# Hopf Galois structures, skew braces for groups of size $p^{n}q$: The cyclic
Sylow subgroup case

Namrata Arvind<EMAIL_ADDRESS>The Institute of Mathematical Sciences, 4th
Cross St, CIT Campus, Tharamani, Chennai, Tamil Nadu 600113, India  and
Saikat Panja<EMAIL_ADDRESS>Harish-Chandra Research Institute- Main
Building, Chhatnag Road, Jhusi, Uttar Pradesh 211019, India

###### Abstract.

Let $n\geq 1$ be an integer, $p$, $q$ be distinct odd primes. Let ${G}$, $N$
be two groups of order $p^{n}q$ with their Sylow-$p$-subgroups being cyclic.
We enumerate the Hopf-Galois structures on a Galois ${G}$-extension, with type
$N$. This also computes the number of skew braces with additive group
isomorphic to $G$ and multiplicative group isomorphic to $N$. Further when
$q<p$, we give a complete classification of the Hopf-Galois structures on
Galois-$G$-extensions.

###### Key words and phrases:

Hopf-Galois structures; Field extensions; Holomorph

###### 2020 Mathematics Subject Classification:

12F10, 16T05.

The first named author is partially supported by the IMSc postdoctoral
fellowship and the second author has been partially supported by HRI
postdoctoral fellowship.

##  1\. Introduction

The study of Hopf-Galois structures comes under the realm of group theory and
number theory. These structures were first studied by S. Chase and M. Sweedler
in $1969$, in their work [10]. Subsequently in [12], C. Greither and B.
Pareigis defined a Hopf-Galois structure for separable extensions. In recent
times, algebraic objects called Skew braces were introduced in the PhD thesis
of D. Bachiller. They have been studied by various mathematicians like W.
Rump, D. Bachiller, F. Cedo in [9], [3] _etc._. Skew braces are known to give
set-theoretic solutions to the Yang-Baxter equations. Subsequently, A.
Smoktunowicz and L. Vendramin noticed a connection between the study of skew
braces and that of Hopf-Galois structures in their work [15]. For more
details, and background on this topic, and the interplay between skew braces
and Hopf-Galois structures, one can refer to the book [11] of L. N. Childs and
the Ph.D. thesis of K. N. Zenouz [22].

A Hopf-Galois structure on a finite field extension is defined in the
following way. Assume $K/F$ is a finite Galois field extension. An $F$-Hopf
algebra $\mathcal{H}$, with an action on $K$ such that $K$ is an $H$-module
algebra and the action makes $K$ into an $\mathcal{H}$-Galois extension, will
be called a Hopf-Galois structure on $K/F$.

A _left skew brace_ is a triple $(\Gamma,+,\times)$ where
$(\Gamma,+),(\Gamma,\times)$ are groups and satisfy $a\times(b+c)=(a\times
b)+a^{-1}+(a\times c),$ for all $a,b,c\in\Gamma$.

Given a group $G$, the $Holomorph$ of $G$ is defined as
$G\rtimes\textup{Aut}(G)$, via the identity map. It is denoted by
$\textup{Hol}(G)$. Let $G$ and $N$ be two finite groups of the same order. By
$e(G,N)$ we mean the number of Hopf-Galois structures on a finite Galois field
extension $L/K$ with Galois group isomorphic to $G$, and the type isomorphic
to $N$. In [12], the authors gave a bijection between Hopf-Galois structures
on a finite Galois extension with Galois group $G$ and regular subgroups in
$\mathrm{Perm}(G)$, which are normalised by $\lambda(G)$. Further in [7], N.
Byott showed that

$e(G,N)=\dfrac{|\textup{Aut}(G)|}{|\textup{Aut}(N)|}\cdot e^{\prime}(G,N),$
(1.1)

where $e^{\prime}(G,N)$ is the number of regular subgroups of
$\textup{Hol}(N)$ isomorphic to $G$. Here a subgroup $\Gamma$ of
$\textup{Hol}(N)$ is called regular if it has exactly one element
$(e_{G},\zeta)\in\Gamma$ with $\zeta=I$, the identity automorphism. We will
use this condition to check regular embeddings of the concerned groups in the
article. It turns out that $e^{\prime}(G,N)$ also gives the number of Skew-
Braces with the additive group isomorphic to $N$ and the multiplicative group
isomorphic to $G$. The number $e(G,N)$ has been computed for several groups.
For example, N. Byott determined $e(G,N)$ when $G$ is isomorphic to a cyclic
group [6]; C. Tsang determined $e(G,N)$ when $G$ is a quasisimple group [17];
N. K. Zenouz consider the groups of order $p^{3}$ [22] to determine $e(G,N)$ ;
T. Kohl determined $e(G,N)$ when $G$ is a dihedral group [13].

Previously in [2], the authors computed $e(G,N)$ whenever $G$ and $N$ are
isomorphic to $\mathbb{Z}_{n}\rtimes\mathbb{Z}_{2}$, where $n$ is odd and its
radical is a Burnside number. Groups of order $p^{2}q$ with cyclic Sylow
subgroups have been considered in [8]. We can show that any group of order
$p^{n}q$ with cyclic Sylow subgroups, when $p$ and $q$ are distinct primes, is
a semidirect product of two cyclic groups (see Section 2). In this article, we
compute $e(G,N)$ (and $e^{\prime}(G,N)$), whenever $G$ and $N$ are groups of
order $p^{n}q$ with cyclic Sylow-$p$ subgroup, where $p$ and $q$ are distinct
odd primes. We do this by looking at the number of regular subgroups of
$\textup{Hol}(N)$ which are isomorphic to $G$. Finally whenever $q<p$ we give
a necessary and sufficient condition on when the pair $(G,N)$ is realizable.

We now fix some notations. For a ring $R$, we will use $R^{\times}$ to denote
the set of multiplicative units of $R$. For a group $G$, the identity element
will be sometimes denoted by $e_{G}$ and mostly by $1$, when the context is
clear. The automorphism group of a group $G$ will be denoted by
$\textup{Aut}(G)$, and the holomorph $G\rtimes_{\mathrm{id}}\textup{Aut}(G)$
will be denoted by $\textup{Hol}(G)$. The binomial coefficients will be
denoted by ${l\choose m}$. The Euler totient function will be denoted by
$\varphi$. We will use $\mathbb{Z}_{m}$ to denote the cyclic group of order
$m.$ We will use $\mathbb{Z}_{m}$ as a group as well as a ring, which will be
clear from the context. Now, we state the two main results of this article. To
state the second result we use notations from Section 2.

######  Theorem 1.1.

Let $p>q$ be odd primes and $q|p-1$. Let $G$ denote the nonabelian group of
the form $\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$ and $C$ denote the cyclic
group of order $p^{n}q$. Then the following are true:

  1. (1)

$e^{\prime}(G,G)=e(G,G)=2+2p^{n}(q-2)$,

  2. (2)

$e^{\prime}(G,C)=q-1$, and $e(G,C)=p^{n}$,

  3. (3)

$e^{\prime}(C,G)=p^{2n-1}$, and $e(C,G)=2p^{n-1}(q-1)$.

######  Theorem 1.2.

Let $p<q$ be odd primes and $p^{a}||q-1$. For $1\leq
b\leq\mathrm{min}\\{n,a\\}$, let $G_{b}$ denote the unique nonabelian group of
the form $\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$ determined by $b$, and $C$
denote the cyclic group of $p^{n}q$. Then the following results hold;

  1. (1)

$e^{\prime}(G_{b},G_{b})=e(G_{b},G_{b})=2\left(p^{n-b}+q\left(\varphi(p^{n})-p^{n-b}\right)\right)$,

  2. (2)

$e^{\prime}(G_{b_{1}},G_{b_{2}})=2qp^{n+b_{1}-b_{2}-1}(p-1)$, and
$e(G_{b_{1}},G_{b_{2}})=2qp^{n-1}(p-1)$ for $b_{1}\neq b_{2}$,

  3. (3)

$e^{\prime}(C,G_{b})=2p^{n-b}q$, and $e(C,G_{b})=2(p-1)p^{n-1}$,

  4. (4)

$e^{\prime}(G_{b},C)=p^{n+b-2}(p-1)$, and $e(G-b,C)=p^{n-1}b$.

The rest of the article is organised as follows. In Section 2, we give a
detailed description of the groups under consideration and determine their
automorphism groups. Next, in Section 3 and Section 4 we will prove Theorem
1.1 and Theorem 1.2 respectively. Lastly, in Section 5 we discuss the
realizability problem and solve them for some of the groups mentioned in this
article.

##  2\. Preliminaries

###  2.1. The groups under consideration

In this subsection we will describe the groups under consideration and fix
some notations. Let $p$ and $q$ be distinct odd primes. We look at groups of
order $p^{n}q$ whose Sylow-$p$-subgroups are cyclic. These come under two
families, depending on whether $p>q$ or $p<q$.

In case $p>q$, the groups are isomorphic to
$\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$, since $\mathbb{Z}_{p^{n}}$ is
normal. Indeed all these groups $G$ fits into the short exact sequence
$1\longrightarrow\mathbb{Z}_{p^{n}}\longrightarrow
G\longrightarrow\mathbb{Z}_{q}\longrightarrow 1$. Thus by the well known
Schur–Zassenhaus theorem $G$ is isomorphic to
$\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$. Since
$\textup{Aut}(\mathbb{Z}_{p^{n}})$ is cyclic, the semidirect product is either
trivial (in this case the group is cyclic) or uniquely nontrivial. Let
$G\cong\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$. If $q\nmid p-1$ then $G$ is
cyclic. In case $q|p-1$, let
$\phi:\mathbb{Z}_{q}\rightarrow\textup{Aut}(\mathbb{Z}_{p^{n}})$ be the
homomorphism defined as $\phi(1)=k$. Here $k$ is an element of
$\textup{Aut}(\mathbb{Z}_{p^{n}})$ of order $q$. Hereafter, we denote
$\mathbb{Z}_{p^{n}}\rtimes_{\phi}\mathbb{Z}_{q}$ by
$\mathbb{Z}_{p^{n}}\rtimes_{k}\mathbb{Z}_{q}$. Let

$\\{x,y|x^{p^{n}}=y^{q}=1,yxy^{-1}=x^{k}\\}$

be a presentation of $\mathbb{Z}_{p^{n}}\rtimes_{k}\mathbb{Z}_{q}$. Note that
since $e(G,G)$ is already known whenever $G$ is cyclic, we will assume $q|p-1$
for our calculations.

Now if $p<q$ we need to use a result of W. Burnside from [5], which states
that for a finite group $G$, all the Sylow subgroups are cyclic if and only if
$G$ is a semidirect product of two cyclic groups of coprime order. Applying
this to our situation, we get that $G$ is either a cyclic group or a non-
trivial semidirect product of the form
$\mathbb{Z}_{q}\rtimes\mathbb{Z}_{p^{n}}$. Next, we elaborate on different
possible semidirect products of the form
$\mathbb{Z}_{q}\rtimes\mathbb{Z}_{p^{n}}$ . Once again in this case we assume
that $p|q-1$. Let $p^{a}||q-1$ and for $b\leq\mathrm{min}\\{n,a\\}$ fix
$\psi_{b}:\mathbb{Z}_{p^{n}}\longrightarrow\textup{Aut}(\mathbb{Z}_{q})$ to be
a homomorphism, such that $|\mathrm{Im}~{}\psi_{b}|=p^{b}$. Take
${G}_{b}=\mathbb{Z}_{q}\rtimes_{\psi_{b}}\mathbb{Z}_{p^{n}}$. The group
${G}_{b}$ is unique up to isomorphism. The presentation of this group can be
taken to be

$\langle x,y|x^{p^{n}}=1,y^{q}=1,xyx^{-1}=y^{k}\rangle,$

where $k$ is an element of order $p^{b}$ in
$\textup{Aut}(\mathbb{Z}_{q})=\mathbb{Z}_{q}^{\times}$. From now on we denote
$\mathbb{Z}_{q}\rtimes_{\psi_{b}}\mathbb{Z}_{p^{n}}$ by
$\mathbb{Z}_{q}\rtimes_{k}\mathbb{Z}_{p^{n}}$.

###  2.2. The basic lemmas

In this subsection we note down the basic group-theoretic results, which will
be used throughout the article.

######  Lemma 2.1.

Let $p$ be a positive odd integer. Take $a=bp^{c}$ where $p\nmid b$. Then we
have that $(1+p)^{a}\equiv 1+dp^{c+1}\pmod{p^{c+2}}$ for some $p\nmid d$, for
all integer $c\geq 0$.

###### Proof.

We prove it by induction on $c$. If $c=0$, then
$(1+p)^{a}=1+ap+a^{\prime}p^{2}$, for some $a^{\prime}\in\mathbb{Z}$. Hence
$(1+p)^{a}\equiv 1+ap\pmod{p^{2}}$ with $d=a$. Next, assume it to be true for
all $l\leq c$ and in particular for $l=c$. Hence
$(1+p)^{bp^{c}}=1+dp^{c+1}+d^{\prime}p^{c+2}$ for some
$d^{\prime}\in\mathbb{Z}$. Then we have

$\displaystyle(1+p)^{bp^{c+1}}=\left(1+dp^{c+1}+d^{\prime}p^{c+2}\right)^{p}=(1+d^{\prime\prime}p^{c+1})^{p},$

for some $d^{\prime\prime}\in Z$ and $(d^{\prime\prime},p)=1$. Hence it
follows that $(1+p)^{bp^{c+1}}\equiv 1+d^{\prime\prime}p^{c+2}\pmod{p^{c+3}}$,
which also finishes the induction, and hence the proof. ∎

######  Lemma 2.2.

Let $G$ be the non-abelian group isomorphic to
$\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$. We have
$\textup{Aut}(G)\cong\textup{Hol}(\mathbb{Z}_{p^{n}})$.

###### Proof.

We first embed $G$ as a normal subgroup of $\textup{Hol}(\mathbb{Z}_{p^{n}})$.
Take the homomorphism $\psi$ defined as

$\displaystyle\psi(x)$ $\displaystyle=\begin{pmatrix}1&k\\\
0&1\end{pmatrix},~{}\psi(y)=\begin{pmatrix}k&1\\\ 0&1\end{pmatrix}.$

This embedding can be shown to be injective. Now consider the following map

$\displaystyle\Phi:\textup{Hol}(\mathbb{Z}_{p^{n}})\longrightarrow\textup{Aut}(G)\text{
defined as }\Phi(z)(w)=zwz^{-1}$

for all $z\in\textup{Hol}(\mathbb{Z}_{p^{n}})$ and $y\in G$ is an injective
group homomorphism, since $\ker\Phi$ consists only of the identity matrix.
From [21, Theorem B] we have
$|\textup{Aut}(G)|=|\textup{Hol}(\mathbb{Z}_{p^{n}})|$. Thus $\Phi$ is an
isomorphism. ∎

######  Lemma 2.3.

Let $p,q$ be primes such that $(p,q)=1$ and $q|p-1$. Let $k$ be a
multiplicative unit in $\mathbb{Z}_{p^{n}}$, of multiplicative order $q$. Then
$k-1$ is a multiplicative unit in $\mathbb{Z}_{p^{n}}$.

###### Proof.

Suppose $k-1$ is not a unit in $\mathbb{Z}_{p^{n}}$, then $k-1=mp$ for some
$m\in\mathbb{Z}_{p^{n}}$. Since $(k)^{q}\equiv 1\pmod{p^{n}}$, we get

$\displaystyle(mp+1)^{q}\equiv 1+{q\choose 1}mp+{q\choose
2}(mp)^{2}+\cdots+(mp)^{q}\equiv 1\pmod{p^{n}},$

which in turn implies that

$\displaystyle mp\cdot\left(q+{q\choose 2}mp+{q\choose
3}(mp)^{2}+\cdots+(mp)^{q-1}\right)\equiv 0\pmod{p^{n}}.$

We note that

$\displaystyle t=q+{q\choose 2}mp+{q\choose 3}(mp)^{2}+\cdots+(mp)^{q-1}$

is a unit since $q$ is a unit and $t-q$ is a nilpotent element. Thus $mp\equiv
0\pmod{p^{n}},$ which implies $k-1\equiv 0\pmod{p^{n}}$. This is a
contradiction since $k$ is an element of order $q$. ∎

######  Lemma 2.4.

Let $G_{b}\cong\mathbb{Z}_{q}\rtimes_{k}\mathbb{Z}_{p^{n}}$, where $k$ is an
element of order $p^{b}$ in $\mathbb{Z}_{q}^{\times}$. Assume $p|q-1$, then
for $b>0$, we have that
$\textup{Aut}({G}_{b})\cong\mathbb{Z}_{p^{n-b}}\times\textup{Hol}(\mathbb{Z}_{q})$.

###### Proof.

The proof will be divided into two steps. First, we calculate the size of the
automorphism group. In the next step, we will determine the group’s
description in terms of generators and relations, from which the result will
follow.

Let us take an automorphism $\Psi$ of ${G}_{b}$. Since an automorphism is
determined by its value on the generator, assume that
$\Psi(x)=y^{\alpha}x^{\gamma}$ and $\Psi(y)=y^{\beta}x^{\delta}$, where
$0\leq\alpha,\beta\leq q-1$ and $0\leq\gamma,\delta\leq p^{n}-1$. Note that we
have
$\Psi(y)^{q}=y^{\beta(1+k^{\delta}+k^{2\delta}+k^{(q-1)\delta})}x^{q\delta}$.
Since $\Psi(y)^{q}=1$, we must have $\delta=0$. Thus $\beta$ should be a unit
in $\mathbb{Z}_{q}$. Now consider the equation
$\Psi(x)\Psi(y)=\Psi(y)^{k}\Psi(x)$. This imposes the condition that
$y^{\alpha+\beta k^{\gamma}}x^{\gamma}=y^{\beta k+\alpha}x^{\gamma}$. Hence we
should have $\beta k^{\gamma}\equiv\beta k\pmod{q}$, whence
$k^{\gamma-1}\equiv 1\pmod{q}$, as $\beta$ is a unit in $\mathbb{Z}_{q}$.
Since $k$ is an element of order $p^{b}$, we get that
$\gamma\in\\{Rp^{b}+1:0\leq R<p^{n-b}\\}$. Next considering the equation
$\Psi(x)^{p^{n}}=1$, we have that
$y^{\alpha(1+k^{\gamma}+k^{2\gamma}+\ldots+k^{(p^{n}-1)\gamma})}x^{p^{n}\gamma}=1$.
Since $x^{p^{n}\gamma}=1$, we have that
$\alpha(1+k^{\gamma}+k^{2\gamma}+\ldots+k^{(p^{n}-1)\gamma})=0\pmod{q}$.
Regardless of the value of $k$, any $0\leq\alpha\leq q$ satisfies the last
congruence. Hence the group is of order $p^{n-b}q(q-1)$.

Hereafter we denote $\Psi$ by $(\gamma,\beta,\alpha)$. Consider the following
three elements of the group given by

$\displaystyle\Psi_{1}=\left((1+p)^{p^{b-1}},1,0\right),\Psi_{2}=\left(1,t,0\right),\Psi_{3}=(1,1,1),$

where $1\leq t\leq q-1$ satisfies
$\mathbb{Z}_{q}^{\times}=\langle\overline{t}\rangle$. Since
$\overline{(1+p)}\in\mathbb{Z}_{p^{n}}^{\times}$ is of order $p^{n-1}$, we get
that $\Psi_{1}$ is an element of order $p^{n-b}$. Given that, $\overline{t}$
is an element of order $q-1$, the element $\Psi_{2}$ is of order $q-1$.
Lastly, $\Psi_{3}$ is an element of order $q$. Note that
$\Psi_{1}\Psi_{2}=\Psi_{2}\Psi_{1}$, follows from an easy calculation. Now,
$\Psi_{1}\Psi_{3}(x)=yx^{(1+p)^{p^{b-1}}}$. Further, we have,

$\displaystyle\Psi_{3}\Psi_{1}(x)=(yx)^{(1+p)^{p^{b-1}}}=y^{1+k+\ldots+k^{\left(1+p\right)^{p^{b-1}}}}x^{(1+p)^{p^{b-1}}}=y^{1+\left(\frac{k^{(1+p)^{p^{b-1}}-1}}{k-1}\right)}x^{k^{(1+p)^{p^{b-1}}-1}}.$

Since ${(1+p)^{p^{b-1}}-1}\equiv 1\pmod{p^{b}}$ and $\overline{k-1}$ is a unit
in $\mathbb{Z}_{q}$, we conclude that
$\Psi_{1}\Psi_{3}(x)=\Psi_{3}\Psi_{1}(x)$. Since
$\Psi_{1}\Psi_{3}(y)=\Psi_{3}\Psi_{1}(y)$, we conclude that
$\Psi_{1}\Psi_{3}=\Psi_{3}\Psi_{1}$. We now take the subgroup generated by
$\Psi_{2}$ and $\Psi_{3}$. In this group $\langle\Psi_{3}\rangle$ is normal as
$\Psi_{2}\Psi_{3}\Psi_{2}^{-1}=\Psi_{3}^{t}\in\langle\Psi_{3}\rangle$. Also
$\langle\Psi_{2}\rangle\cap\langle\Psi_{3}\rangle$ contains only identity.
Hence $|\langle\Psi_{2},\Psi_{3}\rangle|=q(q-1)$. Take the map
$T:\langle\Psi_{2},\Psi_{3}\rangle\longrightarrow\textup{Hol}(\mathbb{Z}_{q})$,
defined as

$\displaystyle T(\Psi_{2})=\begin{pmatrix}t&0\\\ 0&1\end{pmatrix}\text{ and
}T(\Psi_{3})=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}.$

This determines a homomorphism since
$T(\Psi_{2})T(\Psi_{3})T(\Psi_{2})^{-1}=T(\Psi_{3})^{t}$. For any
$\begin{pmatrix}u&v\\\ 0&1\end{pmatrix}\in\textup{Hol}(\mathbb{Z}_{q})$, we
have that $T(\Psi_{2}^{w_{1}}\Psi_{3}^{w_{2}})=\begin{pmatrix}u&v\\\
0&1\end{pmatrix}$, where $w_{1}$ satisfies $t^{w_{1}}=u$ and $w_{2}=v/u$.
Since the order of the groups are the same, we conclude that
$\langle\Psi_{2},\Psi_{3}\rangle\cong\textup{Hol}(\mathbb{Z}_{q})$. Now we
will show that $\langle\Psi_{1}\rangle\cap\langle\Psi_{2},\Psi_{3}\rangle$ has
only the identity element. Indeed, if $\Psi_{1}^{d}=\Psi_{2}^{e}\Psi_{3}^{f}$
(for some $0\leq d<p^{n-b}$, $0\leq e<q-1$ and $0\leq f<q$), then $e=0$,
comparing the evaluation of both the functions at $y$. Finally, if we consider
$\Psi_{1}^{d}(x)=\Psi_{3}^{f}(x)$, we get that $x^{p^{\prime d}}=y^{f}x$ where
$p^{\prime}=(1+p)^{p^{b-1}}$. This forces us to have $f=0$, consequently
$d=0$. Thus
$\langle\Psi_{1},\Psi_{2},\Psi_{3}\rangle\cong\langle\Psi_{1}\rangle\times\langle\Psi_{2},\Psi_{3}\rangle$
and is of order $p^{n-b}q(q-1)$. Hence we have proved that
$\textup{Aut}({G}_{b})\cong\mathbb{Z}_{p^{n-b}}\times\textup{Hol}(\mathbb{Z}_{q})$.
∎

We denote the elements of $\textup{Aut}(G_{b})$ by
$\left(\gamma,\begin{pmatrix}\beta&\alpha\\\
0&1\end{pmatrix}\right)\in\mathbb{Z}_{p^{n}}^{\times}\times\textup{Hol}(\mathbb{Z}_{q})$,
such that $\gamma^{p^{n-b}}=1$.

######  Remark 2.5.

We note down the action of the automorphism group of ${G}_{b}$ on the group
${G}_{b}$, by means of generators. This will be necessary for counting the
Hopf-Galois structures concerning ${G}_{b}$’s. For $b>0$, the action is as
follows.

$\displaystyle\left(\gamma,\begin{pmatrix}\beta&\alpha\\\
0&1\end{pmatrix}\right)\cdot x=y^{\alpha}x^{\gamma}\text{ and,
}\left(\gamma,\begin{pmatrix}\beta&\alpha\\\ 0&1\end{pmatrix}\right)\cdot
y=y^{\beta}.$

######  Remark 2.6.

For $b=0$, the group ${G}_{b}\cong\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{q}$.
Since $(p,q)=1$ and both are abelian groups, it follows from [4, Theorem 3.2]
that
$\textup{Aut}({G}_{b})\cong\mathbb{Z}_{p^{n-1}(p-1)}\times\mathbb{Z}_{q-1}$ in
this case. The action is defined to be component-wise.

##  3\. The case $p>q$

This section is devoted to the proof of Theorem 1.1. As discussed in Section
2, upto isomorphism there are precisely two groups of order $p^{n}q$ whenever
their Sylow subgroups are cyclic. Counting the number of skew braces with
multiplicative group $G$ and additive group $N$ is equivalent to (up to
multiplication by a constant; see [2, Proof of Proposition 3.2]) counting the
number of regular embedding of $G$ in $\textup{Hol}(N)$. Then using Eq. 1.1,
we are able to conclude about the number of Hopf-Galois structures on
$G$-extensions of type $N$. We will use the regularity criterion given in
Section 1. This section will be divided into three subsections, depending on
the isomorphism types of $G$ and $N$. From Lemma 2.2, we have that
$\textup{Aut}(\mathbb{Z}_{p^{n}}\rtimes_{k}\mathbb{Z}_{q})\cong\textup{Hol}(\mathbb{Z}_{p^{n}})$,
where the action is given by,

$\displaystyle\begin{pmatrix}\beta&\alpha\\\ 0&1\end{pmatrix}\cdot
x^{i}y^{j}=x^{\beta i+\alpha k^{-1}-\alpha k^{j-1}}y^{j}.$

###  3.1. Embedding of $\mathbb{Z}_{p^{n}}\rtimes_{k}\mathbb{Z}_{q}$ into
$\textup{Hol}(\mathbb{Z}_{p^{n}}\rtimes_{k}\mathbb{Z}_{q})$

Let
$\Phi:\mathbb{Z}_{p^{n}}\rtimes_{k}\mathbb{Z}_{q}\longrightarrow\textup{Hol}(\mathbb{Z}_{p^{n}}\rtimes_{k}\mathbb{Z}_{q})$
be a regular embedding. Let

$\displaystyle\Phi(x)=\left(x^{i_{1}}y^{j_{1}},\begin{pmatrix}\beta_{1}&\alpha_{1}\\\
0&1\end{pmatrix}\right),\Phi(y)=\left(x^{i_{2}}y^{j_{2}},\begin{pmatrix}\beta_{2}&\alpha_{2}\\\
0&1\end{pmatrix}\right).$

From $(\Phi(x))^{p^{n}}=1$ we get

$\displaystyle j_{1}\equiv 0\pmod{q},$ (3.1)

since $p^{n}j_{1}\equiv 0\pmod{q}$ and $(p,q)=1$,

$\displaystyle\beta_{1}^{p^{n}}\equiv 1$ $\displaystyle\pmod{p^{n}},$ (3.2)
$\displaystyle
i_{1}(1+\beta_{1}+\beta_{1}^{2}+\ldots+\beta_{1}^{p^{n}-1})\equiv 0$
$\displaystyle\pmod{p^{n}},$ (3.3)
$\displaystyle\alpha_{1}(1+\beta_{1}+\beta_{1}^{2}+\ldots+\beta_{1}^{p^{n}-1})\equiv
0$ $\displaystyle\pmod{p^{n}}.$ (3.4)

Similarly from $\Phi(yxy^{-1})=\Phi(x^{k})$ we get

$\displaystyle\beta_{1}^{k-1}$ $\displaystyle\equiv 1\pmod{p^{n}},$ (3.5)

which implies $\beta_{1}=1$ from Eq. 3.2, Eq. 3.5 and using Lemma 2.3;
furthermore,

$\displaystyle\beta_{2}\alpha_{1}+\alpha_{2}$
$\displaystyle\equiv\beta_{1}^{k}\alpha_{2}+\alpha_{1}\pmod{p^{n}},$ (3.6)
$\displaystyle ki_{1}\left(k^{j_{2}-1}\beta_{2}-1\right)$
$\displaystyle\equiv\alpha_{1}\left(1-k^{j_{2}}\right)\pmod{p^{n}}.$ (3.7)

Further taking $\beta_{1}=1$ in Eq. 3.6 and Eq. 3.7 we get that,

$\displaystyle\alpha_{1}\cdot(k-\beta_{2})\equiv 0\pmod{p^{n}},$ (3.8)
$\displaystyle
ki_{1}\cdot(k^{j_{2}-1}\beta_{2}-1)\equiv\alpha_{1}\cdot(1-k^{j_{2}})\pmod{p^{n}}.$
(3.9)

We note that in general,

$\displaystyle\Phi(y)^{k}=\left(x^{\ell_{k}}y^{kj_{2}},\begin{pmatrix}\beta_{2}^{k}&\alpha_{2}(1+\beta_{2}+\beta_{2}^{2}+\cdots+\beta_{2}^{k-1})\\\
&1\end{pmatrix}\right),$

where

$\displaystyle\ell_{k}=i_{2}\left(\sum\limits_{t=0}^{k-1}\left(\beta_{2}k^{j_{2}}\right)^{t}\right)+\left(\alpha_{2}k^{j_{2}-1}-\alpha_{2}k^{2j_{2}-1}\right)\left(1+\sum\limits_{u=1}^{k-2}\left(\sum\limits_{v=0}^{u}\beta_{2}^{v}\right)k^{uj_{2}}\right).$
(3.10)

Using $\Phi(y)^{q}=1$ we get

$\displaystyle\beta_{2}^{q}$ $\displaystyle\equiv 1\pmod{p^{n}},$ (3.11)
$\displaystyle\alpha_{2}(1+\beta_{2}+\beta_{2}^{2}+\ldots+\beta_{2}^{q-1})$
$\displaystyle\equiv 0\pmod{p^{n}}$ $\displaystyle j_{2}\neq 0,$ (3.12)
$\displaystyle\ell_{q}$ $\displaystyle\equiv 0\pmod{p^{n}}.$ (3.13)

From Eq. 3.11 we get $\beta_{2}=k^{a}$, for some $0\leq a\leq q-1$, since
$\mathbb{Z}_{p^{n}}^{*}$ has a unique subgroup of order $q$ and is generated
by $k$. First let us show that, in any regular embedding $j_{2}\neq 0$. If
possible let $j_{2}=0$. Then we get that $\beta_{2}=k$. This forces that for
any $0\leq\omega_{1}\leq p^{n}-1$ and $0\leq\omega_{2}\leq q-1$

$\displaystyle\Phi(x)^{\omega_{1}}\Phi(y)^{\omega_{2}}=\left(x^{\omega_{1}i_{1}+i_{2}\left(1+k+\cdots+k^{\omega_{2}-1}\right)},\begin{pmatrix}k^{\omega_{2}}&\star\\\
0&1\end{pmatrix}\right).$ (3.14)

Since $i_{1}$ is a unit, making a suitable choice of $\omega_{1}$ and
$\omega_{2}$ we get that this embedding will not be regular. Indeed note that
$1-k$ and $1-k^{\omega_{2}}$ are both units and so is
$1+k+\cdots+k^{\omega_{2}-1}$. We now divide the possibilities of $a$ into $3$
cases.

####  3.1.1. Case I: $a=0$

Using Eq. 3.7 and Eq. 3.8, we conclude that $\alpha_{1}\equiv 0\pmod{p^{n}}$,
$j_{2}\equiv 1\pmod{q}$ and, $\alpha_{2}\equiv 0\pmod{p^{n}}$. Since $i_{1}$
is a unit in $\mathbb{Z}_{p^{n}}$ and $i_{2}\in\mathbb{Z}_{p^{n}}$ can take
any value, the total number of embeddings in this case is given by
$p^{n}\varphi(p^{n})$. Moreover, all of these embeddings are regular. We
remark that all the above embedding corresponds to the canonical Hopf-Galois
structure.

####  3.1.2. Case II: $a=1$

Note that using Eq. 3.9 we get that $ki_{1}\equiv-\alpha_{1}\pmod{p^{n}}$. We
deal with this in two subcases depending on the value of $j_{2}$. First, we
consider the case $j_{2}$ being equal to $q-1$. In this case using
$\ell_{q}=0$, we get that $i_{2}$ gets determined by the value of $\alpha_{2}$
since
$\left(\sum\limits_{t=0}^{k-1}\left(\beta_{2}k^{j_{2}}\right)^{t}\right)=q$ is
a unit in $\mathbb{Z}_{p^{n}}$. Hence the number of embedding in this subcase
is given by $p^{n}\varphi(p^{n})$.

For the other case, since the element $k^{j_{2}}(1-k^{a})$ is a unit and
$j_{2}+a\neq 0\pmod{q}$ we get

$\displaystyle
1+\sum\limits_{s=1}^{q-2}\left(\sum\limits_{t=0}^{s}k^{ta}\right)k^{sj_{2}}=\dfrac{1}{k^{j_{2}}(1-k^{a})}\left\\{\sum\limits_{t=1}^{q-1}\left(1-k^{ta}\right)k^{tj_{2}}\right\\}=\dfrac{1}{k^{j_{2}}(1-k^{a})}\cdot(1-1)=0,$

Thus $\Phi(y)^{q}=1$ does not impose any conditions on $i_{2}$ and
$\alpha_{2}$. Hence, in this subcase, the total number of possibilities is
$p^{2n}\varphi(p^{n})(q-2)$. Since $j_{2}\neq 0$, we conclude that all the
embeddings are regular.

####  3.1.3. Case III: $a\geq 2$

This conditions together with Eq. 3.8 and Eq. 3.9, imply that $\alpha_{1}=0$
and $j_{2}\equiv a-1\pmod{q}$. Since $a+j_{2}\not\equiv 0\pmod{q}$, a mutatis
mutandis of Case II gives that $i_{2}$ and $\alpha_{2}$ can be chosen
independently, whence each of them has $p^{n}$ possibilities. Thus, in this
case, the total number of possibilities is given by
$p^{2n}\varphi(p^{n})(q-2)$. Similar to the previous case, all the embeddings
are regular.

Summarizing the above cases we get the following result.

######  Lemma 3.1.

The total number of regular embeddings of
$\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$ inside
$\textup{Hol}(\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q})$ is given by
$2p^{n}\varphi(p^{n})+2p^{2n}\varphi(p^{n})(q-2)$.

######  Proposition 3.2.

Let ${G}$ be a non-abelian groups of the form
$\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$, where $p$ and $q$ are primes
satisfying $q|p-1$. Then $e({G},{G})$ is given by $2+2p^{n}(q-2)$.

###### Proof.

From Lemma 3.1 we get the total number of regular embeddings. Dividing this
number by the Automorphism of $G$ will give us the total number of Hopf-Galois
structures. ∎

###  3.2. Embedding of $G=\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$ in the
$\textup{Hol}(\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{q})$

Next we consider the case of regular embedding of
$G=\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$ in the
$\textup{Hol}(\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{q})$. Let us fix the
presentation of $C=\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{q}$ to be $\langle
r,s|r^{p^{n}}=s^{q}=1,rs=sr\rangle.$ Then it can be shown that
$\textup{Hol}(C)\equiv\textup{Hol}(\mathbb{Z}_{p^{n}})\times\textup{Hol}(\mathbb{Z}_{q})$.
We take a typical element of $\textup{Hol}(C)$ to be
$\left(\begin{pmatrix}b&a\\\ 0&1\end{pmatrix},\begin{pmatrix}d&c\\\
0&1\end{pmatrix}\right)$, where $a$, $c$ are elements of $\mathbb{Z}_{p^{n}}$,
$\mathbb{Z}_{q}$ respectively and $b$, $d$ are elements of
$\mathbb{Z}_{p^{n}}^{\times}$, $\mathbb{Z}_{q}^{\times}$ respectively.
Starting with an embedding $\Phi$ of $G$ inside $\textup{Hol}(C)$ and assuming
that

$\displaystyle\Phi(x)=\left(\begin{pmatrix}b_{1}&a_{1}\\\
0&1\end{pmatrix},\begin{pmatrix}d_{1}&c_{1}\\\
0&1\end{pmatrix}\right),\Phi(y)=\left(\begin{pmatrix}b_{2}&a_{2}\\\
0&1\end{pmatrix},\begin{pmatrix}d_{2}&c_{2}\\\ 0&1\end{pmatrix}\right).$

From $\Phi(x)^{p^{n}}=e_{\textup{Hol}(C)}$ we get the equations

$\displaystyle b_{1}^{p^{n}}$ $\displaystyle\equiv 1\pmod{p^{n}},$ (3.15)
$\displaystyle a_{1}\left(1+b_{1}+\cdots+b_{1}^{p^{n}-1}\right)$
$\displaystyle\equiv 0\pmod{p^{n}},$ (3.16) $\displaystyle d_{1}^{p^{n}}$
$\displaystyle\equiv 1\pmod{q},$ (3.17) $\displaystyle
c_{1}\left(1+d_{1}+\cdots+d_{1}^{p^{n}-1}\right)$ $\displaystyle\equiv
0\pmod{q}.$ (3.18)

Note that $d_{1}^{q-1}\equiv 1\pmod{q}$ and $(q-1,p^{n})=1$. Combining this
with Eq. 3.17, we get that $d_{1}=1$. Then plugging $d_{1}=1$ in Eq. 3.18,
conclude that $c_{1}=0$. For ensuring regularity, we need to take $a_{1}$ is a
unit in $\mathbb{Z}_{p^{n}}$. Using the equation $\Phi(y)^{q}=1$ we get the
equations

$\displaystyle b_{2}^{q}$ $\displaystyle\equiv 1\pmod{p^{n}},$ (3.19)
$\displaystyle a_{2}\left(1+b_{2}+\cdots+b_{2}^{q-1}\right)$
$\displaystyle\equiv 0\pmod{p^{n}},$ (3.20) $\displaystyle d_{2}^{q}$
$\displaystyle\equiv 1\pmod{q},$ (3.21) $\displaystyle
c_{2}\left(1+d_{2}+\cdots+d_{2}^{q-1}\right)$ $\displaystyle\equiv 0\pmod{q}.$
(3.22)

Since the order of $d_{2}$ divides $q-1$, we get $d_{2}=1$ from Eq. 3.21.
Finally comparing both sides of the equation
$\Phi(x)^{k}\Phi(y)=\Phi(y)\Phi(x)$ we get (using the conclusions of the
preceding discussions)

$\displaystyle b_{1}^{k-1}\equiv 1$ $\displaystyle\pmod{p^{n}}$ (3.23)
$\displaystyle b_{2}a_{1}+a_{2}\equiv
b_{1}^{k}a_{2}+a_{1}\left(1+b_{1}+\cdots+b_{1}^{k-1}\right)$
$\displaystyle\pmod{p^{n}}.$ (3.24)

Using Lemma 2.3, Eq. 3.15 and Eq. 3.23 we conclude that $b_{1}=1$. Putting the
value of $b_{1}$ in Eq. 3.24 we get that $b_{2}=k$. Further to ensure
regularity we need to impose $c_{2}\neq 0$ (using a similar argument in the
discussion after Eq. 3.14). Thus the total number of regular embeddings in
this case is given by $\varphi(p^{n})p^{n}(q-1)$.

######  Proposition 3.3.

Let ${C}$ be the cyclic group of order $p^{n}q$ and ${G}$ be the nonabelian
group isomorphic to $\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$, where $p$ and
$q$ are primes. Then $e({G},{C})=p^{n}$ and $e^{\prime}({G},{C})=q-1$.

###  3.3. Embedding of ${C}=\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{q}$ in the
$\textup{Hol}(\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q})$

Recall the description of $\textup{Hol}(G)$ from Section 3.1 and the
presentation for $C$ from Section 3.2. Consider a homomorphism
$\Phi:{C}\longrightarrow\textup{Hol}(G)$ determined by

$\displaystyle\Phi(r)=\left(x^{i_{1}}y^{j_{1}},\begin{pmatrix}\beta_{1}&\alpha_{1}\\\
0&1\end{pmatrix}\right),\Phi(s)=\left(x^{i_{2}}y^{j_{2}},\begin{pmatrix}\beta_{2}&\alpha_{2}\\\
0&1\end{pmatrix}\right).$

Given that $\Phi(r)$ has to be an element of order $p^{n}$ and the embedding
is regular, using a similar argument as in Section 3.1 we conclude that
$j_{1}=0$, $i_{1}$ is a unit in $\mathbb{Z}_{p_{n}}$ and, $j_{2}$ is a unit in
$\mathbb{Z}_{q}$. From $\Phi(r)^{p^{n}}=1$, we get that

$\displaystyle i_{1}\left(1+\beta_{1}+\cdots+\beta^{p^{n}-1}\right)$
$\displaystyle\equiv 0\pmod{p^{n}},$
$\displaystyle\alpha_{1}\left(1+\beta_{1}+\cdots+\beta^{p^{n}-1}\right)$
$\displaystyle\equiv 0\pmod{p^{n}},$ $\displaystyle\beta_{1}^{p^{n}}$
$\displaystyle\equiv 1\pmod{p^{n}}.$

From the last equation above and [2, Corollary 2.2] we get that
$\beta_{1}\equiv 1\pmod{p}$. Hence the first two equations will always be
satisfied irrespective of choices of $i_{1}$ and $\alpha_{1}$. From the
equation $\Phi(s)^{q}=1$, we get

$\displaystyle\beta_{2}^{q}$ $\displaystyle\equiv 1\pmod{p^{n}},$ (3.25)
$\displaystyle\alpha_{2}(1+\beta_{2}+\beta_{2}^{2}+\ldots+\beta_{2}^{q-1})$
$\displaystyle\equiv 0\pmod{p^{n}}$ , (3.26) $\displaystyle\ell_{q}$
$\displaystyle\equiv 0\pmod{p^{n}},$ (3.27)

where $\ell_{q}$ is as defined in Section 3.1. Furthermore
$\Phi(r)\Phi(s)=\Phi(s)\Phi(r)$ gives that

$\displaystyle\alpha_{2}(\beta_{1}-1)$
$\displaystyle\equiv\alpha_{1}(\beta_{2}-1),\pmod{p^{n}}$ (3.28)
$\displaystyle i_{1}+\beta_{1}i_{2}+\alpha_{1}k^{-1}\left(1-k^{j_{2}}\right)$
$\displaystyle\equiv i_{2}+k^{j_{2}}\beta_{2}i_{1}\pmod{p^{n}}.$ (3.29)

Let $\beta_{2}=k^{a}$ for some $a\geq 0$. We divide this into two cases $a=0$
and $a\neq 0$.

####  3.3.1. Case I: a=0

In this case we get $\alpha_{2}=0$ from Eq. 3.26. Hence Eq. 3.28 is always
satisfied. Note that Eq. 3.27 holds true, since $j_{2}+a\neq q$ by using
similar arguments as of Section 3.1. Putting $\beta_{2}=1$ in Eq. 3.29 we get
$\left(i_{1}+\alpha_{1}k^{-1}\right)\left(1-k^{j_{2}}\right)\equiv
i_{2}\left(1-\beta_{2}\right)\pmod{p^{n}}$. Hence the choice of $\alpha_{1}$
gets determined by those of $i_{1}$, $i_{2}$, $\beta_{1}$ and, $j_{2}$. Hence
the total number of embedding in this case becomes
$\varphi(p^{n})p^{2n-1}(q-1)$.

####  3.3.2. Case II: $a\neq 0$

From Eq. 3.28, substituting
$\alpha_{1}={\alpha_{2}(\beta_{1}-1)}{(k^{a}-1)^{-1}}$ in Eq. 3.29 we get

$\displaystyle
i_{1}\left(k^{a}-1\right)\left(1-k^{j_{2}+a}\right)\equiv\left(1-\beta_{1}\right)\left(i_{2}\left(k^{a}-1\right)+\alpha_{2}k^{-1}\left(1-k^{j_{2}}\right)\right)\pmod{p^{n}}.$
(3.30)

We claim that $j_{2}+a=q$. Indeed, if $j_{2}+a\neq q$, we have that the LHS of
Eq. 3.30 is a unit in $\mathbb{Z}_{p^{n}}$, whereas $(1-\beta_{1})$ is never a
unit (since $\beta_{1}\equiv 1\pmod{p^{n}}$). Next, putting $j_{2}+a=q$ in Eq.
3.30, the LHS becomes $0$. Substituting $j_{2}+a=q$ in Eq. 3.27 we get
$i_{2}\equiv-\alpha_{2}k^{-1}(1-k^{j_{2}})(k^{j_{2}}q^{-1})(1+(1+k^{a})k^{j_{2}}+\cdots$$+(1+k^{a}+\cdots+k^{(q-2)a})k^{(q-2)j_{2}})$
$\pmod{p^{n}}$. Further substituting this value of $i_{2}$ to Eq. 3.30, we get
that both sides of the equation become zero. Hence we get that in this case,
the total number of regular embedding of ${C}$ in $\textup{Hol}(G)$ is given
by $\varphi(p^{n})p^{2n-1}(q-1)$.

######  Proposition 3.4.

Let ${C}$ be the cyclic group of order $p^{n}q$ and ${G}$ be the nonabelian
group isomorphic to $\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$. Then
$e({C},{G})=2p^{n-1}(q-1)$ and $e^{\prime}({C},{G})=p^{2n-1}$.

Now Theorem 1.1 follows from Proposition 3.2, Proposition 3.3, and Proposition
3.4.

##  4\. The case $p<q$

In this section, we prove Theorem 1.2. We use methods, described in the
beginning of Section 3. In this case, there are exactly $b+1$ types of groups
up to isomorphism, where $b=\mathrm{min}\\{a,n\\}$ with $p^{a}||q-1$. This
section will be divided into four subsections, depending on the isomorphism
types of $G=G_{b_{1}}$ and $N=G_{b_{2}}$, where $0\leq b_{1},b_{2}\leq n$.

###  4.1. Isomorphic type

First, we consider the isomorphic case. Let
$G=\mathbb{Z}_{q}\rtimes_{k}\mathbb{Z}_{p}^{n}$, where $k$ is an element of
order $p^{b}$. We are looking at $e(G,G)$.

####  4.1.1. The case $b=0$

In this case, the groups are cyclic and $e^{\prime}(G,G)$, $e(G,G)$ have been
enumerated in [6, Theorem 2].

####  4.1.2. The case when $0<b\leq n$

Let us take a group homomorphism
$\Phi:G_{b}\longrightarrow\textup{Hol}(G_{b})$ defined by

$\displaystyle\Phi(x)=\left(y^{j_{1}}x^{i_{1}},\left(\gamma_{1},\begin{pmatrix}\beta_{1}&\alpha_{1}\\\
0&1\end{pmatrix}\right)\right),\text{ and
}\Phi(y)=\left(y^{j_{2}}x^{i_{2}},\left(\gamma_{2},\begin{pmatrix}\beta_{2}&\alpha_{2}\\\
0&1\end{pmatrix}\right)\right).$

From $\Phi(y)^{q}=1$ and from $\Phi(xy)=\Phi(y^{k}x)$, we get the relations
$i_{2}=0$, $\beta_{2}=1$, $\gamma_{2}=1$ and

$\displaystyle\alpha_{2}(k-\beta_{1})\equiv 0$ $\displaystyle\pmod{q},$ (4.1)
$\displaystyle j_{2}(k^{i_{1}-1}\beta_{1}-1)\equiv\alpha_{2}(1+k+k^{2}\cdots
k^{i_{1}-1})$ $\displaystyle\pmod{q}.$ (4.2)

Thus if $\alpha_{2}=0$, then $\beta_{1}=k^{1-i_{1}}$. If $\alpha_{2}\neq 0$,
then $\beta_{1}=k$ and $\alpha_{2}=j_{2}(k-1)$. From $\Phi(x)^{p^{n}}=1$, we
get the following equivalences in $\mathbb{Z}_{q}$.

$\displaystyle{\beta_{1}}^{p^{n}}=1$ (4.3)
$\displaystyle\alpha_{1}(1+\beta_{1}+{\beta_{1}}^{2}\cdots{\beta_{1}^{p^{n}-1}})=0.$
(4.4)

By explicit calculations, we can show that, the exponent of $y$ in
$\Phi(x)^{p^{n}}$ is given by

$\displaystyle{\color[rgb]{0,0,1}\mathrm{Exp}_{y}\left(\Phi(x)^{p^{n}}\right)=j_{1}\left(\sum\limits_{u=0}^{p^{n}-1}m^{u}\right)+\dfrac{\alpha_{1}}{m(k^{\gamma_{1}}-1)}\left\\{\sum\limits_{v=1}^{p^{n}-1}m^{p^{n}-v}\left(k^{i_{1}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}\right)}-k^{i_{1}}\right)\right\\},}$

where $m=\beta_{1}k^{i_{1}}$. Using Eq. 4.1 and Eq. 4.2, we can show that
$m\in\\{k,k^{i_{1}+1}\\}$ First, let us take $m=k$. Then
$\sum\limits_{u=0}^{p^{n}-1}m^{u}\equiv 0\pmod{q}$. We aim to show that the
other summand is also zero in $\mathbb{Z}_{q}$. We have in $\mathbb{Z}_{q}$

$\displaystyle\sum\limits_{v=1}^{p^{n}-1}m^{p^{n}-v}\left(k^{i_{1}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}\right)}-k^{i}\right)=\sum\limits_{v=1}^{p^{n}}k^{i_{1}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}\right)-v}.$

Note that here $i_{1}$ and $\gamma_{1}$ are fixed. Denote by
$\Gamma(v)=i_{1}(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1})-v\pmod{p^{n}}$. Suppose
for $1\leq v_{1}\neq v_{2}\leq p^{n}$ we have
$\Gamma(v_{1})\equiv\Gamma(v_{2})\pmod{p^{n}}$. Then we have
$i(\gamma_{1}^{v_{1}}-\gamma_{1}^{v_{2}})\equiv(v_{1}-v_{2})(\gamma_{1}-1)\pmod{p^{n}}$.
Since the Sylow-$p$-subgroup of $\mathbb{Z}_{p^{n}}^{\times}$ is generated by
$(1+p)$ and $\gamma_{1}$ is an element having $p$-power order, say an element
of order $p^{g}$. Then $p^{n-g}||\gamma_{1}-1$. Thus $v_{1}-v_{2}\equiv
0\pmod{p^{g}}$, using Lemma 2.1. Conversely if $v_{1}-v_{2}\equiv
0\pmod{\mathrm{ord}{\gamma_{1}}}$, then
$i(\gamma_{1}^{v_{1}}-\gamma_{1}{v_{2}})\equiv(v_{1}-v_{2})(\gamma_{1}-1)\pmod{p^{n}}$.
Thus $\Gamma$ gives rise to a function from $\mathbb{Z}_{p^{n}}$ to the subset
$\\{p^{g},2p^{g},3p^{g},\ldots,p^{n}\\}$. Thus the sum is reduced to
$p^{g}\sum\limits_{t=1}^{p^{n-g}}k^{tp^{g}}$. If $k^{p^{g}}=1$, we get the sum
to be zero. Otherwise this sum is $p^{g}\dfrac{k^{p^{n}}-1}{k^{p^{g}}-1}=0$.
This finishes the proof.

Now, take the case when $m=k^{i_{1}+1}$. Then again the multiplier of $j_{1}$
is zero in $\mathbb{Z}_{q}$. We claim that the other summand is also zero in
the above expression. We have in this case,

$\displaystyle\sum\limits_{v=1}^{p^{n}-1}m^{p^{n}-v}\left(k^{i_{1}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}\right)}-k^{i_{1}}\right)=\sum\limits\limits_{v=1}^{p^{n}}k^{i_{1}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}-v\right)-v}-\sum\limits_{v=1}^{p^{n}}k^{i_{1}(1-v)-v}$
$\displaystyle=$
$\displaystyle\begin{cases}\sum\limits_{v=1}^{p^{n}}k^{i_{1}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}\right)-\left(i_{1}+1\right)v}&\text{
when }i_{1}+1\neq 0\pmod{p}\\\
\sum\limits_{v=1}^{p^{n}}k^{i_{1}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}-v\right)-v}-\sum\limits_{v=1}^{p^{n}}k^{i_{1}(1-v)-v}&\text{otherwise}.\end{cases}$

We start by considering the first subcase, i.e. $i_{1}+1$ being a unit in
$\mathbb{Z}_{p^{n}}$. Again denote by
$\Gamma(v)=i_{1}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}-v\right)-(i_{1}+1)v$.
Then $\Gamma(v_{1})\equiv\Gamma(v_{2})\pmod{p^{n}}$ implies that
$i_{1}(\gamma_{1}^{v_{1}}-\gamma_{1}^{v_{2}})\equiv(i_{1}+1)(\gamma_{1}-1)(v_{1}-v_{2})\pmod{p^{n}}$.
Then proceeding as before, we get the result. Next, consider the second
subcase. In this case, we show that both of the sums are zero. Take
$\Gamma_{1}(v)=i_{1}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}-v\right)-(i_{1}+1)v$
and $\Gamma_{2}(v)=i_{1}(1-v)-v$. Assume $p^{h}||i_{1}+1$, then
$\Gamma_{2}(v^{\prime})=\Gamma_{2}(v^{\prime\prime})$ iff $v^{\prime}\equiv
v^{\prime\prime}\pmod{p^{n-h}}$, using Lemma 2.1. Thus $\Gamma_{2}$ determines
a function to the subset $\\{p^{n-h},2p^{n-h},\ldots,p^{n}\\}$ and hence the
second term of the expression above vanishes. An argument similar to the
previous cases of $\Gamma(v)$, shows that the first term is $0$ as well in
$\mathbb{Z}_{q}$. Thus we have proved the following lemma.

######  Lemma 4.1.

In $\mathrm{Exp}_{y}\left(\Phi(x)^{p^{n}}\right)$, if the coefficient of
$j_{1}$ is zero in $\mathbb{Z}_{q}$, then so is the coefficient of
$\alpha_{1}$.

We claim that $i_{1}$ is a unit. Suppose $i_{1}$ is not a unit. We note that
$\Phi(x)^{p^{n-1}}=\left(y^{J},\left(1,\begin{pmatrix}1&0\\\
0&1\end{pmatrix}\right)\right)$, for some $J$. Note that if $\beta_{1}=0$ then
$\alpha_{1}=0$, otherwise $1+\beta_{1}+\ldots+\beta_{1}^{p^{n}-1}=0$, whence
the matrix entry is justified. Now, if $J=0$ then this map is not regular.
Otherwise when $J\neq 0$, we get $J$ is a unit in $\mathbb{Z}_{q}$. Since $p$
is a unit is $\mathbb{Z}_{q}$, we get that $\Phi(x_{1})^{p^{n}}$ is not
identity element. This proves claim. Now we are ready to count the number of
Hopf-Galois structures on extensions, whose group is of the form ${G}_{b}$ for
some $0<b<n$. This will be divided into four cases. Before proceeding, we note
that none of the cases, impose any condition on $j_{2}$ and $\gamma_{1}$.

Case $1$: The case $\beta_{1}=1$. This implies $\alpha_{2}=0$. Since if
$\alpha_{2}\neq 0$, then $\beta_{1}=k\neq 1$. From $\alpha_{2}=0$ we get that
$i_{1}\equiv 1\pmod{p^{b}}$, from which we get that $i_{1}$ has $p^{n-b}$
possibilities. Further $\alpha_{1}=0$ from Eq. 4.4. In this case, $j_{1}$ has
$q$ possibilities since $m\neq 1$, using Lemma 4.1. Thus in this case we get
$\varphi(q)qp^{2(n-b)}$ embedding.

Case $2$: The case $\beta_{1}\neq 1$, and $\alpha_{2}=0$. Note that
$\alpha_{2}=0$ implies that $\beta_{1}=k^{1-i_{1}}$. Also, $\beta_{1}\neq 1$
imposes the condition that $i_{1}$ has $\varphi(p^{n})-p^{n-b}$ possibilities.
In this case, $j_{1}$ and $\alpha_{1}$ have $q$ possibilities each. Thus in
this case we have $\varphi(q)\left(\varphi(p^{n})-p^{n-b}\right)q^{2}p^{n-b}$
embeddings.

Case $3$: The case $\beta_{1}\neq 1$, $\alpha_{2}\neq 0$, and $1+i_{1}\equiv
0\pmod{p^{b}}$. Since $1+i_{1}\equiv 0\pmod{p^{b}}$, we get $m=1$. Hence the
value of $j_{1}$ gets fixed. Thus in this case, we have
$\varphi(q)qp^{2(n-b)}$ embeddings.

Case $4$: The case $\beta_{1}\neq 1$, $\alpha_{2}\neq 0$, and
$1+i_{1}\not\equiv 0\pmod{p^{b}}$. In this case $i_{1}$ has
$\varphi(p^{n})-p^{n-b}$ possibilities. Similar to Case $2$, $j_{1}$ has $q$
possible values. Thus in this case, we have
$\varphi(q)\left(\varphi(p^{n})-p^{n-b}\right)q^{2}p^{n-b}$ embeddings.

In all of the cases above, the embeddings are regular, which is guaranteed by
the conditions that $i_{1}$ and $j_{2}$ are units. Furthermore, In conclusion,
we have proved the following result.

######  Proposition 4.2.

Let ${G}_{b}=\mathbb{Z}_{q}\rtimes_{k}\mathbb{Z}_{p^{n}}$, where
$k\in\mathbb{Z}_{q}$ is of order $p^{b}$ for some $0<b\leq n$. Then
$e^{\prime}\left({G}_{b},{G}_{b}\right)=e\left({G}_{b},{G}_{b}\right)=2\left(p^{n-b}+q\left(\varphi(p^{n})-p^{n-b}\right)\right)$.

###  4.2. Non-isomorphic type

This case will be divided into three cases, depending on the values of $b_{1}$
and $b_{2}$.

####  4.2.1. The case $1\leq b_{1}\neq b_{\leq}n$

We will need a variation of Lemma 4.1, for dealing with this case. We start
with a presentation of these two groups. For $t=1$ and $2$, let us fix

$\displaystyle{G}_{b_{t}}=\left\langle
x_{t},y_{t}\middle|x_{t}^{p^{n}}=y_{t}^{q}=1,x_{t}y_{t}x_{t}^{-1}=y_{t}^{k_{t}}\right\rangle,$

where $k_{t}$ is an element of order $p^{b_{t}}$. Now we consider
$\Phi:{G}_{b_{1}}\longrightarrow\textup{Hol}\left({G}_{b_{2}}\right)$ is an
regular embedding and
$\Phi(x_{1})=\left(y_{2}^{j_{1}}x_{2}^{i_{1}},\left(\gamma_{1},\begin{pmatrix}\beta_{1}&\alpha_{1}\\\
0&1\end{pmatrix}\right)\right)$, then it can be proved that,

$\displaystyle{\color[rgb]{0,0,1}\mathrm{Exp}_{y}\left(\Phi(x)^{p^{n}}\right)=j\left(\sum\limits_{u=0}^{p^{n}-1}m^{u}\right)+\dfrac{\alpha_{1}}{m(k_{2}^{\gamma_{1}}-1)}\left\\{\sum\limits_{v=1}^{p^{n}-1}m^{p^{n}-v}\left(k_{2}^{i_{2}\left(1+\gamma_{1}+\ldots+\gamma_{1}^{v-1}\right)}-k_{2}^{i_{2}}\right)\right\\},}$

where $m=\beta_{1}k_{2}^{i_{1}}$. It can be shown that
$m\in\left\\{k_{1},k_{1}k_{2}^{i_{1}}\right\\}$ modulo $q$, using Eq. 4.1 and
Eq. 4.2. Note that in any of the cases $b_{1}<b_{2}$ or $b_{2}<b_{1}$, $m$ is
purely a power of $k_{1}$ or $k_{2}$, since $\mathbb{Z}_{p^{n}}^{\times}$ is
cyclic. Then a variation of the argument before Lemma 4.1, proves the
following result.

######  Lemma 4.3.

In $\mathrm{Exp}_{y}\left(\Phi(x_{1})^{p^{n}}\right)$ if the coefficient of
$j_{1}$ is $0$ in $\mathbb{Z}_{q}$, then so is the coefficients of
$\alpha_{1}$.

Hoping that the reader is now familiar with the flow of arguments, without
loss of generality in this case we will assume that the embedding is given by,

$\displaystyle\Phi(x_{1})=\left(y_{2}^{j_{1}}x_{2}^{i_{1}}\left(\gamma_{1},\begin{pmatrix}\beta_{1}&\alpha_{1}\\\
0&1\end{pmatrix}\right)\right),\Phi(y_{1})=\left(y_{2}^{j_{2}}\left(1,\begin{pmatrix}1&\alpha_{2}\\\
0&1\end{pmatrix}\right)\right),$

where $i_{1}$ is a unit in $\mathbb{Z}_{p^{n}}$ (using same argument as in
Section 4.1), $\gamma_{1}$ is a unit in $\mathbb{Z}_{p^{n}}$ satisfying
$\gamma_{1}^{p^{n-b_{2}}}=1$, and $j_{2}$ is a unit in $\mathbb{Z}_{q}$.
Comparing the both sides of the equation
$\Phi(x_{1})\Phi(y_{1})=\Phi(y_{1})^{k_{1}}\Phi(x_{1})$, we get

$\displaystyle\alpha_{2}(k_{1}-\beta_{1})\equiv 0$ $\displaystyle\pmod{q},$
(4.5) $\displaystyle k_{2}^{i_{1}}\beta_{1}j_{2}\equiv
j_{2}k_{1}+j_{2}\left(1+k_{2}+\ldots+k_{2}^{i_{1}-1}\right)$
$\displaystyle\pmod{q}.$ (4.6)

From Eq. 4.5 either $\alpha_{2}=0$ or $\beta_{1}=k_{1}$. Irrespective of the
cases $\beta_{1}k_{1}^{i_{1}}\neq 1$. Thus from Lemma 4.3 $j_{1}$ can take any
value from $\mathbb{Z}_{q}$. Now, in the first case,
$\beta_{1}=k_{1}k_{2}^{-i_{1}}$ (from Eq. 4.6). Also $\gamma_{1}$ and
$\alpha_{1}$ have $p^{n-b_{2}}$ and $q$ many choices respectively. This gives
that total number of embeddings in this case is given by
$\varphi(q)\varphi(p^{n})q^{2}p^{n-b_{2}}$. In the second case,
$\alpha_{2}=(k_{2}-1)j_{2}$ and $\gamma_{1}$, $\alpha_{1}$ have $p^{n-b_{2}}$,
$q$ many choices respectively. Thus the total number of embeddings arising
from this case is given by $\varphi(q)\varphi(p^{n})q^{2}p^{n-b_{2}}$. Given
that $i_{1}$ and $j_{2}$ are units, we get that the constructed map is
regular. We now have the following result.

######  Proposition 4.4.

Let ${G}_{b_{t}}=\mathbb{Z}_{q}\rtimes_{k_{t}}\mathbb{Z}_{p^{n}}$, where
$k_{t}$ is an element of $\mathbb{Z}_{p^{n}}$ of order $p^{b_{t}}$, for $t=1$,
$2$. Let $0<b_{1}\neq b_{2}\leq n$. Then

$\displaystyle
e^{\prime}\left({G}_{b_{1}},{G}_{b_{2}}\right)=2qp^{n+b_{1}-b_{2}-1}(p-1),~{}e\left({G}_{b_{1}},{G}_{b_{2}}\right)=2qp^{n-1}(p-1).$

####  4.2.2. The case $0=b_{1}<b_{2}\leq n$

In this case ${G}_{b_{1}}$ is cyclic and hence the presentations of the groups
${G}_{b_{1}}$ and ${G}_{b_{2}}$ are chosen to be

$\displaystyle{G}_{b_{1}}=\left\langle
x_{1},y_{1}\middle|x_{1}^{p^{n}}=y_{1}^{q}=1,x_{1}y_{1}x_{1}^{-1}=y_{1}\right\rangle,{G}_{b_{2}}=\left\langle
x_{2},y_{2}\middle|x_{2}^{p^{n}}=y_{2}^{q}=1,x_{2}y_{2}x_{2}^{-1}=y_{2}^{k_{2}}\right\rangle,$

with $k_{2}\in\mathbb{Z}_{p^{n}}$ being an element of multiplicative order
$p^{b_{2}}$. Fix a homomorphism
$\Phi:{G}_{b_{1}}\longrightarrow\textup{Hol}({G}_{b_{2}})$ given by

$\displaystyle\Phi(x_{1})=\left(y_{2}^{j_{1}}x_{2}^{i_{1}}\left(\gamma_{1},\begin{pmatrix}\beta_{1}&\alpha_{1}\\\
0&1\end{pmatrix}\right)\right),\Phi(y_{1})=\left(y_{2}^{j_{2}}x_{2}^{i_{2}}\left(\gamma_{2},\begin{pmatrix}\beta_{2}&\alpha_{2}\\\
0&1\end{pmatrix}\right)\right).$

From the condition $\Phi(y_{1})^{q}$, we get the conditions that $i_{2}=0$,
$\gamma_{2}=0$ and $\beta_{2}=1$. To ensure the regularity of the maps, we
will need $i_{1}$ and $j_{2}$ to be units in $\mathbb{Z}_{p^{n}}$ and
$\mathbb{Z}_{q}$ respectively (see Section 4.1). Equating the two sides of the
equality $\Phi(x_{1})\Phi(y_{1})=\Phi(y_{1})\Phi(x_{1})$, we get that

$\displaystyle\alpha_{2}(1-\beta_{1})\equiv 0$ $\displaystyle\pmod{q},$ (4.7)
$\displaystyle\beta_{1}k_{2}^{i_{1}}j_{2}\equiv
j_{2}+\alpha_{2}\left(1+k_{2}+\ldots+k_{2}^{i_{1}-1}\right)$
$\displaystyle\pmod{q}.$ (4.8)

Hence from Eq. 4.7 we have either $\alpha_{2}=0$ or $\beta_{1}=1$. In case
$\alpha_{2}=0$, plugging the value in Eq. 4.8 we get that
$\beta_{1}k_{2}^{i_{1}}=1$, whence $j_{1}$ has fixed choice, once $\alpha_{1}$
is fixed. Furthermore $\alpha_{1}$, $\gamma_{1}$ have $q$, $p^{n-b_{2}}$
choices. In the case $\beta+1=1$, from Eq. 4.8 we get that
$\alpha_{2}=j_{2}(k_{2}-1)$ and $\beta_{1}k_{2}^{i_{1}}\neq 1$. Hence Lemma
4.3 applies. Thus $j_{1}$, and $\gamma_{1}$ have $q$, and $p^{n-b_{2}}$
possibilities. We conclude that in both cases the number of regular embedding
of the cyclic group of order $p^{n}q$ in
$\textup{Hol}\left({G}_{b_{2}}\right)$ is given by
$q\varphi(q)p^{n-b_{2}}\varphi(p^{n})$. We have the following result.

######  Proposition 4.5.

Let ${C}$ denotes the cyclic group of order $p^{n}q$ and
${G}_{b}\cong\mathbb{Z}_{q}\rtimes_{k_{b}}\mathbb{Z}_{p^{n}}$, where
$k_{b}\in\mathbb{Z}_{q}$ is an element of multiplicative order $p^{b}$. Then

$\displaystyle e^{\prime}({C},{G}_{b})=2p^{n-b}q,\text{ and
}e({C},{G}_{b})=2(p-1)p^{n-1}$

####  4.2.3. The case $0=b_{2}<b_{1}\leq n$

Here we count the number $e^{\prime}({G}_{b_{1}},{G}_{b_{2}})$ (equivalently
$e({G}_{b_{1}},{G}_{b_{2}})$). Here ${G}_{b_{2}}$ is a cyclic group of order
$p^{n}q$. In this case, we have,

$\displaystyle\textup{Hol}({G}_{b_{2}})\cong\left\\{\left(y_{2}^{j}x_{2}^{i},(\omega,\delta)\middle|\begin{subarray}{c}(j,i)\in\mathbb{Z}_{q}\times\mathbb{Z}_{p^{n}}\\\
(\omega,\delta)\in\mathbb{Z}_{q}^{\times}\times\mathbb{Z}_{p^{n}}^{\times}\end{subarray}\right)\right\\}.$

We fix an embedding
$\Phi:{G}_{b_{1}}\longrightarrow\textup{Hol}\left({G}_{b_{2}}\right)$
determined by

$\displaystyle\Phi(x_{1})=\left(y_{2}^{j_{1}}x_{2}^{i_{1}},\left(\omega_{1},\delta_{1}\right)\right),~{}\Phi(x_{1})=\left(y_{2}^{j_{2}}x_{2}^{i_{2}},\left(\omega_{2},\delta_{2}\right)\right).$

From $\Phi(y_{1})^{q}=1$, we get $\omega_{2}=1,\delta_{2}=1$ and $i_{2}=0$.
Considering $\Phi(x_{1})^{p^{n}}=1$ we get that $\omega_{1}^{p^{n}}=1$,
$\delta_{1}^{p^{n}}=1$, and

$\displaystyle
j_{1}\left(1+\omega_{1}+\ldots+\omega_{1}^{p^{n}-1}\right)\equiv 0$
$\displaystyle\pmod{q},$ (4.9) $\displaystyle
i_{1}\left(1+\delta_{1}+\ldots+\delta_{1}^{p^{n}-1}\right)\equiv 0$
$\displaystyle\pmod{p^{n}}..$ (4.10)

Finally comparing both sides of the equation
$\Phi(x_{1})\Phi(y_{1})=\Phi(y_{1})^{k_{1}}\Phi(x_{1})$, we get that
$\omega_{1}=k_{1}$ and hence Eq. 4.9 gets satisfied automatically. To ensure
that the embedding is regular, we will need that $i_{1}$ and $j_{2}$ are
units. Any choice of $\delta_{1}$ satisfies Eq. 4.10. Thus $j_{1}$, $j_{2}$,
$i_{1}$, and $\delta_{1}$ have $\varphi(q)$, $q$, $\varphi(p^{n})$, and
$p^{n-1}$ possibilities respectively. We conclude with the following result.

######  Proposition 4.6.

Let ${G}_{b}\cong\mathbb{Z}_{q}\rtimes_{k_{b}}\mathbb{Z}_{p^{n}}$, where
$k_{b}$ is an element of $\mathbb{Z}_{q}$ of order $p^{b}$, $1\leq b\leq n$,
and ${C}$ denote the cyclic group of order $p^{n}q$. Then we have

$\displaystyle
e^{\prime}\left({G}_{b},{C}\right)=p^{n+b-2}(p-1),~{}e\left({G}_{b},{C}\right)=p^{n-1}q.$

The Theorem 1.2 follows from Proposition 4.2, Proposition 4.4, Proposition
4.5, and Proposition 4.6.

##  5\. Realizable pair of groups

Given two finite groups $G$ and $N$ of the same order, we say that the pair
$(G,N)$ is _realizable_ if there exists a Hopf-Galois structure on a Galois
$G$-extension, of type $N$. In other words a pair $(G,N)$ is realizable if
$e(G,N)\neq 0$. This is equivalent to saying there exists a skew brace with
its multiplicative group isomorphic to $G$ and its additive group isomorphic
to $N$. This problem is not well studied since given an integer $n$, the
classification of all the groups of size $n$ is not known. However, they have
been studied for a variety of groups. When $G$ is a cyclic group of odd order
and the pair $(G,N)$ is realizable then in [6], the author showed that if $N$
is abelian then it is cyclic. If $N$ is a non ableian simple group and $G$ is
a solvable group with the pair $(G,N)$ being realizable, then in [18] $N$ was
completely classified. Whenever $N$ or $G$ is isomorphic to
$\mathbb{Z}_{n}\rtimes\mathbb{Z}_{2}$ for an odd $n$ then their
realizabilities were studied in [1].

Among a few available techniques, the notion of bijective crossed homomorphism
to study realizability problems for a pair of groups of the same order, was
introduced by Tsang in the work [19]. Given an element
$\mathfrak{f}\in\mathrm{Hom}(G,\textup{Aut}(N))$, a map
$\mathfrak{g}\in\mathrm{Map}(G,N)$ is said to be a _crossed homomorphism_ with
respect to $\mathfrak{f}$ if
$\mathfrak{g}(ab)=\mathfrak{g}(a)\mathfrak{f}(a)(\mathfrak{g}(b))\text{ for
all }a,b\in G$. Setting
$Z_{\mathfrak{f}}^{1}(G,N)=\\{\mathfrak{g}:\mathfrak{g}$ is bijective crossed
homomorphism with respect to $\mathfrak{f}\\}$, we have the following two
results.

######  Proposition 5.1.

[19, Proposition 2.1] The regular subgroups of $\textup{Hol}(N)$ which are
isomorphic to $G$ are precisely the subsets of $\textup{Hol}(N)$ of the form
$\\{(\mathfrak{g}(a),\mathfrak{f}(a)):a\in G\\},$ where
$\mathfrak{f}\in\mathrm{Hom}(G,\textup{Aut}(N)),\mathfrak{g}\in
Z_{\mathfrak{f}}^{1}(G,N)$.

######  Proposition 5.2.

[20, Proposition 3.3] Let $G,N$ be two groups such that $|G|=|N|$. Let
$\mathfrak{f}\in\mathrm{Hom}(G,\textup{Aut}(N))$ and $\mathfrak{g}\in
Z_{\mathfrak{f}}^{1}(G,N)$ be a bijective crossed homomorphism (i.e. $(G,N)$
is realizable). Then if $M$ is a characteristic subgroup of $N$ and
$H=\mathfrak{g}^{-1}(M)$, we have that the pair $(H,M)$ is realizable.

We will need the following two results, where the realizability of cyclic
groups have been characterized. We will use modifications of these
characterizations towards proving the realizability of groups of the form
$\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$.

######  Proposition 5.3.

[16, Theorem 3.1] Let $N$ be a group of odd order $n$ such that the pair
$(\mathbb{Z}_{n},N)$ is realizable. Then $N$ is a $C$-group (i.e. all the
Sylow subgroups are cyclic).

######  Proposition 5.4.

[14, Theorem 1] Let $G$ be a group of order $n$ such that $(G,\mathbb{Z}_{n})$
is realizable. Then $G$ is solvable and almost Sylow-cyclic (i.e. its Sylow
subgroups of odd order are cyclic, and every Sylow-$2$ subgroup of G has a
cyclic subgroup of index at most $2$).

######  Theorem 5.5.

Let ${N}$ be a group of order $qp^{n}$, where $q$ is a prime, $q<p$ and
$(q,p)=1$. Then the pair $(\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q},{N})$ (or
$(N,\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q})$) is realizable if and only if
${N}\cong\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$.

###### Proof.

Let $(\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q},{N})$ be realizable. By
Proposition 5.1 there exists a bijective crossed homomorphism $\mathfrak{g}\in
Z^{1}_{\mathfrak{f}}(\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q},N)$ for some
$\mathfrak{f}\in\mathrm{Hom}(\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q},\textup{Aut}(N))$.
Let $H_{p}$ be the Sylow-$p$ subgroup of $N$ (it is unique since $q<p$). Then
using Proposition 5.2 the pair $(\mathfrak{g}^{-1}H_{p},H_{p})$ is realizable.
Note that $\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$ has unique subgroup of
order $p^{n}$, which is cyclic. This implies that $(\mathbb{Z}_{p^{n}},H_{p})$
is realizable. Hence by Proposition 5.3 we get that $H_{p}$ is isomorphic to
$\mathbb{Z}_{p^{n}}$ and therefore
$N\cong\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$. Conversely if
$N\cong\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$ then the pair
$(\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q},{N})$ is realizable since
$e(\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q},{N})$ is non-zero from Section 3.

Now if the pair $(N,\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q})$ is realizable,
by Proposition 5.1 there exists a bijective crossed homomorphism
$\mathfrak{g}\in
Z^{1}_{\mathfrak{f}}(N,\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q})$ for some
$\mathfrak{f}\in\mathrm{Hom}(G,\textup{Aut}(\mathbb{Z}_{n}\rtimes\mathbb{Z}_{2}))$.
Since $\mathbb{Z}_{p^{n}}$ is a characteristic subgroups of
$\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$, we get that
$\mathfrak{g}^{-1}(\mathbb{Z}_{p^{n}})$ is a subgroup of $N$ and
$(\mathfrak{g}^{-1}(\mathbb{Z}_{p^{n}}),\mathbb{Z}_{p^{n}})$ is realizable.
Then by Proposition 5.4, we have that $\mathfrak{g}^{-1}(\mathbb{Z}_{p^{n}})$
is almost Sylow-cylic and therefore isomorphic to $\mathbb{Z}_{p^{n}}$. Hence
$N\cong\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$. Conversely if
$N\cong\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q}$, then by Section 3 we have the
pair $(N,\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{q})$ is realizable. ∎

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