Finite Groups All of Whose Subgroups are $\mathbb {P}$-Subnormal or ${{\,\textrm{TI}\,}}$-Subgroups

Abstract

Let $\mathbb {P}$ be the set of all prime numbers. A subgroup H of a finite group G is said to be $\mathbb {P}$-subnormal in G if there exists a chain of subgroups

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

such that either $H_{i-1}$ is normal in $H_i$ or $|H_i{:}\, H_{i-1}|$ is a prime number for every $i = 1, 2, \ldots , n$. A subgroup H of G is called a ${{\,\textrm{TI}\,}}$-subgroup if every pair of distinct conjugates of H has trivial intersection. The aim of this paper is to give a complete description of all finite groups in which every non-$\mathbb {P}$-subnormal subgroup is a ${{\,\textrm{TI}\,}}$-subgroup.

1 Introduction

All groups considered in this paper are finite.

A subgroup H of a group G is called a ${{\,\textrm{TI}\,}}$-subgroup or trivial intersection subgroup if $H \cap H^x \in \{1,H \}$ for every $x \in G$. Normal subgroups, subgroups of prime order and Frobenius complements of Frobenius groups are examples of ${{\,\textrm{TI}\,}}$-subgroups. Using the classification of simple groups it is proved in [2] that every cyclic Sylow subgroup of a non-abelian simple group is also a ${{\,\textrm{TI}\,}}$-subgroup. Structural description of groups in which some distinguished families of subgroups are ${{\,\textrm{TI}\,}}$-subgroups is an interesting research topic in group theory (see [4, 7,8,9,10,11]).

An important subgroup embedding property in which we also are interested in this paper is the $\mathbb {P}$-subnormality, where $\mathbb {P}$ is the set of all prime numbers. It is an extension of the subnormality, and is closely related to the K-$\mathfrak {U}$-subnormality, where $\mathfrak {U}$ is the saturated formation of all supersoluble groups [1, Chapter 6]. In fact, in the soluble universe, $\mathbb {P}$-subnormality coincides with K-$\mathfrak {U}$-subnormality.

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

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

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 $H_i$.

Trivial intersection subgroups and $\mathbb {P}$-subnormal subgroups played an important role in the structural study of the groups and so it is natural and interesting to ask about groups in which every subgroup is either ${{\,\textrm{TI}\,}}$-subgroup or $\mathbb {P}$-subnormal ($\mathbb {P}{{\,\textrm{STI}\,}}$-groups for short). The main result of this paper is an answer to this question and extends the main result of [8].

In the following, if p is a prime, $\mathfrak {A}_{(p-1)}$ will denote the class of all abelian groups of exponent dividing $p-1$ and if n is a positive integer, $n_p$ will denote the largest power of p dividing n.

Theorem A

A group G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group if, and only if, either G is a supersoluble group or G is a Frobenius group with non-cyclic p-elementary abelian kernel N, for some prime p, which is complemented in G by a maximal subgroup M. Furthermore, every subgroup F of M with $F \notin \mathfrak {A}_{(p-1)}$ acts irreducibly on N and one of the following statements holds:

• 1. $p = 2$, M is cyclic;

• 2. $p \ne 2$, M is either a cyclic group or the direct product of a quaternion group ${{\,\textrm{Q}\,}}_8$ with a cyclic group of odd order; 

• 3. $(p-1)_2 > 2$, M is the direct product of a generalised quaternion group ${{\,\textrm{Q}\,}}_{2^m}$ $(m > 3)$ with a cyclic group of odd order. Every subgroup of M is either normal in M or belongs to $\mathfrak {A}_{p-1}$.

• 4. $p \ne 2$, M is the semidirect product $M = LC$, where L is a normal subgroup of prime order r, $C \in \mathfrak {A} _{(p-1)}$.

2 Proof of Theorem A

We work toward a proof of Theorem A.

The following result describes the structure of a Frobenius group.

Lemma 1

[6, Theorem 8.5.5, Theorem 10.5.6] Let G be a Frobenius group and M be a Frobenius complement of G. Then,

• 1. $N = G {\setminus } \bigcup _{x \in G} (M {\setminus } 1)^x$ is a normal subgroup of G such that $G = NM$ and $M \cap N = 1$. N is called the Frobenius kernel of G and M is a complement of N.

• 2. ${{\,\textrm{C}\,}}_G(x) \subseteq N$ for every $1 \ne x \in N$,

• 3. |M| divides $|N| - 1$; in particular, $(|M|, |N|) = 1$,

• 4. N is nilpotent; in particular, $N = {{\,\textrm{F}\,}}(G)$,

• 5. the Sylow p-subgroups of M are cyclic if $p > 2$ and cyclic or a generalised quaternion if $p = 2$,

• 6. all complements of N are conjugate in G.

Our second lemma collects some basic properties of $\mathbb {P}$-subnormal subgroups. Its proof is rather straightforward.

Lemma 2

Let G be a group and A be a subgroup of G. Then,

• 1. if N is a normal subgroup of G and A is $\mathbb {P}$-subnormal in G, then AN/N is $\mathbb {P}$-subnormal in G/N;

• 2. if N is a normal subgroup in G contained in A and A/N is $\mathbb {P}$-subnormal in G/N, then A is $\mathbb {P}$-subnormal in G;

• 3. if A contains $G^{\mathfrak {U}}$, the supersoluble residual of G, then A is $\mathbb {P}$-subnormal in G;

• 4. if A is $\mathbb {P}$-subnormal and G is soluble, then A is $\mathbb {P}$-subnormal in B for every subgroup B of G containing A;

• 5. G is supersoluble if, and only if, every subgroup of G is $\mathbb {P}$-subnormal in G.

Proof

The proofs of the Statements (1)–(3) are rather straightforward. Statement (4) follows from [1, Lemma 6.1.7]. Assume that G is a supersoluble group. Then, $G^{\mathfrak {U}} =1$ and so every subgroup of G is $\mathbb {P}$-subnormal in G by Statement (3). Assume that every subgroup of G is $\mathbb {P}$-subnormal in G. Then, every maximal subgroup of G has a prime index in G. Applying [5, Theorem 6], it follows that G is supersoluble. $\square$

Proof of Theorem A

Let G be $\mathbb {P}{{\,\textrm{STI}\,}}$-group. Assume that all maximal subgroups of G are $\mathbb {P}$-subnormal in G. Then, every maximal subgroup has prime index in G. By Lemma 2, G is supersoluble.

Assume that G has a non-$\mathbb {P}$-subnormal maximal subgroup, M say. Then M is not normal in G and $|G{:}\, M|$ is a composite number. We split the rest of the proof into several steps.

Step 1. $G = [N]M$ is a Frobenius group with kernel N and complement M, where N is an abelian minimal normal subgroup of G and $|N| = p^n$, for some prime p and $n \ge 2$.

By hypothesis, M is a ${{\,\textrm{TI}\,}}$-subgroup of G. Since M is maximal and not normal in G, it follows that G is a Frobenius group with complement M. Let N be the Frobenius kernel of G. Then, by Lemma 1, $G = NM$ is a primitive group, $N = {{\,\textrm{Soc}\,}}(G)$ is the unique minimal normal subgroup of G and M is core-free in G. By Lemma 1, N is an elementary abelian p-group for some prime p, $|N| = p^n$ say. Since $|G{:}\, M|$ is a composite number, it follows that $n \ge 2$.

Applying Lemma 1, we have that

Step 2. $(p, |M|) = 1$. In particular, M is a Hall $p'$-subgroup of G.

Step 3. M is supersoluble and $N = G^{\mathfrak {U}}$ is the supersoluble residual of G. In particular, G is a soluble group.

Let E be a maximal subgroup of M. Then $NE$ is a maximal subgroup of G. Suppose that $NE$ is not $\mathbb {P}$-subnormal in G. Then, $NE$ is a non-normal ${{\,\textrm{TI}\,}}$-subgroup of G by hypothesis. Therefore, $NE$ is a Frobenius complement. Since, by Lemma 1, all Frobenius complements are conjugate, we have that $NE$ is a conjugate of M. This contradiction yields that $NE$ is $\mathbb {P}$-subnormal in G. Consequently, every maximal subgroup of M is $\mathbb {P}$-subnormal in M. By Lemma 2, M is supersoluble. Since G is not supersoluble and N is a minimal normal subgroup containing $G^{\mathfrak {U}}$, it follows that $N = G^{\mathfrak {U}}$.

Step 4. If H is a $\mathbb {P}$-subnormal subgroup of G, then either it has a Hall $p'$-subgroup $H_{p'} \in \mathfrak {A}_{(p-1)}$ or $N \subseteq H$.

Assume that $H_{p'}$ does not belong to $\mathfrak {A}_{(p-1)}$ and H does not contain N. Let

$$\begin{aligned} H = H_s \subset H_{s-1} \subset \cdots \subset H_i \subset H_{i-1} \subset \cdots \subset H_1 \subset H_0 = G \end{aligned}$$

be a chain of subgroups such that for every $i = 1, 2, \ldots , s$, $H_i$ is maximal in $H_{i-1}$ and $|H_{i-1}{:}\, H_i|$ is a prime. Working by induction on s, we can assume that either $H_{s-1}$ has a Hall $p'$-subgroup $(H_{s-1})_{p'} \in \mathfrak {A} _{(p-1)}$ or $N \subseteq H_{s-1}$.

Suppose that $(H_{s-1})_{p'} \in \mathfrak {A} _{(p-1)}$. Since the class $\mathfrak {A} _{(p-1)}$ is subgroup-closed, $H_{p'} = (H_s)_{p'} \in \mathfrak {A} _{(p-1)}$, contrary to assumption.

Suppose that $N \subseteq H_{s-1}$ and assume that $|H_{s-1}{:}\, H| = q$, q a prime. If $q \ne p$, then $N \subseteq H$, contrary to supposition. Hence, $q = p$. Note that $H_{s-1} / {{\,\textrm{Core}\,}}_{H_{s-1}}(H)$ is a supersoluble group and so $H_{s-1}^{\mathfrak {U}} \subseteq {{\,\textrm{Core}\,}}_{H_{s-1}}(H) \cap N$. Consequently, in both cases, we have that $H_{s-1}^{\mathfrak {U}} \subseteq {{\,\textrm{Core}\,}}_{H_{s-1}}(H) \cap N$.

Assume that $H_{s-1}$ is supersoluble. Then, $H_{p'} = (H_{s-1})_{p'}$ is isomorphic to $H_{s-1}/{{\,\textrm{O}\,}}_{p'p}(H_{s-1}) = H_{s-1}/N$ and so $H_{p'}$ belongs to $\mathfrak {A} _{(p-1)}$, contrary to assumption. Consequently, $H_{s-1}^{\mathfrak {U}} \ne 1$ and if S is a supersoluble projector of $H_{s-1}$, we have that S cannot be $\mathbb {P}$-subnormal in $H_{s-1}$ because otherwise S would be contained in a maximal subgroup of $H_{s-1}$ containing $H_{s-1}^{\mathfrak {U}}$. By Lemma 2, S is not $\mathbb {P}$-subnormal in G. Hence, S is a ${{\,\textrm{TI}\,}}$-subgroup of G. Note that ${{\,\textrm{N}\,}}_{H_{s-1}}(S) = S \times {{\,\textrm{C}\,}}_N(S)$. Since S is $\mathfrak {U}$-maximal in $H_{s-1}$, we conclude that ${{\,\textrm{N}\,}}_{H_{s-1}}(S) = S$. Thus, $H_{s-1}$ is a Frobenius group with complement S. Let K be the Frobenius kernel of $H_{s-1}$. By Lemma 1, K is a normal nilpotent subgroup of $H_{s-1}$ and so K is contained in $N = {{\,\textrm{F}\,}}(H_{s-1})$. By Step 1, K is a p-group. Moreover, by Lemma 1, $(|K|,|S|) = 1$. Hence, S is a $p'$-subgroup and $H_{s-1} = NS$. Thus, $N = H_{s-1}^{\mathfrak {U}} \subseteq H$, a contradiction.

Hence, Step 4 holds.

Step 5. Let $\pi = \pi (p - 1)$. If $q \notin \pi$, then every Sylow q-subgroup Q of M is normal in M. In particular, ${{\,\textrm{F}\,}}(M)$ contains a Hall $\pi '$-subgroup of M.

Assume that Q is a non-normal Sylow q-subgroup of M, and suppose that ${{\,\textrm{N}\,}}_M (Q)$ is a proper subgroup of ${{\,\textrm{N}\,}}_G (Q)$. Then, p divides $|{{\,\textrm{N}\,}}_G (Q)|$ and so there is an element $1\ne x \in {{\,\textrm{N}\,}}_G(Q) \cap N$. Then, $Q \subseteq {{\,\textrm{C}\,}}_G(x) \subseteq N$ by Lemma 1, which is a contradiction. Hence, ${{\,\textrm{N}\,}}_M (Q) = {{\,\textrm{N}\,}}_G (Q)$. Since ${{\,\textrm{N}\,}}_G (Q) \notin \mathfrak {A} _{p-1}$, it follows by Step 4 that ${{\,\textrm{N}\,}}_G (Q)$ is a ${{\,\textrm{TI}\,}}$-subgroup of G. Since ${{\,\textrm{N}\,}}_G({{\,\textrm{N}\,}}_G (Q)) = {{\,\textrm{N}\,}}_G (Q)$, we have that ${{\,\textrm{N}\,}}_G (Q)$ is a Frobenius complement of G. By Lemma 1, ${{\,\textrm{N}\,}}_G (Q)$ is a conjugate of M, a contradiction. Therefore, Q is normal in M.

Step 6. Let $\pi = \pi (p - 1)$. If M has a non-trivial Hall $\pi '$-subgroup $M_{\pi '}$, then either M is nilpotent or $M_{\pi '}$ is a normal cyclic group of prime order r, and a Hall $\pi$-subgroup $M_\pi$ of M is a nilpotent subgroup of M that belongs to $\mathfrak {A} _{(p-1)}$.

By Step 5, $M_{\pi '}$ is a normal nilpotent subgroup of M. We show that the Hall $\pi$-subgroup $M_\pi$ of M is nilpotent. Assume that $q \in \pi$ and let $M_q$ be a Sylow q-subgroup of M contained in $M_\pi$. Write $D = M_qM_{\pi '}$. Then D is a Hall $(\pi ' \cup \{q \})$-subgroup of M. Arguing as in Step 5, we conclude that ${{\,\textrm{N}\,}}_G(D) = M$, and so $M_q = M_\pi \cap D \unlhd M_\pi$. Hence, every Sylow subgroup of $M_\pi$ is normal in $M_\pi$, and $M_\pi$ is nilpotent. Since $M = {{\,\textrm{F}\,}}(M)M_\pi$, it follows that $M_\pi$ is contained in a Carter subgroup C of M by [1, Proposition 2.3.17]. If C were normal in G, then we would have that M is nilpotent. Therefore, we may assume that C is not normal in M.

Assume that $M_\pi \notin \mathfrak {A} _{(p-1)}$. Then, $C \notin \mathfrak {A} _{(p-1)}$. Thus, by Step 4, C cannot be $\mathbb {P}$-subnormal in G. Thus, C is a non-normal ${{\,\textrm{TI}\,}}$-subgroup of G. Since ${{\,\textrm{N}\,}}_G(C) = C$, it follows that C is a Frobenius complement of G. By Lemma 1, C is a conjugate of M, and so M is nilpotent.

Suppose now that $M_\pi \in \mathfrak {A} _{(p-1)}$. If $M_\pi$ were a proper subgroup of C, then C would not belong to $\mathfrak {A} _{(p-1)}$ and so C would be a Frobenius complement of G. This would yield the nilpotency of M.

Assume that $M_\pi = C$. Let L be a minimal normal subgroup of M contained in $M_{\pi '}$. Since M is supersoluble, L is cyclic of prime order $r \in \pi '$. Then, $LM_\pi \notin \mathfrak {A} _{(p-1)}$, and so $LM_\pi$ is not $\mathbb {P}$-subnormal in G by Step 4. Then, $LM_\pi$ is a ${{\,\textrm{TI}\,}}$-subgroup of G. Since ${{\,\textrm{N}\,}}_G(LM_\pi ) = LM_\pi$, we have that $LM_\pi$ is a Frobenius complement of G. By Lemma 1, $M = LM_\pi$.

Step 7. Let $\pi = \pi (p - 1)$. If $M_{\pi '} = 1$, then either M is nilpotent or $M = LC$, where $C \in \mathfrak {A} _{(p-1)}$ is a Carter subgroup of M and L is a normal subgroup of M of prime order.

Assume that M is not nilpotent. Then, a Carter subgroup C of M is a proper subgroup of M. Assume that $C \notin \mathfrak {A} _{(p-1)}$. Then, by Step 4, C is not $\mathbb {P}$-subnormal in G. Hence, C is a ${{\,\textrm{TI}\,}}$-subgroup of G. Since ${{\,\textrm{N}\,}}_G(C) = C$, C is not normal in M and so C is a Frobenius complement of G. By Lemma 1, we have that C is a conjugate of M, a contradiction. Consequently, $C \in \mathfrak {A} _{(p-1)}$. Since M is supersoluble, M has a normal Hall $2'$-subgroup and so it contains the nilpotent residual A of M. By Lemma 1, the Sylow q-subgroups of M for $q \ne 2$ are cyclic. Furthermore, $A \subseteq M'$, which is nilpotent because M is supersoluble. Consequently, A is a non-trivial cyclic normal subgroup of M, which is complemented in M by C [3, Theorem IV.5.18]. Let L be a minimal normal subgroup of M contained in A. Then, $|L| = r$ for some prime r and $LC \notin \mathfrak {A} _{(p-1)}$. By Step 4, LC is not $\mathbb {P}$-subnormal in G. Since ${{\,\textrm{N}\,}}_G(LC) = LC$, it follows that LC is a Frobenius complement of G. By Lemma 1, $M = LC$.

Step 8. Every subgroup F of M with $F \notin \mathfrak {A} _{(p-1)}$ acts irreducibly on N. In particular, if $p = 2$, then every subgroup of M acts irreducibly on N.

Let F be a subgroup of M with $F \notin \mathfrak {A} _{(p-1)}$ and suppose that K is a non-trivial subgroup of N normalised by F. Let L be a maximal subgroup of G such that $KF \subseteq L$.

If L is not $\mathbb {P}$-subnormal in G, then $G^\mathfrak {U}L = NL = G$. Thus, L is a Frobenius complement of G and $M = L^x$ for some $x \in G$. Hence, $1 \ne K^x \subseteq (N \cap L)^x = N \cap L^x = N \cap M$, a contradiction. Hence, L is $\mathbb {P}$-subnormal in G and so $N \subseteq L$. In particular, $KF$ is a proper subgroup of L. Let

$$\begin{aligned} KF = L_t \subset L_{t-1} \subset \cdots \subset L_i \subset L_{i-1} \subset \cdots \subset L_1 \subset L_0 = L \end{aligned}$$

be a chain of subgroups of L, such that $L_i$ is maximal in $L_{i-1}$ for every $i = 1, 2, \ldots , t$. We will show that N is contained in $L_i$ for every $i = 1, 2, \ldots , t$. We may assume that $N \subseteq L_{t-1}$. By Lemma 2, $L_{t-1}$ is $\mathbb {P}$-subnormal in G. Since, by Step 4, $KF$ is not $\mathbb {P}$-subnormal in G, we have that $KF$ is not $\mathbb {P}$-subnormal in $L_{t-1}$ by Lemma 2. Then, $KF$ is a non-normal maximal ${{\,\textrm{TI}\,}}$-subgroup of $L_{t-1}$. Hence, $L_{t-1}$ is a Frobenius group with complement $KF$. By Lemma 1, $(|L_{t-1}{:}\, KF |, | KF |) = 1$. Therefore p does not divide $|L_{t-1}{:}\, KF |$ and so N is contained in KL. Thus $K = N$ and F acts irreducibly on N.

Step 9. If M is a direct product of a generalised quaternion group ${\text {Q}}_{2^m}$ ($m \ge 3$) with a cyclic group of odd order, then every subgroup of M is either normal in M or belongs to $\mathfrak {A}_{p-1}$.

Let D be a subgroup of M. If D is not normal in M, then D contains a Sylow 2-subgroup $D_2$ which is not normal in M. Thus, there exists $x\in M$ such that $D_2 \ne D_2^x$. Note that $1\ne D_2 \cap D_2^x$, as ${{\,\textrm{Z}\,}}({{\,\textrm{Q}\,}}_{2^m}) \subseteq D\cap D^x$. Therefore, D is not a ${{\,\textrm{TI}\,}}$-subgroup of G and so, D is $\mathbb {P}$-subnormal. According to Step 4, $D = D_{p'} \in \mathfrak {A}_{p-1}$.

Conclusion.

Assume that N is a 2-group. Then, by Step 7 and Lemma 1, M is a cyclic group of odd order. Furthermore, by Step 8, every subgroup of M acts irreducibly on N.

Suppose that $p > 2$. Then, by Steps 6 and 7, one of the following statements holds:

• 1. M is nilpotent,

• 2. $M = LM_{\pi (p-1)}$, where $M_{\pi (p-1)}$ is a nilpotent Hall $\pi (p-1)$-subgroup of M such that $M_{\pi (p-1)}\in \mathfrak {A} _{(p-1)}$, and L is a normal subgroup of G of order r, for a prime $r \notin \pi (p-1)$,

• 3. $M = LC$, where $C \in \mathfrak {A} _{(p-1)}$ is a Carter subgroup of M and L is a normal subgroup of G of order r, for a prime $r \in \pi (p-1)$.

Observe that cases 2 and 3 correspond to Statement (4) of the theorem.

Assume that M is nilpotent. By Lemma 1, M is either a cyclic group or the direct product of a generalised quaternion group ${{\,\textrm{Q}\,}}_{2^m}$ ($m \ge 3$) with a cyclic group of odd order.

Assume that M is not cyclic. If $(p-1)_2 = 2$, by Lemma 1, $M_2 \cong Q_{2^m}$, $m\ge 3$, for $M_2$ the Sylow 2-subgroup $M_2$ of M. Let T be the normal Hall $2'$-subgroup of M and let S be a subgroup of $M_2$ with $|S| \ge 4$. By Step 4, ST is not $\mathbb {P}$-subnormal in G. Hence, ST is a ${{\,\textrm{TI}\,}}$-subgroup of G. Since ${{\,\textrm{Z}\,}}(M_2) \subseteq S$, we have that ST is normal in M and $S \unlhd M_2$. Recall that if $|S| = 2$, then $S = {{\,\textrm{Z}\,}}(M_2)$ and $S \unlhd M_2$. This means that $M_2$ is a Dedekind group. Therefore, $m = 3$ and $M_2 \cong Q_8$. Then, by Step 9, M satisfies Statements (2) and (3) of the theorem.

In all cases, by Step 8, every subgroup F of M with $F \notin \mathfrak {A} _{(p-1)}$ acts irreducibly on N.

Consequently, G is one of the groups described in Statements (1)–(4) of the theorem.

Conversely, assume that G is either a supersoluble group or a Frobenius group satisfying one of the Statements (1)–(4) of the theorem. We show that G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group.

If G is supersoluble, every subgroup of G is $\mathbb {P}$-subnormal in G by Lemma 2. Then, G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group.

Assume that $G = NM$ is a Frobenius group with kernel N, a non-cyclic elementary abelian p-group for some prime p, which is complemented in G by a maximal subgroup M. Note that M is supersoluble and $N = G^\mathfrak {U}$. Moreover, every subgroup F of M with $F \notin \mathfrak {A} _{(p-1)}$ acts irreducibly on N.

Let D be a non-trivial proper subgroup of G. If $D \subseteq N$ or $N \subseteq D$, then, by Lemma 2, D is $\mathbb {P}$-subnormal in G. Therefore, we may assume that D is not contained in N and D does not contain N. Assume that D has a Hall $p'$-subgroup $D_{p'} \in \mathfrak {A} _{(p-1)}$. Then, every chief factor of ND below N is cyclic by [3, Theorem B.9.8] and ND is supersoluble. By Lemma 2, D is $\mathbb {P}$-subnormal in G. If $D_{p'} \notin \mathfrak {A} _{(p-1)}$, then D acts irreducibly on N. Therefore, $N \cap D = 1$ and so D is a $p'$-subgroup. Without loss of generality we may assume that D is contained in M, and $D \notin \mathfrak {A} _{(p-1)}$.

Assume that M is a Dedekind group. Then, ${{\,\textrm{N}\,}}_G(D) = M$ and D is a ${{\,\textrm{TI}\,}}$-subgroup of G. Consequently, if G satisfies Statements (1) and (2) of the theorem, G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group.

Assume that G satisfies Statement (3) of the theorem. Since $D\notin \mathfrak {A}_{p-1}$, then $D \unlhd M$. Thus, D is a ${{\,\textrm{TI}\,}}$-subgroup of G and G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group.

Suppose that G satisfies Statement (4) of the theorem. Then, M is the semidirect product $M = LC$, where L is a normal subgroup of prime order r, and $C \in \mathfrak {A} _{(p-1)}$. Since $D \notin \mathfrak {A} _{(p-1)}$, $L \subseteq D$. But then, $D \unlhd M$ and D is a ${{\,\textrm{TI}\,}}$-subgroup of G. Thus, G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group.

The proof of Theorem A is complete. $\square$

3 Examples

The construction of $\mathbb {P}{{\,\textrm{STI}\,}}$-groups satisfying one of the sufficient conditions of Theorem A will follow a common pattern: assume that M is one of the groups in Cases (1)–(4) of Theorem A. Then, ${{\,\textrm{Soc}\,}}(M)$ is a product of pairwise non-G-isomorphic minimal normal subgroups of G. Then, M has a faithful and irreducible module N over a field of p elements for every prime p not dividing the order of ${{\,\textrm{Soc}\,}}(M)$ by [3, Corollary B.11.8]. We consider the semidirect product $G = [N]M$ of N with M.

Example 1

Assume that M is isomorphic to the cyclic group of order 3. Then, N is an M-module of dimension 2 over the field of 2 elements, and G is isomorphic to the alternating group of degree 4. G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group satisfying Statement (1) of Theorem A.

Example 2

Assume that $M = A$, where A is isomorphic to the quaternion group ${{\,\textrm{Q}\,}}_8$ of order 8. Then, N is an M-module of dimension 2 over the field of 5 elements. G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group satisfying Statement (2) of Theorem A.

Example 3

Let $Q=\langle a,b\mid a^4=1, b^4=a^2, b^a=b^{-1}\rangle$, isomorphic to a generalised quaternion group of order 16, and $C=\langle c\rangle$, a cyclic group of order 5. Then, $M = Q\times C$ acts on $N=\langle d, e\mid d^{41}=e^{41}=1, d^e=d\rangle$ by means of

$$\begin{aligned} d^a&=e^{40},&d^b&=d^{27},&d^c&=d^{10},\\ e^a&=d,&e^b&=e^{38},&e^c&=e^{10}. \end{aligned}$$

With this action, N is a faithful and irreducible M-module over ${\text {GF}}(41)$, the field of 41 elements. Every subgroup F of M such that $F \notin \mathfrak {A} _{(40)}$ is normal in M and N is an irreducible F-module. Hence, G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group satisfying Statement (3) of Theorem A.

Example 4

Let $M = [A]B$ be a group isomorphic to the symmetric group of degree 3. Then, N is a module of dimension 2 over the field of 5 elements such that N is irreducible for the Sylow 3-subgroup A of M. Therefore, G is a $\mathbb {P}{{\,\textrm{STI}\,}}$-group satisfying Statement (4) of Theorem A.