\begin{document}


\title{On p-groups of maximal class}

\author{Noureddine Snanou$^{1}$}

\thanks{{\scriptsize
\hskip -0.4 true cm \textbf{Keywords:}  p-groups of maximal class, fundamental subgroup, nilpotency class.
\newline \textbf{Mathematics subject classification:} 20D15. }}


\maketitle
\begin{abstract} Recall that a p-group of order $p^ {n} >p^ {3} $ is of maximal
class, if its nilpotency class is $n-1$. In this paper, we recall
some preliminaries of nilpotent groups and give some basic material
about p-groups of maximal class. Furthermore, we introduce the
fundamental subgroup of a p-group of maximal class by specifying its
importance in the general theory of p-groups of maximal class.
\end{abstract}
\vskip 0.2 true cm


\pagestyle{myheadings}

\markboth{\rightline {\sl \hskip 7 cm N. Snanou }}
         {\leftline{\sl On p-groups of maximal class}}


\section{\bf Introduction}
A group of order $p^{n}>p^{3}$ and nilpotency class $n-1$, is said
to be of maximal class. These groups have been studied by various
authors, the basic material about these groups is to be found in
Blackburn's paper \cite{Bla1}. A more modern reference for the
theory of p-groups of maximal class is section $14$ of Chapter III
in Huppert's book \cite{Hup67}. The p-groups of maximal class with
an abelian maximal subgroup were completely classified by A. Wiman
in \cite{W.A46}. The most results about these groups that we present
in this paper can be found in \cite{Ber1, Ber2, Ber3}. These results play a fundamental role in finite
p-group theory. Throughout this paper, we use the standard notation,
such as in \cite{Ber1}.

The paper is organized as follows: In the second section, we recall
some preliminaries of nilpotent groups. The third section covers the
basic material about p-groups of maximal class. In section 4, we
introduce a subgroup of a p-group of maximal class called the
fundamental subgroup. This group plays a fundamental role in the
development of the general theory of p-groups of maximal class.

\section{\bf Preliminaries}
\vskip 0.4 true cm

 Let $G$ be a group and $H$ be a subgroup of $G$. An element $g\in G$ normalizes $H$ if $gHg^{-1}= H$. We call
$N_{G}(H)=\{g\in G | gHg^{-1}=H\}$ the normalizer of $H$ in $G$. An
element $g\in G$ centralizes $H$ if $ghg^{-1}=h$ for any $h\in H$.
We call $C_{G}(H) =\{g\in G | ghg^{-1}=h, \forall h\in H\}$ the
centralizer of $H$ in $G$. If $H=G$, then $Z(G)=C_{G}(G)$ is called
the center of $G$.

 Note that for a subgroup $H$ of $G$, $C_{G}(H)=G$
if and only if $H\leq Z(G)$. It is easy to see that, for any
subgroup $H$, $N_{G}(H)$ and $C_{G}(H)$ are subgroups of G with
$C_{G}(H)\leq N_{G}(H)$, and that if $H\leq G$ then $H\leq
N_{G}(H)$. The N/C-theorem \cite[Introduction, Proposition
12]{Ber1} asserts that if $H\leq G$, then the quotient group
$N_{G}(H)/C_{G}(H)$ is isomorphic to a subgroup of $Aut(H)$. In
particular, $G/Z(G)$ is isomorphic to a subgroup of $Aut(G)$.

 Let $G$ be a group. Set $Z_{0}(G)=\{1\},
Z_{1}(G)=Z(G)$. Suppose that $Z_{i}(G)$ has been defined for $i\leq
k$. Define $Z_{k+1}(G)$ as follows: $
Z_{k+1}(G)/Z_{k}(G)=Z(G/Z_{k}(G))$. The chain $\{1\}=Z_{0}(G)\leq
Z_{1}(G)\leq...\leq Z_{k}(G)\leq...$ is said to be the upper central
series of $G$. All members of that series are characteristic in $G$.

\begin{defn} For elements $x,y\in G$, their commutator $x^{-1}y^{-1}xy$ is written
as $[x,y]$. If $X,Y\subseteq G$, then $[X, Y]$ is the subgroup
generated by all commutators $[x,y]$ with $x\in X, y\in Y$.
\end{defn}

The lower central series $G=K_{1}(G)\geq K_{1}(G)\geq
K_{2}(G)\geq...$ of $G$ is defined as follows: $K_{1}(G)=G,
K_{i+1}(G)=[K_{i}(G),G], i>0$. All members of that series are
characteristic in $G$. We have $K_{i}(G)/K_{i+1}(G)\leq
Z(G/K_{i}(G))$. If $H\leq G$, then $K_{i}(H)\leq K_{i}(G)$ for all
$i$.

Since $[y,x]=[x,y]^{-1}$, we have $[Y,X]=[X,Y]$. We write
$[G,G]=G'$, the subgroup $G'$ is called the commutator (or derived
subgroup) of $G$. We also write $G^{0}=G$, $G'=G^{1}$. Then the
subgroup $G^{i+1}=[G^{i},G^{i}]$ is called the (i+1)th derived
subgroup of $G$, $i\geq 0$. The chain $G=G^{0}\geq G^{1}\geq ...\geq
G^{n}\geq...$ is called the derived series of $G$. All members of
this series are characteristic in $G$ and all factors
$G^{i}/G^{i+1}$ are abelian. The group $G$ is said to be solvable if
$G^{n}={1}$ for some $n$.

\begin{defn}
A group $G$ is said to be nilpotent if the upper central series of
$G$ contains $G$. In other words, $G$ is nilpotent of class $c$, if
$Z_{c}(G)=G$ but $Z_{c-1}(G)<G$, we write $c=cl(G)$. In particular,
the class of the identity group is $0$ and the class of a
nonidentity abelian group is $1$.
\end{defn}

\begin{lem} The following are equivalent:
\begin{enumerate}
  \item $G$ is nilpotent of class $c$.
  \item $G=K_{1}(G)>K_{2}(G)>K_{3}(G)>...>K_{c+1}(G)=\langle1\rangle$.
  \item $\langle1\rangle=Z_{0}(G)<Z_{1}(G)<...<Z_{c-1}(G)<Z_{c}(G)=G$.
\end{enumerate}
\end{lem}

\begin{thm}\label{NIL}
Let $G$ be a p-group of order $p^{m}\geq p^{2}$. Then:
\begin{enumerate}
\item $G$ is nilpotent of class at most $m-1$.
\item If $G$ has nilpotency class $c$ then $|G:Z_{c-1}(G)|\geq p^{2}$.
\item The maximal subgroups of $G$ are normal and of index $p$.
\item $|G:G'|\geq p^{2}$.
\end{enumerate}
\end{thm}

\begin{cor}\label{CORNIL} Let $G$ be a p-group and let $N$ be a normal subgroup of $G$
of index $p^{i}\geq p^{2}$. Then $K_{i}(G)\leq N$.
\end{cor}

\begin{proof} The group $G/N$ has order $p^{i}\geq p^{2}$. It follows from
part $(1)$ of the Theorem \ref{NIL} that $G/N$ has class $\leq i-1$
and consequently $K_{i}(G/N)=\overline{1}$. Since
$K_{i}(G/N)=K_{i}(G)N/N$, this proves that $K_{i}(G)\leq N$.
\end{proof}

\section{P-group of maximal class }
\vskip 0.4 true cm

Recall that a group of order $p^{m}$ is of maximal class, if
$cl(G)=m-1>1$. Of course, any group of order $p^{2}$ and any
nonabelian p-group of order $p^{3}$ have maximal class. If $G$ is a
group of maximal class and of order $p^{m}$, then the lower and
upper central series of $G$ are:
$$G=K_{1}(G)>K_{2}(G)>K_{3}(G)>...>K_{m}(G)=\langle1\rangle,$$
and
$$\langle1\rangle=Z_{0}(G)<Z_{1}(G)<...<Z_{m-2}(G)<Z_{m-1}(G)=G.$$
The two series coincide, and the sections are all of order $p$,
except the first which has order $p^{2}$ and is not cyclic. Clearly
$G/K_{i}(G)$ is also of maximal class for $4\leq i\leq m-1$.

\begin{prop}\label{maxi}
Let $G$ be a p-group of maximal class and order $p^{m}$. Then:
\begin{enumerate}
  \item $|G:G'|=p^{2}$, $|Z(G)|=p$ and $|K_{i}(G):K_{i+1}(G)|=p$ for $2\leq i\leq m-1$.

  \item \label{maxim}If $1\leq i<m-1$, then $G$ has only one normal subgroup of order
$p^{i}$. More precisely, if $N$ is a normal subgroup of $G$ of index
$p^{i}\geq p^{2}$, then $N=K_{i}(G)$.
  \item $G$ has $p+1$ maximal subgroups.
\end{enumerate}
\end{prop}
\begin{proof}
(1) We have that
$$p^{m}=|G|=|G:G'|\prod_{i=2}^{m-1}|K_{i}(G):K_{i+1}(G)|$$. Now it
suffices to observe that $|G:G'|\geq p^{2}$, by theorem \ref{NIL},
and that $|K_{i}(G):K_{i+1}(G)|\geq p$ for $2\leq i\leq
m-1$. \\

(2) Let $N$ be any normal subgroup of $G$ of index $p^{i}$ with
$0\leq i\leq m$.If $i=0$ or $1$ then $N=K_{1}(G)$ or $N$ is maximal
in $G$. Otherwise $i\geq 2$ and $K_{i}(G)\leq N$ by Corollary
\ref{CORNIL}. Since $|G:K_{i}(G)|=p^{i}$, we conclude that
$N=K_{i}(G)$. \\

(3) As $G/K_{2}(G)$ has exponent $p$, the Frattini subgroup
$\Phi(G)=K_{2}(G)$. Hence $G/\Phi(G)$ has order $p^{2}$ and can be
regarded as a vector space over $\mathbb{F}_{p}$ of dimension $2$.
This vector space has $p+1$ subspaces of dimension $1$ and these
correspond to the maximal subgroups of $G$.
\end{proof}

\begin{rem}\
\begin{enumerate}
\item (The converse of the proposition \ref{maxi}(2)) If $G$ is a noncyclic group of order
$p^{m} > p^{2}$ containing only one normal subgroup of order $p^{i}$
for each $1\leq i<m-1$, then it is of maximal class \cite[Lemma 9.1]{Ber1}.
\item A p-group $G$ of order $p^{n}$ has exactly $n+p-1$ nontrivial normal subgroups if and only if it is of maximal class \cite[exercie 0.30]{Ber1}.
\end{enumerate}
\end{rem}

Suppose that a p-group $G$ has only one normal subgroup $T$ of index
$\geq p^{p + 1}$. If $G/T$ is of maximal class so is $G$
(see \cite[Theorem 12.9]{Ber1}). Conversely, let $G$ be a p-group of
maximal class. If $N$ is a normal subgroup of $G$ of index $\geq
p^{2}$, then $G/N$ has also maximal class. Indeed, since the class
of $G/K_{i}(G)$ is $i-1$ Whenever $2\leq i\leq m$, then the result
follows immediately from proposition \ref{maxi}.

\begin{prop} Let $G$ be a nonabelian group of order $p^{m} > p^{p}$. If $G$ contains only
one normal subgroup of index $p^{k}$ for any $k\in\{2, . . ., p +
1\} $, then it is of maximal class.
\end{prop}
\begin{proof} Obviously, $|G : G'|= p^{2}$ hence $ d(G) = 2 $. Assume that $G$
is not of maximal class, then $m > p + 1$. Let $T < G'$ be
G-invariant of index $p^{p + 1}$ in $G$, then $G/T$ is of maximal
class. As, by hypothesis, $T$ is the unique normal subgroup of index
$p^{p + 1}$ in $G$, then $G$ is of maximal class.
\end{proof}

\begin{prop}\label{M.Suzuki}(M.Suzuki). Let $G$ be nonabelian p-group. If $A<G$ of order
$p^{2}$ is such that $C_{G}(A)=A$, then $G$ is of maximal class.
\end{prop}
\begin{proof} We use induction on $|G|$. Since $p^{2}\nmid|Aut(A)|$ then, by
N/C-theorem, $N_{G}(A)$ is nonabelian of order $p^{3}$. As $A<G$,
then $ Z(G)<A $ and $|Z(G)|=p$. Obviously, $C_{G/Z(G)}(A/Z(G))=
N_{G/Z(G)}(A/Z(G))$. Since $ C_{G/Z(G)}(N_{G}(A)/Z(G))\leq
C_{G/Z(G)}(A/Z(G))=N_{G}(A)/Z(G)$ is of order $p^{2}$ then, by
induction, $G/Z(G)$ is of maximal class so $G$ is also of maximal
class since $|Z(G)|=p$.
\end{proof}

 \begin{cor} A p-group $G$ is of maximal class if and only if $G$ has an element with centralizer of
order $p^{2}$.
 \end{cor}

\begin{prop} \label{CGB}Let $G$ be a p-group, $B\leq G$ nonabelian of order $p^{3}$ and $C_{G}(B)< B$. Then
$G$ is of maximal class.
\end{prop}
\begin{proof} Assume that $|G|\geq p^{4}$ and the proposition has been proved
for groups of order $<|G|$ . It is known that a Sylow p-subgroup of
$Aut(B)$ is nonabelian of order $p^{3}$. Now, $C_{G}(B)=Z(B)=Z(G)$.
Therefore, by N/C-Theorem, $N_{G}(B)/Z(G)$ is nonabelian of order
$p^{3}$. If $x\in G-C_{G}(B)$ centralizes $N_{G}(B)/Z(G)$, then $x$
normalizes $B$ so $x\in N_{G}(B)$, a contradiction. Thus,
$C_{G}(N_{G}(B)/Z(G))<N_{G}(B)/Z(G)$ so, by induction, $G/Z(G)$ is
of maximal class. Since $|Z(G)|=p$, we are done.\end{proof}

\begin{prop}\label{NORM} Let $G$ be a p-group. If $G$ has a subgroup $H$ such
that $N_{G}(H)$ is of maximal class, then it is of maximal class.
\end{prop}
\begin{proof} Assume that $|N_{G}(H)| > p^{3}$ (otherwise, $C_{G}(H)=H$ and
$G$ is of maximal class, by Proposition \ref{M.Suzuki}). We use
induction on $|G|$. One may assume that $N_{G}(H)< G$, then $H$ is
not characteristic in $N_{G}(H)$ so by Proposition \ref{maxi}, we
have $|N_{G}(H):H|=p$ hence $|H|>p$. As $|Z(N_{G}(H))|=p$ and
$Z(G)<N_{G}(H)$, we get $Z(G)=Z(N_{G}(H))$ so $|Z(G)|=p$ and
$Z(G)<H$. Then $N_{G/Z(G)}(H/Z(G))=N_{G}(H)/Z(G)$ is of maximal
class, so $G/Z(G)$ is also of maximal class by induction. Since
$|Z(G)|=p$, then $G$ is of maximal class.
\end{proof}

\begin{lem}
Let $G$ be a p-group and let $N\lhd G$ be of order $> p$. Suppose
that $G/N$ of order $> p$ has cyclic center. If $R/N\lhd G/N$ is of
order $p$ in $G/N$, then $R$ is not of maximal class.
\end{lem}
\begin{proof} Let $T$ be a G-invariant subgroup of index $p^{2}$ in $N$. Then
$R\leq C_{G}(N/T)$ so $R/T$ is abelian of order $p^{3}$, and we
conclude that $R$ is not of maximal class.
\end{proof}

\begin{prop}\label{index}Let $A < G$ be of order $> p$. If all subgroups of $G$ containing $A$
as a subgroup of index $p$ are of maximal class, then $G$ is also of
maximal class.
\end{prop}
\begin{proof} Set $N=N_{G}(A)$. In view of proposition \ref{NORM} and
hypothesis, one may assume that $|N:A|>p$ (otherwise, there is
nothing to prove). Let $D<A$ be N-invariant of index $p^{2}$ ( $D$
exists since $|A| > p$). Set $C=C_{N}(A/D)$, then $C > A$. Let $F/A
\leq C/A$ be of order $p$, then $F$ is not of maximal class, a
contradiction.
\end{proof}

Let $H < G$ be of index $> p^{k}$, $k > 1$. If all subgroups of $G$
of order $p^{k}|H|$, containing $H$, are of maximal class, then $G$
is also of maximal class. Indeed, let $H < M < G$, where $|M :H|=
p^{k}-1$. Then all subgroups of $G$ containing $M$ as a subgroup of
index $p$, are of maximal class. Now the result follows from
Proposition \ref{index}.

\begin{prop}
Let $G$ be a p-group of maximal class and order $p^{m}$, $p > 2$, $m
> 3$, and let $N\lhd G$ be of index $p^{3}$. Then $exp(G/N)=p$.
\end{prop}
\begin{proof} Assume that this is false. Let $T$ be a G-invariant subgroup of
index $p$ in $N$. By hypothesis, $G/N$ has two distinct cyclic
subgroups $C/N$ and $Z/N$ of order $p^{2}$. Then $C/N\cap
Z/N=Z(G/T)$ and $G/T$ is not of maximal class, a
contradiction.
\end{proof}

Let $A$ be an abelian subgroup of index $p$ of a nonabelian p-group
$G$. By \cite[lemma 1.1]{Ber1}, we have $|G|=p|G'||Z(G)|$. Hence,
we have the following proposition.

\begin{prop}\label{AGG}
Suppose that a nonabelian p-group $G$ has an abelian subgroup $A$ of
index $p$. If $|G:G'|=p^{2}$, then $G$ is of maximal class.
\end{prop}
\begin{proof} We proceed by induction on $|G|$. One may assume that
$|G|>p^{3}$. We have $|Z(G)|=\frac{1}{p}|G:G'|=p$, so $Z(G)$ is a
unique minimal normal subgroup of $G$. Since $Z(G)<G'$, then wa have
$|G/Z(G):G'/Z(G)|=|G:G'|=p^{2}$. Hence, the quotient group $G/Z(G)$
is of maximal class by induction, and the result follows since
$|Z(G)|=p$.
\end{proof}

The previous result also holds if $G$ contains a subgroup of maximal
class and index $p$ (see \cite[Theorem 9.10]{Ber1}).

\begin{exer}
Let $G$ be a p-group of order $p^{4}$. Show that $G$ is of maximal
class if and only if $|G:G'|=p^{2}$.
\end{exer}
\textit{Solution}. Let $|G:G'|=p^{2}$. We have to prove that $G$ is
of class $3$. Assume that $cl(G)=2$. It follows from \cite[Lemma 65.1]{Ber2} that $G$ contains a nonabelian subgroup $B$ of order
$p^{3}$. By proposition \ref{CGB}, we have $G=BZ(G)$, then
$|G:G'|=p^{3}$, a contradiction.

\begin{exer}
Let $G$ be a p-group of order $p^{4}$ and exponent $p$. Prove that
if $G$ has no nontrivial direct factors then it is of maximal class.
\end{exer}
\textit{Solution}. Let $H$ be an $A_{1}$-subgroup of $G$; then
$|H|=p^{3}$. If $Z(G)\nleq H$, then $G=H\times C$, where $C<Z(G)$ is
of order $p$ such that $C\nleq H$. Thus, $Z(G)<H$ so $Z(G)$ is of
order $p$. Since $C_{G}(H)=Z(G)$, we get $C_{G}(H)<H$. Then, by
proposition \ref{CGB}, $G$ is of maximal class.

\begin{exer}
If $G$ is of order $p^{4}$ and exponent $p$. Prove that if $d(G)=2$
then $G$ is of maximal class.
\end{exer}
\textit{Solution}. Obviously, the group $G$ has an abelian subgroup
of index $p$. Since $G'=\Phi(G)$ and, by hypothesis, $|G:G'|=p^{2}$,
the result follows from Proposition \ref{AGG}.

\section{Fundamental subgroup of a p-group of maximal class}
\vskip 0.4 true cm

Let $G$ be a $p$-group of maximal class and order $
p^{n} $, $n>3$. The structure of $G$ is largely determined by the structure of a certain subgroup $
G_{1}$ called the fundamental subgroup of $ G $ and defined as
follows: $G_{1}\diagup K_{4}(G)$ is the
centralizer of $K_{2}(G)\diagup K_{4}(G)$ in $G\diagup K_{4}(G)$, where $
G > K_{2}(G)$ $ > ... > K_{n}(G)=1$ is the lower central series of
$G$. For each $ i=2,...,m-2 $, $ K_{i}(G)/K_{i+2}(G) $ is a
noncentral normal subgroup of order $ p^{2} $ in ${G}/{K_{i+2}(G)}$
so $ M_{i}=C_{G}(K_{i}(G)/K_{i+2}(G)) $ is a maximal subgroup of $
G$. Indeed, by N/C-theorem, the quotient group $G\diagup M_{i}$ is
isomorphic to a subgroup of $\ Aut(K_{i}(G)\diagup K_{i+2}(G))$.
But, a p-sylow subgroup of $\ Aut(K_{i}(G)\diagup K_{i+2}(G))$ has order $p$. Thus, $
\left\vert G,M_{i}\right\vert =p$.

\begin{prop}
Let $G$ be a p-group of maximal class. Then $G_{1}$ is a
characteristic subgroup of $G$.
\end{prop}
\begin{proof} Let $\varphi\in Aut(G)$. The subgroup $G_{1}$ is composed of the
elements $x\in G$ such that $[x,G_{2}]\leq G_{4}$. The subgroups
$G_{2}$ and $G_{4}$ are characteristic in $G$, then we have that
$$[\varphi(x),G_{2}]=[\varphi(x),\varphi(G_{2})]=\varphi([x,G_{2}])\leq \varphi(G_{4})=G_{4}.$$
This proves that $G_{1}$ is characteristic in $G$.
\end{proof}

We shall write $G_{i}$ instead $K_{i}(G)$ when there is no possible
confusion. We have the following main properties:
\begin{enumerate}
\item $G\diagup G_{2}$ is elementary abelian of order $p^{2}$ and $\left[
G_{i},G_{i+1}\right] =p$ for $1\leq i\leq n-1.$ Hence $\left\vert
G:G_{i}\right\vert =p^{i}$ for $1\leq i\leq n.$

\item For $i\geq 2$, $G_{i}$ is the unique normal subgroup of index $p^{i}$.

\item $G_{i}=Z_{n-i}(G)$, $i=2,...,n$.
\end{enumerate}

\begin{rem}\label{Fund}\
\begin{enumerate}
  \item If $N$ is a normal subgroup of $G$ such that $|G/N|\geq p^{4}$, it
is clear from the definition that $(G/N)_{1}=G_{1}/N$.
  \item Let $G$ be a p-group of maximal class, $|G|> p^{4}$. Let $M$ be a maximal subgroup of $G$ and let $M_{1}$ be the
fundamental subgroup of $M$. Then $|G:M_{1}|=p^{2}$ and $M_{1}\lhd
G$ so $M_{1}=\Phi(G)<G_{1}$, and we get $M_{1}=G_{1}\cap M$.
\end{enumerate}
\end{rem}

\begin{lem}\cite[Theorem 9.6(e)]{Ber1}\label{Fund1}
Let $G$ be a group of maximal class and order $p^{m}$, $p > 2$, $m >
p+1$. Then the set of all maximal subgroups of $G$ is
$$\Gamma_{1}=\{M_{1}=G_{1}; M_{2}; ... ; M_{p+1}\}$$
Where $G_{1}$ is the fundamental subgroup of $G$, and the subgroups
$M_{2}; ... ; M_{p+1}$ are of maximal class.
\end{lem}

\begin{prop}
Let a p-group $G$ of maximal class have order $p^{m} > p^{p+1}$. If
$H<G$ is of order $> p^{p}$ and $H\nleq G_{1}$, then $H$ is of
maximal class.
\end{prop}
\begin{proof} We proceed by induction on $m$. If $m=p+ 2$, the result follows
from Lemma \ref{Fund1} (indeed, then all members of the set
$\Gamma_{1}-{G_{1}}$ are of maximal class). Now let $m > p + 2$ and
let $H\leq M\in \Gamma_{1}$, then $M$ is of maximal class (Lemma
\ref{Fund1}). The subgroup $M_{1}=M\cap G_{1}$ is the fundamental
subgroup of $M$ (Remark \ref{Fund}(2)). As $H\nleq M_{1}$, then $H$
is of maximal class, by induction, applied to the pair $H<M$.
\end{proof}

\begin{prop}
Let $G$ be of maximal class and order $>p^{p+1}$. If $H<G$ is of
order $>p^{2}$, then either $H\leq G_{1}$ or $H$ is of maximal.
class.
\end{prop}
\begin{proof} Let $|H|=p^{k}$. The result is known if $k=3$ \cite{Bla1}.
Assuming that $k>3$, we use induction on $k$. Then all maximal
subgroups of $H$ which $\neq H\cap G_{1}$ are of maximal class, by
induction. Then the set $\Gamma_{1}(H)$ contains exactly
$|\Gamma_{1}(H)|-1\neq 0(mod p^{2})$ members of maximal class so $H$
is of maximal class, by \cite[Theorem 12.12(c)]{Ber1}.
\end{proof}

\begin{exer}\cite[Exercise 9.28]{Ber1}
Let $G$ be a p-group of maximal class and order $>p^{p+1}$. Show the
following:
\begin{enumerate}
  \item $exp(G_{1})=exp(G)$.
  \item If $x\in G$ is of order $\geq p^{3}$, then $x\in G_{1}$.
\end{enumerate}
\end{exer}
{"config": "arxiv", "file": "2006.15737.tex"}