,\textrm{PSL}\,}}_3(3)$.

Assume that $G \cong {{\,\textrm{PSL}\,}}_3(3)$. By [6], we know that G contains a maximal subgroup $L \cong {{\,\textrm{S}\,}}_4$ and a maximal subgroup $A\cong 13:3$. By hypothesis, L permutes with some conjugate C of F in G. Suppose that $G = CL$. Then, by order considerations, neither any conjugate of F can be contained in A nor any conjugate of F can permute with A. This contradicts our supposition. Consequently, C is contained in L.

Let D be a conjugate of F permuting with A. Then, DA must be a proper subgroup of G, because |D| divides 24. Since A is maximal in G, it follows that $D \subseteq A$, and so, D is a Sylow 3-subgroup of A.

Then, C is a Sylow 3-subgroup of L and L has a subgroup R containing C, such that $R \cong {{\,\textrm{A}\,}}_4$. Since C is R-permutable in R, it follows that R has a subgroup of order 6, which is not possible.

(5) G is not ${{\,\textrm{Sz}\,}}(q)$, where $q = 2^{p}$, where p is an odd prime.

Assume that $G \cong {{\,\textrm{Sz}\,}}(q)$. By [5, Theorem 7.3.3], $|G| = q^2(q-1)(q^2+s+1)(q^2-s+1)$, with $s = \sqrt{2q}$, and G has four conjugacy classes of maximal subgroups. Let $M_1$, $M_2$, $M_3$, and $M_4$ be representatives of these classes. Then, $M_1$ is isomorphic to the dihedral group of order $2(q -1)$, $M_2 \cong (q + s +1):4$, $M_3 \cong (q - s +1): 4$ and $M_4$ is a Frobenius group of order $q^2(q-1)$ (see also [17]).

Assume that $M_4$ does not contain any conjugate of F in G. Then, $G = M_4C$ for a conjugate C of F in G. Thus, both $q + s +1$ and $q - s +1$ divide |C|, which is not possible.

Therefore, we may assume that F is a subgroup of $M = M_4$. Thus, |F| divides $q^2(q-1)$, and then, for every conjugate C of F, $G\ne M_iC$ for each $i\in \{1,2,3\}$. Therefore, |F| divides $(|M_1|,|M_2|, |M_3|)$. Since $(|M_1|,|M_2|, |M_3|)$ divides 2, the only possibility for |F| is 2. Then, we may assume that F is contained in the kernel R of M, which is a Sylow 2-subgroup of G. Since F is M-permutable in M, there exists $y\in M$, such that $L = F^yS = SF^y$, where S is a cyclic complement of R in M. Since $F^y$ is normal in L and $|L:S| = 2$, it follows that $L = F^y \times S$ and ${{\,\textrm{C}\,}}_R(S) \ne 1$, a contradiction.

The proof of the lemma is complete.

$\square$

3 Proof of Theorems A and B

Proof of Theorem A

We assume that G is a simple non-abelian group and derive a contradiction.

(1) G is a group of Lie type over a field of positive characteristic p.

Suppose that G is a simple group of Lie type. Assume that

$$\begin{aligned} G \notin \{{{\,\textrm{PSL}\,}}_6(2), {{\,\textrm{PSp}\,}}_6(2), \Omega _8^{+}(2), {{\,\textrm{PSU}\,}}_4(2)\}. \end{aligned}$$

Then, by Lemma 7, there is a prime $r \in \pi (G)$, such that r does not divide the order of any proper parabolic subgroup of G. According to Lemma 5, it follows that there is an r-closed Schmidt subgroup [R]T of G. Since [R]T is G-permutable in G, there exists a Sylow p-subgroup U of G permuting with [R]T. It implies that U([R]T) is a subgroup of G. Assume that U([R]T) is a proper subgroup of G. Since U is a unipotent subgroup of G, we can apply Lemma 8 to conclude that U([R]T) is contained in some proper maximal parabolic subgroup L of G. But then, r divides |L|, which contradicts Lemma 7. Therefore, $G = U([R]T)$. By Lemma 4, $|\pi (G)| = 3$ and [R]T is a Hall $p'$-subgroup of G. Then, by [8], G is isomorphic to one of the following simple groups:

$$\begin{aligned} \{{{\,\textrm{PSL}\,}}_2(5), {{\,\textrm{PSL}\,}}_2(7), {{\,\textrm{PSL}\,}}_2(8), {{\,\textrm{PSL}\,}}_2(9), {{\,\textrm{PSL}\,}}_2(17), {{\,\textrm{PSL}\,}}_3(3), {{\,\textrm{PSU}\,}}_3(3).\} \end{aligned}$$

We appeal to [6] to derive a contradiction. Assume that $G \cong {{\,\textrm{PSL}\,}}_2(5)$. Then, G contains a Schmidt subgroup $H \cong 3:2$. Thus, G has a subgroup of order 30, which is not possible.

Suppose that $G \cong {{\,\textrm{PSL}\,}}_2(7)$. Then, G contains a Schmidt subgroup $H \cong 7:3$. Thus, G has a subgroup of order 42, which is not possible.

Suppose that $G \cong {{\,\textrm{PSL}\,}}_2(8)$. Then, G contains a Schmidt subgroup $H \cong ({{\,\textrm{C}\,}}_2 \times {{\,\textrm{C}\,}}_2\times {{\,\textrm{C}\,}}_2):7$ of index 9. Thus, G has a subgroup of index 3, which is not possible.

Assume that $G \cong {{\,\textrm{PSL}\,}}_2(9)$. Then, G contains a Schmidt subgroup $H \cong 3:2$. Thus, G has a subgroup of order 30, which is not possible.

Assume that $G \cong {{\,\textrm{PSL}\,}}_2(17)$. Then, G contains a Schmidt subgroup $H \cong 3:2$. Thus, G has a subgroup of order $2\cdot 3\cdot 17$, which is not possible.

Suppose that $G \cong {{\,\textrm{PSL}\,}}_3(3)$. Then, G contains a Schmidt subgroup $H \cong 13:3$. Thus, G has a subgroup of order $2\cdot 3\cdot 13$, which is not possible.

Suppose that $G \cong {{\,\textrm{PSU}\,}}_3(3)$. Then, G contains a Schmidt subgroup $H \cong 7:3$. Thus, G has a subgroup of order $2\cdot 3\cdot 7$, which is not possible.

Therefore, G should be isomorphic to one of the groups in the set

$$\begin{aligned} \{{{\,\textrm{PSL}\,}}_6(2), {{\,\textrm{PSp}\,}}_6(2), \Omega _8^{+}(2), {{\,\textrm{PSU}\,}}_4(2)\}. \end{aligned}$$

In this case, we get a contradiction by finding a maximal subgroup M of G and a Schmidt subgroup H of G which neither permutes with any conjugate of M in G nor is contained in any conjugate of M in G.

(1.1) $G\cong {{\,\textrm{PSL}\,}}_6(2)$.

By [6], we know that G has a maximal subgroup M that is isomorphic to $2^9:({{\,\textrm{PSL}\,}}_3(2)\times {{\,\textrm{PSL}\,}}_3(2))$ and G has a Schmidt subgroup $H \cong 31:5$. Since H permutes with a conjugate A of M in G and H is not contained in A, it follows that $G=HA$. This is not possible by order considerations.

(1.2) $G\cong {{\,\textrm{PSp}\,}}_6(2)$.

We can apply [6] to conclude that G has a maximal subgroup M that is isomorphic to $L \cong {{\,\textrm{S}\,}}_3 \times {{\,\textrm{S}\,}}_6$ and a Schmidt subgroup $H \cong 7:3$. It is clear that neither H permutes with any conjugate of M in G not H is contained in any conjugate of M in G. This is a contradiction.

(1.3) $G\cong \Omega _8^{+}(2)$.

By [6], G has a maximal subgroup $M \cong 2_{+}^{1+8}:({{\,\textrm{S}\,}}_3 \times {{\,\textrm{S}\,}}_3 \times {{\,\textrm{S}\,}}_3)$ and a Schmidt subgroup $H \cong 7:3$, such that $H^g$ is not in M for all $g \in G$. Since H permutes with some $M^t$ for some $t \in G$, it follows that $G = HM^t$, which is a contradiction.

(1.4) $G\cong {{\,\textrm{PSU}\,}}_4(2)$.

By [6], G has a maximal subgroup M that is isomorphic to $2^{.}({{\,\textrm{A}\,}}_4 \times {{\,\textrm{A}\,}}_4).2$ and a Schmidt subgroup $H \cong 5:2$. It is clear that H is not contained in any conjugate of M in G. Since H permutes with $M^g$ for some $g \in G$, it follows that $G = HM^g$, a contradiction.

(2) G is a sporadic simple group.

It follows from [1, Theorem] that $\text {J}_1$ is the unique sporadic simple group having a proper $\text {J}_1$-permutable subgroup $H \ne 1$, and |H| must be 2. Hence, any Schmidt subgroup of any sporadic simple group G is not G-permutable in G.

(3) $G \cong {{\,\textrm{A}\,}}_n$, $n \ge 5$.

Suppose that $n \le 12$ first. Since

$$\begin{aligned} {{\,\textrm{A}\,}}_5 \cong {{\,\textrm{PSL}\,}}_2(5), \quad {{\,\textrm{A}\,}}_6 \cong {{\,\textrm{PSL}\,}}_2(9),\quad {{\,\textrm{A}\,}}_8 \cong {{\,\textrm{PSL}\,}}_4(2), \end{aligned}$$

we have to assume that $n \in \{7, 9, 10, 11, 12\}$. In all these cases, G has a Schmidt subgroup $H \cong 7:3$ (see [6]).

(3.1) $G\cong {{\,\textrm{A}\,}}_7$

According to [6], we know that the maximal subgroups of G of order divisible by 7 are isomorphic to ${{\,\textrm{PSL}\,}}_2(7)$. Since H permutes with a Sylow 3-subgroup S of G, it follows that $G = HS$, a contradiction.

(3.2) $G\cong {{\,\textrm{A}\,}}_9$.

G has a maximal subgroup $M \cong ({{\,\textrm{A}\,}}_5 \times {{\,\textrm{A}\,}}_4):2$ permuting with H ( [6]). Since M does not contain H, it follows that $G = HM$, which is a contradiction.

(3.3) $G\cong {{\,\textrm{A}\,}}_{10}$.

G has a maximal subgroup $M \cong ({{\,\textrm{A}\,}}_6 \times {{\,\textrm{A}\,}}_4):2$ permuting with H ( [6]). Since M does not contain H, it follows that $G = HM$, which is a contradiction.

(3.4) $G\cong {{\,\textrm{A}\,}}_{11}$

G has a maximal subgroup $M \cong ({{\,\textrm{A}\,}}_6 \times {{\,\textrm{A}\,}}_5):2$ permuting with H ( [6]). Since M does not contain H, it follows that $G = HM$, which is a contradiction.

(3.5) $G \cong {{\,\textrm{A}\,}}_{12}$.

G has a maximal subgroup $M \cong ({{\,\textrm{A}\,}}_6 \times {{\,\textrm{A}\,}}_6):2^2$ permuting with H ( [6]). Since M does not contain H, it follows that $G = HM$, which is a contradiction.

(3.6) Let $n \ge 13$. If $n = 2l+1$, then put $k = l$. If $n = 2l$, then put $k = l-1$. Then, $k \ne n-k$ and $A = ({{\,\textrm{S}\,}}_k \times {{\,\textrm{S}\,}}_{n-k}) \cap {{\,\textrm{A}\,}}_n$ is a maximal subgroup of ${{\,\textrm{A}\,}}_n$. By the choice of k, we have that $k \ge 6$ and $n-k \ge 6$.

For every $13 \le n \le 18$, we can find a 3-tuple $\{k, n-k, r\}$, where $r \in \pi ({{\,\textrm{A}\,}}_n)$ and r does not divide k! and $(n-k)!$. For example

$$\begin{aligned} \{ 6,7,13\}, \quad \{6,8,13\}, \quad \{7,8,13\}, \quad \{7,9,13\}, \quad \{8,9,17\},\quad \{8,10,17\}. \end{aligned}$$

If $n\ge 19$, we have $k \ge 9$. Since $(n-k) - k \le 2$, by Lemma 9, there is $r \in \pi ({{\,\textrm{A}\,}}_n)$, such that $r > n-k$.

Hence, for $n \ge 13$, there is $r \in \pi ({{\,\textrm{A}\,}}_n)$, such that $r \notin \pi (A)$.

By Lemma 5, there exists an r-closed Schmidt subgroup T of ${{\,\textrm{A}\,}}_n$. By hypothesis, T permutes with a conjugate $A^g$, for some $g\in G$, of A in G. Since T is not contained in $A^g$, it follows that $G = A^gT$. However, since $k \ge 6$ and $n-k \ge 6$, we can apply Lemma 10 to conclude that A is not a factor in any proper factorization of ${{\,\textrm{A}\,}}_n$. This is a contradiction. The proof of the theorem is complete. $\square$

Proof of Theorem B

Suppose that the result is false and let G be a counterexample of the smallest possible order. It is clear that the hypotheses of the theorem hold in every subgroup of G. The minimal choice of G yields that every maximal subgroup of G is soluble. Hence, G is a minimal non-soluble group. By Theorem A, $\Phi (G) \ne 1$ and $G/\Phi (G)$ is a minimal simple group.

Let $K/\Phi (G)$ be a Schmidt subgroup of $G/\Phi (G)$ and L be a minimal supplement to $\Phi (G)$ in K. Assume that $K/\Phi (G)$ is a $\text {S}_{\langle p,q\rangle }$-group, for some primes $p,q \in \pi (G)$. By Lemma 6, L contains an $\text {S}_{\langle p,q\rangle }$-subgroup [P]Q, such that Q is not contained in $\Phi (G)$. Thus, $[P]Q\Phi (G)/\Phi (G)$ is a proper non-trivial subgroup of $G/\Phi (G)$. Since [P]Q is G-h-permutable in G, we can apply Lemma 3 to conclude that $[P]Q\Phi (G)/\Phi (G)$ is $G/\Phi (G)$-h-permutable in $G/\Phi (G)$. This contradicts Lemma 11. $\square$