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     {\huge СИБИРСКИЕ \ ЭЛЕКТРОННЫЕ} \\[2mm]
     {\huge МАТЕМАТИЧЕСКИЕ ИЗВЕСТИЯ} \\[2mm]
     {\large Siberian Electronic Mathematical Reports} \\[1mm]
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     {\flushleft\it Том 16, стр. 144--144 (2019) \hspace{65mm}{\rm\small УДК \udcs}}
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     	{\rm\small DOI~10.33048/semi.2019.16.xxx}\hphantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}{\rm\small MSC\ \ \mscs }
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\title[Finite groups with formational subnormal \ldots]{Finite groups with formational subnormal primary subgroups of bounded exponent}
\author{{V.S. Monakhov, I.L. Sokhor}}%
\address{Victor Stepanovich Monakhov, Irina Leonidovna Sokhor  % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Francisk Skorina Gomel State University,
\newline\hphantom{iii} Kirova Str. 119,
\newline\hphantom{iii} 246019, Gomel, Belarus}%
\email{victor.monakhov@gmail.com, irina.sokhor@gmail.com}%

\thanks{\sc Monakhov, V.S., Sokhor, I.L.
Submodularity and formational subnormality}
\thanks{\copyright \ 2023 Monakhov V.S., Sokhor I.L.}
%\thanks{\rm Работа поддержана Белорусским республиканским фондом фундаментальных исследований (грант Ф23РНФ-237)}
\thanks{\rm This work is supported by the Belarusian Republican Foundation for Fundamental Research (Grant F23RNF-237)}
\thanks{\it Поступила 1 января 2015 г., опубликована 31 декабря 2015 г.}%


\semrtop \vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } Let $\mathfrak{U}_k$ be the class of
all supersoluble groups in which exponents are not divided by
the $(k+1)$-th powers of primes. We investigate the classes
$\mathrm{w}\mathfrak{U}_k$ and $\mathrm{v}\mathfrak{U}_k$
that contain all finite groups in which every Sylow and, respectively,
every cyclic primary subgroup is $\mathfrak{U}_k$-subnormal.
We prove that $\mathrm{w}\mathfrak{U}_k$ and $\mathrm{v}\mathfrak{U}_k$
are subgroup-closed saturated formations and obtain the characterizations
of these formations.
\medskip

\noindent{\bf Keywords:} finite group, primary subgroup, subnormal subgroup.
 \end{quote}
}


\section{Introduction}

All groups in this paper are finite. A primary group is a group of prime power order.
All fragments of the theory of group classes that we used
correspond to~\cite{ballcl}.

Let $\mathfrak F$ be a non-empty formation. A subgroup~$H$ of a group~$G$ is called
$\mathfrak F$-subnormal in~$G$ if either~$G=H$ or there is a subgroup chain
\begin{equation}\label{equ}
H=H_0\lessdot\ldots\lessdot H_i\lessdot H_{i+1}\lessdot \ldots
\lessdot H_n=G
\end{equation}
such that   $H_{i+1}/{(H_i)}_{H_{i+1}}\in \mathfrak F$ for all~$i$,
\cite[Definition~6.1.2]{ballcl}.
We write $X\lessdot Y$ if $X$ is a maximal subgroup of a group~$Y$,
and $X_Y=\cap_{y\in Y}X^y$ is the core of a subgroup~$X$ in a group~$Y$.
If~$\mathfrak X$ and~$\mathfrak Y$ are formations and
$\mathfrak X\subseteq \mathfrak Y$, then, clearly,
every $\mathfrak X$-subnormal subgroup is $\mathfrak Y$-subnormal.
If $\mathfrak F$ is a soluble formation (i.\,e. all groups in $\mathfrak F$
are soluble) and $H$ is a soluble $\mathfrak F$-subnormal subgroup of a group~$G$,
then~$G$ is soluble,~\cite[Lemma~2.13]{ms18}.

Let $\mathbb P$ be the set of all primes. If $|H_{i+1}:H_i|\in \mathbb P$ for every~$i$
in~\eqref{equ}, then $H$ is $\mathbb P$-subnormal in $G$,~\cite[Definition~1]{vvt2010}.

The class of groups with all Sylow subgroups (all cyclic primary subgroups)
$\mathfrak F$-subnormal is denoted by $\mathrm{w}\mathfrak{F}$
($\mathrm{v}\mathfrak{F}$, respectively).
If $\mathfrak{F}=\mathfrak U$ is the formation of all supersoluble groups,
then the class $\mathrm{w}\mathfrak{U}$ ($\mathrm{v}\mathfrak{U}$)
coincides with the class of all groups in which every Sylow subgroup
(every cyclic primary subgroup, respectively) is $\mathbb P$-subnormal,
\cite[lemma 1.12]{mon_smj16}. The classes  $\mathrm{w}\mathfrak{U}$ and $\mathrm{v}\mathfrak{U}$
are quite well investigated~\cite{vvt2010}--\cite{mon21}.
In particular, these classes are subgroup-closed saturated formations,
$\mathrm{w}\mathfrak U\subset \mathrm{v}\mathfrak U$ and every group from
$\mathrm{v}\mathfrak{U}$ has a Sylow tower of supersoluble type.
The inclusion $\mathrm{w}\mathfrak U\subset\mathrm{v}\mathfrak U$ is proper,
every biprimary minimal non-supersoluble group with non-cyclic non-normal Sylow subgroup
belongs to $\mathrm{v}\mathfrak{U}\setminus \mathrm{w}\mathfrak{U}$, see~\cite[{Example~2, Example~3}]{mon21}.

The exponent of a group $G$ is the least common multiple of the orders of
all elements of~$G$ and denoted by~$\exp(G)$. The set of all positive integers is
denoted by $\mathbb{N}$ and the set of all positive integers not divided by the $(k+1)$-th
powers of primes for $k\in\mathbb{N}$ is denoted by $\mathbb{N}_k$.
If $\mathfrak{X}$ is a formation, then $\mathfrak{X}_k$ is the class of
all groups from $\mathfrak{X}$ with exponents from~$\mathbb{N}_k$.
It is clear that $\mathfrak{X}_k=\mathfrak{X}\cap\mathfrak{E}_k$, where
$\mathfrak{E}$ is the formation of all finite groups.

Introduce the following classes of groups:
\begin{itemize}
  \item $\mathfrak{U}_k$ is the class of all supersoluble groups
        with exponents from~$\mathbb{N}_k$;
  \item $\mathrm{w}\mathfrak{U}_k=\mathrm{w}(\mathfrak{U}_k)$ is
        the class of all groups in which every Sylow subgroup is
        $\mathfrak{U}_k$-subnormal;
  \item $\mathrm{v}\mathfrak{U}_k=\mathrm{v}(\mathfrak{U}_k)$ is
        the class of all groups in which every cyclic primary subgroup
        is $\mathfrak{U}_k$-subnormal.
\end{itemize}

Since $\mathfrak{U}_k \subset \mathfrak{U}$, we have
$\mathrm{w}\mathfrak{U}_k\subset \mathrm{w}\mathfrak{U}$ and
$\mathrm{v}\mathfrak{U}_k\subset \mathrm{v}\mathfrak{U}$. Hence groups in
$\mathrm{w}\mathfrak{U}_k$ and $\mathrm{v}\mathfrak{U}_k$ possess the properties
of groups from $\mathrm{w}\mathfrak{U}$ and $\mathrm{v}\mathfrak{U}$,
respectively. In particular, groups in $\mathrm{w}\mathfrak{U}_k$
and $\mathrm{v}\mathfrak{U}_k$ have Sylow towers of supersoluble type.
In addition, $\mathrm{w}\mathfrak{U}_k\subset \mathrm{v}\mathfrak{U}_k$
(Lemma~\ref{lwv}) and this inclusion is proper for every~$k$ (Example~\ref{evw}).

Although $\mathfrak{U}_k$ is a subgroup-closed non-saturated formation,
$\mathrm{w}\mathfrak{U}_k$ and $\mathrm{v}\mathfrak{U}_k$ are subgroup-closed saturated
formations (Proposition~\ref{p1} and Proposition~\ref{pv}). The following theorems contain
the characterizations of groups from these formations.


\begin{theorem}\label{tw}
For a group~$G$, the following statements are equivalent.

$(1)$~Every Sylow subgroup of~$G$ is $\mathfrak{U}_k$-subnormal in~$G$,
i.\,e. $G\in\mathrm{w}\mathfrak{U}_k$.

$(2)$~$G/\Phi(G)\in (\mathrm{w}\mathfrak{U}_k)_k$;

$(3)$~$A/\Phi (A)\in \mathfrak{U}_k$
for every metanilpotent subgroup $A$ of~$G$;

$(4)$~$B/\Phi (B)\in \mathfrak{U}_k$
for every biprimary subgroup $B$ of~$G$.
\end{theorem}

\begin{corollary}\label{cw}
If $G\in\mathrm{w}\mathfrak{U}_k$, then $G/F(G)\in \mathcal A_k$.
\end{corollary}

Here $\mathcal A_k$ is the class of all groups with
abelian Sylow subgroups of exponent from~$\mathbb{N}_k$.

\begin{corollary}\label{ccw}
For a metanilpotent group~$G$, the following statements are equivalent.

$(1)$~Every Sylow subgroup of~$G$ is $\mathfrak{U}_k$-subnormal in~$G$.

$(2)$~$G/\Phi (G)\in \mathfrak U_k$.

$(3)$~$G\in \mathfrak{U}$ and $G/F(G)\in \mathfrak{A}_k$.
\end{corollary}

Here $\mathfrak{A}_k$ is the class of all abelian groups
with exponents from~$\mathbb{N}_k$.

\begin{theorem}\label{tkv}
For a group~$G$, the following statements are equivalent.

$(1)$~Every cyclic primary subgroup of~$G$ is $\mathfrak{U}_k$-subnormal
in~$G$, i.\,e. $G\in\mathrm{v}\mathfrak{U}_k$.

$(2)$~$G/\Phi(G)\in (\mathrm{v}\mathfrak{U}_k)_k$.

$(3)$~$A/\Phi (A)\in \mathfrak{U}_k$ for every subgroup~$A$
with nilpotent derived subgroup.

$(4)$~$B/\Phi (B)\in \mathfrak{U}_k$ for every biprimary subgroup~$B$
with cyclic Sylow subgroup.
\end{theorem}

\begin{corollary}\label{cvw}
$\mathfrak{U}  \cap \mathrm{w}\mathfrak{U}_k=\mathfrak{N}^2\cap \mathrm{w}\mathfrak{U}_k=
\mathfrak{U}  \cap \mathrm{v}\mathfrak{U}_k=\mathfrak{NA} \cap \mathrm{v}\mathfrak{U}_k$.
In particular, every Sylow subgroup of a supersoluble group $G$ is
$\mathfrak{U}_k$-subnormal in $G$ if and only if every cyclic primary
subgroup of $G$ is $\mathfrak{U}_k$-subnormal in~$G$.
\end{corollary}

Here $\mathfrak{N}^2$ is the class of all metanilpotent groups and
$\mathfrak{NA}$ is the class of all groups with nilpotent derived subgroup.
Both of these classes are subgroup-closed saturated formations.

\section{Preliminaries}

Throughout this paper, %$G$ always denotes a finite group and
$k$ denotes an  positive integer.
We write $H\le G$ ($H<G$) if $H$ is a (proper) subgroup of~$G$. A subgroup $H$ of
$G$ is non-trivial if $H\neq 1$ and $H\neq G$. By $\pi (k)$ we denote the set of
all primes dividing~$k$. For a group~$G$, $\pi (G)=\pi(|G|)$, where $|G|$
is the order of~$G$. If
$$
|G|={p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_n^{\alpha_n}},\  {p_1<p_2<\ldots < p_n},
$$
and $G$ has a normal series $G=G_0\ge G_1\ge \ldots \ge G_{n-1}\ge G_n=1$
such that $G_{i-1}/G_i$ is isomorphic to a Sylow $p_i$"~subgroup of~$G$ for every~$i$,
then we say that $G$ has \textit{a Sylow tower of supersoluble type}.
It is easy to check that the class $\mathfrak D$ of all groups with Sylow tower
of supersoluble type is a subgroup-closed saturated Fitting formation.
The class $\mathcal A$ of all groups with abelian Sylow subgroups is
a subgroup-closed formation, but it is not a saturated formation and
it is not a Fitting formation.

The greatest common divisor (gcd) and the least common multiple (lcm) of
integers $a$ and $b$ are denoted by~$(a,b)$ and~$[a,b]$, respectively.
We repeatedly use the following simplest properties of~$\mathbb{N}_k$.

\begin{lemma}\label{lnk}
$(1)$~If $n\in\mathbb{N}_k$ and $d$ divides $n$,
then $d\in\mathbb{N}_k$ and~$n/d\in\mathbb{N}_k$.

$(2)$~If $a,b\in\mathbb{N}_k$,
then $(a,b)\in\mathbb{N}_k$ and $[a,b]\in \mathbb{N}_k$.
\end{lemma}

\begin{lemma}\label{lexpg}
$(1)$~$\pi (G)=\pi (\exp(G))$ and $\exp(G)$ divides~$|G|$.

$(2)$~The exponent of~$G$ is equal to \emph{lcm} of orders of
primary elements of $G$.

$(3)$~If~$H$ is a subgroup of~$G$ and~$N$ is a normal subgroup of~$G$,
then~$\exp(H)$ and~$\exp(G/N)$ divide~$\exp(G)$.

$(4)$~If~$G=G_1\times G_2$, where $G_1\le G$ and $G_2\le G$,
then~$\exp(G)=[\exp(G_1),\exp (G_2)]$.
\end{lemma}

\begin{proof}[Proof]
$(1)$~For every $p\in \pi(G)$, there is an element of order~$p$ in~$G$
by Sylow's Theorem. Hence  $\pi (G)=\pi (\exp(G))$. In view of Lagrange's Theorem,
the order of every element of $G$ divides $|G|$. Therefore $\exp(G)$
divides~$|G|$ by Lemma~\ref{lnk}\,(2).

$(2)$~Assume that $\pi (G)=\{p_1,p_2,\ldots ,p_m\}$, $p_1<p_2<\ldots <p_m$,
and $\exp(G)=p_1^{n_1}p_2^{n_2}\ldots p_m^{n_m}$.
For every $i=1,2,\ldots ,m$, there is an element $x_i$ in $G$
such that $|x_i|=p_i^{n_i}t_i$ and $p_i$ does not divide~$t_i$.
It is clear that $x_i^{t_i}$ is a primary element of order~$p_i^{n_i}$
and $[x_1^{t_i},x_2^{t_2},\ldots ,x_m^{t_m}]=\exp (G)$.

$(3)$~This statement true in view of Lagrange's Theorem.


$(4)$~According to Statement~$(3)$, $\exp(G_1)$ and $\exp (G_2)$ divide~$\exp(G)$.
Hence $[\exp(G_1),\exp (G_2)]$ divides~$\exp(G)$.
Since any element $g\in G$ can be represented as~$g=g_1g_2$,
where $g_1\in G_1$, $g_2\in G_2$ and $|g|=[|g_1|,|g_2|]$,
we conclude that~$\exp(G)$ divides $[\exp(G_1),\exp (G_2)]$.
Consequently, $[\exp(G_1),\exp (G_2)]=\exp(G)$.
\end{proof}

A class $\mathfrak X$ is saturated if $G\in \mathfrak X$
whenever~$G/\Phi (G)\in \mathfrak X$.
Here $\Phi (G)$ is the Frattini subgroup of a group~$G$.
If $H\in \mathfrak X$ whenever~$H\le G$ and $G\in \mathfrak X$,
then~$\mathfrak X$ is a subgroup-closed class.

\begin{lemma}\label{lek}
$(1)$~$\mathfrak{E}_k$ is a subgroup-closed formation.

$(2)$~If $\mathfrak{X}$ is a (subgroup-closed) formation,
then $\mathfrak{X}_k=\mathfrak{X}\cap\mathfrak{E}_k$ is
a (subgroup-closed) formation.

$(3)$~If $\mathfrak X$ and $\mathfrak Y$ are formations, then
$(\mathfrak X\cap \mathfrak Y)_k=\mathfrak X_k\cap \mathfrak Y_k$
and $(\mathfrak X\mathfrak Y)_k\subset \mathfrak X_k\mathfrak Y_k$.
\end{lemma}

\begin{proof}[Proof]
$(1)$~Assume that $G\in\mathfrak{E}_k$ and $N$ is a normal subgroup of~$G$.
Then $\exp(G)\in\mathbb{N}_k$ and $\exp(G/N)$ divides $\exp(G)$
by Lemma~\ref{lexpg}\,(3). Hence $\exp(G/N)\in \mathbb{N}_k$
by Lemma~\ref{lnk}\,(1), and $G/N\in\mathfrak{E}_k$. Consequently,
$\mathfrak{E}_k$ is a homomorph.

Let $N_1$ and $N_2$ be normal subgroups of~$G$ and let
$G/N_1, G/N_2\in \mathfrak{E}_k$. By Remak's Lemma, $G/(N_1\cap N_2)$
is isomorphic to a subgroup which is a subdirect product of the direct product
$G/N_1\times G/N_2$. Since $\exp(G/N_i)\in \mathbb{N}_k$ for $i=1,2$
and $\exp (G/N_1\times G/N_2)=[\exp(G/N_1),\exp(G/N_2)]$ by Lemma~\ref{lexpg}\,(4),
we get $\exp (G/N_1\times G/N_2)\in \mathbb{N}_k$ by Lemma~\ref{lnk}\,(2).
Consequently, $\mathfrak{E}_k$ is a formation.

Assume that $G\in\mathfrak{E}_k$ and $H$ is a subgroup of~$G$.
In that case, $\exp(G)\in\mathbb{N}_k$ and $\exp(H)$ divides $\exp(G)$.
Consequently, $\exp(H)\in\mathbb{N}_k$ by Lemma~\ref{lnk}\,(1)
and $H\in\mathfrak{E}_k$. Thus $\mathfrak{E}_k$ is a subgroup-closed formation.

$(2)$~Since the intersection of (subgroup-closed) formations is
a (subgroup-closed) formation and in view of Statement~$(1)$,
Statement~$(2)$ is true.

$(3)$~Let $G\in (\mathfrak X\cap \mathfrak Y)_k$.
In that case, $G\in (\mathfrak X\cap \mathfrak Y)$ and $\exp(G)\in\mathbb{N}_k$.
Hence $G\in \mathfrak {X}_k$ и $G\in \mathfrak {Y}_k$. It follows that
$G\in \mathfrak {X}_k \cap \mathfrak {Y}_k$ and
$(\mathfrak X\cap \mathfrak Y)_k\subseteq\mathfrak X_k\cap \mathfrak Y_k$.
Now assume that $G\in \mathfrak {X}_k \cap \mathfrak {Y}_k$.
Then $G\in \mathfrak {X}_k\subseteq\mathfrak{X}$ and $G\in \mathfrak {Y}_k\subseteq\mathfrak{Y}$,
$\exp(G)\in\mathbb{N}_k$. Therefore $G\in (\mathfrak X\cap \mathfrak Y)_k$
and $(\mathfrak X\cap \mathfrak Y)_k=\mathfrak X_k\cap \mathfrak Y_k$.

Let $G\in (\mathfrak X\mathfrak Y)_k$. In that case,
$G\in \mathfrak X\mathfrak Y$ и $\exp(G)\in\mathbb{N}_k$.
Since $G\in \mathfrak X\mathfrak Y$, we get $G^\mathfrak Y\in \mathfrak X$.
From $\exp(G)\in\mathbb{N}_k$ it follows that
$\exp(G^\mathfrak Y)\in\mathbb{N}_k$ and $\exp(G/G^\mathfrak Y)\in\mathbb{N}_k$.
Hence $G^\mathfrak{Y}\in\mathfrak{X}_k$, $G/G^\mathfrak{Y}\in \mathfrak Y_k$
and $G^{\mathfrak{Y}_k}\le G^\mathfrak{Y}$. But $\mathfrak{Y}_k\subseteq \mathfrak{Y}$.
Therefore $G^\mathfrak{Y}\le G^{\mathfrak{Y}_k}$. Consequently,
$G^{\mathfrak{Y}_k}=G^\mathfrak{Y}$ and $G\in \mathfrak X_k\mathfrak Y_k$.
Thus, $(\mathfrak X\mathfrak Y)_k\subseteq \mathfrak X_k\mathfrak Y_k$.
\end{proof}

\begin{example}
Note that the reverse inclusion in Lemma~\ref{lek}\,(3) does not hold,
an example is $D_8\in \mathfrak N_1\mathfrak N_1\setminus (\mathfrak N\mathfrak N)_1$.
Here $D_8$ is the dihedral group of order $8$.
\end{example}

Note that $\mathrm{w}\mathfrak{U}_k$ and $(\mathrm{w}\mathfrak{U})_k$ are distinct classes:
$\mathrm{w}\mathfrak{U}_k=\mathrm{w}(\mathfrak{U}_k)$ is the class of all groups
in which every Sylow subgroup is $\mathfrak{U}_k$-subnormal;
the class $(\mathrm{w}\mathfrak{U})_k=\mathrm{w}\mathfrak{U}\cap\mathfrak{E}_k$
consists of all groups with $\mathfrak{U}$-subnormal Sylow subgroups and
exponent that is not divided by the $(k+1)$-th powers of primes.

Similarly, $\mathrm{v}\mathfrak{U}_k$ and $(\mathrm{v}\mathfrak{U})_k$ are also distinct classes:
$\mathrm{v}\mathfrak{U}_k=\mathrm{v}(\mathfrak{U}_k)$ is the class of all groups
in which every primary cyclic subgroup is $\mathfrak{U}_k$-subnormal;
$(\mathrm{v}\mathfrak{U})_k=\mathrm{v}\mathfrak{U}\cap \mathfrak{E}_k$
consists of all groups with $\mathfrak{U}$-subnormal primary cyclic subgroups and
exponent that is not divided by the $(k+1)$-th powers of primes.

\begin{lemma}\label{l10}
$(1)$~$(\mathrm{w}\mathfrak{U})_k$ and $(\mathrm{v}\mathfrak{U})_k$ are
subgroup-closed formations for any~$k$.

$(2)$~$(\mathrm{w}\mathfrak{U})_k\subset \mathrm{w}\mathfrak{U}_k$
and $(\mathrm{v}\mathfrak{U})_k\subset \mathrm{v}\mathfrak{U}_k$
for any~$k$.
\end{lemma}

\begin{proof}[Proof]
$(1)$~Since $\mathrm{w}\mathfrak{U}$ and $\mathrm{v}\mathfrak{U}$ are
subgroup-closed saturated formations,
$(\mathrm{w}\mathfrak{U})_k=\mathrm{w}\mathfrak{U}\cap\mathfrak{E}_k$ and $(\mathrm{v}\mathfrak{U})_k=\mathrm{v}\mathfrak{U}\cap\mathfrak{E}_k$,
we deduce that $(\mathrm{w}\mathfrak{U})_k$ and $(\mathrm{v}\mathfrak{U})_k$
are subgroup-closed formations by Lemma~\ref{lek}\,(2).

$(2)$~Let $G\in (\mathrm{w}\mathfrak{U})_k$. In that case, $\exp(G)\in\mathbb{N}_k$
and every Sylow subgroup of~$G$ is $\mathfrak{U}$-subnormal in~$G$. Assume that $R$
is a Sylow subgroup of~$G$. By hypothesis, there is a subgroup chain
$$
R=H_0\lessdot\ldots\lessdot H_i\lessdot H_{i+1}\lessdot \ldots \lessdot H_n=G
$$
such that   $H_{i+1}/{(H_i)}_{H_{i+1}}\in\mathfrak{U}$ for every~$i$.
By Lemma~\ref{lnk}, $\exp\left(H_{i+1}/{(H_i)}_{H_{i+1}}\right)\in\mathbb{N}_k$.
Hence $H_{i+1}/{(H_i)}_{H_{i+1}}\in \mathfrak{U}_k$ for every~$i$.
Thus $R$ is $\mathfrak{U}_k$-subnormal in~$G$ and $G\in \mathrm{w}\mathfrak{U}_k$.

Let $G\in (\mathrm{v}\mathfrak{U})_k$. Then $\exp(G)\in\mathbb{N}_k$
and every cyclic primary subgroup of~$G$ is $\mathfrak{U}$-subnormal in~$G$.
Assume that $A$ is a cyclic primary subgroup of~$G$. By hypothesis,
there is a subgroup chain
$A=H_0\lessdot\ldots\lessdot H_i\lessdot H_{i+1}\lessdot \ldots \lessdot H_n=G$
such that   $H_{i+1}/{(H_i)}_{H_{i+1}}\in\mathfrak{U}$ for every~$i$.
Since $\exp\left(H_{i+1}/{(H_i)}_{H_{i+1}}\right)\in\mathbb{N}_k$,
we get $H_{i+1}/{(H_i)}_{H_{i+1}}\in \mathfrak{U}_k$ for every~$i$.
Therefore $A$ is $\mathfrak{U}_k$-subnormal in~$G$ and $G\in \mathrm{v}\mathfrak{U}_k$.
\end{proof}

\begin{example}
In Lemma~~\ref{l10}\,(2), the inclusion is proper. In the non-cyclic group
$G=C_3\rtimes C_{2^{k+1}}=\langle a,b\mid a^3=b^{2^{k+1}}=1,\ a^b=a^2\rangle$, a Sylow $3$-subgroup $C_3$ is normal.
Therefore $C_3$ is $\mathfrak{U}_1$-subnormal in $G$.
A Sylow $2$-subgroup $C_{2^{k+1}}$ is also $\mathfrak{U}_1$-subnormal in~$G$,
since
$$
{(C_{2^{k+1}})}_G\cong C_{2^{k}}, \ \
G/{(C_{2^{k+1}})}_G\cong S_3\in\mathfrak{U}_1.
$$
Thus, $G\in \mathrm{w}\mathfrak{U}_1\subset\mathrm{w}\mathfrak{U}_k$,
but $G\notin(\mathrm{w}\mathfrak{U})_k$ in view of~$\exp(G)=3\cdot 2^{k+1}$.
\end{example}

Remind the properties of $\mathfrak{F}$-subnormal subgroups that we use.

\begin{lemma}\label{fsn}
Let $\mathfrak{F}$ be a formation, let $H$ and $K$ be subgroups of~$G$ and
let $N$ be a normal subgroup of~$G$. The following statement hold.

$(1)$~If $K$ is $\mathfrak{F}$-subnormal in $H$ and $H$ is $\mathfrak{F}$-subnormal in $G$,
then $K$ is $\mathfrak{F}$-subnormal in $G$~\textup{\cite[6.1.6(1)]{ballcl}}.

$(2)$~If $K/N$ is $\mathfrak{F}$-subnormal in $G/N$,
then $K$ is $\mathfrak{F}$-subnormal in $G$~\textup{\cite[6.1.6(2)]{ballcl}}.

$(3)$~If $H$ is $\mathfrak{F}$-subnormal in $G$,
then $HN/N$ is $\mathfrak{F}$-subnormal in $G/N$~\textup{\cite[6.1.6(3)]{ballcl}}.

$(4)$~If $\mathfrak{F}$ is a subgroup-closed formation and
$H$ is $\mathfrak{F}$-subnormal in $G$, then $H\cap K$ is
$\mathfrak{F}$-subnormal in $K$~\textup{\cite[6.1.7(2)]{ballcl}}.

$(5)$~If $\mathfrak{F}$ is a subgroup-closed formation and
$H$ and $K$ are $\mathfrak{F}$-subnormal in $G$,
then $H \cap K$ is $\mathfrak{F}$-subnormal in $G$~\textup{\cite[6.1.7(3)]{ballcl}}.
\end{lemma}

\begin{lemma}\label{hf}
If $\mathfrak{F}$ is a subgroup-closed formation and $H$ is $\mathfrak{F}$-subnormal
in~$G$, then $H^\mathfrak{F}$ is subnormal in~$G$.
\end{lemma}

\begin{proof}[Proof]
Use induction on $|G|$. If $H=G$, then $H^\mathfrak{F} = G^\mathfrak{F}$
is normal in $G$. Let $H$ be a proper subgroup of~$G$. In that case, there is
a maximal subgroup $M$ of~$G$ such that $M$ contains $H$ and $G^\mathfrak{F}$.
By induction, $H^\mathfrak{F}$ is subnormal in $M$. Since
$H^\mathfrak{F} \leq G^\mathfrak{F} \leq M$, we deduce that $H^\mathfrak{F}$ is
subnormal in $G^\mathfrak{F}$. But $G^\mathfrak{F}$ is normal in $G$.
Therefore $H^\mathfrak{F}$ is subnormal in $G$.
\end{proof}

\begin{lemma}\label{lnet}
If $H$ is a subnormal subgroup of a soluble group $G$,
then $H$ is $\mathfrak{U}_1$"~subnormal in $G$.
\end{lemma}

\begin{proof}[Proof]
Assume that $H$ is a subnormal subgroup of a soluble group~$G$.
In that case, there is a composition series such that
$$
1=H_0\leq H_1\leq\ldots\leq H_j=H\leq H_{j+1}\leq\ldots\leq H_m=G.
$$
Since $G$ is soluble, we get $|H_{j+1}/{(H_j)}_{H_{j+1}}|=|H_{j+1}/H_j|\in\mathbb{P}$
and $H_{j+1}/H_j\in\mathfrak{U}_k$. Therefore $H$ is $\mathfrak{U}_k$-subnormal in~$G$.
\end{proof}

\begin{example}
In the Frobenius group~$F_{5}=C_5\rtimes C_4$,
a Sylow subgroup~$C_4$ is $\mathfrak{U}$"~subnormal,
but $C_4$ is not $\mathfrak{U}_1$-subnormal and not subnormal.
\end{example}

We repeatedly use the following lemma.

\begin{lemma}\label{ind}
If $H$ is a non-normal subgroup of a soluble group $G$
and $|G:H|=r\in\mathbb{P}$, then $G/H_G \cong C_r\rtimes C_t$,
where $t$ divides $r-1$. In particular, $G/H_G$ is supersoluble.
\end{lemma}

\begin{proof}[Proof]
According to $|G:H|\in\mathbb{P}$, we deduce that $H$ is a maximal subgroup of $G$
and $\overline{G}=G/H_G$ is a soluble primitive group.
Therefore $\overline{G}=\overline{N}\rtimes \overline{H}$,
where $\overline{N}=F(\overline{G})=C_{\overline{G}}(\overline{N})$
is the unique minimal normal subgroup of $\overline{G}$,
$\overline{H}=H/H_G$ is a maximal subgroup of $\overline{G}$.
Hence
$$
|\overline{N}|=|\overline{G}:\overline{H}|=|G:H|=r, \
\overline{N}\cong C_r, \
N_{\overline{G}}(\overline{N})/C_{\overline{G}}(\overline{N})=
\overline{G}/\overline{N}\cong \overline{H}
$$
and $\overline{H}$ is isomorphic to a subgroup of the automorphism group of~$\overline{N}$.
Therefore $\overline{H}\cong C_t$ and $t$ divides $r-1$.
Thus, $G/H_G\cong C_r\rtimes C_t$, in particular, $G/H_G$ is supersoluble.
\end{proof}



\section{Groups with $\mathfrak{U}_k$-subnormal Sylow subgroups}

We repeatedly use the following properties of groups with
$\mathfrak{U}$-subnormal Sylow subgroups.

\begin{lemma}\label{lwu}
$(1)$~A group $G\in\mathrm{w}\mathfrak{U}$ if and only if
every metanilpotent subgroup of~$G$ is supersoluble,~\textup{\cite[Theorem~2.6\,(2)]{mon_smj16}}.
In particular, $\mathfrak{U}=\mathrm{w}\mathfrak{U}\cap \mathfrak {N}^2$.

$(2)$~A group $G\in \mathrm{w}\mathfrak{U}$ if and only if
every biprimary subgroup of~$G$ is supersoluble,~\textup{~\cite[Theorem~B\,(1)]{mk_rm13},
\cite[Theorem~1\,(2)]{mon21}}.

$(3)$~If $G\in \mathrm{w}\mathfrak{U}$, then $G$ has
a Sylow tower of supersoluble type and every Sylow subgroup of~$G/F(G)$
is abelian,~\textup{~\cite[Proposition~2.8; Theorem~2.13\,(3)]{vvt2010}}.

$(4)$~Every minimal non-supersoluble subgroup of~$G$ is threeprimary
if and only if $G\in \mathrm{w}\mathfrak{U}$,~\textup{\cite[Corollary 1\,(2)]{mon21}}.
\end{lemma}


\begin{proposition}\label{p1}
$\mathrm{w}\mathfrak{U}_k$ is a subgroup-closed saturated formation.
\end{proposition}

\begin{proof}[Proof]
By Lemma~\ref{lek}\,(2), $\mathfrak{U}_k$ is a subgroup-closed formation.
Therefore $\mathrm{w}\mathfrak{U}_k$ is a subgroup-closed formation by~\cite[Theorem~3.1\,(5)]{vvv}.
%Since $\mathrm{w}\mathfrak{U}_k\subseteq \mathrm{w}\mathfrak{U}$,
%every group in $\mathrm{w}\mathfrak{U}_k$ has a Sylow tower
%of supersoluble type in view of Lemma~\ref{lwu}\,(3).

%Assume that $G\in \mathrm{w}\mathfrak{U}_k$, $N$ is a normal subgroup of~$G$
%and $H/N$ is a Sylow subgroup of~$G/N$. In that case, there is a Sylow subgroup~$R$
%of~$H$ such that $H=RN$. Since $R$ is a Sylow subgroup of~$G$, $R$ is
%$\mathfrak{U}_k$-subnormal in~$G$ by hypothesis. In view of Lemma~\ref{fsn}\,(3),
%$H/N$ is $\mathfrak{U}_k$-subnormal in~$G/N$. Consequently,
%$G/N\in\mathrm{w}\mathfrak{U}_k$ and $\mathrm{w}\mathfrak{U}_k$ is a homomorph.

%Assume that $N_1$ and $N_2$ are normal subgroups of~$G$,
%$G/N_i\in\mathrm{w}\mathfrak{U}_k$, $i=1,2$, $N_1\cap N_2=1$,
%and $R$ is a Sylow subgroup of~$G$. In that case, $RN_i/N_i$ is
%a Sylow subgroup of~$G/N_i$ and $RN_i/N_i$ is $\mathfrak{U}_k$-subnormal
%in~$G/N_i$, $i=1,2$. According to Lemma~\ref{fsn}\,(2), $RN_i$ is
%$\mathfrak{U}_k$-subnormal in~$G$, $i=1,2$. Since $RN_1\cap RN_2=R$,
%we deduce that $R$ is $\mathfrak{U}_k$-subnormal in~$G$
%by Lemma~\ref{fsn}\,(5). Thus $\mathrm{w}\mathfrak{U}_k$ is a formation.

%Let $H\leq G\in \mathrm{w}\mathfrak{U}_k$ and let $Q$ be a Sylow subgroup of~$H$.
%By Sylow's Theorem, there is a Sylow subgroup $Q^*$ of~$G$ such that
%$Q=Q^* \cap H$. By hypothesis, $Q^*$ is $\mathfrak{U}_k$"~subnormal in~$G$.
%Since $\mathfrak{U}_k$ is a subgroup-closed formation, we get $Q$ is
%$\mathfrak{U}_k$"~subnormal in~$H$ by Lemma~\ref{fsn}\,(4).
%Thus $\mathrm{w}\mathfrak{U}_k$ is a subgroup-closed  formation.

Now we prove that $\mathrm{w}\mathfrak{U}_k$ is a saturated formation.
Assume the contrary and let $G$ be a group of least order such that
$G/\Phi(G)\in\mathrm{w}\mathfrak{U}_k$ and $G\notin\mathrm{w}\mathfrak{U}_k$.

Assume that $N\neq 1$ is a normal subgroup of~$G$ and $\Phi(G/N)=K/N$.
Since
$$
\Phi(G)N/N =\left(\cap_{M\lessdot G}M\right)N/N\le
\left(\cap_{N\leq H\lessdot G} H\right)/N=\Phi(G/N)=K/N,
$$
we get $\Phi(G)N\leq K$. Since
$$
G/K\cong (G/\Phi(G))/(K/\Phi(G)), \ G/\Phi(G)\in\mathrm{w}\mathfrak{U}_k
$$
and $\mathrm{w}\mathfrak{U}_k$ is a homomorph,
we have $G/K\in \mathrm{w}\mathfrak{U}_k$. Hence
$$
(G/N)/(\Phi(G/N))=(G/N)/(K/N)\cong G/K\in \mathrm{w}\mathfrak{U}_k.
$$
Since $|G/N|<|G|$, we get $G/N\in \mathrm{w}\mathfrak{U}_k$.
Thus $G/N\in \mathrm{w}\mathfrak{U}_k$ for every non-identity
normal subgroup~$N$ of~$G$. Since $\mathrm{w}\mathfrak{U}_k$ is
a formation, $G$ has the unique minimal normal subgroup.

Since~$G$ has a Sylow tower of supersoluble type, a Sylow $r$-subgroup
$R$ of $G$ is normal in $G$ for $r=\max \pi(G)$. It is clear that $R=F(G)$
and $O_p(G)=1$ for all $p\in \pi (G)\setminus \{r\}$.
In view of Lemma~\ref{lnet}, $R$ is $\mathfrak{U}_k$-subnormal in $G$.

Let $Q$ be a Sylow $q$-subgroup of~$G$ for $q\neq r$.
Since $G/\Phi(G)\in\mathrm{w}\mathfrak{U}_k$, we deduce that
$Q\Phi(G)/\Phi(G)$ is $\mathfrak{U}_k$-subnormal in~$G/\Phi(G)$.
By Lemma~\ref{hf},
$$
(Q\Phi(G)/\Phi(G))^{\mathfrak{U}_k}=Q^{\mathfrak{U}_k}\Phi(G)/\Phi(G)
$$
is subnormal in $G/\Phi(G)$. Consequently,
$$
Q^{\mathfrak{U}_k}\Phi(G)/\Phi(G)\leq F(G/\Phi(G))=F(G)/\Phi(G), \ Q^{\mathfrak{U}_k}=1.
$$
Therefore exponents of all Sylow $r^\prime$-subgroup of $G$
belong to~$\mathbb{N}_k$. Since $QR/R$ is a Sylow $q$-subgroup
of~$G/R\in\mathrm{w}\mathfrak{U}_k$, $QR/R$ is $\mathfrak{U}_k$-subnormal in $G/R$.
According to Lemma~\ref{fsn}\,(2), $QR$ is $\mathfrak{U}_k$-subnormal in $G$.
In view of $QR\leq G\in\mathrm{w}\mathfrak{U}$, we have $Q$ is
$\mathfrak{U}$-subnormal in $QR$. Therefore there is a subgroup chain
$$
Q=M_0 \lessdot M_1\lessdot \ldots \lessdot M_i\lessdot M_{i+1}\lessdot \ldots \lessdot M_n=QR
$$
such that $|M_{i+1}:M_i|\in\mathbb{P}$ for every $i$.
Denote $M_i=A$ and $M_{i+1}=B$. Clearly, $|B:A|=r$.
In view of Lemma~\ref{ind}, $B/A_B\cong C_r\rtimes C_t$, where $t$ divides $r-1$.
Since $\exp(Q)\in\mathbb{N}_k$, we deduce that $\exp(B/A_B)\in\mathbb{N}_k$ and
$B/A_B\in\mathfrak{U}_k$. Hence $Q$ is $\mathfrak{U}_k$"~subnormal in $QR$.
Consequently, $Q$ is $\mathfrak{U}_k$"~subnormal in $G$ by Lemma~\ref{fsn}\,(1).
Thus all Sylow subgroups of $G$ are $\mathfrak{U}_k$"~subnormal in $G$ and
$G\in\mathrm{w}\mathfrak{U}_k$.
\end{proof}

\begin{proof}[Proof of Theorem~\textup{\ref{tw}}]
$(1)\Rightarrow (2)$:
Assume that every Sylow subgroup of~$G$ is $\mathfrak{U}_k$"~subnormal in~$G$,
i.\,e. $G\in \mathrm{w}\mathfrak{U}_k$. Use induction on $|G|$ to prove
$G/\Phi (G)\in (\mathrm{w}\mathfrak{U}_k)_k$. Suppose that there is
a maximal subgroup $M$ of~$G$ such that $M_G=1$. In that case, $G$ is
a primitive group, $\Phi (G)=1$, $G=F(G)\rtimes M$, where $F(G)$ is
the unique minimal normal subgroup of $G$. Since~$G$ has a Sylow tower
of supersoluble type, a Sylow $r$-subgroup $R$ is normal in $G$ for $r=\max \pi(G)$.
Hence $R=F(G)$ and $R$ is an elementary abelian $r$-subgroup. If $Q$ is
a Sylow $q$-subgroup of~$G$ for $q\neq r$, $Q$ is $\mathfrak{U}_k$-subnormal in $G$
and $Q^{{\mathfrak{U}}_k}$ is subnormal in $G$ by Lemma~\ref{hf}. Therefore
$Q^{{\mathfrak{U}}_k}\leq F(G)=R$ in view of~\cite[Theorem~2.2]{isaacs}.
Consequently, $Q^{{\mathfrak{U}}_k}=1$ and the exponent of every Sylow
$r^\prime$"~subgroup of~$G$ belongs to~$\mathbb{N}_k$.
Thus all Sylow subgroups of~$G$ have exponents from~$\mathbb{N}_k$
and $G\in (\mathrm{w}\mathfrak{U}_k)_k$ by Lemma~\ref{lexpg}\,(2).

Now assume that $M_G\neq 1$ for every maximal subgroup $M$ of~$G$.
Since $G/M_G\in\mathrm{w}\mathfrak{U}_k$, by induction,
$$
(G/M_G)/\Phi(G/M_G))\in (\mathrm{w}\mathfrak{U}_k)_k.
$$
But $G/M_G$ is a primitive group, hence $\Phi(G/M_G)=1$
and $G/M_G\in (\mathrm{w}\mathfrak{U}_k)_k$ for every maximal subgroup~$M$ of~$G$.
Since $\Phi(G)=\bigcap_{M\lessdot G}M_G$ and $(\mathrm{w}\mathfrak{U}_k)_k$ is
a formation, we get $G/\Phi(G)\in (\mathrm{w}\mathfrak{U}_k)_k$.

$(1)\Leftarrow (2)$:
Let $G/\Phi(G)\in (\mathrm{w}\mathfrak{U}_k)_k$. Since
$(\mathrm{w}\mathfrak{U}_k)_k\subseteq \mathrm{w}\mathfrak{U}_k$
and $\mathrm{w}\mathfrak{U}_k$ is a saturated formation in view of
Proposition~\ref{p1}, we get~$G\in\mathrm{w}\mathfrak{U}_k$.

Thus, $(1)\Leftrightarrow (2)$ is proved.

$(1)\Rightarrow (3)$:
Assume that $G\in\mathrm{w}\mathfrak{U}_k$ and $A$ is a metanilpotent subgroup of~$G$.
In that case, $G\in\mathrm{w}\mathfrak{U}$, and by Lemma~\ref{lwu}\,(1),
$A\in \mathfrak{U}$. Since $\mathrm{w}\mathfrak{U}_k$ is a subgroup-closed formation
in view of Proposition~\ref{p1}, we get $A\in \mathrm{w}\mathfrak{U}_k$.
According proved Statement $(1)\Rightarrow (2)$,
$A/\Phi (A)\in (\mathrm{w}\mathfrak{U}_k)_k$. Consequently,
$A/\Phi(A)\in\mathfrak{U}\cap(\mathrm{w}\mathfrak{U}_k)_k\subseteq \mathfrak{U}_k$.

$(1)\Leftarrow (3)$:
Let $A/\Phi (A)\in \mathfrak{U}_k$ for every metanilpotent subgroup $A$ of~$G$.
Since $\mathfrak{U}_k\subseteq\mathfrak{U}$, every metanilpotent subgroup $A$ of~$G$
is supersoluble. In view of Lemma~\ref{lwu}\,(1), $G\in\mathrm{w}\mathfrak{U}$. Choose $G$
of least order such that $G\in\mathrm{w}\mathfrak{U}\setminus \mathrm{w}\mathfrak{U}_k$.
Since $G\in\mathrm{w}\mathfrak{U}$, a Sylow $r$-subgroup $R$ of~$G$ is normal in $G$
for $r=\max\pi(G)$. In view of Lemma~~\ref{lnet}, $R$ is $\mathfrak{U}_k$-subnormal in $G$.
Assume that $Q$ is a Sylow $q$-subgroup of $G$ for  $q\neq r$. In that case,
$R\rtimes Q$ is metanilpotent and
$R\rtimes Q/\Phi(R\rtimes Q)\in \mathfrak{U}_k \subseteq \mathrm{w}\mathfrak{U}_k$
by the choice of $G$.  Since $\mathrm{w}\mathfrak{U}_k$ is a saturated formation
by Proposition~\ref{p1}, we get  $R\rtimes Q\in \mathrm{w}\mathfrak{U}_k$. Hence $QR$
is a proper subgroup of~$G$ and $Q$ is $\mathfrak{U}_k$-subnormal in $QR$.
Let $U_1/R$ be a metanilpotent subgroup of~$G/R$. Since $(|U_1/R|,|R|)=1$,
by the Schur-Zassenhaus Theorem, there is a subgroup $U$ such that $U_1=R\rtimes U$
and $U_1/R\cong U$ is metanilpotent. By the choice of $G$, $U/\Phi (U)\in \mathfrak{U}_k$. Hence
$$
(U_1/R)/\Phi (U_1/R)\cong U/\Phi (U)\in \mathfrak{U}_k.
$$
Thus $G/R$ satisfies Statement~(3) and $G/R\in \mathrm{w}\mathfrak{U}_k$ by
the choice of~$G$. Hence a Sylow subgroup $QR/R$ is $\mathfrak{U}_k$-subnormal in $G/R$.
According to Lemma~\ref{fsn}\,(2), $QR$ is $\mathfrak{U}_k$-subnormal in $G$, and $Q$ is
$\mathfrak{U}_k$-subnormal in $G$ by Lemma~\ref{fsn}\,(1). Thus all Sylow subgroups of $G$
is $\mathfrak{U}_k$-subnormal in $G$ and~$G\in \mathrm{w}\mathfrak{U}_k$.

Statement~$(1)\Leftrightarrow (3)$ is proved.

$(1)\Rightarrow (4)$:
Assume that $G\in\mathrm{w}\mathfrak{U}_k$ and $B$ is a biprimary subgroup of~$G$.
In that case, $G\in\mathrm{w}\mathfrak{U}$, and by Lemma~\ref{lwu}\,(2),
$B$ is supersoluble. Since $\mathrm{w}\mathfrak{U}_k$ is a subgroup-closed formation
by Proposition~\ref{p1}, we have $B\in \mathrm{w}\mathfrak{U}_k$.
By proved Statement~$(1)\Rightarrow (2)$, $B/\Phi (B)\in (\mathrm{w}\mathfrak{U}_k)_k$.
Consequently, $B/\Phi (B)\in \mathfrak{U}\cap (\mathrm{w}\mathfrak{U}_k)_k\subseteq \mathfrak{U}_k$.

$(1)\Leftarrow (4)$:
Let $G$ be a group of least order such that $B/\Phi (B)\in \mathfrak{U}_k$ for
every biprimary subgroup $B$ of $G$ and $G\notin\mathrm{w}\mathfrak{U}_k$.
In that case, $G$ has a Sylow $q$-subgroup $Q$ for a prime $q\in\pi(G)$ that is not
$\mathfrak{U}_k$-subnormal in~$G$. Since $\mathfrak{U}_k\subseteq\mathfrak{U}$,
every biprimary subgroup of~$G$ is supersoluble. By Lemma~\ref{lwu}\,(2),
$G\in \mathrm{w}\mathfrak{U}$, in particular, $G$ has a Sylow tower of supersoluble type.
Consequently, for $r=\max \pi(G)$, a Sylow $r$-subgroup $R$ of~$G$ is normal in~$G$.
In view of Lemma~\ref{lnet}, $R$ is $\mathfrak{U}_k$-subnormal in $G$ and $r>q$.
By the choice of $G$, $QR/\Phi(QR)\in\mathfrak{U}_k\subseteq\mathrm{w}\mathfrak{U}_k$.
Hence $QR\in\mathrm{w}\mathfrak{U}_k$ by Proposition~\ref{p1}, in particular, $Q$ is
$\mathfrak{U}_k$-subnormal in~$QR$ and~$QR<G$. Assume that $H/R$ is a biprimary subgroup
of~$G/R$. By the Schur-Zassenhaus Theorem, there is a biprimary subgroup $B$ of~$H$
such that $H=R\rtimes B$, $H/R\cong B$. By the choice of $G$, $B/\Phi (B)\in \mathfrak{U}_k$.
Therefore
$$
(H/R)/\Phi(H/R)\cong B/\Phi (B)\in \mathfrak{U}_k.
$$
By induction, $G/R\in \mathrm{w}\mathfrak{U}_k$, hence $QR/R$ is
$\mathfrak{U}_k$-subnormal in~$G/R$. It follows that $QR$ is
$\mathfrak{U}_k$-subnormal in $G$ by Lemma~\ref{fsn}\,(2),
and $Q$ is $\mathfrak{U}_k$-subnormal in $G$ by Lemma~\ref{fsn}\,(1), a contradiction.

Statement~$(1)\Leftrightarrow (4)$ is proved.
\end{proof}

\begin{proof}[Proof of Corollary~\textup{\ref{cw}}]
Since $G\in \mathrm{w}\mathfrak{U}_k \subset \mathrm{w}\mathfrak{U}$,
we get $G/F(G)\in \mathcal A$ by Lemma~\ref{lwu}\,(3).
In view of theorem~\ref{tw}\,($(1)\Rightarrow(2)$)
$G/\Phi (G)\in (\mathrm{w}\mathfrak{U}_k)_k$. Therefore
\[
G/F(G)\cong (G/\Phi (G))/(F(G)/\Phi (G))\in \mathcal A\cap
(\mathrm{w}\mathfrak{U}_k)_k\subseteq \mathcal A_k.\qedhere
\]
\end{proof}

\begin{proof}[Proof of Corollary~\textup{\ref{ccw}}]
$(1)\Leftrightarrow (2)$:
If $G\in\mathfrak{N}^2$ and every Sylow subgroup of~$G$ is
$\mathfrak{U}_k$-subnormal in~$G$, then $G/\Phi (G)\in \mathfrak{U}_k$
by Theorem~\ref{tw}\,($(1)\Rightarrow (3)$).
Conversely, if $G/\Phi (G)\in \mathfrak{U}_k$, then $G\in \mathrm{w}\mathfrak{U}_k$
by Theorem~\ref{tw}\,($(1)\Leftarrow (2)$).

$(1)\Leftrightarrow (3)$:
If $G\in \mathfrak{N}^2\cap \mathrm{w}\mathfrak{U}_k$, then
$G/\Phi (G)\in \mathfrak{U}_k$ by proved Statement~$(1)\Rightarrow(2)$.
Since $G/F(G)$ is abelian, we get
$G/F(G)\cong (G/\Phi (G))/(F(G)/\Phi (G))\in \mathfrak A\cap \mathfrak U_k=\mathfrak A_k$.
Conversely, let $G/F(G)\in \mathfrak A_k$ and let $G\in \mathfrak U$.
Use induction on $|G|$ to prove that every Sylow subgroup of~$G$ is
$\mathfrak U_k$-subnormal in~$G$. Assume that $P$ is a Sylow $p$-subgroup and
$N$ is a minimal normal subgroup of~$G$ such that $|N|=r$ and $r=\max \pi (G)$.
By induction, $PN/N$ is $\mathfrak U_k$-subnormal in~$G/N$. Hence $PN$ is
$\mathfrak U_k$-subnormal in~$G$ and~$p<r$. Since
$$
F(G)\le C_G(N), \ PN/C_{PN}(N)=PN/(PN\cap C_G(N))\cong PC_G(N)/C_G(N)\le
$$
$$
\le G/C_G(N)\cong (G/F(G))/(C_G(N)/F(G))\in \mathfrak A_k,
$$
and $G/C_G(N)$ is cyclic, we deduce that $PN/C_{PN}(N)$ is cyclic and $|PN/C_{PN}(N)|=p^t\le p^k$.
Next,
$$
C_{PN}(N)=P_1\times N, \  P_1=P_{PN}\le P\lessdot PN, \
PN/P_1\cong C_r\rtimes C_{p^t}\in \mathfrak U_k,
$$
therefore $P$ $\mathfrak U_k$-subnormal in~$PN$. By Lemma~\ref{fsn}\,(1),
$P$ is $\mathfrak U_k$-subnormal in~$G$.
\end{proof}

\section{Groups with $\mathfrak{U}_k$-subnormal cyclic primary subgroups}

Groups with $\mathfrak{U}$-subnormal cyclic primary subgroups
were first considered in~\cite{mk_rm13}. The class of such groups was later
denoted by $\mathrm{v}\mathfrak{U}$. In Introduction, we indicate that
$\mathrm{w}\mathfrak{U}\subset \mathrm{v}\mathfrak{U}$
and this inclusion is proper.

\begin{lemma}\label{lwv}
$\mathrm{w}\mathfrak{U}_k\subset \mathrm{v}\mathfrak{U}_k$.
\end{lemma}

\begin{proof}[Proof]
Let $G\in \mathrm{w}\mathfrak{U}_k$. Then every Sylow subgroup of~$G$
is $\mathfrak{U}_k$-subnormal in~$G$. In view of Lemma~\ref{lnet},
every $p$-subgroup is $\mathfrak{U}_1$-subnormal in a Sylow $p$-subgroup.
Hence every primary subgroup of~$G$ is $\mathfrak{U}_k$-subnormal in~$G$
and $G\in \mathrm{v}\mathfrak{U}_k$.
\end{proof}

\begin{example}\label{evw}
In $GL(3,7)$, there is  a non-abelian subgroup $Q$
of order $3^3$ and exponent~$3$ that acts irreducibly on
an elementary abelian group~$P$ of order~$7^3$~\cite{bloom}.
The semidirect product $G=P\rtimes Q$ is a minimal non-supersoluble group
and $G\in \mathrm{v}\mathfrak{U}$ according to~\cite[Corollary~2\,(2)]{mon21}.
It corresponds to the group from~\cite[Theorem~9\,(Type~$10$)]{be}.
Since $\exp (G)=3\cdot 7$, we have $G\in \mathrm{v}\mathfrak{U}_1$.
Biprimary groups in~$\mathrm{w}\mathfrak{U}$ are supersoluble,
therefore $G\notin \mathrm{w}\mathfrak{U}$, and $G\notin \mathrm{w}\mathfrak{U}_1$.
Clearly, $G\in \mathrm{v}\mathfrak{U}_k\setminus \mathrm{w}\mathfrak{U}_k$
for any~$k$.
\end{example}

We repeatedly use the following properties of groups with $\mathfrak{U}$-subnormal
primary cyclic subgroups.

\begin{lemma}\label{lvu}
$(1)$~A group $G\in\mathrm{v}\mathfrak{U}$ if and only if
every subgroup of~$G$ with nilpotent derived subgroup is
supersoluble,~\textup{\cite[Theorem 2.6\,(1)]{mon_smj16},\cite[Theorem 2\,(1)]{mon21}}.
In particular, $\mathfrak{U}=\mathrm{v}\mathfrak{U}\cap \mathfrak{N}\mathfrak A$.

$(2)$~A group $G\in \mathrm{v}\mathfrak{U}$ if and only if
every biprimary subgroup of~$G$ with cyclic Sylow subgroup
is supersoluble,~\textup{\cite[Theorem~B\,(3)]{mk_rm13},\cite[Theorem 2\,(2)]{mon21}}.

$(3)$~The quotient group~$H/H^\mathfrak{U}$ is non-cyclic for
every minimal non-supersoluble subgroup $H$ of $G$ if and only if
$G\in \mathrm{v}\mathfrak{U}$, \textup{\cite[Corollary~2\,(2)]{mon21}}.

$(4)$~$\mathrm{w}\mathfrak{U}=\mathrm{v}\mathfrak{U}\cap\mathfrak{N}\mathcal A$
and every group of $\mathrm{v}\mathfrak{U}$ has a Sylow tower of supersoluble
type,~\textup{\cite[Theorem~3\,(1)]{mon21}}.
\end{lemma}

\begin{proposition}\label{pv}
$\mathrm{v}\mathfrak{U}_k$ is a subgroup-closed saturated formation.
\end{proposition}

\begin{proof}[Proof]
By Lemma~\ref{lek}\,(2), $\mathfrak{U}_k$ is a subgroup-closed formation.
Therefore $\mathrm{v}\mathfrak{U}_k$ is a subgroup-closed  formation by~\cite[Theorem~A\,(3)]{mur2014}.
%Since $\mathrm{v}\mathfrak{U}_k\subseteq \mathrm{v}\mathfrak{U}$,
%every group in $\mathrm{v}\mathfrak{U}_k$ has a Sylow tower of
%supersoluble type in view of Lemma~\ref{lvu}\,(4).

%Let $H\leq G\in\mathrm{v}\mathfrak{U}_k$ and let $A$ be a  primary cyclic subgroup of $H$.
%Then $A$ is $\mathfrak{U}_k$-subnormal in $G$. Since  $\mathfrak{U}_k$ is a subgroup-closed
%formation, $A$ is $\mathfrak{U}_k$-subnormal in~$H$ by Lemma~\ref{fsn}\,(4).
%Consequently, $\mathrm{v}\mathfrak{U}_k$ is a subgroup-closed class.

%Assume that $N$ is a normal subgroup of~$G\in\mathrm{v}\mathfrak{U}_k$
%and $A/N$ is a cyclic $p$-subgroup of $G/N$ for a prime $p\in\pi(G/N)$.
%If $B$ is a minimal supplement for $N$ in $A$, then $A=BN$ and $B\cap N\leq \Phi(B)$.
%Since $A/N=BN/N\cong B/(B\cap N)$, we deduce that $B$ is a primary cyclic subgroup.
%By the choice of~$G$, $B$ is $\mathfrak{U}_k$-subnormal in $G$. Consequently,
%$A/N$ is $\mathfrak{U}_k$-subnormal in~$G/N$ by Lemma~\ref{fsn}\,(3).
%Thus $\mathrm{v}\mathfrak{U}_k$ is a subgroup-closed homomorph.

%Let $N_1$ and $N_2$ be normal subgroups of~$G$ such that
%$$
%N_1\cap N_2=1, \ G/N_i\in\mathrm{v}\mathfrak{U}_k, \ i=1,2,
%$$
%and let $A$ be a cyclic $p$-subgroup of~$G$ for a prime~$p\in\pi(G)$.
%Assume that $A\leq N_1$. By Lemma~\ref{lnet}, $N_1$ is
%$\mathfrak{U}_k$-subnormal in~$G$. Since $A\cap N_2=1$ and
%$AN_2/N_2 \leq G/N_2\in\mathrm{v}\mathfrak{U}_k$, we deduce that
%$AN_2$ is $\mathfrak{U}_k$-subnormal in $G$ and~$AN_2\cap N_1=A(N_2\cap N_1)=A$
%is $\mathfrak{U}_k$-subnormal in~$N_1$ by Lemma~\ref{fsn}\,(4).
%Now $A$ is $\mathfrak{U}_k$-subnormal in~$G$ by Lemma~\ref{fsn}\,(1).
%Therefore we can assume that $A$ is contained neither in~$N_1$ nor in~$N_2$.

%Since $G/N_i\in\mathrm{v}\mathfrak{U}_k$, we get $AN_i/N_i$ is
%$\mathfrak{U}_k$-subnormal in~$G/N_i$. Hence by Lemma~\ref{fsn}\,(2),
%$AN_i$ is $\mathfrak{U}_k$-subnormal in $G$, and by Lemma~\ref{fsn}\,(5),
%$H=AN_1\cap AN_2$ is $\mathfrak{U}_k$-subnormal in~$G$. Let $P$ be
%a Sylow subgroup such that $A\leq P$. Since
%$$
%A\le H=AN_1\cap AN_2\le PN_1\cap PN_2=P,
%$$
%we have $A$ is $\mathfrak{U}_k$-subnormal in~$H$ by Lemma~\ref{lnet}.
%Now $A$ is $\mathfrak{U}_k$-subnormal in $G$ by Lemma~\ref{fsn}\,(1).
%Thus $\mathrm{v}\mathfrak{U}_k$ is a subgroup-closed formation.

Now we prove that $\mathrm{v}\mathfrak{U}_k$ is a saturated formation.
Assume the contrary and let $G$ be a group of least order such that
$G/\Phi(G)\in\mathrm{v}\mathfrak{U}_k$ and $G\notin\mathrm{v}\mathfrak{U}_k$.
By analogy with the proof of Proposition~\ref{p1}, we can easily prove that
$G$ has the unique minimal normal subgroup.
%Assume that $N\neq 1$ is a normal subgroup of $G$ and $\Phi(G/N)=K/N$. Since
%$$
%\Phi(G)N/N =\left(\cap_{M\lessdot G}M\right)N/N\le
%\left(\cap_{N\leq H\lessdot G} H\right)/N=\Phi(G/N)=K/N,
%$$
%we get $\Phi(G)N\leq K$. Since
%$$
%G/K\cong (G/\Phi(G))/(K/\Phi(G)), \ G/\Phi(G)\in \mathrm{v}\mathfrak{U}_k
%$$
%and $\mathrm{v}\mathfrak{U}_k$ is a homomorph, we have $G/K\in \mathrm{v}\mathfrak{U}_k$.
%Hence
%$$
%(G/N)/(\Phi(G/N))=(G/N)/(K/N)\cong G/K\in \mathrm{v}\mathfrak{U}_k.
%$$
%From $|G/N|<|G|$ it follows that $G/N\in \mathrm{v}\mathfrak{U}_k$.
%Thus $G/N\in \mathrm{v}\mathfrak{U}_k$ for every non-identity normal subgroup~$N$ of~$G$.
%Since $\mathrm{v}\mathfrak{U}_k$ is a formation, we deduce that $G$ has
%the unique minimal normal subgroup.
Since~$G$ has a Sylow tower of supersoluble type, a Sylow $r$-subgroup $R$
is normal in~$G$ for $r=\max \pi(G)$. It is clear that $R=F(G)$ and $O_p(G)=1$
for all $p\in \pi (G)\setminus \{r\}$.

Let $A$ be a cyclic $q$-subgroup for a prime $q\in\pi(G)$. If $q=r$, then $A$ is
$\mathfrak{U}_k$-subnormal in $G$ in view of Lemma~\ref{lnet}. Analogously,
if $A\leq \Phi(G)$, then $A$ is $\mathfrak{U}_k$-subnormal in~$G$ by Lemma~\ref{lnet}.
Assume that $q\neq r$ and $A$ is not contained in $\Phi(G)$. Since
$G/\Phi(G)\in\mathrm{v}\mathfrak{U}_k$, we deduce that $A\Phi(G)/\Phi(G)$
is $\mathfrak{U}_k$-subnormal in~$G/\Phi(G)$.
By Lemma~\ref{hf},
$$
(A\Phi(G)/\Phi(G))^{\mathfrak{U}_k}=A^{\mathfrak{U}_k}\Phi(G)/\Phi(G)
$$
is subnormal in~$G/\Phi(G)$. Consequently,
$$
A^{\mathfrak{U}_k}\Phi(G)/\Phi(G)\leq F(G/\Phi(G))=F(G)/\Phi(G)=R/\Phi(G), \
A^{\mathfrak{U}_k}=1.
$$
Therefore $A\in\mathfrak{U}_k$. Since $AR/R$ is a cyclic $q$-subgroup
of~$G/R\in\mathrm{v}\mathfrak{U}_k$, we deduce that $AR/R$ is $\mathfrak{U}_k$-subnormal
in $G/R$. By Lemma~\ref{fsn}\,(2), $AR$ is $\mathfrak{U}_k$-subnormal in $G$.
Since $AR\leq G\in\mathrm{v}\mathfrak{U}$, we get $A$ is $\mathfrak{U}$-subnormal in $AR$.
Hence there is a subgroup chain
$$
A=M_0 \lessdot M_1\lessdot \ldots \lessdot M_i\lessdot M_{i+1}\lessdot \ldots \lessdot M_n=AR
$$
such that $|M_{i+1}:M_i|\in\mathbb{P}$ for every $i$.
Denote $M_i=H$ and $M_{i+1}=K$. Clearly, $|K:H|=r$. It follows that
$K/H_K\cong C_r\rtimes C_t$, where $t$ divides $r-1$ in view of Lemma~\ref{ind}.
Since $\exp(A)\in\mathbb{N}_k$, we have $\exp(K/H_K)\in\mathbb{N}_k$ and
$K/H_K\in\mathfrak{U}_k$. Hence $A$ is $\mathfrak{U}_k$-subnormal in~$AR$.
Consequently, $A$ is $\mathfrak{U}_k$-subnormal in $G$ by Lemma~\ref{fsn}\,(1).
Thus all primary cyclic subgroups of~$G$ are $\mathfrak{U}_k$-subnormal in~$G$
and $G\in\mathrm{v}\mathfrak{U}_k$.
\end{proof}

\begin{proof}[Proof of Theorem~\textup{\ref{tkv}}]
$(1)\Rightarrow (2)$:
Let $G\in \mathrm{v}\mathfrak{U}_k$. Use induction on $|G|$ to prove
$G/\Phi (G)\in (\mathrm{v}\mathfrak{U}_k)_k$. Suppose that there is
a maximal subgroup $M$ of~$G$ such that $M_G=1$. In that case, $G$ is
a primitive group, $\Phi (G)=1$, $G=F(G)\rtimes M$, where $F(G)$
is the unique minimal normal subgroup of~$G$. In view of Lemma~\ref{lvu}\,(1),
a Sylow $r$-subgroup $R$ is normal in~$G$  for $r=\max \pi(G)$. Hence
$R=F(G)$ and $R$ is an elementary abelian $r$-subgroup.

Let $A$ be a cyclic $q$-subgroup for a prime $q\in\pi(G)$, $q\neq r$.
In that case, $A$ is $\mathfrak{U}_k$-subnormal in $G$, and by Lemma~\ref{hf},
$A^{{\mathfrak{U}}_k}$ is subnormal in $G$. Hence $A^{{\mathfrak{U}}_k}\leq F(G)=R$
by~\cite[Theorem~2.2]{isaacs}. Consequently, $A^{{\mathfrak{U}}_k}=1$ and
the exponent of every primary cyclic $r'$-subgroup belongs to~$\mathbb{N}_k$.
Thus all primary cyclic subgroups of~$G$ have exponents from~$\mathbb{N}_k$
and $G\in (\mathrm{v}\mathfrak{U}_k)_k$ by Lemma~\ref{lexpg}\,(2).

Now assume that $M_G\neq 1$ for every maximal subgroup $M$ of~$G$.
Since $G/M_G\in\mathrm{v}\mathfrak{U}_k$, we get
$(G/M_G)/\Phi(G/M_G))\in (\mathrm{v}\mathfrak{U}_k)_k$ by induction.
But $G/M_G$ is a primitive group, therefore $\Phi(G/M_G)=1$ and
$G/M_G\in (\mathrm{v}\mathfrak{U}_k)_k$ for every maximal subgroup~$M$
of~$G$.  Since $\Phi(G)=\bigcap_{M\lessdot G}M_G$ and
$(\mathrm{v}\mathfrak{U}_k)_k$ is a formation, we conclude
that $G/\Phi(G)\in (\mathrm{v}\mathfrak{U}_k)_k$.

$(1)\Leftarrow (2)$:
Let $G/\Phi(G)\in (\mathrm{v}\mathfrak{U}_k)_k$.
Since $(\mathrm{v}\mathfrak{U}_k)_k\subseteq \mathrm{v}\mathfrak{U}_k$
and $\mathrm{v}\mathfrak{U}_k$ is a saturated formation by Proposition~\ref{pv},
we get  $G\in\mathrm{v}\mathfrak{U}_k$.

Statement $(1)\Leftrightarrow (2)$ is proved.

$(1)\Rightarrow (3)$:
Assume that $G\in\mathrm{v}\mathfrak{U}_k$ and $A$ is a subgroup of~$G$
with nilpotent derived subgroup. In that case, $G\in\mathrm{v}\mathfrak{U}$,
and by Lemma~\ref{lvu}\,(1), $A\in\mathfrak{U}$. By proved Statement $(1)\Rightarrow (2)$,
$A/\Phi (A)\in (\mathrm{v}\mathfrak{U}_k)_k$. Consequently,
$A/\Phi (A)\in \mathfrak{U}\cap(\mathrm{v}\mathfrak{U}_k)_k\subseteq\mathfrak{U}_k$.

$(1)\Leftarrow (3)$:
Let $A/\Phi (A)\in \mathfrak{U}_k$ for every subgroup~$A$ of $G$
with  nilpotent derived subgroup. Since $\mathfrak{U}_k\subseteq\mathfrak{U}$,
every subgroup $A$ of $G$ with nilpotent derived subgroup is supersoluble.
In view of Lemma~\ref{lvu}\,(1), $G\in\mathrm{v}\mathfrak{U}$.
Choose a group $G$ of least order such that
$G\in\mathrm{v}\mathfrak{U}\setminus \mathrm{v}\mathfrak{U}_k$.
Since $G\in\mathrm{v}\mathfrak{U}$, a Sylow $r$-subgroup $R$ of~$G$
is normal in~$G$ for $r=\max\pi(G)$. In view of Lemma~\ref{lnet},
every cyclic $r$-subgroup of $G$ is $\mathfrak{U}_k$-subnormal in~$G$.
Let $H$ be a cyclic $q$-subgroup of~$G$ for a prime $q\in\pi(G)$, $q\neq r$.
The derived subgroup $(R\rtimes H)'\leq R\in\mathfrak{N}$.
Therefore by the choice of $G$,
$R\rtimes H/\Phi(R\rtimes H)\in\mathfrak{U}_k\subseteq\mathrm{w}\mathfrak{U}_k$.
By Proposition~\ref{pv}, we get $R\rtimes H\in\mathrm{w}\mathfrak{U}_k$.
Hence  $HR$ is a  proper subgroup of $G$ and $H$ is $\mathfrak{U}_k$-subnormal in $HR$.

Let $U_1/R$ be a subgroup with nilpotent derived subgroup in~$G/R$.
Since $(|U_1/R|,|R|)=1$, by the Schur-Zassenhaus theorem,
there is a subgroup $U$ such that $U_1=R\rtimes U$ and $U_1/R\cong U$
has the derived subgroup. By the choice of $G$,
$U/\Phi (U)\in \mathfrak{U}_k$. Hence
$$
(U_1/R)/\Phi (U_1/R)\cong U/\Phi (U)\in \mathfrak{U}_k.
$$
Thus $G/R$ satisfies Statement~(3) and $G/R\in \mathrm{v}\mathfrak{U}_k$
by the choice of~$G$. Therefore $HR$ is $\mathfrak{U}_k$-subnormal in~$G$
by Lemma~\ref{fsn}\,(2), and $H$ is $\mathfrak{U}_k$-subnormal in $G$
by Lemma~\ref{fsn}\,(1). Thus, $G\in\mathrm{v}\mathfrak{U}_k$.

Statement $(1)\Leftrightarrow (3)$ is proved.

$(1)\Rightarrow (4)$:
Assume that $G\in\mathrm{v}\mathfrak{U}_k$ and $B$ is a biprimary subgroup
with cyclic Sylow subgroup in~$G$. In that case, $G\in\mathrm{v}\mathfrak{U}$,
and by Lemma~\ref{lvu}\,(2), $B$ is supersoluble. Since $\mathrm{v}\mathfrak{U}_k$
is a subgroup-closed formation by Proposition~\ref{pv}, we get
$B\in\mathrm{v}\mathfrak{U}_k$. According to proved Statement $(1)\Rightarrow (3)$,
we have $B/\Phi (B)\in \mathfrak{U}\cap(\mathrm{w}\mathfrak{U}_k)_k\subseteq\mathfrak{U}_k$.

$(1)\Leftarrow (4)$:
Let $G$ be a group of least order such that $B/\Phi (B)\in \mathfrak{U}_k$
for every biprimary~$B$ with cyclic Sylow subgroup and $G\notin\mathrm{v}\mathfrak{U}_k$.
In that case, $G$ contains a cyclic $q$-subgroup $H$ for a prime $q\in\pi(G)$
that is not $\mathfrak{U}_k$-subnormal in~$G$. Since $\mathfrak{U}_k\subseteq\mathfrak{U}$,
every biprimary subgroup with cyclic Sylow subgroup in~$G$ is supersoluble.
By Lemma~\ref{lvu}\,(2), $G\in\mathrm{v}\mathfrak{U}$, in particular,
$G$ has a Sylow tower of supersoluble type. Consequently, a Sylow $r$-subgroup $R$
of~$G$ is normal in~$G$ for $r=\max \pi(G)$.  In view of Lemma~\ref{lnet}, $R$ is
$\mathfrak{U}_k$-subnormal in $G$ and $r>q$. By the choice of $G$,
$HR/\Phi(HR)\in\mathfrak{U}_k\subseteq\mathrm{v}\mathfrak{U}_k$. Hence
$HR\in\mathrm{v}\mathfrak{U}_k$ by Proposition~\ref{pv}. Consequently,
$HR$ is a proper subgroup of~$G$ and $H$ is $\mathfrak{U}_k$-subnormal in $HR$.
Let $K_1/R$ be a biprimary subgroup with cyclic Sylow subgroup in~$G/R$.
By the Schur-Zassenhaus theorem, there is a biprimary subgroup $K$
with cyclic Sylow subgroup in $K_1$ such that $K_1=R\rtimes K$ and $K_1/R\cong K$.
By the choice of $G$, $K/\Phi (K)\in \mathfrak{U}_k$. Therefore
$$
(K_1/R)/\Phi(K_1/R)\cong K/\Phi (K)\in \mathfrak{U}_k.
$$
By induction, $G/R\in \mathrm{v}\mathfrak{U}_k$. It follows that $HR/R$ is
$\mathfrak{U}_k$-subnormal in $G/R$. Hence $HR$ is $\mathfrak{U}_k$-subnormal in $G$
by Lemma~\ref{fsn}\,(2), and $H$ is $\mathfrak{U}_k$-subnormal in $G$
by Lemma~\ref{fsn}\,(1), a contradiction.

Statement $(1)\Leftrightarrow (4)$ is proved.
\end{proof}


\begin{proof}[Proof of Corollary~\textup{\ref{cvw}}]
Since every supersoluble group is metanilpotent, we have
$\mathfrak{U}\cap \mathrm{w}\mathfrak{U}_k\subseteq \mathfrak {N}^2\cap \mathrm{w}\mathfrak{U}_k$.
If $G\in \mathfrak {N}^2\cap \mathrm{w}\mathfrak{U}_k$, then
$G/\Phi (G)\in \mathfrak{U}_k$ by Theorem~\ref{tw}\,($(1)\Rightarrow (3)$). Now
$G\in \mathfrak{U}$ and $\mathfrak{U}\cap \mathrm{w}\mathfrak{U}_k\supseteq \mathfrak {N}^2\cap \mathrm{w}\mathfrak{U}_k$.
Hence $\mathfrak{U}\cap \mathrm{w}\mathfrak{U}_k=\mathfrak {N}^2\cap \mathrm{w}\mathfrak{U}_k$.

Since the derived subgroup of a supersoluble group is nilpotent, we get
$\mathfrak{U}\cap\mathrm{v}\mathfrak{U}_k\subseteq \mathfrak {NA}\cap \mathrm{v}\mathfrak{U}_k$.
If $G\in \mathfrak {NA}\cap \mathrm{v}\mathfrak{U}_k$, then
$G/\Phi (G)\in \mathfrak{U}_k$ by Theorem~\ref{tkv}\,($(1)\Rightarrow (3)$). Now
$G\in \mathfrak{U}$ and $\mathfrak{U}\cap \mathrm{v}\mathfrak{U}_k\supseteq \mathfrak {NA}\cap \mathrm{v}\mathfrak{U}_k$.
Hence $\mathfrak{U}\cap \mathrm{v}\mathfrak{U}_k=\mathfrak {NA}\cap \mathrm{v}\mathfrak{U}_k$.

In view of Lemma~\ref{lwv}, $\mathrm{w}\mathfrak{U}_k\subset \mathrm{v}\mathfrak{U}_k$.
Therefore
$\mathfrak{U}\cap \mathrm{w}\mathfrak{U}_k\subseteq \mathfrak{U}\cap \mathrm{v}\mathfrak{U}_k$.

Conversely, let $G\in \mathfrak{U}\cap \mathrm{v}\mathfrak{U}_k$.
By Theorem~\ref{tkv}\,($(1)\Rightarrow (2)$), $G/\Phi(G)\in \mathfrak{U}\cap (\mathrm{v}\mathfrak{U}_k)_k\subseteq
\mathfrak{U}_k\subseteq \mathfrak{U}\cap \mathrm{w}\mathfrak{U}_k$, and
$G\in \mathfrak{U}\cap \mathrm{w}\mathfrak{U}_k$ since
$\mathfrak{U}\cap \mathrm{w}\mathfrak{U}_k$ is a saturated formation.

Since $\mathfrak{U}\cap \mathrm{w}\mathfrak{U}_k=\mathfrak{U}\cap \mathrm{v}\mathfrak{U}_k$,
it follows that every Sylow subgroup of a supersoluble group $G$ is
$\mathfrak{U}_k$-subnormal in $G$ if and only if every cyclic primary
subgroup of $G$ is $\mathfrak{U}_k$-subnormal in~$G$.
\end{proof}


\bigskip
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\end{document}
