Characterizations of Finite $\sigma$-Nilpotent and $\sigma$-Quasinilpotent Groups

Abstract

Let $\sigma =\{\sigma _{i} | i\in I\}$ be a partition of the set $\mathbb {P}$ of all primes and G a finite group. A chief factor H / K of G is said to be $\sigma$-central (in G) if the semidirect product $(H/K) \rtimes (G/C_{G}(H/K))$ is a $\sigma _{i}$-group for some $i=i(H/K)$; otherwise, it is called $\sigma$-eccentric. We say that G is: $\sigma$-nilpotent if every chief factor of G is $\sigma$-central; $\sigma$-quasinilpotent if for every $\sigma$-eccentric chief factor H / K of G, every automorphism of H / K induced by an element of G is inner. In this paper, we study properties of $\sigma$-nilpotent and $\sigma$-quasinilpotent subgroups of finite groups.

1 Introduction

Throughout this paper, all groups are finite and G always denotes a finite group. Moreover, ${\mathbb {P}}$ is the set of all primes, $\pi \subseteq \mathbb {P}$ and $\pi ' = \mathbb {P} \setminus \pi$. If n is an integer, the symbol $\pi (n)$ denotes the set of all primes dividing n; as usual, $\pi (G)=\pi (|G|)$, the set of all primes dividing the order of G.

In what follows, $\sigma =\{\sigma _{i} | i\in I\}$ is some partition of $\mathbb {P}$, that is, $\mathbb {P}=\bigcup _{i\in I} \sigma _{i}$ and $\sigma _{i}\cap \sigma _{j}= \emptyset$ for all $i\ne j$. We say that G is $\sigma$-primary [1] provided it is a $\sigma _{i}$-group for some i. A chief factor H / K of G is said to be $\sigma$-central (in G) if the semidirect product $(H/K) \rtimes (G/C_{G}(H/K))$ is $\sigma$-primary; otherwise, it is called $\sigma$-eccentric. A normal subgroup E of G is said to be ${\sigma }$-hypercentral (in G) if either $E=1$ or every chief factor of G below E is $\sigma$-central in G.

The group G is called: $\sigma$-soluble [1] if every chief factor of G is $\sigma$-primary; $\sigma$-decomposable (Shemetkov [2]) or $\sigma$-nilpotent (Guo and Skiba [3]) if $G= G_{1} \times \cdots \times G_{n}$ for some $\sigma$-primary groups $G_{1}, \ldots , G_{n}$.

In fact, $\sigma$-nilpotent groups are exactly the groups whose chief factors are $\sigma$-central [1], and such groups have proved to be very useful in the formation theory (see [4, 5] and the books [6, Ch. 6], [2, Ch. IV]). In the recent years, the $\sigma$-nilpotent groups have found new and to some extent unexpected applications in the theories of permutable and generalized subnormal subgroups (see, for example, the recent papers [1, 3, 7,8,9,10] and the survey [11]).

In this paper, we consider the following generalization of $\sigma$-nilpotency.

Definition 1.1

We say that G is $\sigma$-quasinilpotent if for every $\sigma$-eccentric chief factor H / K of G, every automorphism of H / K induced by an element of G is inner (cf. [12, Ch.X, Definition 13.2]).

We say that G is $\sigma$-semisimple if either $G=1$ or $G=A_{1}\times \cdots \times A_{t}$ is the direct product of simple non-$\sigma$-primary groups $A_{1}, \ldots , A_{t}$.

Example 1.2

Let $G=(A_{5}\wr A_{5})\times (A_{7}\times A_{11})$ and $\sigma = \{\{2, 3, 5\}, \{2, 3, 5\}'\}$. Then G is $\sigma$-quasinilpotent but G is not $\sigma$-nilpotent. The group $A_{7}\times A_{11}$ is $\sigma$-semisimple.

Let $Z_{\sigma }(G)$ be the product of all normal ${\sigma }$-hypercentral subgroups of G. It is not difficult to show (see Proposition 2.5(i) below) that $Z_{\sigma }(G)$ is also ${\sigma }$-hypercentral in G. We call the subgroup $Z_{\sigma }(G)$ the ${\sigma }$-hypercenter of G.

The product of all normal $\sigma$-nilpotent (respectively $\sigma$-quasinilpotent) subgroups of G is said to be the $\sigma$-Fitting subgroup [1] (respectively the generalized $\sigma$-Fitting subgroup) of G and denoted by $F_{\sigma }(G)$ (respectively by $F^{*}_{\sigma }(G)$).

Note that the classical case, when $\sigma =\{\{2\}, \{3\}, \ldots \}$, a chief factor H / K of G is central in G (that is, $C_{G}(H/K)=G$) if and only if it is ${\sigma }$-hypercentral in G. Thus in this case the subgroups $Z_{\sigma }(G)$, $F_{\sigma }(G)$ and $F^{*}_{\sigma }(G)$ coincide with $Z_{\infty }(G)$, F(G) and $F^{*}(G)$, respectively.

In this paper, we study the influence of the subgroups $Z_{\sigma }(G)$, $F_{\sigma }(G)$ and $F^{*}_{\sigma }(G)$ the structure of G. In particular, using such subgroups, we give some characterizations of $\sigma$-nilpotent and $\sigma$-quasinilpotent groups.

A set ${{{\mathcal {H}}}}$ of subgroups of G is said to be a complete Hall $\sigma$-set of G [11] if every member $\ne 1$ of ${{{\mathcal {H}}}}$ is a Hall $\sigma _{i}$-subgroup of G for some $\sigma _{i}\in \sigma$ and ${{{\mathcal {H}}}}$ contains exactly one Hall $\sigma _{i}$-subgroup of G for every i such that $\sigma _{i}\cap \pi (G)\ne \emptyset$.

A subgroup H of G is said to be a maximal $\sigma$-nilpotent subgroup of G if H is $\sigma$-nilpotent but every subgroup E of G such that $H < E$ is not $\sigma$-nilpotent.

In Sect. 2, we study some properties of the subgroup $Z_{\sigma }(G)$. In particular, we prove in the section the following

Theorem A

• (i) The subgroup $Z_{\sigma }(G)$ coincides with the intersection of all maximal $\sigma$-nilpotent subgroups of G.

• (ii) If G possesses a complete Hall $\sigma$-set ${{{\mathcal {H}}}}=\{H_{1}, \ldots , H_{t}\}$, then

  $$\begin{aligned} Z_{\sigma }(G)=\bigcap _{x\in G}(N_{G}(H_{1}^{x})\cap \cdots \cap N_{G}(H_{t}^{x})). \end{aligned}$$

G is said to be $\pi$-decomposable if $G=O_{\pi }(G)\times O_{\pi '}(G)$. In the case when $\sigma =\{\pi , \pi '\}$, we get from Theorem A the following result.

Corollary 1.3

Suppose that G possesses a Hall $\pi$-subgroup and a Hall $\pi '$-subgroup. Then the intersection of all maximal $\pi$-decomposable subgroups coincides with the intersection of the normalizers of all Hall $\pi$-subgroups and all Hall $\pi '$-subgroups of G.

In the case when $\sigma =\{\{2\}, \{3\}, \ldots \}$ we get from Theorem A the following well-know results.

Corollary 1.4

(Baer) The hypercenter $Z_{{\infty }}(G)$ of G coincides with the intersection of all maximal nilpotent subgroups of G.

Corollary 1.5

(Baer) The hypercenter $Z_{{\infty }}(G)$ of G coincides with the intersection of the normalizers of all Sylow subgroups of G.

In Sect. 3, we obtain the following characterization of $\sigma$-quasinilpotent groups.

Theorem B

The following are equivalent:

• (i) G is $\sigma$-quasinilpotent.

• (ii) $G/Z_{\sigma }(G)$ is $\sigma$-semisimple.

• (iii) $G/F_{\sigma }(G)$ is $\sigma$-semisimple and $G=F_{\sigma }(G)C_{G}(F_{\sigma }(G))$.

Corollary 1.6

G is quasinilpotent if and only if G / F(G) is semisimple and $G=F(G)C_{G}(F(G))$.

Corollary 1.7

(See Theorem 13.6 in [12, Ch.X]) G is quasinilpotent if and only if $G/Z_{\infty }(G)$ is semisimple.

A formation is a class ${\mathfrak {F}}$ of groups which is closed under taking subdirect products and homomorphic images. The formation ${\mathfrak {F}}$ is said to be: hereditary if $H\in {\mathfrak {F}}$ whenever $H\le G \in {\mathfrak {F}}$, (solubly) saturated if $G\in {\mathfrak {F}}$ whenever $G/\Phi (N) \in {\mathfrak {F}}$ for some (soluble) normal subgroup N of G; a Fitting formation if ${\mathfrak {F}}$ is closed under taking normal subgroups and products of normal subgroups.

As another application of Theorem B, we prove the following result.

Theorem C

The class ${\mathfrak {N}}^{*}_{\sigma }$, of all $\sigma$-quasinilpotent groups, is a solubly saturated Fitting formation.

Corollary 1.8

(Shemetkov [13]) The class ${\mathfrak {N}}^{*}$, of all quasinilpotent groups, is a solubly saturated formation.

Corollary 1.9

(See Lemma 13.4 and Corollary 13.11 in [12, Ch.X]) The class ${\mathfrak {N}}^{*}$ is a Fitting formation.

Remark 1.10

Let $\sigma =\{\sigma _{1}, \sigma _{2}, \ldots \}$ be any partition of $\mathbb {P}$ with $|\sigma | > 1$. We show that the formation ${\mathfrak {N}}_{\sigma }^{*}$ is not saturated. We can assume without loss of generality that $2\in \sigma _{1}$. Let q be the largest prime in $\sigma _{2}$, and let p be a prime such that $p=q$ if $q > 3$ and p is any odd prime in $\sigma _{1}$ in the case when $q=3$. Finally, let $A_{p}$ be the alternating group of degree p. Then $A_{p}$ is a simple non-$\sigma$-primary group.

Let $G=V\rtimes A_{p}$, where V is a projective envelope of a trivial $\mathbb {F}_{p}A_{p}$-module. Let C be the intersection of the centralizers in $A_{p}$ of all chief factors of G below V. Then $\Phi (G) \cap V= \text {Rad}(V)$ by Lemma B.3.14 in [14], and $C = O_{p', p}(A_{p})=1$ by Theorem VII.14.6 in [15]. Hence, since $V/\text {Rad}(V)=V/\Phi (G) \cap V$ is a central chief factor of G (that is, $C_{G}(V/\text {Rad}(V))=G$) by [14, B, 4.8], G has a Frattini chief factor K / L (that is, $K/L\le \Phi (G/L)$) such that $C_{G}(K/L)=V$ and for every chief factor M / N of G between K and V we have $C_{G}(M/N)=G$. Then G / K is $\sigma$-quasinilpotent by Theorem B. On the other hand, Theorem B implies that G / L is not $\sigma$-quasinilpotent. Thus, the formation ${\mathfrak {N}}_{\sigma }^{*}$ is not saturated.

Finally, being based on Theorems B and C, we prove also the following

Theorem D

If G / E is $\sigma$-nilpotent and every cyclic subgroup of $F_{\sigma }^{*}(E)$ of prime order or order 4 is contained in $Z_{{\sigma }}(G)$, then G is $\sigma$-nilpotent.

Corollary 1.11

(Derr et al. [16]) If G / E is nilpotent and every cyclic subgroup of E of prime order or order 4 is contained in the hypercenter $Z_{\infty }(G)$ of G, then G is nilpotent.

Corollary 1.12

(N. Ito) If every cyclic subgroup of G of prime order or order 4 is contained in the center Z(G) of G, then G is nilpotent.

2 Proof of Theorem A

The following lemma is evident.

Lemma 2.1

If H / K and T / L are G-isomorphic chief factors of G, then

$$\begin{aligned} (H/K)\rtimes (G/C_{G}(H/K))\simeq (T/L)\rtimes (G/C_{G}(T/L)). \end{aligned}$$

Lemma 2.2

(see Proposition 2.3 in [1]) The following are equivalent:

• (i) G is $\sigma$-nilpotent.

• (ii) G has a complete Hall $\sigma$-set ${{{\mathcal {H}}}}=\{H_{1}, \ldots , H_{t} \}$ such that $G=H_{1}\times \cdots \times H_{t}$.

• (iii) Every chief factor of G is $\sigma$-central in G.

Lemma 2.3

(See Corollary 2.4 and Lemma 2.5 in [1]) The class ${\mathfrak {N}}_{\sigma }$, of all $\sigma$-nilpotent groups, is a hereditary saturated Fitting formation.

Lemma 2.4

Let N be a normal $\sigma _{i}$-subgroup of G. Then $N\le Z_{\sigma }(G)$ if and only if $O^{\sigma _{i}}(G)\le C_{G}(N)$.

Proof

If $O^{\sigma _{i}}(G)\le C_{G}(N)$, then for every chief factor H / K of G below N both groups H / K and $G/C_{G}(H/K)$ are ${\sigma _{i}}$-groups since $G/O^{\sigma _{i}}(G)$ is a ${\sigma _{i}}$-group. Hence, $(H/K)\rtimes (G/C_{G}(H/K))$ is $\sigma$-primary. Thus $N\le Z_{\sigma }(G)$.

Now assume that $N\le Z_{\sigma }(G)$. Let $1= Z_{0}< Z_{1}< \ldots < Z_{t} = N$ be a chief series of G below N and $C_{i}= C_{G}(Z_{i}/Z_{i-1})$. Let $C= C_{1} \cap \cdots \cap C_{t}$. Then G / C is a ${\sigma _{i}}$-group. On the other hand, $C/C_{G}(N)\simeq A\le \text {Aut}(N)$ stabilizes the series $1= Z_{0}< Z_{1}< \ldots < Z_{t} = N$, so $C/C_{G}(N)$ is a $\pi (N)$-group by Theorem 0.1 in [17]. Hence, $G/C_{G}(N)$ is a ${\sigma _{i}}$-group and so $O^{\sigma _{i}}(G)\le C_{G}(N)$.

The lemma is proved.

We write $\sigma (G) =\{\sigma _{i} |\sigma _{i}\cap \pi (G)\ne \emptyset \}$, and we say that G is a $\Pi$-group provided $\sigma (G)\subseteq \Pi \subseteq \sigma$.

Proposition 2.5

Let $Z=Z_{\sigma }(G)$. Let A, B and N be subgroups of G, where N is normal in G.

• (i) Z is ${\sigma }$-hypercentral in G.

• (ii) $Z_{\sigma }(A)N/N\le Z_{\sigma }(AN/N)$.

• (iii) $Z_{\sigma }(B)\cap A\le Z_{\sigma }(B\cap A)$.

• (iv) If $N\le Z$ and N is a $\Pi$-group, then N is $\sigma$-nilpotent and $G/C_{G}(N)$ is a $\sigma$-nilpotent $\Pi$-group.

• (v) If G / Z is $\sigma$-nilpotent, then G is also $\sigma$-nilpotent.

• (vi) If $N\le Z$, then $Z/N= Z_{\sigma }(G/N)$.

• (vii) If A is $\sigma$-nilpotent, then ZA is also $\sigma$-nilpotent. Hence, Z is contained in each maximal $\sigma$-nilpotent subgroup of G. Moreover, if A is a Hall $\sigma _{i}$-subgroup of G, for some $i\in I$, then $Z\le N_{G}(A)$.

• (viii) If $G=A\times B$, then $Z=Z_{\sigma }(A)\times Z_{\sigma }(B)$.

Proof

• (i) It is enough to consider the case when $Z=A_{1}A_{2}$, where $A_{1}$ and $A_{2}$ are normal ${\sigma }$-hypercentral subgroups of G. Moreover, in view of the Jordan–Hölder theorem, it is enough to show that if $A_{1}\le K < H \le A_{1}A_{2}$, then H / K is $\sigma$-central. But in this case we have $H=A_{1}(H\cap A_{2})$, where evidently $H\cap A_{2}\nleq K$, so we have the G-isomorphism $(H\cap A_{2})/(K\cap A_{2})\simeq (H\cap A_{2})K/K=H/K$, and hence H / K is $\sigma$-central in G by Lemma 2.1.

• (ii) First assume that $A=G$, and let H / K be a chief factor of G such that $N\le K < H\le NZ$. Then H / K is G-isomorphic to the chief factor $(H\cap Z)/(K\cap Z)$ of G below Z. Therefore, H / K is $\sigma$-central in G by Assertion (i) and Lemma 2.1. Consequently, $ZN/N\le Z_{\sigma }(G/N)$.

  Now let A be any subgroup of G, and let $f:A/A\cap N \rightarrow AN/N$ be the canonical isomorphism from $A/A\cap N$ onto AN / N. Then $f(Z_{\sigma }(A/A\cap N))=Z_{\sigma }(AN/N)$ and

  $$\begin{aligned}f(Z_{\sigma }(A) (A\cap N)/(A\cap N))=Z_{\sigma }(A)N/N. \end{aligned}$$

  Hence, in view of the preceding paragraph, we have

  $$\begin{aligned} Z_{\sigma }(A)(A\cap N)/(A\cap N)\le Z_{\sigma }(A/A\cap N). \end{aligned}$$

  Hence, $Z_{\sigma }(A)N/N\le Z_{\sigma }(AN/N)$.

• (iii) First assume that $B=G$, and let $1= Z_{0}< Z_{1}< \ldots < Z_{t} = Z$ be a chief series of G below Z and $C_{i}= C_{G}(Z_{i}/Z_{i-1})$. Now consider the series

  $$\begin{aligned} 1= Z_{0}\cap A \le Z_{1}\cap A \le \ldots \le Z_{t} \cap A= Z\cap A. \end{aligned}$$

  We can assume without loss of generality that this series is a chief series of A below $Z\cap A$.

  Let $i\in \{1, \ldots , t \}$. Then, by Assertion (i), $Z_{i}/Z_{i-1}$ is $\sigma$-central in G, $(Z_{i}/Z_{i-1})\rtimes (G/C_{i})$ is a $\sigma _{k}$-group say. Hence, $(Z_{i}\cap A)/(Z_{i-1}\cap A)$ is a $\sigma _{k}$-group. On the other hand, $A/A\cap C_{i}\simeq C_{i}A/C_{i}$ is a $\sigma _{k}$-group and

  $$\begin{aligned} A\cap C_{i}\le C_{A}((Z_{i}\cap A)/(Z_{i-1}\cap A)). \end{aligned}$$

  Thus $(Z_{i}\cap A)/(Z_{i-1}\cap A)$ is $\sigma$-central in A. Therefore, in view of the Jordan–Hölder theorem for the chief series, we have $Z\cap A\le Z_{\sigma }(A)$.

  Now assume that B is any subgroup of G. Then, in view of the preceding paragraph, we have

  $$\begin{aligned} Z_{\sigma }(B) \cap A = Z_{\sigma }(B) \cap (B\cap A)\le Z_{\sigma }(B\cap A). \end{aligned}$$

• (iv) By Assertion (iii) and Lemma 2.2, N is ${\sigma }$-nilpotent, and it has a complete Hall $\sigma$-set $\{H_{1}, \ldots , H_{t} \}$ such that $N= H_{1} \times \cdots \times H_{t}$. Then

  $$\begin{aligned} C_{G}(N)= C_{G}(H_{1}) \cap \cdots \cap C_{G}(H_{t}). \end{aligned}$$

  It is clear that $H_{1}, \ldots , H_{t}$ are normal in G. We can assume without loss of generality that $H_{i}$ is a $\sigma _{i}$-group. Then , by Assertion (i) and Lemma 2.4, $G/C_{G}(H_{i})$ is a $\sigma _{i}$-group. Hence,

  $$\begin{aligned} G/C_{G}(N)=G/(C_{G}(H_{1}) \cap \cdots \cap C_{G}(H_{t})) \end{aligned}$$

  is a $\sigma$-nilpotent $\Pi$-group.

• (v), (vi) These assertions are corollaries of Assertion (i) and the Jordan–Hölder theorem.

• (vii) Since A is $\sigma$-nilpotent, $ZA/Z\simeq A/A\cap Z$ is $\sigma$-nilpotent by Lemma 2.3. On the other hand, $Z\le Z_{\sigma }(ZA)$ by Assertion (iii). Hence, ZA is $\sigma$-nilpotent by Assertion (v).

  Finally, let A be a Hall $\sigma _{i}$-subgroup of G. Then A is $\sigma$-nilpotent and so ZA is also $\sigma$-nilpotent. Therefore, $Z\le N_{G}(A)$ by Lemma 2.2.

• (viii) Let $Z_{1}=Z_{\sigma }(A)$ and $Z_{2}= Z_{\sigma }(B)$. Since $Z_{1}$ is characteristic in A, it is normal in G.

  First assume that $Z_{1}\ne 1$ and let R be a minimal normal subgroup of G contained in $Z_{1}$. Then R is $\sigma$-primary, R is a $\sigma _{i}$-group say, by Assertion (iv). Hence, $A/C_{A}(R)$ is a $\sigma _{i}$-group by Lemma 2.4. But $C_{G}(R)=B(C_{G}(R)\cap A)=BC_{A}(R)$, so

  $$\begin{aligned} G/C_{G}(R)=AB/C_{A}(R)B\simeq A/(A\cap C_{A}(R)B)=A/C_{A}(R)(A\cap B)=A/C_{A}(R) \end{aligned}$$

  is a $\sigma _{i}$-group, and hence R is $\sigma$-central in G. Then $R\le Z_{\sigma }(G)$, so $Z_{\sigma }(G)/R=Z_{\sigma }(G/R)$ by Assertion (vi). On the other hand, we have $Z_{1}/R=Z_{\sigma }(A/R)$ and $Z_{2}R/R=Z_{\sigma }(BR/R)$, so by induction we have

  $$\begin{aligned} Z_{\sigma }(G/R)= & {} Z_{\sigma }((A/R)\times (BR/R))=Z_{\sigma }(A/R)\times Z_{\sigma }(BR/R) \\= & {} (Z_{1}/R)\times (Z_{2}R/R)= Z_{1}Z_{2}/R= Z/R. \end{aligned}$$

  Hence $Z=Z_{1}\times Z_{2}$.

Finally, suppose that $Z_{1}=1=Z_{2}$. Assume that $Z_{\sigma }(G)\ne 1$ and let R be a minimal normal subgroup of G contained in $Z_{\sigma }(G)$. Then, in view of Assertions (i) and (iii), $R\cap A=1=R\cap B$ and hence $G=A\times B\le C_{G}(R)$. Thus $R\le Z(G)=Z(A)\times Z(B)=1$, a contradiction. Hence we have (viii).

The proposition is proved.

Temporarily, we write $I_{\sigma }(A)$ to denote the intersection of all maximal $\sigma$-nilpotent subgroups of a group A; if A possesses a complete Hall $\sigma$-set ${{{\mathcal {H}}}} = \{H_{1}, \ldots , H_{t}\}$, then we use $I_{{{\mathcal {H}}}}(A)$ to denote the intersection

$$\begin{aligned} \bigcap _{x\in A}(N_{A}(H_{1}^{x})\cap \cdots \cap N_{A}(H_{t}^{x})). \end{aligned}$$

Proof of Theorem A

Let $Z=Z_{\sigma }(G)$. (i) Suppose that this is false and let G be a counterexample of minimal order. Let $I= I_{\sigma }(G)$. Then $Z < I$ by Proposition 2.5(vii) and the choice of G, so $I\ne 1$ and G is not $\sigma$-nilpotent. Let N be a minimal normal subgroup of G and let L be a minimal normal subgroup of G contained in I. Then L is a $\sigma _{i}$-group for some $i\in I$.

(1) $IN/N\le I_{\sigma }(G/N)$.

Let U / N be a maximal $\sigma$-nilpotent subgroup of G / N, and let V be a minimal supplement to N in U. Then $V\cap N\le \Phi (V)$ and, by Lemma 2.3, $V/V\cap N\simeq VN/N=U/N$ is $\sigma$-nilpotent. Hence V is $\sigma$-nilpotent by Lemma 2.3. Let $U_{0}$ be a maximal $\sigma$-nilpotent subgroup of G such that $V\le U_{0}$. Then $U_{0}N/N\simeq U_{0}/U_{0}\cap N$ is $\sigma$-nilpotent and $U/N\le U_{0}N/N$. But then $U/N=U_{0}N/N$, so $IN/N\le I_{\sigma }(G/N)$.

(2) $G/I\, is \,not\, \sigma -nilpotent$.

Indeed, suppose that G / I is $\sigma$-nilpotent, and let V be a minimal supplement to I in G. Then V is $\sigma$-nilpotent, so for a maximal $\sigma$-nilpotent subgroup U of G such that $V\le U$ we have $G=IU=U$, a contradiction.

(3) $IN/N\le Z_{\sigma }(G/N)= \text {I}_{\sigma }(G/N)$.

Indeed, $IN/N\le I_{\sigma }(G/N)$ by Claim (1). On the other hand, by the choice of G, $Z_{\sigma }(G/N)=I_{\sigma }(G/N)$.

(4) $L\nleq Z$.

Suppose that $L\le Z$. Then $Z/L=Z_{\sigma }(G/L)$ by Proposition 2.5(vi), so $I/L\le I_{\sigma }(G/L) =Z_{\sigma }(G/L)=Z/L$ by Claim (3). Hence $I \le Z$ and so $I=Z$, a contradiction.

(5) $\textit{If } L\le M < G, then L\le Z_{\sigma }(M)$.

Let V be any maximal $\sigma$-nilpotent subgroup of M, and let H be a maximal $\sigma$-nilpotent subgroup of G such that $V\le H$. Then $V\le H\cap M$, where $H\cap M$ is $\sigma$-nilpotent by Lemma 2.3, which implies $L\le V=H\cap M$. Hence $L \le I_{\sigma }(M)$. But $|M| < |G|$, so $I_{\sigma }(M)=Z_{\sigma }(M)$ by the choice of G. Hence $L\le Z_{\sigma }(M)$.

(6) $L=N$ is a unique minimal normal subgroup of G.

Suppose that $L\ne N$. From Claim (3) we deduce that $NL/N\le Z_{\sigma }(G/N)$, so from the G-isomorphism $NL/N\simeq L$ and Lemma 2.1 we obtain $L\le Z$, which contradicts to Claim (4).

(7) $L\nleq \Phi (G)$.

Suppose that $L\le \Phi (G)$. Let $C=C_{G}(L)$ and M be any maximal subgroup of G. Then $L\le M$, so $L\le Z_{\sigma }(M)$ by Claim (5). Hence $M/M\cap C$ is a $\sigma _{i}$-group by Lemma 2.4. If $C\nleq M$, then $G/C=CM/C\simeq M/M\cap C$ is a $\sigma _{i}$-group, so L is $\sigma$-central in G and hence $L\le Z$, contrary to Claim (4). Hence $C\le M$ for all maximal subgroups M of G, so C is nilpotent. Therefore, in view of Claim (6), C is a p-group for some $p\in \sigma _{i}$ since C is normal in G. Thus M is a $\sigma _{i}$-group. But then G is either a $\sigma _{i}$-group or a group of prime order q for some $q\in \sigma _{i}'$, so G is $\sigma$-nilpotent. This contradiction shows that we have (7).

(8) L is not abelian.

Suppose that L is abelian. Then from Claims (6) and (7) we deduce that $G=L\rtimes M$ for some maximal subgroup M of G and, by [14, Ch.A, 15.6], $C=C_{G}(L)=L.$ Let E be a maximal subgroup of M, $V=LE$. Then, by Claim (5), $L\le Z_{\sigma }(V)$, so $E\simeq V/L=V/C_{V}(L)$ is a $\sigma _{i}$-group by Lemma 2.4. Hence M is either a $\sigma _{i}$-group or a group of prime order, contrary to Claim (2). Hence we have (8).

(9) $\textit{If } L\le M < G, \textit{then } M \textit{ is a } \sigma _{i}-\textit{group}$.

By Claim (5), $L\le Z_{\sigma }(M)$. On the other hand, by Claims (6) and (8), $C_{G}(L)=1$. Hence $M\simeq M/1=M/C_{M}(L)$ is a $\sigma _{i}$-group by Lemma 2.4.

Final contradiction for (i). Let U be a minimal supplement to L in G. Let V be any maximal subgroup of U. Then $LV\ne G$, so LV is a $\sigma _{i}$-group by Claim (9). Hence V is a $\sigma _{i}$-group. Therefore, every maximal subgroup of U is a $\sigma _{i}$-group. Then U is either a $\sigma _{i}$-group or a group of prime order. Hence U is $\sigma$-nilpotent and so $G/L=UL/L\simeq U/U\cap L$ is $\sigma$-nilpotent. But then $G/I\simeq (G/L)/(I/L)$ is $\sigma$-nilpotent by Lemma 2.3, contrary to Claim (2). Thus Assertion (i) is proved.

(ii) Assume that this is false and let G be a counterexample with minimal order. Let $I=I_{{{\mathcal {H}}}}(G)$.

First note that if A is a Hall $\sigma _{i}$-subgroup of G, then $Z\le N_{G}(A)$ by Proposition 2.5(vii) and so $Z\le I$. Thus the choice of G implies that $I\ne 1$. Moreover, since $(N_{G}(A))^{x}=N_{G}(A^{x})$, I is a normal $\sigma$-nilpotent subgroup of G. Let R be a minimal normal subgroup of G contained in I. Then R is a $\sigma _{i}$-group for some $i\in I$. Therefore, for each $j\ne i$ we have $R\le C_{G}(H_{j})$ since $R\le N_{G}(H_{j})$, so $G/C_{G}(R)$ is a $\sigma _{i}$-group. Hence R is $\sigma$-central in G.

It is clear that ${{{\mathcal {H}}}}_{0}=\{H_{1}R/R, \ldots , H_{t}R/R\}$ is a complete Hall $\sigma$-set of G / R. Since

$$\begin{aligned} N_{G}(H_{i}^{x})R/R\le N_{G/R}(H_{i}^{x}R/R)=N_{G/R}((H_{i}R/R)^{xR}), \end{aligned}$$

$RI/R\le I_{{{{\mathcal {H}}}}_{0}}(G/R)$. The choice of G implies that $RI/R\le I_{{{{\mathcal {H}}}}_{0}}(G/R)=Z_{\sigma }(G/R)$. But since R is $\sigma$-central in G, $R\le Z$ and so $Z/R=Z_{\sigma }(G/R)$ by Proposition 2.5(vi). Thus $RI/R\le Z/R$, so $I\le Z$. Thus $I=Z$, as required.

The theorem is proved.

3 Proof of Theorem B

Lemma 3.1

• (i) If G is a $\sigma$-quasinilpotent group and N is a normal subgroup of G, then N and G / N are $\sigma$-quasinilpotent.

• (ii) If G / N and G / L are $\sigma$-quasinilpotent, then $G/(N\cap L)$ is also $\sigma$-quasinilpotent.

Proof

See the proof of Lemma 13.3 in [12, Ch.X].

Lemma 3.2

Let N be a minimal normal subgroup of the group G. Then every automorphism of N induced by an element of G is inner if and only if $G=NC_G(N)$.

Proof

See the proof of Lemma 13.4 in [12, Ch.X].

Proof of Theorem B

Let $Z= Z_{\sigma }(G)$.

• (i) $\Rightarrow$ (ii) Assume that this is false and let G be a counterexample of minimal order. Then the hypothesis holds for G / Z by Lemma 3.1(i). On the other hand, $Z_{\sigma }(G/Z)=1$ by Proposition 2.5(vi). Hence in the case when $Z\ne 1$, $G/Z_{\sigma }(G)$ is $\sigma$-semisimple by the choice of G.

  Now assume that $Z=1$ and let R be any minimal normal subgroup of G. Then R / 1 is a $\sigma$-eccentric chief factor of G, so $G=RC_{G}(R)$ by Lemma 3.2. Therefore, since $Z(G)\le Z =1$, $C_{G}(R)\ne G$ and hence R is $\sigma$-semisimple. Thus $G=R\times C_{G}(R)$. Therefore, $Z_{\sigma }(R)\times Z_{\sigma }(C_{G}(R))=Z_{\sigma }(G)= 1$ by Proposition 2.5(viii). Moreover, the choice of G implies that $C_{G}(R)$ is $\sigma$-semisimple, so $G\simeq G/Z=G/1$ is $\sigma$-semisimple and hence Assertion (ii) is true, a contradiction.

• (ii) $\Rightarrow$ (iii) First note that $Z\le F_{\sigma }(G)$ by Proposition 2.5(iv), so $Z= F_{\sigma }(G)$ since G / Z is $\sigma$-semisimple by hypothesis. But then $G/C_{G}(F_{\sigma }(G))$ is $\sigma$-nilpotent by Proposition 2.5(iv). Hence $G=F_{\sigma }(G)C_{G}(F_{\sigma }(G))$ since $G/F_{\sigma }(G)=G/Z$ is $\sigma$-semisimple.

• (iii) $\Rightarrow$ (i) Let H / K be a chief factor of G. If $F_{\sigma }(G)\le K$, then every automorphism of H / K induced by an element of G is inner by Lemma 3.2 since $G/F_{\sigma }(G)$ is $\sigma$-semisimple by hypothesis. Now suppose that $H\le F_{\sigma }(G)$. Then

  $$\begin{aligned} C_{G}(H/K)=C_{G}(H/K)\cap F_{\sigma }(G)C_{G}(F_{\sigma }(G)) =C_{G}(F_{\sigma }(G))C_{F_{\sigma }(G)}(H/K), \end{aligned}$$

  so

  $$\begin{aligned} G/C_{G}(H/K)= & {} F_{\sigma }(G)C_{G}(F_{\sigma }(G))/C_{G}(F_{\sigma }(G))C_{F_{\sigma }(G)}(H/K) \\\simeq & {} F_{\sigma }(G)/F_{\sigma }(G)\cap C_{G}(F_{\sigma }(G))C_{F_{\sigma }(G)}(H/K)\\= & {} F_{\sigma }(G)/C_{F_{\sigma }(G)}(H/K)Z(F_{\sigma }(G)) \\\simeq & {} (F_{\sigma }(G)/C_{F_{\sigma }(G)}(H/K)) /(C_{F_{\sigma }(G)}(H/K)Z\\&(F_{\sigma }(G))/C_{F_{\sigma }(G)}(H/K)) \end{aligned}$$

  is $\sigma$-primary by Lemma 2.4. Therefore, H / K is $\sigma$-central in G. Now applying the Jordan–Hölder theorem, we get that for every $\sigma$-eccentric chief factor H / K of G, every automorphism of H / K induced by an element of G is inner. Hence G is $\sigma$-quasinilpotent.

The theorem is proved.

Corollary 3.3

If a $\sigma$-quasinilpotent group $G\ne 1$ is $\sigma$-soluble, then $G= O_{\sigma _{1}}(G)\times \cdots \times O_{\sigma _{t}}(G)$, where $\{\sigma _{1}, \ldots , \sigma _{t}\}= \sigma (G)$.

Proof

This directly follows from Theorem B and Lemma 2.2.

We say that G is $\sigma$-perfect if $O^{\sigma _{i}}(G)=G$ for all i.

Corollary 3.4

Let G be $\sigma$-quasinilpotent.

• (i) If G is $\sigma$-perfect, then $Z_{\sigma }(G)=Z(G)$.

• (ii) If H is a normal $\sigma$-soluble subgroup of G, then $H\le Z_{\sigma }(G)$.

Proof

• (i) This assertion follows from Theorem B and Proposition 2.5(iv).

• (ii) This directly follows from Theorem B.

4 Proof of Theorem C

For any function f of the form

$$\begin{aligned} f:\mathbb {P}\cup \{0\}\rightarrow \{\text {group formations}\}, \end{aligned}$$

(*)

we put, following [18],

$$\begin{aligned}&CF(f)=\{G\text { is a group}\mid G/C_{G}(H/K)\in f(0)\\&\text { for each non-abelian chief factor} \ H/K \text { of} \ G \\&\text {\ and}\ G/C_{G}(H/K)\in f(p)\text {\ for any abelian } \text {{ p}-chief factor } \ H/K \text { of }\ G \}. \end{aligned}$$

In the paper [18], the following useful fact is proved.

Lemma 4.1

For any function f of the form $(*)$, the class CF(f) is a solubly saturated formation.

Proof of Theorem C

Let $\mathfrak {M}=CF(f)$, where $f(p)= {\mathfrak {G}}_{\sigma _{i}}$ is the class of all $\sigma _{i}$-groups for all $p\in \sigma _{i}$, and $f(0)={\mathfrak {N}}^{*}_{\sigma }$. We show that $\mathfrak {M}= {\mathfrak {N}}^{*}_{\sigma }$. First assume that $\mathfrak {M} \nsubseteq {\mathfrak {N}}^{*}_{\sigma }$ and G be a group of minimal order in $\mathfrak {M} \setminus {\mathfrak {N}}^{*}_{\sigma }$ with a minimal normal subgroup R. Then $G/R\in \mathfrak {M}$ by Lemma 4.1, so G / R is $\sigma$-quasinilpotent. Hence R is a unique minimal normal subgroup of G by Lemma 3.1(ii). Therefore, in view of Theorem B, R is not $\sigma$-central in G. Hence R is non-abelian. But then $C_{G}(R)=1$ and so $G\simeq G/C_{G}(R)\in f(0)={\mathfrak {N}}^{*}_{\sigma }$, a contradiction. Thus $\mathfrak {M}\subseteq {\mathfrak {N}}^{*}_{\sigma }$.

Now, assume that ${\mathfrak {N}}^{*}_{\sigma }\nsubseteq \mathfrak {M}$ and G be a group of minimal order in ${\mathfrak {N}}^{*}_{\sigma } \setminus \mathfrak {M}$ with a minimal normal subgroup R. Then $G/R\in {\mathfrak {N}}^{*}_{\sigma }$ by Lemma 3.1(i), so $G/R\in \mathfrak {M}$. Hence R is a unique minimal normal subgroup of G by Lemma 4.1. If R is non-abelian, then $G\simeq G/1=G/C_{G}(R)\in f(0)={\mathfrak {N}}^{*}_{\sigma }.$ Hence $G\in \mathfrak {M}$ since $G/R\in \mathfrak {M}$, a contradiction. Hence R is a p-group for some $p\in \sigma _{i}$, so $R \rtimes (G/C_{G}(R))$ is a $\sigma _{i}$-group by Theorem B. Therefore, $G/C_{G}(R)\in f(p)$ and so $G\in \mathfrak {M}$. Thus $\mathfrak {M}= {\mathfrak {N}}^{*}_{\sigma }$. Therefore, ${\mathfrak {N}}^{*}_{\sigma }$ is a solubly saturated formation by Lemma 4.1. Lemma 3.1(i) implies that this formation is normally hereditary.

Therefore, in order to complete the proof of the theorem it is enough to show that if $G=AB$, where A and B are normal $\sigma$-quasinilpotent subgroups of G, then G is $\sigma$-quasinilpotent. Suppose that this is false and let G be a counterexample of minimal order. Let R be a minimal normal subgroup of G and $C=C_{G}(R)$. By Lemma 3.1(i), the hypothesis holds for G / R and so the choice of G implies that G / R is $\sigma$-quasinilpotent. Therefore, in view of Lemma 3.1(ii), R is a unique minimal normal subgroup of G.

Let $Z_{1}=Z_{\sigma }(A)$ and $Z_{2}= Z_{\sigma }(B)$. If $A\cap B=1$, then $Z_{\sigma }(G)=Z_{1}\times Z_{2}$ by Proposition 2.5(viii). On the other hand, $A/Z_{1}$ and $B/Z_{2}$ are $\sigma$-semisimple by Theorem B, so

$$\begin{aligned} G/Z=(A\times B)/(Z_{1}\times Z_{2})\simeq (A/Z_{1})\times (B/Z_{2}) \end{aligned}$$

is $\sigma$-semisimple. Hence G is $\sigma$-quasinilpotent by Theorem B. Therefore, $A\cap B\ne 1$, so $R\le A\cap B$. First assume that R is $\sigma$-primary, R is a $\sigma _{i}$-group say. Then by Theorem B, $R\le Z_{1}\cap Z_{2}$ and so $AC/C\simeq A/A\cap C$ and $BC/C\simeq B/B\cap C$ are $\sigma _{i}$-groups. Hence $G/C=(AC/C)(BC/C)$ is a $\sigma _{i}$-group. Thus R is $\sigma$-central in G. Therefore, $R\le Z_{\sigma }(G)$ and so $Z_{\sigma }(G/R)=Z_{\sigma }(G)/R$ by Proposition 2.5(vi). Thus G is $\sigma$-quasinilpotent by Theorem B.

Therefore, R is not $\sigma$-primary. Hence R is non-abelian, so $C=1$. Then $R=R_{1} \times \cdots \times R_{t}$, where $R_{1}, \ldots , R_{t}$ are minimal normal subgroups of A. Let $C_{i}= C_{A}(R_{i})$ ($i=1, \ldots , t$). Then $C=1= C_{1} \cap \cdots \cap C_{t}$. Since A is $\sigma$-quasinilpotent by hypothesis, $A=R_{i}C_{i}$ for all $i=1, \ldots , t$ by Lemma 3.2. Hence

$$\begin{aligned} R= & {} RC=R_{1}\ldots R_{t}(C_{t} \cap \cdots \cap C_{1})=R_{1}\ldots R_{t-1}(R_{t}C_{t} \cap C_{t-1} \cap \cdots \cap C_{1})\\= & {} R_{1}\ldots R_{t-1}(A\cap C_{t-1} \cap \cdots \cap C_{1})=R_{1}\ldots R_{t-1}(C_{t-1} \cap \cdots \cap C_{1})\\= & {} \cdots =R_{1}C_{1}=A. \end{aligned}$$

Similarly one can get that $B=R$, so $G=R$ is $\sigma$-semisimple. Hence G is $\sigma -\hbox {quasinilpotent}$.

The theorem is proved.$\square$

5 Proof of Theorem D

Recall that G is said to be a Schmidt group if G is not nilpotent but every proper subgroup of G is nilpotent.

Lemma 5.1

(See Proposition 1.6 in [8]) Let G be $\sigma$-soluble. If G is not $\sigma$-nilpotent but all proper subgroups of G are $\sigma$-nilpotent, then G is a Schmidt group.

Lemma 5.2

(See [19, Ch.III, 5.2] and [19, Ch.IV, 5.4]) If G is not p-nilpotent but every proper subgroup of G is p-nilpotent, then G is a p-closed Schmidt group and so $G=P\rtimes Q$, where P is a Sylow p-subgroup of G and Q is a Sylow q-subgroup of G for some primes $p\ne q$. Moreover, P is of exponent p or exponent 4 (if P is a non-abelian 2-group).

Lemma 5.3

(see [19, Ch.IV, 5.12]) Let P be a p-group, a a $p'$-automorphism of P.

• (1) If $[a, \Omega _{2} (P)]=1$, then $a=1$.

• (2) If $[a, \Omega _{1} (P)]=1$ and either p is odd or P is abelian, then $a=1$.

Proof of Theorem D

Let $Z= Z_{{\sigma }}(G)$ and $F^{*}=F^{*}_{\sigma }(E)$. Let $F=F_{\sigma }(E)$ and $C=C_{G}(F)$. It is enough to show that if every cyclic subgroup of $F^{*}$ of prime order or order 4 is contained in Z, then G is $\sigma$-nilpotent. Assume that this is false and let G be a counterexample with $|G| +|E|$ minimal.

(1) $F\ne E. Hence E=G.$

Assume that $F=E$. Then G is $\sigma$-soluble since G / E is $\sigma$-nilpotent by hypothesis. Let M be any maximal subgroup of G. Then $M/M\cap E\simeq EM/E$ is $\sigma$-nilpotent and $M\cap E$ is a normal $\sigma$-nilpotent subgroup of M by Lemma 2.3. Hence $F^{*}_{\sigma }(M\cap E)=F_{\sigma }(M\cap E)$ by Corollary 3.3. If A is a cyclic subgroup of $M\cap E$ of prime order or order 4, then $A\le Z\cap M\le Z_{{\sigma }}(M)$ by Proposition 2.5(iii). Therefore, the hypothesis holds for $(M, M\cap E)$, so the choice of G implies that M is $\sigma$-nilpotent. Hence G is a Schmidt group by Lemma 5.1, and so by Lemma 5.2, $G=P\rtimes Q$, where P is a Sylow p-subgroup of G and Q is a Sylow q-subgroup of G for some primes $p\ne q$ dividing |G|. Moreover, P is of exponent p or exponent 4 (if P is a non-abelian 2-group). Thus $P\le Z$, which implies that G is $\sigma$-nilpotent by Proposition 2.5(v). This contradiction shows that $F\ne E$. Therefore, since the hypothesis holds for (E, E) by Proposition 2.5(iii), the choice of G and Theorem B imply that $E=G$.

(2) If N is a normal subgroup of G, then the hypothesis holds for (N, N).

Indeed, $F^{*}_{\sigma }(N)\le F^{*}_{\sigma }(E)$ by Theorem C since $F^{*}_{\sigma }(N)$ is characteristic in N. Hence the hypothesis holds for (N, N) by Proposition 2.5(iii).

(3) $F=F^{*}$. Hence G / F is not $\sigma$-nilpotent.

Assume that $F < F^{*}$. Then $F^{*}$ is not $\sigma$-soluble by Theorem B. Moreover, the hypothesis holds for $(F^{*}, F^{*})$ by Claim (2), so the choice of G implies that $F^{*}=G$. Then, by Theorem B, $G=FC$. Hence the choice of G and Lemma 2.3 imply that C is not $\sigma$-nilpotent. But the hypothesis holds for (C, C) by Claim (2). Hence $C=G$, so $Z=F= Z(G)$. Since $G=F^{*}$ is not $\sigma$-soluble by Theorem B, G is not p-nilpotent for some odd prime p and so G has a p-closed Schmidt subgroup $H=P\rtimes Q$, where P is of exponent p by Lemma 5.2. Hence $P\le Z(G)\cap H$ and so H is nilpotent. This contradiction shows that $F = F^{*}$. Finally, note that if G / F is a $\sigma$-nilpotent, the hypothesis holds for (G, F), so the choice of G and Claim (1) imply that G is $\sigma$-nilpotent. This contradiction shows that we have (3).

(4) G / F is a simple non-$\sigma$-primary group.

Let $F\le M < G$, where M is a normal subgroup of G with simple quotient G / M. Then the hypothesis holds for (M, M) by Claim (2). Therefore, $M=F$ by the choice of G.

(5) $Z=Z(G)$. Hence F is nilpotent.

The hypothesis holds for $(C_{G}(Z), C_{G}(Z))$ by Claim (2). Assume that $C_{G}(Z) < G$. Then the choice of G implies that $C_{G}(Z)$ is $\sigma$-nilpotent. But $G/C_{}(Z)$ is $\sigma$-nilpotent by Proposition 2.5(iv). Hence G is $\sigma$-soluble, contrary to Claim (4). Therefore, $C_{G}(Z)=G$. Arguing now as in the proof of Claim (1), one can show that F is nilpotent.

(6) $F=Z(G)$.

In view of Claim (5), it is enough to show that for any prime p dividing |F| we have $P\le Z(G)$, where P is the Sylow p-subgroup of F. Let $\Omega =\Omega _{1}(P)$ if $p > 2$ and $\Omega = \Omega _{2}(P)$ if $p=2$. Then $\Omega (P)\le Z(G)$ by Claims (1) and (5), so $P\le Z(G)$ by Lemma 5.3.

Final contradiction From Claims (4) and (6), it follows that $G=F^{*}$ is $\sigma$-quasinilpotent by Theorem B, contrary to Claim (3).

The theorem is proved.