On Boundary Factors and Traces of Subgroups of Finite Groups

Abstract

A subgroup $E$ of a finite group $G$ is called hypercyclically embedded in $G$ if every chief factor of $G$ below $E$ is cyclic. Let $A$ be a subgroup of a group $G$. Then we call any chief factor $H/A_{G}$ of $G$ a $G$-boundary factor of $A$. For any $G$-boundary factor $H/A_{G}$ of $A$, we call the subgroup $(A\cap H)/A_{G}$ of $G/A_{G}$ a $G$-trace of $A$. On the basis of these notions, we give some new characterizations of hypercyclically embedded subgroups.

1 Introduction

Throughout this paper, all groups are finite and $G$ always denotes a finite group, $p$ is supposed to be a prime.

It is obvious that a maximal subgroup $M$ of $G$ cannot be written as a proper intersection of subgroups of $G$. Proper subgroups of $G$ with this property are called primitive [1] or meet-irreducible [2] since, in fact, they are just the meet-irreducible elements in the sense of Birkhoff [3, p. 93] of the lattice $L(G)$ of all subgroups of G.

Recall that a normal subgroup of $G$ is called hypercyclically embedded [4, p. 217] in $G$ if either $E=1$ or $E\ne 1$ and every chief factor of $G$ below $E$ is cyclic. With study and applications of hypercyclically embedded subgroup are connected a large number of researches. Some results related to such subgroups are discussed in the books [2, 4, 5]. Among recent papers in this line of researches, see for example [6–13].

Our goal here is to give new characterizations of hypercyclically embedded subgroups on the basis of the following new notion.

Definition 1.1

Let $A$ be a proper subgroup of $G$. We call any chief factor $H/A_{G}$ of $G/H_{G}$ a $G$-boundary factor or simply boundary factor of $A$. For any $G$-boundary factor $H/A_{G}$ of $A$, we call the subgroup $(A\cap H)/A_{G}$ of $G/A_{G}$ a $G$-trace of $A$ or simply a trace of $A$.

Example 1.2

A subgroup $A$ of $G$ is said to be a $CAP$-subgroup [14, p. 37] if $A$ either covers or avoids each chief factor of $G$; a partial $CAP$-subgroup [15, 16] or semi $CAP$-subgroup [17] if $H$ either covers or avoids each factor of some chief series of $G$. It is clear that if $A$ is a $CAP$-subgroup of $G$, then every $G$-trace of $A$ is identity. If $A$ is a proper partial $CAP$-subgroup, then some $G$-trace of $A$ is identity. Indeed, let $1=G_{0} < G_{1} < \cdots < G_{t}=G$ be a chief series of $G$ such that $H$ either covers or avoids each factor of this series. Then there is an index $i$ such that $G_{i} \le A$ and $G_{i+1} \nleq A$. Hence $A$ does not cover the factor $G_{i+1}/G_{i}$, so $A$ avoids it, that is, $A\cap G_{i+1}\le G_{i}$. Thus the trace $(A\cap G_{i+1})/G_{i}$ is identity.

Example 1.3

Let $M$ be a maximal subgroup of $G$ and $H/M_{G}$ and $H_{1}/M_{G}$ be chief factors of $G$. Then $|H/M_{G}|=|H_{1}/M_{G}|$ by Baer’s theorem [14, Ch.A, 15.2(3)], and the number $|H/M_{G}|$ is called the normal index of $M$ (Deskins). In general, the chief trace $(H\cap M)/M_{G}$ may be non-trivial (for example, if $G$ is a simple non-abelian group). But if $G$ is soluble, then every chief factor of $G$ is abelian and so any trace of $M$ is trivial by [14, Ch.A,15.2(1)(2)]. Hence in this case, the normal index of $M$ coincides with its index $|G:M|$ (Deskins).

Example 1.4

Let $G$ be $p$-soluble (respectively, $p$-supersoluble). Let $A$ be a proper subgroup of $G$ and $H/A_{G}$ be any $G$-boundary factor of $A$ such that $p$ divides $|H/A_{G}|$. Then $H/A_{G}$ is an abelian $p$-group (respectively, $H/A_{G}$ is a group of order $p$). Hence the trace $(H\cap A)/A_{G}$ is subnormal in $G/A_G$, and it is trivial in the case where $G$ is $p$-supersoluble. Therefore, if $G$ is soluble, then every trace of any proper subgroup of $G$ is subnormal; if $G$ is supersoluble, then every trace of any proper subgroup of $G$ is trivial.

Example 1.5

Every subgroup $A$ of order 2 of the alternative group $A_{4}$ of degree 4 is meet-irreducible and $A$ has a unique boundary factor which coincides with the Sylow 2-subgroup of $A_{4}.$

Recall that a subgroup $A$ of $G$ is said to permute with a subgroup $B$ if $AB=BA$.

Now we can state our first result.

Theorem A

A non-identity normal subgroup $E$ of $G$ is hypercyclically embedded in $G$ if and only if for every meet-irreducible subgroup $X$ of $E$ and some $G$-boundary factor $H/X_{G}$ of $X$, where $H\le E$, the trace $(H\cap X)/X_{G}$ permutes with some Sylow $q$-subgroup of $G/X_{G}$ for every prime $q$ dividing $|G/X_{G}|$.

Let $\mathfrak {F}$ be a class of groups. If $1\in \mathfrak {F}$, then we write $G^{\mathfrak {F}}$ to denote the intersection of all normal subgroups $N$ of $G$ with $G/N\in \mathfrak {F}$. A non-empty class $\mathfrak {F}$ of groups is said to be a formation provided, for every group $G$, every homomorphic image of $G/G^{\mathfrak {F}}$ belongs to $\mathfrak {F}$. A formation $\mathfrak {F}$ is said to be solubly saturated if $G\in \mathfrak {F}$ whenever $G^{\mathfrak {F}} \le \Phi (S)$ for some normal soluble subgroup $S$ of $G$.

It is not difficult to show that if $G/E\in \mathfrak {F}$, where $\mathfrak {F}$ is a solubly saturated formation, and $E$ is hypercyclically embedded in $G$, then $G\in \mathfrak {F}$ (see [13, Lemma 2.16]). Hence, the following result follows directly from Theorem A.

Corollary 1.6

Let $\mathfrak {F}$ be a solubly saturated formation containing all supersoluble groups and $E$ a non-identity normal subgroup of $G$ with $G/E\in \mathfrak {F}$. Suppose that for every meet-irreducible subgroup $X$ of $E$ and some $G$-boundary factor $H/X_{G}$ of $X$, where $H\le E$, the trace $(H\cap X)/X_{G}$ permutes with some Sylow $q$-subgroup of $G/X_{G}$ for every prime $q$ dividing $|G/X_{G}|$. Then $G \in \mathfrak {F}$.

Corollary 1.7

$G$ is supersoluble if and only if for every meet-irreducible subgroup $X$ of $G$ and some $G$-boundary factor $H/X_{G}$ of $X$, the trace $(H\cap X)/X_{G}$ permutes with some Sylow $q$-subgroup of $G/X_{G}$ for every prime $q$ dividing $|G/X_{G}|$.

It is clear that if $|G:A|=p^{n}$, then for every prime $q\in \pi (G)$ there is a Sylow $q$-subgroup $Q$ of $G$ such that $A$ permutes with $Q$. Thus from Corollary 1.7 we get

Corollary 1.8

(See Johnson [1]). If the index $|G:X|$ is a prime power for every meet-irreducible subgroup $X$ of $G$, then $G$ is supersoluble.

The following example shows that in general the inverse of Corollary 1.8 is not true.

Example 1.9

Let $p$ and $q$ be primes such that $q$ divides $p-1$. Let $G= (\langle a\rangle \rtimes \langle b\rangle )\times \langle c\rangle$, where $|a|=|c|=p, |b|=q$ and $A=\langle a\rangle \rtimes \langle b \rangle$ is a non-abelian group of order $pq$. Then $G$ is a supersoluble group, so any $G$-trace of the subgroup $E=\langle ac\rangle$ is trivial. Moreover, $|G:E|=pq$ and $E$ does not permute with any Sylow $q$-subgroup of $G$. Hence $E$ is a meet-irreducible subgroup of $G$.

The product of all normal quasinilpotent subgroups of $G$ is denoted by $F^{*}(G)$ and it is called the generalized Fitting subgroup of $G$.

Theorem B

A non-identity normal subgroup $E$ of $G$ is hypercyclically embedded in $G$ if and only if for every two meet-irreducible subgroups $X$ and $Y$ of $E$ with $X_{G}=Y_{G}$, there are $G$-boundary factors $H/X_{G}$ and $H_{1}/X_{G}$ of $X$ and $Y$, respectively, where $H\le E$ and $H_1\le E$, such that the traces $(X\cap H)/X_{G}$ and $(Y\cap H_{1})/X_{G}$ are conjugated by an element of the generalized Fitting subgroup $F^{*}(G/X_{G})$.

Corollary 1.10

Let $\mathfrak {F}$ be a solubly saturated formation containing all supersoluble groups and $E$ a non-identity normal subgroup of $G$ with $G/E\in \mathfrak {F}$. Suppose that for every two meet-irreducible subgroups $X$ and $Y$ of $E$ with $X_{G}=Y_{G}$, there are $G$-boundary factors $H/X_{G}$ and $H_{1}/X_{G}$ of $X$ and $Y$, respectively, where $H\le E$ and $H_1\le E$, such that the traces $(X\cap H)/X_{G}$ and $(Y\cap H_{1})/X_{G}$ are conjugated by an element of the generalized Fitting subgroup $F^{*}(G/X_{G})$. Then $G\in \mathfrak {F}$.

It is well known [18, II, 3.2] that in a soluble group $G$ any two maximal subgroups $M$ and $L$ with $M_{G}=L_{G}$ are conjugated. It is not difficult to show that the inverse statement is also true: If any two maximal subgroups $M$ and $L$ with $M_{G}=L_{G}$ are conjugated, then $G$ is soluble (see the proof of Theorem B*). These observations are the motivation for our next result.

Corollary 1.11

$G$ is supersoluble if and only if for every two meet-irreducible subgroups $X$ and $Y$ of $G$ with $X_{G}=Y_{G}$, there are $G$-boundary factors $H/X_{G}$ and $H_{1}/X_{G}$ of $X$ and $Y$, respectively, such that the traces $(X\cap H)/X_{G}$ and $(Y\cap H_{1})/X_{G}$ are conjugated by an element of the generalized Fitting subgroup $F^{*}(G/X_{G})$.

It is clear that if $A$ is a meet-irreducible subgroup of $G$, then there is a unique subgroup $A \le A_{0}$ of G such that $A$ is a maximal subgroup of $A_{0}$. We say that $A_0$ is the covering subgroup for $A, |A_{0} : A|$ is the small index of $A$ and denote it by $|G : A|_{0}$.

Theorem C

A non-identity normal subgroup $E$ of $G$ is hypercyclically embedded in $G$ if and only if for every meet-irreducible subgroup $X$ of $E$ and some of its $G$-boundary factor $H/X_{G}$, where $H\le E$, we have $|E:X|_{0} =|H/X_{G}|$.

Corollary 1.12

Let $\mathfrak {F}$ be a solubly saturated formation containing all supersoluble groups and $E$ a non-identity normal subgroup of $G$ with $G/E\in \mathfrak {F}$. Suppose that for every meet-irreducible subgroup $X$ of $E$ and some of its $G$-boundary factor $H/X_{G}$ of $X$, where $H\le E$, we have $|E:X|_{0} =|H/X_{G}|$. Then $G\in \mathfrak {F}$.

The following result is well known (see Deskins [19]): A group $G$ is soluble if and only if the normal index $|G:M|_{n}$ of any maximal subgroup $M$ of $G$ coincides with the index $|G:M|$. Our next result is an analogue of this result for supersoluble groups.

Corollary 1.13

$G$ is supersoluble if and only if for every meet-irreducible subgroup $X$ of $G$ and some of its $G$-boundary factor $H/X_{G}$, we have $|E:X|_{0} =|H/X_{G}|$.

2 Proofs of Theorems A, B and C

The following results are useful in our proof.

Lemma 2.1

(See [2, Ch. 4, 5.1])). Let $R \le A \le G$, where $R$ is normal in $G$.

• (1) If $A\le B\le G$, where $A$ is a meet-irreducible subgroup of $B$, then there is a meet-irreducible subgroup $X$ of $G$ such that $A=B\cap X$.

• (2) $A$ is a meet-irreducible subgroup of $G$ if and only if $A/R$ is a meet-irreducible subgroup of $G/R$.

Lemma 2.2

(See [20]). Let $H\mathrm{,} K$ and $N$ be pairwise permutable subgroups of $G$ and $H$ is a Hall subgroup of $G$. Then $N\cap HK=(N\cap H)(N\cap K).$

Recall that the largest normal subgroup $E$ of $G$ such that every chief factor of $G$ below $E$ is cyclic is called the $\mathfrak {U}$-hypercentre of $G$ and is denoted by $Z_{\mathfrak {U}}(G)$ (see [14, p. 389]).

Theorem A is a corollary of the following more general result since $\cap _{p_i\in \pi (E)}O_{p_i'}(E)=1$.

Theorem A*

Let $E$ be a non-identity normal subgroup of $G$. Then $E/O_{p'}(E)\le Z_{\mathfrak {U}}(G/O_{p'}(E))$ if and only if for every meet-irreducible subgroup $X$ of $E$ such that $p$ divides the order of some $G$-boundary factor $H/X_{G}$ of $X$, where $H\le E$, the trace $(H\cap X)/X_{G}$ permutes with some Sylow $q$-subgroup of $G/X_{G}$ for all primes $q$ dividing $|G/X_{G}|$.

Proof

Sufficiency. Assume that this is false and let $G$ be a counterexample with $|G| + |E|$ minimal. Let $R$ be a minimal normal subgroup of $G$ contained in $E$. Then

• (1) The hypothesis holds for $(G/R, E/R)$. Hence $(E/R)/O_{p'}(E/R)\le Z_{\mathfrak {U}}((G/R)/O_{p'}(E/R))$. Let $X/R$ be any meet-irreducible subgroup of $G/R$ such that $p$ divides the order of some $(G/R)$-boundary factor $(H/R)/(X/R)_{G/R}$ of $X/R$, where $H/R\le E/R$. Let $q$ be any prime dividing $|(G/R)/(X/R)_{G/R}|=|(G/R)/(X_{G}/R)|=|G/X_{G}|$. Then, by hypothesis, there is a Sylow $q$-subgroup $Q/X_{G}$ of $G/X_{G}$ such that

  $$\begin{aligned} Q(H\cap X)/X_{G}= & {} (Q/X_{G})((H\cap X)/X_{G})\\= & {} ((H\cap X)/X_{G})(Q/X_{G})=(H\cap X)Q/X_{G}, \end{aligned}$$

  so

  $$\begin{aligned} (Q(H\cap X)/R)/(X_{G}/R)=((H\cap X)Q/R)/(X_{G}/R). \end{aligned}$$

  Hence

  $$\begin{aligned}&((Q/R)/(X/R)_{G/R})(((H/R)\cap (X/R))/(X/R)_{G/R}) \\&\quad =(((H/R)\cap (X/R))/(X/R)_{G/R})((Q/R)/(X/R)_{G/R}), \end{aligned}$$

  where $(Q/R)/(X/R)_{G/R}=(Q/R)/(X_{G}/R)$ is a Sylow $q$-subgroup of $(G/R)/(X/R)_{G/R}$. Therefore the hypothesis holds for $(G/R, E/R)$. The choice of $G$ implies that $(E/R)/O_{p'}(E/R)\le Z_{\mathfrak {U}}((G/R)/O_{p'}(E/R))$.

• (2) $O_{p'}(E)=1$. Hence $p$ divides $|R|$. Assume that $O_{p'}(E)\ne 1.$ Without loss of generality, we can assume that $R\le O_{p'}(E)$. Then $O_{p'}(E/R)=O_{p'}(E)/R$. Claim (1) implies that $(E/R)/O_{p'}(E/R)\le Z_{\mathfrak {U}}((G/R)/O_{p'}(E/R))$. On the other hand, from the $G$-isomorphism $E/O_{p'}(E) \simeq (E/R)/(O_{p'}(E)/R)$ it follows that every chief factor of $G$ between $E$ and $O_{p'}(E)$ is cyclic. Hence $E/O_{p'}(E)\le Z_{\mathfrak {U}}(G/O_{p'}(E))$, contrary to the choice of $G$. Therefore $O_{p'}(E)=1$ and so $p$ divides $|R|$.

• (3) If $R$ is the only minimal normal subgroup of $G$ contained in $E$ and $X$ is a meet-irreducible subgroup of $E$, then $X\cap R$ permutes with some Sylow $q$ -subgroup $Q$ of $G$ for every prime $q$ dividing $|G|$. If $R\le X$, it is evident. Now assume that $R\nleq X$. Then $X_G=1.$ Since $R$ is the unique minimal normal subgroup of $G$ contained in $E, R/1$ is the unique $G$-boundary of $X$. Therefore, by hypothesis, there is a Sylow $q$-subgroup $Q$ of $G$ such that $(X\cap R)Q=Q(X\cap R)$.

• (4) $C_{E}(R)\ne 1$. Hence $E$ is $p$ -soluble by (1) and $R$ is a $p$ -group. If $G$ has two different minimal normal subgroups contained in $E$, then it follows from Claim (1). We may, therefore, assume that $R$ is the unique minimal normal subgroup of $G$ contained in $E$. Now, in view of Claims (1) and (2), it is enough to show that $R$ is abelian. Suppose that this is false. Then by the Feit–Thompson theorem, 2 divides $|R|$ and $R$ is not a 2-group. Let $R_{0}=P_{1}\cap R$ be a Sylow 2-subgroup of $R$, where $P_1$ is some Sylow $2$-subgroup of $G$, and $V$ a maximal subgroup of $R_{0}$. Then $N_{R}(V)/V$ is 2-nilpotent by [18, IV,2.8], so $N_{R}(V)$ has a subgroup $W$ such that $|N_{R}(V):W|=2$ and $V\le W$. Then $W$ is a meet-irreducible subgroup of $N_{R}(V)$. Hence by Lemma 2.1, there is a meet-irreducible subgroup $X_{0}$ of $R$ such that $W= X_{0}\cap N_{R}(V)$. Clearly, 2 divides $|R:X_{0}|$ and 4 does not divide $|R:X_{0}|$. Again applying Lemma 2.1, we obtain that $X_{0}=X\cap R$ for some meet-irreducible subgroup $X$ of $G$. Claim (3) implies that $G$ has a Sylow 2-subgroup $P$ such that $X_{0}P=PX_{0}$. Let $R_{2}=P\cap R$ be a Sylow 2-subgroup of $R$. Then $X_{0}P\cap R=X_{0}(P\cap R)=X_{0}R_{2}$ is a subgroup of $R$, and $X_{0} < X_{0}R_{2}$ since 2 divides $|R:X_{0}|$. Suppose that some prime $q\ne 2$ divides $|R:X_{0}|$. Since $R$ is the unique minimal normal subgroup of $G$ contained in $E$, Claim (3) implies that there is a Sylow $q$-subgroup $Q$ of $G$ such that $X_0Q=QX_0$. Then $X_0Q\cap R=X_{0}(Q\cap R)$ is a subgroup of $R$ and $X_{0} < X_{0}R_q$, where $R_q=Q\cap R$ is a Sylow $q$-subgroup of $R$. It is also clear that $X_0Q\cap X_{0}R_{2}=X_{0}$. Hence $X_{0}$ is not a meet-irreducible subgroup of $R$. This contradiction shows that $|R:X_{0}|=2$, so $X_{0}$ is normal in $R$. It follows that $1\ne O^{2}(R)\ne R$. Since $O^{2}(R)$ is characteristic in $R$, it is normal in $G$, which contradicts the minimality of $R$. Hence $R$ is an abelian $p$-group.

• (5) If $R$ is the only minimal normal subgroup of $G$ contained in $E$, then $|R|=p$. In view of Claim (4), $R$ is a $p$-group. Let $V$ be a maximal subgroup of $R$ such that $V$ is normal in a Sylow $p$-subgroup $G_{p}$ of $G$. Let $q\ne p$ be any other prime dividing $|G|$. By Lemma 2.1, there is a meet-irreducible subgroup $X$ of $G$ such that $V=X\cap R$. Claim (3) implies that for some Sylow $q$-subgroup $Q$ of $G$ we have $VQ=QV$. It is also clear that $V= VQ\cap R$ is normal in $VQ$, so $q$ does not divide $|G:N_{G}(V)|$. Hence $V$ is normal in $G$ and therefore $|R|=p$.

• (6) If $R\le \Phi (G)$, then $O_{p'}(E/R)=1$ and so $E/R\le Z_{\mathfrak {U}}(G/R)$. Claim (1) implies that $(E/R)/O_{p'}(E/R)\le Z_{\mathfrak {U}}((G/R)/O_{p'}(E/R))$. Let $V/R=O_{p'}(E/R)$. Clearly, $V$ is $p$-soluble, so by the Frattini argument we have $G=VN_{G}(S)=RSN_{G}(S)= N_{G}(S)$, where $S$ is a Hall $p'$-subgroup of $V$. But this contradicts Claim (2). Hence $O_{p'}(E/R)=1$, and so $E/R\le Z_{\mathfrak {U}}(G/R)$.

• (7) If for some minimal normal subgroup $L$ of $G$ we have $L\le E$ and $L\ne R$, then $R$ and $L$ are cyclic $p$ -groups. First note that $p$ divides $|L|$ by Claim (2). On the other hand, by Claim (1), $(E/R)/O_{p'}(E/R)\le Z_{\mathfrak {U}}((G/R)/O_{p'}(E/R))$, so from the $G$-isomorphism $L\simeq LR/R$ we see that $L$ is either a $p'$-group or a cyclic group. Hence $L$ must be a cyclic $p$-group. Similarly, $R$ is cyclic $p$-groups.

• (8) $\Phi (G)\cap O_{p}(E)=1$. Assume that $\Phi (G)\cap O_{p}(E)\ne 1$. Then $E$ has a minimal normal subgroup $R\le \Phi (G).$ Hence $E/R\le Z_{\mathfrak {U}}(G/R)$ by Claim (6). But by Claims (5) and (7), $R$ is cyclic. It follows that $E\le Z_{\mathfrak {U}}(G)$. This contradiction shows that we have (8). Final contradiction for the sufficiency. Since $\Phi (O_{p}(E))\le \Phi (G)$, Claim (8) implies that $\Phi (O_{p}(E))=1$, so $O_{p}(E)$ is abelian. Therefore, again by Claim (8), for every minimal normal subgroup $N$ of $G$ contained in $O_{p}(E)$, there is a maximal subgroup $M$ of $G$ such that $O_{p}(E)=N\times (O_{p}(E)\cap M)$, where $O_{p}(E)\cap M$ is normal in $G$. Hence $O_{p}(E)= N_{1}\times \cdots \times N_{t}$ for some minimal normal subgroups $N_{1}, \ldots , N_{t}$ of $G$. Let $C_{i}=C_{G}(N_{i})$. Then $C=C_{G}(O_{p}(E))= C_{1}\cap \cdots \cap C_{t}$. Since $E$ is $p$-soluble by Claim (4) and $O_{p'}(E)=1$ by Claim (2), $O_{p}(E)=C\cap E$ by [18, VI,6.5]. Note that $|N_{i}|=p$ for all $i=1, \ldots t$. Indeed, if $t > 1$, then it follows from Claim (7). Suppose that $t=1$. Then Claims (2), (4) and (7) imply that $N_{1}=R$ is a unique minimal normal subgroup of $G$ contained in $E$, so $|N_{1}|=p$ by Claim (5). Hence $G/C_{i}$ is cyclic for all $i=1, \ldots , t$. This implies that $G/C$ is an abelian group. Therefore from the $G$-isomorphism $CE/C\simeq E/C\cap E=E/O_{p}(E)$, we see that every chief factor of $G$ between $E$ and $O_{p}(E)$ is cyclic. Thus $E\le Z_{\mathfrak {U}}(G)$ by the Jordan-H$\ddot{o}$lder theorem [14, Ch.A,3.2]. The final contradiction completes the proof of the sufficiency. Necessity. Let $X$ be any meet-irreducible subgroup of $E$ such that $p$ divides the order of some $G$-boundary factor $H/X_{G}$ of $X$, where $H\le E$. Since $E/O_{p'}(E)\le Z_{\mathfrak {U}}(G/O_{p'}(E)), E$ is $p$-supersoluble, so $|H/X_{G}|=p.$ Hence $(H\cap X) /X_{G}=1$. The theorem is proved. $\square$

A group $G$ is called semisimple if either $G=1$ or $G$ is the direct product of some simple non-abelian groups.

Theorem B is a corollary of the following our result.

Theorem B*

Let $E$ be a non-identity normal subgroup of $G$. Then $E/O_{p'}(E)\le Z_{\mathfrak {U}}(G/O_{p'}(E))$ if and only if for every two meet-irreducible subgroups $X$ and $Y$ of $E$ with $X_{G}=Y_{G}$ such that $p$ divides the order of some $G$-boundary factors $H/X_{G}$ and $H_{1}/X_{G}$ of $X$ and $Y$ respectively, where $H\le E$ and $H_1\le E$, the traces $(X\cap H)X_{G}/X_{G}$ and $(Y\cap H_{1})X_{G}/X_{G}$ are conjugated by an element of the generalized Fitting subgroup $F^{*}(G/X_{G})$.

Proof

Sufficiency. Assume that this is false and let $G$ be a counterexample with $|G| + |E|$ minimal. Let $R$ be a minimal normal subgroup of $G$ contained in $E$. We first show that the hypothesis holds for $(G/R, E/R)$.

Let $X/R$ and $Y/R$ be meet-irreducible subgroups of $E/R$ with $(X/R)_{G/R}=(Y/R)_{G/R}$ such that $p$ divides the orders of some $(G/R)$-boundary factors $(H/R)/(X/R)_{G/R}$ and $(H_{1}/R)/(X/R)_{G/R}$ of $X/R$ and $Y/R$ respectively, where $H/R\le E/R$ and $H_1/R\le E/R$. Then $X$ and $Y$ are meet-irreducible subgroups of $E$ with $X_{G}=Y_{G}$ such that $p$ divides the orders of the $G$-boundary factors $H/X_{G}$ and $H_{1}/X_{G}$ of $X$ and $Y$ respectively, where $H\le E$ and $H_1\le E$. By hypothesis, for some element $xX_{G}\in F^{*}(G/X_{G})$ we have

$$\begin{aligned} ((X\cap H)/X_{G})^{xX_{G}}=(Y\cap H_{1})/X_{G}. \end{aligned}$$

Let $f: G/X_{G} \rightarrow (G/R)/(X_{G}/R)$ be the natural isomorphism. Then

$$\begin{aligned} f(xX_{G})=xR(X_{G}/R)\in F^{*}((G/R)/(X_{G}/R)) \end{aligned}$$

and

$$\begin{aligned} f(((X\cap H)/X_{G}))^{xX_{G}})= & {} (((X\cap H)/R)/(X_{G}/R))^{xR(X_{G}/R)} \\= & {} ((X/R)\cap (H/R))/(X/R)_{G/R})^{xR(X_{G}/R)}\\= & {} f((Y\cap H_{1})/X_{G})= ((Y/R)\cap (H_{1}/R))/(X/R)_{G/R}. \end{aligned}$$

Therefore the hypothesis holds for $(G/R, E/R)$.

Then $O_{p'}(E)=1$ and so $p$ divides $|R|$ (see Claim (2) in the proof of Theorem $A^{*}$). Moreover, if for some minimal normal subgroup $L$ of $G$ we have $L\le E$ and $L\ne R$, then $R$ and $L$ are cyclic $p$-groups (see Claim (6) in the proof of Theorem $A^{*}$).

Now assume that $R$ is the only minimal normal subgroup of $G$ contained in $E$. We now show that $|R|=p$. First we claim that $R$ is a $p$-group. Assume that this is false. Then $R$ is a non-abelian group, so the Frattini argument implies that, for any prime $q$ dividing $|R|$, there is a maximal subgroup $L$ of $E$ such that $R\nleq L$ and a Sylow $q$-subgroup $R_{q}$ of $R$ is contained in $R\cap L$. On the other hand, if $M$ is a maximal subgroup of $E$ such that $R \nleq M$, then $R/1=R/M_G$ is the only $G$-boundary factor of $M$. Hence for any two maximal subgroups $M$ and $L$ not containing $R$, we have $|M\cap R|=|L\cap R|$ by hypothesis. It follows that $R\cap M=R$, a contradiction. Thus $R$ is a $p$-group.

Next we show that $|R|=p$. Indeed, assume that $|R| > p$. Then it has two different maximal subgroups $V$ and $W$. Hence $V$ and $W$ are meet-irreducible subgroups of $R$. By Lemma 2.1, there are meet-irreducible subgroups $X$ and $Y$ of $E$ such that $V=X\cap R$ and $W=Y\cap R$. Since $R$ is the only minimal normal subgroup of $G$ contained in $E, R/1=R/M_G$ is the only $G$-boundary factor of $V$ and $W$ satisfying $R\le E$. Therefore by hypothesis there is an element $x\in F^{*}(G)$ such that $W=V^{x}$. If $F^{*}(G)\cap E=1$, then $F^{*}(G)\le C_G(E)$ and so $W=V^{x}=V$, a contradiction. Hence $F^{*}(G)\cap E\ne 1$. But then $R\le F^{*}(G)\cap E$. By [21, X, 13.6], $F^{*}(G)/Z_{\infty }(F^{*}(G))$ is a semisimple group. Since $R$ is a $p$-group, it follows that $R\le Z_{\infty }(F^{*}(G))$. Thus $F^{*}(G)/C_{F^{*}(G)}(R)$ is a $p$-group by [22, Ch.5, 3.2]. On the other hand, $Z_{\infty }(F^{*}(G))=F(G)$ by [21, X, 13.7]. Hence $Z_{\infty }(F^{*}(G))\le C_{G}(R)$ by [2, AppendixC, 3.2]. But then we have $C_{F^{*}(G)}(R)=F^{*}(G)$ and so again we get that $W=V$. This contradiction shows that $|R|=p$.

The above shows that every minimal normal subgroup of $G$ contained in $E$ is a cyclic $p$-group. Now, the final part of the proof of the sufficiency can be proved similarly as the final part of the proof of the sufficiency in Theorem $A^{*}$.

Necessity. Let $X$ and $Y$ be any two meet-irreducible subgroups with $X_{G}=Y_{G}$ such that $p$ divides the order of some $G$-boundary factors $H/X_{G}$ and $H_{1}/X_{G}$ of $X$ and $Y$, respectively, where $H\le E$ and $H_1\le E$. Then, since $E/O_{p'}(E)\le Z_{\mathfrak {U}}(G/O_{p'}(E)), |H/X_{G}|= p = |H_{1}/X_{G}|$ and so the traces $(X\cap H)/X_{G}$ and $(Y\cap H_{1})/X_{G}$ are the identities.

The theorem is proved. $\square$

Lemma 2.3

Let $R$ be a normal subgroup and $A$ a meet-irreducible subgroup of $G$.

• (1) If $R \le A$, then $|G : A|_{0} = |G/R : A/R|_{0}$.

• (2) If $R \nleq A$, then $A_{0} = A(A_{0} \cap R)$ and $|G : A|_{0} = |(A_{0} \cap R) : (A \cap R)|$.

Proof

Both assertions are evident. Indeed, if, for example, $R \nleq A$, then $A < AR$ and so $A_{0} \le AR$. Hence $A_{0} = A(A_{0} \cap R)$. It follows that $|G : A|_{0} = |A_{0} : A| = |(A_{0} \cap R) : (A \cap A_{0} \cap R)| = |(A_{0} \cap R) : (A \cap R)|$. $\square$

Theorem C is a corollary of the following result.

Theorem C*

Let $E$ be a non-identity normal subgroup of $G$. Then $E/O_{p'}(E)\le Z_{\mathfrak {U}}(G/O_{p'}(E)$ if and only if for every meet-irreducible subgroup $X$ of $E$ such that $p$ divides the order of some $G$-boundary factor $H/X_{G}$ of $X$, where $H\le E$, we have $|E:X|_{0} =|H/X_{G}|$.

Proof

Sufficiency. Assume that this is false and let $G$ be a counterexample with $|G| + |E|$ minimal. Let $R$ be a minimal normal subgroup of $G$ contained in $E$.

We first show that the hypothesis holds for $(G/R, E/R)$. Indeed, let $X/R$ be a meet-irreducible subgroup of $E/R$ such that $p$ divides the order of a $(G/R)$-boundary factor $(H/R)/(X/R)_{G/R}$ of $X/R$ satisfying $H/R\le E/R$. Then $X$ is a meet-irreducible subgroup of $E$ and $H/X_{G}$ is a $G$-boundary factor of $X$ such that $H\le E$ and $p$ divides $|H/X_{G}|$. Hence $|E:X|_{0} =|H/X_{G}|$ by hypothesis, so

$$\begin{aligned} |E/R:X/R|_{0}=|E:X|_{0} =|(H/R)/(X_{G}/R)|=|(H/R)/(X/R)_{G/R}|. \end{aligned}$$

This shows that the hypothesis holds for $(G/R, E/R)$. It follows that $O_{p'}(E)=1$ and so $p$ divides $|R|$ (see Claim (2) in the proof of Theorem $A^{*}$).

If for some minimal normal subgroup $L$ of $G$ we have $L\le E$ and $L\ne R$, then $R$ and $L$ are cyclic $p$-groups (see Claim (7) in the proof of Theorem $A^{*}$).

Now assume that $R$ is the only minimal normal subgroup of $G$ contained in $E$. If $R$ is a non-abelian group, then for some Sylow $p$-subgroup $R_{p}$ of $R$ we have $N_{E}(R_{p})\ne E$. Let $M$ be a maximal subgroup of $E$ such that $N_{E}(R_{p})\le M$. The Frattini argument implies that $R\nleq M$. Then, in view of the $G$-isomorphism $RM_{G}/M_{G}\simeq R$, we have that $RM_{G}/M_{G}$ is a $G$-boundary factor of $M$ such that $p$ divides $|RM_{G}/M_{G}|$ and $RM_G\le E$. The hypothesis implies that $|RM_{G}/M_{G}|=|E:M|_{0}=|E:M|$. Hence $p$ divides $|E:M|$. But for a Sylow $p$-subgroup $E_{p}$ of $E$ containing $R_{p}$ we have $E_{p}\cap R=R_{p}$, so $E_{p}\le N_{E}(R_{p})\le M.$ Hence $p$ does not divide $|E:M|$. This contradiction show that $R$ is a $p$-group.

Now we show that $|R|=p$. Assume that this is false. Let $V$ be a maximal subgroup of $R$. Then $V\ne 1$ and, by Lemma 2.1, for some meet-irreducible subgroup $X$ of $E$ we have $V=X\cap R$. It is clear that $R\nleq X$, so $X_{G}=1$ since $R$ is the only minimal normal subgroup of $G$ contained in $E$. Hence $R/1=R/M_G$ is the only $G$-boundary factor of $X$ satisfying $R\le E$. Thus $|E:X|_{0} =|R|$ by hypothesis. But $|E:X|_{0}=|R/V|=p$, so $V=1$. This contradiction shows that $|R|=p$.

The above shows that every minimal normal subgroup of $G$ contained in $E$ is a cyclic $p$-group. Now, the final part of the proof of the sufficiency can be proved similarly as the final proof of the sufficiency in Theorem $A^{*}$.

Necessity. Let $X$ be any meet-irreducible subgroup of $E$ such that $p$ divides the order of some $G$-boundary factor $H/X_{G}$ of $X$ where $H\le E$. Since $E/O_{p'}(E)\le Z_{\mathfrak {U}}(G/O_{p'}(E)), E$ is $p$-supersoluble. This implies that $|H/X_{G}|=p.$ Hence by Lemma 2.3(2), $1\ne |E:X|_0=|H\cap X_0 : X\cap H|=|H\cap X_0 : X_G|.$ Consequently $|E:X|_{0} =|H/X_{G}|$. The theorem is thus proved. $\square$

3 Some Other Applications

In this section, we discuss some other applications of the notions of boundary factor and trace of a subgroup.

1. Recall that a series $M < T < G$, where $T$ is a maximal subgroup of $G$ and $M$ is a maximal subgroup of $T$, is said to be a maximal chain of $G$ of length $2$.

A large number of results are based on a stronger condition for subgroups “cover or avoid chief factors”. For example, it is known that if either every maximal subgroup of $G$ is a (partial) $CAP$-subgroup of $G$ or every second maximal subgroup of $G$ is a (partial) $CAP$-subgroup of $G$ (see [17, 23]), then $G$ is soluble. Note that the subgroup $A$ in Example 1.5 is 2-maximal and it is not a partial $CAP$ subgroup of $A_{4}$.

Nevertheless, we can prove the following theorem which contains the above-mentioned results in [17, 23].

Theorem 3.1

$G$ is soluble if and only if every maximal chain of $G$ of length $2$ contains a proper subgroup $M$ of $G$ such that some $G$-trace of $M$ is subnormal.

In order to prove Theorem 3.1, the following well-known result (see [24, (6.6.3)]) is used.

Lemma 3.2

Let $G=R\rtimes M$. If $M$ is a soluble maximal subgroup of $G$, then $R$ is abelian.

Proof of Theorem 3.1

In view of Example 1.4, it is enough to prove that if every maximal chain of $G$ of length $2$ contains a proper subgroup $M$ of $G$ such that some $G$-trace of $M$ is subnormal, then $G$ is soluble. Assume that this is false and let $G$ be a counterexample of minimal order. Let $R$ be a minimal normal subgroup of $G, q$ the largest prime dividing $|R|$ and $R_{q}$ a Sylow $q$-subgroup of $R$. Then:

• (1) $G/R$ is soluble. Hence $R$ is the unique minimal normal subgroup of $G, R\not \subseteq \Phi (G), C_{G}(R)=1$ and so $q > 3$. Let $M/R < T/R < G/R$ be any maximal chain of $G/R$ of length $2$. Then $M < T < G$ is a maximal chain of $G$ of length $2$, so for one of the subgroups $M$ or $T$ (we denote it by $L$) some $G$-trace $(H_{1}\cap L)/L_{G}$ of $L$ is subnormal in $G/L_{G}$ by hypothesis. But then the $(G/R)$-trace

  $$\begin{aligned} ((H_{1}/R)\cap (L/R))/(L/R)_{G/R}=((H_{1}\cap L)/R)/(L_{G}/R) \end{aligned}$$

  of $L/R$ is subnormal in $(G/R)/(L/R)_{G/R}.$ This shows that the hypothesis holds for $G/R$. The choice of $G$ implies that $G/R$ is soluble, and it follows that $R$ is the unique minimal normal subgroup of $G, R\not \subseteq \Phi (G), C_{G}(R)=1$. It is clear that $2\in \pi (R)$ and $q > 3$ by the well-known Feit–Thompson theorem and Burnside $p^{a}q^{b}$-theorem.

• (2) For some maximal subgroup $M$ of $G$ and some Sylow $q$ -subgroup $G_{q}$ of $G$, we have $R_{q}\le G_{q}\le N_{G}(R_{q})\le M$ and $M_{G}=1$. (This follows from Claim (1) and Frattini argument).

• (3) $R\ne G$, so $D=M\cap R\ne M$. Suppose that $R= G$ is a simple non-abelian group. Then $G/1$ is the only $G$-boundary factor of any proper subgroup of $G$. By Claim (1), $G$ is not $2$-nilpotent and so $G$ has a $2$-closed Schmidt subgroup $H=H_{2}\rtimes H_{t}$ by [18, IV, 5.4]. It is clear that $|H_2|\ne 2$ and $H\ne G$. Hence $G$ has a maximal chain $T < L < G$, where $T\ne 1$. Since $p$ divides $|G|$, at least one of the subgroups $T$ or $L$ is a proper non-identity subnormal subgroup of $G$ by hypothesis, so $G$ is not a simple non-abelian group. This contradiction shows that $R\ne G$. Since $M_{G}=1, MR=G$ and so $D=M\cap R\ne M$.

• (4) $D$ is a normal non-nilpotent subgroup of $M$, so $D\nleq \Phi (M)$. Clearly, $R_ q$ is a Sylow subgroup of $D$. Suppose that $D$ is nilpotent. Then $R_{q}$ is a characteristic subgroup of $D$, so $M\le N_{G}(R_{q})$ since $D$ is normal in $M$. Hence, in view of Claim (1), we have $M=N_{G}(R_{q})$ and so $N_{R}(R_{q})=D$ is nilpotent. Then $N_{R}(R_{q})/C_{R}(R_{q})$ is a $q$-group and hence $O^{q}(R)\ne R$ by [21, X, 8.13] since $q > 3$ by Claim(1). But in view of Claim (1) again, every composition factor of $R$ is non-abelian, a contradiction. Thus we have (4).

• (5) $M$ has a maximal subgroup $T$ such that $M=DT$ and $D\cap T\ne 1$. In view of Claim (4), there is a maximal subgroup $T$ of $M$ such that $M=DT$. Assume that $D\cap T= 1$. Then $D$ is a minimal normal subgroup of $M$. Note also that

  $$\begin{aligned} G/R\simeq MR/R\simeq M/M\cap R=M/D\simeq T \end{aligned}$$

  is soluble by Claim (1). Then $D$ is a abelian group by Lemma 3.2, which contradicts Claim (4). Hence $D\cap T\ne 1$.

• (6) At least one of the subgroups $T\cap R$ or $D=M\cap R$ is subnormal in $G$. By hypothesis, a maximal chain $T < M < G$ contains a proper subgroup $L$ of $G$ such that some $G$-trace of $L$ is subnormal. In view of Claim (1), $R/1$ is the unique $G$-boundary factor of $L$. Therefore, $L \cap R$ is subnormal in $G$ by [14, Ch.A, 14.2] Final contradiction. By [14, A, 14.3], $R$ normalizes every subnormal subgroup of $G$. Hence, in view of Claim (6), either $R\le N_{G}(T\cap R)$ or $R\le N_{G}(D)$. Then since $(T\cap R)^{G}=(T\cap R)^{RM}\le (T\cap R)^M\le D^{M}\le M_{G}$ and $1\ne D\cap T\le T\cap R$, we obtain that $M_G\ne 1$. But this contradicts Claim (2). The final contradiction completes the proof.$\square$

Question 3.3

Is it true that $G$ is $p$-soluble if and only if every maximal chain of $G$ of length $2$ contains a proper subgroup $M$ of $G$ such that either some $G$-trace of $M$ is subnormal or every $G$-boundary factor of $M$ is a $p'$-group?

2. It is clear that every trace of any maximal subgroup of a soluble group is abelian. This fact is a motivation for our next observations, which can be proved similarly as Theorem 3.1.

Theorem 3.4

$G$ is soluble if and only if every maximal subgroup of $G$ has a nilpotent trace.

Corollary 3.5

(O. Yu. Schmidt [14, Ch.A, 10.7]). If every maximal subgroup of $G$ is nilpotent, then $G$ is soluble.

Nevertheless, we do not know the answer to the following

Question 3.6

Suppose that every maximal subgroup of $G$ has a supersoluble trace. Does it true then that $G$ is soluble?

3. It is well known that a $p$-soluble group is $p$-supersoluble if and only if for every its maximal subgroup $M$, we have that $|G:M|$ is either $p$ or a $p'$-number [18, VI,9.2,9.3]. Note that if $G$ is $p$-soluble and $p$ divides the order of some boundary factor $H/M_{G}$ of a maximal subgroup $M$ of $G$, then $H/M_{G}$ is abelian $p$-group and so $|H/M_{G}|=|G:M|$. This elementary observation is a motivation for the following generalization of the Theorems 9.2 and 9.3 in [18].

Theorem 3.7

$G$ is $p$-supersoluble if and only if for every maximal subgroup $M$ of $G$ such that $p$ divides the order of some boundary factor $H/M_{G}$ of $M$, we have $|H/M_{G}|=|G:M|$.

Proof

See the proof of Theorem $C^{*}$. $\square$