On the behavior of $\pi$-submaximal subgroups under homomorphisms

TL;DR

This paper investigates how $\pi$-submaximal subgroups behave under group homomorphisms by providing specific examples that highlight differences from $\pi$-maximal subgroups.

Contribution

It offers new insights and explicit examples demonstrating the distinct behavior of $\pi$-submaximal subgroups under homomorphisms.

Findings

Examples showing divergence in behavior of $\pi$-submaximal subgroups

Differences between $\pi$-maximal and $\pi$-submaximal subgroups under homomorphisms

Clarification of subgroup behavior in group theory contexts

Abstract

We construct examples that show the difference in behavior or $\pi$-maximal and $\pi$-submaximal subgroups under group homomorphisms.

Equations

$$[M,G^{\ast }]⩽U\cap MW=M(U\cap W)=M,$$

$$U=[U,G]⩽U\cap G⩽W\cap G=V,$$

$$M_i^g=(M\cap GW{)}^{g_{i}g}=(M\cap GW{)}^{t_i{g}_{i\sigma }}=(M\cap GW{)}^{g_{i\sigma }}=M_{i\sigma },$$

$$M\cap G⩽M\cap GW=M_1⩽M_X⩽K.$$

Full text

On the behavior of $\pi$-submaximal subgroups

under homomorphisms

Danila O. Revin

Danila O. Revin

iii Sobolev Institute of Mathematics,

iii 4, Koptyug av.

iii 630090, Novosibirsk, Russia

revin@math.nsc.ru

 and 

Andrei V. Zavarnitsine

Andrei V. Zavarnitsine

iii Sobolev Institute of Mathematics,

iii 4, Koptyug av.

iii 630090, Novosibirsk, Russia

zav@math.nsc.ru

Abstract. We construct examples that show the difference in behavior or $\pi$-maximal and $\pi$-submaximal subgroups under group homomorphisms.

Keywords: $\pi$-maximal subgroup, $\pi$-submaximal subgroup, minimal nonsolvable group

MSC2010: 20E28, 20E34, 20D20

1. Introduction

Let $\pi$ be a set of primes and let $G$ be a finite group. A subgroup $H\leqslant G$ is a $\pi$-subgroup if every prime divisor of $|H|$ is contained in $\pi$. We say that $H$ is $\pi$-maximal, if it is a $\pi$-subgroup and is not contained in a strictly larger $\pi$-subgroup. In the particular case $\pi=\{p\}$, the $\pi$-maximal subgroups are precisely the $p$-Sylow subgroups.

The $\pi$-maximal subgroups have been extensively studied in the literature, see, for example, [3, 4, 5, 8, 11, 14, 15, 17], [12, Ch. 5, § 3]. The most well-studied are the $\pi$-maximal subgroups in solvable groups. According to P. Hall’s classical theorem [6], [12, Ch. 4, Theorem 5.6], every two $\pi$-maximal subgroups of a solvable group $G$ are conjugate, or, equivalently, a complete analog of Sylow’s theorem for $\pi$-subgroups holds. More precisely, a solvable group $G$ has a $\pi$-subgroup whose index is not divisible by the primes in $\pi$ (such subgroups are called $\pi$-Hall and they are necessarily $\pi$-maximal), every two $\pi$-Hall subgroups of $G$ are conjugate, and every $\pi$-subgroup is contained in a $\pi$-Hall subgroup. For nonsolvable group, the analog of Hall’s theorem does not hold. However, the study of $\pi$-Hall subgroups in nonsolvable groups is substantially facilitated by the fact that if $A$ is a normal subgroup of $G$ then, for every $\pi$-Hall subgroup $H$ of $G$, the subgroups $H\cap A$ and $HA/A$ are $\pi$-Hall in $A$ and $G/A$, respectively [12, Ch. 4, (5.11)].

The principal difficulty in the study of $\pi$-maximal subgroups of nonsolvable groups is that the $\pi$-maximality agrees poorly with the normal structure of the group. For example [12, Example 2, p. 170], a $2$-Sylow subgroup $H$ in $G=\operatorname{PGL}_{2}(7)$ is maximal, see [1], hence is $\{2,3\}$-maximal. However, the normal subgroup $A=\operatorname{PSL}_{2}(7)\cong\operatorname{GL}_{3}(2)$ of order $2^{3}\cdot 3\cdot 7$ in $G$ is such that the intersection $H\cap A$ is a $2$-Sylow subgroup of $A$ (and $|H\cap A|=2^{3}$), but is not a $\{2,3\}$-maximal subgroup of $A$, because the stabilizers in $\operatorname{GL}_{3}(2)$ of a line and a plane of the natural $3$-dimensional module both have order $2^{3}\cdot 3$, i. e. are $\{2,3\}$-subgroups, and $H\cap A$ is conjugate to a $2$-Sylow subgroup of each of the stabilizers by Sylow’s theorem.

Homomorphic images of $\pi$-maximal subgroups exhibit an even more irregular behavior which we will talk about below. As far as intersections of $\pi$-maximal and normal subgroups are concerned, not every $\pi$-subgroup of a normal subgroup $A$ can be such an intersection. An obstruction here is the Wielandt–Hartley theorem [17, Hauptsatz 13.2], [7, Lemmas 2 and 3], [12, Ch. 5, (3.20)] which states that if $H$ is a $\pi$-maximal subgroup of $G$ and $A\trianglelefteqslant G$ then the order of $N_{A}(H\cap A)/(H\cap A)$ is not divisible by the primes in $\pi$; in particular, $H\cap A\neq 1$ if $|A|$ is divisible by at least one prime in $\pi$. Later Wielandt announced111Unfortunately, a proof of this analog has not been published to this day. In 2003, Shemetkov obtained the following weaker fact [11, Theorem 7]: if $A\trianglelefteqslant\trianglelefteqslant G$ and $H$ is $\pi$-maximal in $G$ then $H\cap A\neq 1$ if $|A|$ is divisible by at least one prime in $\pi$. Recently, Skresanov has found a proof of Wielandt’s claim in its entire form [15, 5.4(a)] and is currently preparing it for publication. an analog [15, 5.4(a)] of this theorem for the case where $A\trianglelefteqslant\trianglelefteqslant G$. It allowed him to introduce the new concept of a $\pi$-submaximal subgroup.

According to Wielandt [15, Definition on p. 170], we say that $H$ is $\pi$-submaximal in $G$ if there exists a finite group $G^{*}$ such that $G\trianglelefteqslant\trianglelefteqslant G^{*}$, i. e. $G$ can be embedded into $G^{*}$ as a subnormal subgroup, and $H=G\cap K$ for some $\pi$-maximal subgroup $K$ of $G^{*}$.

Clearly, every $\pi$-maximal subgroup is $\pi$-submaximal, but the converse does not hold in general. For example, a $2$-Sylow subgroup of $\operatorname{GL}_{3}(2)$ is $\{2,3\}$-submaximal, but not $\{2,3\}$-maximal, see above. Unlike $\pi$-maximal subgroups, $\pi$-submaximal ones have the following inductive property which resembles that of $\pi$-Hall subgroups: if $A\trianglelefteqslant\trianglelefteqslant G$ then $H\cap A$ is $\pi$-submaximal in $A$ for every $\pi$-submaximal subgroup $H$ in $G$. The Wielandt–Hartley theorem can be equivalently reformulated as follows: if $H$ is a $\pi$-submaximal subgroup of $G$ then $|N_{G}(H)/H|$ is not divisible by the primes in $\pi$ [15, 5.4(a)].

The concept of a $\pi$-submaximal subgroup proved to be quite useful. For example, the above-mentioned inductive property, along with the Wielandt–Hartley theorem, shows that every $\pi$-submaximal (and, consequently, every $\pi$-maximal) subgroup of $G$ is uniquely, up to conjugacy, determined by its projections on the quotients of any subnormal series of $G$ [15, 5.4(c)] (cf. [12, Ch. 5, Theorem 3.21]). More detailed information on $\pi$-submaximal subgroups can be found in [4, 5].

In this paper, we compare the behavior of $\pi$-maximal and $\pi$-submaximal subgroups under group homomorphisms.

Wielandt [17, (14.2)], [15, 4.2] pointed out quite a general construction showing that if $\phi:G\rightarrow G_{1}$ is a homomorphism of groups and $H$ is a $\pi$-maximal subgroup of $G$, then $H^{\phi}$ is in general not $\pi$-maximal in $G^{\phi}$. Suppose that $X$ is a finite group that has more than one conjugacy class of $\pi$-maximal subgroups, and let $Y$ be an arbitrary finite group. We consider the regular wreath product $X\wr Y$ and the natural epimorphism $G\rightarrow Y$. Then every (not just $\pi$-maximal) $\pi$-subgroup of $Y=G^{\phi}$ coincides with $H^{\phi}$ for some $\pi$-maximal subgroup $H$ of $G$.

However, there exists a group $A$ with the following property: if some group $G$ contains a normal subgroup $N$ isomorphic to $A$ then the image $HN/H$ of every $\pi$-maximal subgroup $H$ is $\pi$-maximal in $G/N$. Such is every $\pi$-group [17, (12.4)], every $\pi^{\prime}$-group [17, (12.7)], where $\pi^{\prime}$ is the complement to $\pi$ in the set of all primes, and also (by induction) every $\pi$-separable group, i. e. a group having a subnormal series whose every factor is either a $\pi$- or a $\pi^{\prime}$-group, see. [17, (12.9)], [15, 4.3]. In other words,

• $(*)$

if $\phi:G\rightarrow G_{1}$ is a homomorphism of groups whose kernel is $\pi$-separable then $H^{\phi}$ is a $\pi$-maximal subgroup in $G^{\phi}$ for every $\pi$-maximal subgroup $H$ in $G$.

An important remark about the behaviour of $\pi$-maximal subgroups under homomorphisms is the following consequence of [9, Kap. III, Satz 3.8 and Satz 3.9]:

• $(**)$

if $\phi:G\rightarrow G_{1}$ is a homomorphism of groups then every $\pi$-maximal subgroup in $G^{\phi}$ is the image under $\phi$ of some $\pi$-maximal subgroup of $G$.

Properties $(*)$ and $(**)$ imply that if $\phi:G\rightarrow G_{1}$ is a homomorphism of groups whose kernel is $\pi$-separable then $H\mapsto H^{\phi}$ maps surjectively the set of $\pi$-maximal subgroups of $G$ to the similar set of $G^{\phi}$ and induces a bijection between the conjugacy classes of $\pi$-maximal subgroups of $G$ and $G^{\phi}$.

Do the $\pi$-submaximal subgroups have similar properties? In [4, Proposition 9], it was proved that under a group homomorphism $\phi:G\rightarrow G_{1}$ the image of a $\pi$-submaximal subgroup of $G$ is a $\pi$-submaximal subgroup of $G^{\phi}$ if the kernel of $\phi$ coincides with the $\pi$-radical, $\pi^{\prime}$-radical, nilpotent radical of $G$, and, more broadly, with the radical $G_{\mathfrak{F}}$, where $\mathfrak{F}$ is a Fitting class that consists of groups whose all maximal $\pi$-subgroups are conjugate222Recall that a Fitting class $\mathfrak{F}$ is a class of finite groups which contains with every group all its normal subgroups, and if some group $G$ is the product of two of its normal subgroups belonging to $\mathfrak{F}$ then $G$ itself belongs to $\mathfrak{F}$. If $\mathfrak{F}$ is a Fitting class then every finite group has an $\mathfrak{F}$-radical $G_{\mathfrak{F}}$, i. e. the largest normal $\mathfrak{F}$-subgroup, which coincides with the product of all normal $\mathfrak{F}$-subgroups of $G$. Solvable groups, nilpotent groups, $\pi$-groups, $\pi^{\prime}$-groups, $\pi$-separable groups are examples of Fitting classes that consist of groups whose all maximal $\pi$-subgroups are conjugate.. This property plays an important role in [3]. H. Wielandt posed [15, Question (g)] the problem of classifying $\pi$-submaximal subgroups in minimal nonsolvable groups333Recall that a group is minimal nonsolvable if it is not solvable, but all its proper subgroups are solvable. If, additionally, the group is simple then it is minimal simple.. Such a classification for the minimal simple groups was obtained in [3]. The quotient $L=G/\Phi(G)$ of a minimal nonsolvable group $G$ by its Frattini subgroup $\Phi(G)$ (which in this case coincides with the nilpotent radical) is a minimal simple group. As a consequence, the image of every $\pi$-submaximal subgroup of $G$ under the canonical epimorphism $G\to L$ is a $\pi$-submaximal subgroup of $L$ [3, Proposition 1.2]. For every minimal simple group (the list of which was obtained by J.Thompson [13]), all $\pi$-submaximal subgroups are found in [3, Tables 1–11]. The question of whether an analog of $(*)$ is true remains: is a $\pi$-submaximal subgroup of $L$ the image of some $\pi$-submaximal subgroup of $G$? Note that most $\pi$-submaximal subgroups of $L$ are maximal (see [3, Tables 1–11]), hence the above question is answered in the affirmative for such subgroups due to $(*)$.

In the present paper, we construct examples showing that analogs of $(*)$ and $(**)$ for $\pi$-submaximal subgroups do not hold even in the case where the kernel of $\phi$ is an abelian $\pi$-group. A counterexample $G$ to the analog of $(**)$ will be a minimal nonsolvable group and the corresponding homomorphism $\phi$ will be the canonical epimorphism $G\rightarrow L=G/\Phi(G)$. We will therefore also construct an example of a $\pi$-submaximal subgroup in a minimal simple group $L$ that cannot be lifted to a $\pi$-submaximal subgroup of some minimal nonsolvable group covering $L$.

2. Preliminaries

We will require the following result.

Proposition 1**.**

Let $G$ be a finite group, let $\pi$ be a set of primes, and let $V$ be a unique minimal normal subgroup of $G$. Assume that $V$ is a $p$-group for $p\in\pi$, $V\nleqslant\operatorname{Z}(G)$, and $L=G/V$ is nonabelian simple. Let $H$ be a $\pi$-submaximal subgroup of $G$ and let $G^{*}$ be a finite group of minimal order such that $G\trianglelefteqslant\trianglelefteqslant G^{*}$ and there is a $\pi$-maximal subgroup $K$ of $G^{*}$ with $H=G\cap K$. Then $G\trianglelefteqslant G^{*}$ and $C_{G^{*}}(G)=1$.

Proof.

Denote $W=\langle V^{g}\mid g\in G^{*}\rangle$, i. e. the normal closure of $V$ in $G^{*}$, and let $\overline{\phantom{a}}:G^{*}\to G^{*}/W$ be the canonical epimorphism. Note that $W$ is a $p$-group as a subgroup generated by subnormal $p$-subgroups, see [16, Satz 6.5(a)]. Also, $W\leqslant K$, since $p\in\pi$ and $K$ is a $\pi$-maximal subgroup. Moreover, $W\cap G=V$ due to the structure of $G$.

Denote $X=\langle G^{g}\mid g\in G^{*}\rangle$. By the minimality of $G^{*}$, we have $G^{*}=KX$. Note that $\overline{X}$ is a minimal normal subgroup of $\overline{G^{*}}$, because it is the normal closure in $\overline{G^{*}}$ of the simple subnormal subgroup $\overline{G}\cong L$. This also yields $GW\trianglelefteqslant X$.

We show that every minimal normal subgroup of $G^{*}$ is a $\pi$-group and is therefore contained in $K$. Let $U$ be a minimal normal subgroup of $G^{*}$. If $U\leqslant W$, the claim follows, since $W$ is a $p$-group. Otherwise, $U\cap W=1$ and $\overline{U}\cong U$. Then $\overline{U}$ is a minimal normal subgroup of $\overline{G^{*}}$. (Indeed, if $M\leqslant U$ is such that $\overline{M}\trianglelefteqslant\overline{G^{*}}$ then

[TABLE]

i. e. $M\trianglelefteqslant G^{*}$, which implies $M=U$ or $M=1$.) It follows that either $\overline{U}=\overline{X}$ or $\overline{U}\cap\overline{X}=1$. The former case, however, is impossible, since we would have in this case $X=UW$ and $[U,W]\leqslant U\cap W=1$, i. e. $X\cong W\times U$ which would contain no subgroup isomorphic to $G$. Hence, the latter case holds, and because $G^{*}=KX$, we have $U\cong\overline{U}\leqslant\overline{K}$ is a $\pi$-group as claimed.

We now show that $W$ is the unique minimal normal subgroup of $G^{*}$. Again, let $U$ be a minimal normal subgroup of $G^{*}$. Then $U\leqslant K$ by the above. Suppose that $U\cap V=1$. Then $U\cap G=1$ as $V$ is the unique proper normal subgroup of $G$. Let $\widetilde{\phantom{a}}:G^{*}\to G^{*}/U$ be the canonical epimorphism. We have $G\cong\widetilde{G}\trianglelefteqslant\trianglelefteqslant\widetilde{G^{*}}$ and $\widetilde{K}$ is a $\pi$-maximal subgroup of $\widetilde{G^{*}}$, since $U\leqslant K$. We have $GU\cap K=(G\cap K)U=HU$, and so $\widetilde{G}\cap\widetilde{K}=\widetilde{H}$. The minimality of $G^{*}$ excludes this possibility. Therefore, $U\cap V\neq 1$ and thus $V\leqslant U$, because $V$ is minimal normal in $G$. But then $W=\langle V^{g}\mid g\in G^{*}\rangle\leqslant U$, which gives $W=U$. This proves the uniqueness of $W$ as a minimal normal subgroup of $G^{*}$.

We have $W=VC_{W}(G)$. Indeed, Clifford’s theorem and the subnormality of $G$ in $G^{*}$ imply that $W$ is a completely reducible $\mathbb{F}_{p}G$-module. If $U$ is an irreducible submodule of $W$ in and $U$ is not contained in $C_{W}(G)$ then

[TABLE]

where we have used the fact that $W$ normalizes $G$ by [10, Theorem 2.6]. This implies that $W=VC_{W}(G)$.

We also have $C_{G^{*}}(X)=1$. Indeed, assuming the contrary we have $W\leqslant C_{G^{*}}(X)\trianglelefteqslant G^{*}$, since $W$ is the unique minimal normal subgroup of $G^{*}$. Therefore, $V=W\cap G\leqslant C_{G}(G)=\operatorname{Z}(G)$ which contradicts the assumption that $V\nleqslant\operatorname{Z}(G)$.

Henceforth, denote $N=N_{K}(GW)$ and $G^{0}=N_{G^{*}}(GW)$.

We have $G^{0}=NX$. Indeed, $N\leqslant G^{0}$ by definition, and $X\leqslant G^{0}$, because $GW\trianglelefteqslant X$. Conversely, $G^{0}\leqslant NX$, since $G^{*}=KX$.

We also have $H=N\cap G$. Indeed, $N\cap G\leqslant K\cap G=H$. Also, $H$ normalizes both $G$ and $W$, hence $H\leqslant N_{K}(GW)=N$.

We now show that $H=M\cap G$ for some $\pi$-maximal subgroup $M$ of $G^{0}$. Being a $\pi$-group, $N$ is contained in a maximal $\pi$-subgroup, say $M$, of $G^{0}$. We have $G^{0}=NX=MX$ by the above. Let $1=g_{1},\ldots,g_{m}$ be a right transversal of $N$ in $K$. This will also be a right transversal of $G^{0}$ in $G^{*}$, since $G^{0}=NX$, $G^{*}=KX$, and $N\cap X=K\cap X$. Denote $M_{i}=(M\cap GW)^{g_{i}}$. For every $g\in K$, there exist $\sigma\in\operatorname{Sym}_{m}$ and elements $t_{1},\ldots,t_{m}\in N$ such that $g_{i}g=t_{i}g_{i\sigma}$. Therefore,

[TABLE]

where we have used the fact that $t_{i}$ normalizes both $M$ and $GW$, since $t_{i}\in N$. It follows that $K$ normalizes the subgroup $M_{X}=\langle M_{i}\mid i=1,\ldots,m\rangle$.

Since $W\leqslant K$, we have $W\leqslant N\leqslant M$ and $W\leqslant(M\cap GW)^{g_{i}}=M_{i}$ for every $i$. The fact that $g_{1},\ldots,g_{m}$ form a right transversal of $G^{0}$ in $G^{*}$ implies that the factors $M_{i}/W$ lie in distinct components of the minimal normal subgroup $X/W$ of $G^{*}/W$, and therefore, pairwise commute. Consequently, $M_{X}/W$ is a $\pi$-group, hence so is $M_{X}$. The maximality of $K$ implies that $M_{X}\leqslant K$. On the other hand,

[TABLE]

Now, we have $H=N\cap G\leqslant M\cap G\leqslant K\cap G=H$, and so $H=M\cap G$ as claimed.

In view of the minimality of $G^{*}$, we have $G^{*}=G^{0}$, $GW\trianglelefteqslant G^{*}$, $X=GW$, and $X/W\cong L$. It follows that $C_{W}(G)=C_{W}(X)\leqslant C_{G^{*}}(X)$. But we showed that $C_{G^{*}}(X)=1$ and $W=VC_{W}(G)$. Consequently, $W=V$ and $X=G$. This implies $G\trianglelefteqslant G^{*}$. ∎

3. Examples

In this section, we give two examples showing that $\pi$-submaximality is generally not preserved under homomorphic images or preimages. The following consequence of Proposition 1 will be used.

Corollary 2**.**

Let $\pi=\{2,3\}$, let $L=\operatorname{GL}_{3}(2)$, and let $V$ be an elementary abelian group of order $8$. Suppose that $G$ is an upward extension of $V$ by $L$ such that the conjugation action of $G$ induces on $V$ the natural $\mathbb{F}_{2}L$-module structure. Then no $2$-Sylow subgroup of $G$ is $\pi$-submaximal.

Proof.

Let $S$ be a $2$-Sylow subgroup of $G$ and suppose that it is $\pi$-submaximal. Let $\overline{\phantom{a}}:G\to L$ be the canonical epimorphism. Observe that $S$ is not $\pi$-maximal, because neither is the $2$-Sylow subgroup $\overline{S}$ of $L$, inasmuch as $\overline{S}$ is contained in a subgroup $M\leqslant L$ of order $24$. Let $G^{*}$ be a finite group of minimal order such that $G\trianglelefteqslant\trianglelefteqslant G^{*}$ and there is a $\pi$-maximal subgroup $K$ of $G^{*}$ with $S=G\cap K$. By Proposition 1, $G\trianglelefteqslant G^{*}$ and and $C_{G^{*}}(G)=1$, i. e. $G^{*}\leqslant\operatorname{Aut}(G)$. There are only two possibilities for $G$: it is either the natural semidirect product $V\!\!:\!L$, or the unique444The nonsplit extension $V^{\,\textstyle\cdot}L$ is well known [18, p. 120]. It occurs as a maximal subgroup of the simple Chevalley group $G_{2}(p)$ for an odd prime $p$. The uniqueness follows from the fact that $\dim H^{2}(L,V)=1$. nonsplit extension $V^{\,\textstyle\cdot}L$. It can be shown using [2] that in both cases $\operatorname{Aut}(G)$ is an extension of $G$ by an outer automorphism $\alpha$ of order $2$ which acts trivially on both $V$ and $L$. Consequently, we have either $G^{*}=G$ or $G^{*}=\operatorname{Aut}(G)$. Since $S$ is not $\pi$-maximal in $G$, we cannot have $G^{*}=G$. Hence, $G^{*}=\operatorname{Aut}(G)$.

We extend ‘‘\ \overline{\phantom{a}}\’’ to the canonical epimorphism $G^{*}\to G^{*}/V\cong L\times C$, where $C$ is cyclic of order $2$ generated by the image of $\alpha$. Since $K$ is $\pi$-maximal in $G^{*}$ and $V$ is a $\pi$-subgroup, it follows that $\overline{K}$ is $\pi$-maximal in $G^{*}/V$. Also, we have $\overline{K}\cong\overline{S}\times C$ by assumption. But this subgroup is not $\pi$-maximal in $G^{*}/V$ as it is properly contained in the $\pi$-subgroup $M\times C$, a contradiction. ∎

The first example shows that the image of a $\pi$-submaximal subgroup under an epimorphism $\phi$ whose kernel is an abelian $\pi$-group is not $\pi$-submaximal in $\operatorname{Im}\phi$.

Example 1. As in Corollary 2, let $V$ be the natural $3$-dimensional $\mathbb{F}_{2}L$-module for $L=\operatorname{GL}_{3}(2)$, and let $V^{*}$ be its contragredient module. The inverse-transpose automorphism $\gamma$ of $L$ of order $2$ permutes $V$ and $V^{*}$ and so naturally acts on $V\oplus V^{*}$. Consequently, the semidirect product $G=(V\oplus V^{*})\!:\!L$ can be extended to $G^{*}=G\!:\!\langle\gamma\rangle$. The $2$-Sylow subgroup of $G^{*}$ is $\pi$-maximal, where $\pi=\{2,3\}$. Hence, the $2$-Sylow subgroup $S$ of $G$ is $\pi$-submaximal. Let $\overline{\phantom{a}}:G\to G/V^{*}$ be the canonical epimorphism. Corollary 2 implies that $\overline{S}$ is not $\pi$-submaximal in $\overline{G}$, because $\overline{G}$ is an upward extension of $V$ by $L$.

The second example shows that, for a homomorphism $\phi$ whose kernel is an abelian $\pi$-group, a $\pi$-submaximal subgroup in $\operatorname{Im}\phi$ is not the image of any $\pi$-submaximal subgroup. Moreover, this example implies that there exists a minimal nonsolvable group $G$ such that a $\pi$-submaximal subgroup in the minimal simple group $G/\Phi(G)$ is the image of no $\pi$-submaximal subgroup of $G$.

Example 2. Again, let $V$ be the natural module for $L=\operatorname{GL}_{3}(2)$ and let $G$ be the nonsplit extension $V^{\,\textstyle\cdot}L$. Then $G$ is a minimal nonsolvable group. A $2$-Sylow subgroup $H$ of $L$ is $\pi$-submaximal by [3], where $\pi=\{2,3\}$. However, there is no $\pi$-submaximal subgroup of $G$ whose image in $L$ under the canonical epimorphism $G\to L$ would equal $H$, because such a subgroup would necessarily be $2$-Sylow in $G$, but no $2$-Sylow subgroup of $G$ is $\pi$-submaximal by Corollary 2.

Remark. We know of examples where $G$ has a normal abelian $\pi^{\prime}$-subgroup $A$ such that the image under the canonical epimorphism $G\to G/A$ of a $\pi$-submaximal subgroup of $G$ is not $\pi$-submaximal in $G/A$, and conversely a $\pi$-submaximal subgroup of $G/A$ is the image of no $\pi$-submaximal subgroup of $G$. A justification of these examples requires an analog of Proposition 1 for the case $p\not\in\pi$. We have obtained a proof of this analog which, however, uses the entire form of Wielandt’s claim [15, 5.4(a)], and therefore we postpone the publication of our examples until after a proof of this claim has appeared in print, see footnote on p. 1.