# On $\mathbb{Z}_{N}\rtimes\mathbb{Z}_{2}$-Hopf-Galois structures

Namrata Arvind, Saikat Panja<EMAIL_ADDRESS><EMAIL_ADDRESS>IISER Pune, Dr. Homi Bhabha Road, Pashan, Pune 411 008, India

(Date: August 21, 2020)

###### Abstract.

Let $K/F$ be a finite Galois extension of fields with
$\textup{Gal}(K/F)=\Gamma$. In an earlier work of Timothy Kohl, the author
enumerated dihedral Hopf-Galois structures acting on dihedral extensions.
Dihedral group is one particular example of semidirect product of
$\mathbb{Z}_{n}$ and $\mathbb{Z}_{2}$. In this article we count the number of
Hopf-Galois structures with Galois group $\Gamma$ of type $G$, where
$\Gamma,G$ are groups of the form $\mathbb{Z}_{N}\rtimes_{\phi}\mathbb{Z}_{2}$
when $N$ is odd with radical of $N$ being a Burnside number. As an application
we also find the corresponding number of skew braces.

###### 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 IISER Pune research
fellowship and the second author has been partially supported by supported by
NBHM fellowship.

##  1\. Introduction

###  1.1. Hopf-Galois structures

Let $\mathcal{R}$ be a commutative ring with unity and let $\mathcal{H}$ be an
$\mathcal{R}$-bialgebra. Then $\mathcal{H}$ will be called an
$\mathcal{R}$-Hopf algebra if there is an $\mathcal{R}$-module homomoprhism
$\lambda:\mathcal{H}\rightarrow\mathcal{H}$ (the antipode map), which is both
an $\mathcal{R}$-algebra and an $\mathcal{R}$-coalgebra antihomomophism such
that:

$\displaystyle\lambda(h\otimes h^{\prime})$
$\displaystyle=\lambda(h)\otimes\lambda(h^{\prime})$
$\displaystyle\Delta\lambda(h)$
$\displaystyle=(\lambda\otimes\lambda)\tau\Delta,$

where $\Delta$ is the comultiplication map and $\tau$ is the switch map
$\tau(h_{1}\otimes h_{2})=h_{2}\otimes h_{1}$. Now assume that $\mathcal{H}$
is commutative. An $\mathcal{R}$-Hopf algebra $\mathcal{H}$ is called a finite
algebra if it is finitely generated and a projective $\mathcal{R}$-module. Now
if $\mathcal{S}$ is an $\mathcal{R}$-algebra which is an $\mathcal{H}$-module,
then $\mathcal{S}$ is called an $\mathcal{H}$-module algebra if

$\displaystyle h(st)=\sum h_{(1)}(s)h_{(2)}(t)\text{ and }h(1)=\epsilon(h)1$

for all $h\in\mathcal{H},s,t\in\mathcal{S}$, where
$\Delta(h)=\sum\limits_{(h)}h_{(1)}\otimes
h_{(2)}\in\mathcal{H}\otimes\mathcal{H}$ according to Sweedler’s ([Sw69])
notation and $\epsilon:\mathcal{H}\rightarrow\mathcal{R}$ is the co-unit map.

Then $\mathcal{S}$, a finite commutative $\mathcal{R}$-algebra is called an
$\mathcal{H}$-Galois extension over $\mathcal{R}$ if $\mathcal{S}$ is a left
$\mathcal{H}$-module algebra and the $\mathcal{R}$-module homomorphism

$\displaystyle
j:\mathcal{S}\otimes_{\mathcal{R}}\mathcal{H}\rightarrow\textup{End}_{\mathcal{R}}(\mathcal{S}),$

given by $j(s\otimes h)(s^{\prime})=sh(s^{\prime})$ for
$s,s^{\prime}\in\mathcal{S},h\in\mathcal{H}$, is an isomorphism. Now we define
a Hopf-Galois structure on a separable field extension. Assume $K/F$ is a
finite separable 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 $K/F$.

###  1.2. Greither-Pareigis theory [GP87] and Boytt’s translation [B96]

Given a group $G$ we define $Holomorph$ of $G$ as a semidirect product
$G\rtimes_{\psi}\textup{Aut}(G)$, where $\psi$ is the identity map. Holomorph
of a group $G$ (denoted by $\textup{Hol}(G)$) sits inside $\textup{Perm}(G)$
(set of permutations on $G$) as follows

$\textup{Hol}(G)=\\{\eta\in\textup{Perm}(G):\eta\text{ normalizes
}\lambda(G)\\}$

where $\lambda$ is the left regular representation. Now we state some results
which will help us count the number of Hopf-Galois structures on a given field
extension. The following result is due to [GP87].

######  Lemma 1.1.

[C00, Theorem 6.8] Let $K/F$ be a separable extension of fields. Let $L$ be
its normal closure,
$\Gamma=\textup{Gal}(L/F),\Gamma^{\prime}=\textup{Gal}(L/K)$. Define the set
$X=\Gamma/{\Gamma^{\prime}}$. Then there is a bijection between Hopf-Galois
structure on $K/F$ and regular subgroups $G$ of $\textup{Perm}(X)$ normalized
by $\lambda(\Gamma)$ where $\lambda$ is the left regular representation.

In the proof of the above lemma, given a regular subgroup
$G\leq\textup{Perm}(X)$ normalized by $\lambda(\Gamma)$, the Hopf-Galois
structure on $K/F$ corresponding to $G$ is $K[G]^{\Gamma}$. Here $\Gamma$ acts
on $G$ by conjugation inside $\textup{Perm}(X)$ and it acts on $K$ by field
automorphism, which induces an action of $\Gamma$ on $K[G]$. This $G$ is
called the type of the Hopf-Galois extension. Although Greither-Pareigis
theory simplifies the problem of counting the number of Hopf-Galois structure
for a given separable extension, the size of $\textup{Perm}(X)$ is large
($|X|!$) in general. The next lemma (also known as Boytt’s translation)
further simplifies the problem by considering regular embedding in
$\textup{Hol}(X)$, which is comparatively smaller in size. From the proof of
[B96, Proposition 1] we have the following:

Let $\Gamma$ be a finite group and $\Gamma^{\prime}\subseteq\Gamma$ be a
subgroup. Let $X=\Gamma/\Gamma^{\prime}$ and $G$ be group of order $|X|$. Then
there is a bijection between the following sets:

  1. (1)

$\\{\alpha:G\rightarrow\textup{Perm}(X)\text{ a monomorphism, }\alpha(G)\text{
is regular}\\}$

  2. (2)

$\\{\beta:\Gamma\rightarrow\textup{Perm}(G)\text{ a monomorphism,
}\beta(\Gamma^{\prime})\text{ is stabilizer of the identity of }G\\}$

Now assume $K/F$ is a Galois extension with Galois group isomorphic to
$\Gamma$. Let $e(\Gamma,G)$ be the number of regular subgroups in
$\textup{Perm}(\Gamma)$ isomorphic to $G$ which is normalized by
$\lambda(\Gamma)$ i.e. the number of Hopf-Galois structures on $K/F$ of type
$G$. Let $e^{\prime}(\Gamma,G)$ denote the number of subgroups $\Gamma^{*}$ of
$\textup{Hol}(G)$ isomorphic to $\Gamma$, such that the stabilizer in
$\Gamma^{*}$ of $e_{G}$ is trivial. Then we have the following result.

######  Lemma 1.2.

[B96, See Proposition 1] With the notations as above we have,

$e(\Gamma,G)=\dfrac{|\textup{Aut}(\Gamma)|}{|\textup{Aut}(G)|}e^{\prime}(\Gamma,G).$

Note that $\Gamma^{*}$ is a regular subgroup of $\textup{Hol}(G)$ implies
$\Gamma^{*}$ has the same cardinality as $G$. A typical element of
$\textup{Hol}(G)$ is of the form $(g,\zeta)$ where $g\in
G,\zeta\in\textup{Aut}(G)$. Hence to say $\Gamma^{*}$ is a regular subgroup of
$\textup{Hol}(G)$ we have to make sure that $g=e_{G}$ implies that $\zeta$ is
the identity automorphism of $G$ for every $(g,\zeta)\in\Gamma^{*}$. We will
use this condition to check regular embeddings of the concerned groups in the
article.

###  1.3. Notations

For $a,b\in\mathbb{Z}$ we will use $(a,b)$ to denote the g.c.d. of $a,b$. For
a number $n$, we take $\pi(n)=\\{p:p\text{ divides }n,p\text{ prime}\\}$. The
notation $v_{p}(n)$ denotes the $p$-valuation of $n$. For $n\in\mathbb{N}$,
the radical of $n$ is defined to be product of the distinct primes in $\pi(n)$
which will be denoted as $\mathfrak{R}(n)$. The symbol $\varphi(n)$ denotes
the Euler’s totient function at $n\in\mathbb{N}$. A number $n\in\mathbb{N}$ is
called a Burnside number if $(n,\varphi(n))=1$.

###  1.4. Result

In [B96] the author has proved that If $K/F$ is a finite Galois extension of
field of degree $T$, then this extesnion admits a unique Hopf-Galois structure
if and only if $T$ is Burnside number. Since in our case $N>1$ is odd and
hence $2N$ is not Burnside, the extension has at least $2$ Hopf-Galois
structure. The number of Hopf-Galois structure for various groups have been
studied by E. Campedel et al. [CCC20], T. Kohl [Ko20], Carnahan S. et al
[CC99] et cetera. For an extensive literature review one may look at the PhD
thesis of K. N. Zenouz [Ze18]. In [Ko20], T. Kohl has computed $e(G.G)$ when
$G$ is Dihedral group. For $N$ odd we look at groups of order $2N$ of the form
$\mathfrak{M}_{k,l}:=\mathcal{D}_{2k}\times\mathcal{C}_{l}$ where
$kl=N,(k,l)=1$, whenever radical of $N$ is a Burnside number. Our main result
is the following.

######  Theorem 1.3.

Let $K/F$ be a Galois extension of fields with $\textup{Gal}(K/F)\cong\Gamma$
and $N\in\mathbb{N}$ be odd. If $\Gamma=\mathfrak{M}_{k_{1},l_{1}}$ and
$G=\mathfrak{M}_{k_{2},l_{2}}$ where $k_{1}l_{1}=k_{2}l_{2}=N$ and
$\mathfrak{R}(N)$ is a Burnside number, then the number of Hopf-Galois
structure on $K/F$ of type $G$ is given by

$\displaystyle
e(\Gamma,G)=\dfrac{l_{1}l_{2}}{(l_{1},l_{2})\mathfrak{R}(l_{1})}\cdot
2^{|\pi(k_{2})|}.$

Rest of the article is organized as follows. In section $2$ we collect
background materials needed for the proof of the theorem. Section $3$ is
devoted to the proof of the main result. Next in section $4$ we derive some
further consequences of the results found in section $3$. All the necessary
equations which we use in section $3$ have been included in section $5$.

##  2\. Preliminaries

In this section we give complete description of groups of the form
$\mathbb{Z}_{N}\rtimes\mathbb{Z}_{2}$ and state some basic number theoretic
results which will be used in Section $3$ to enumerate regular embeddings.

###  2.1. Groups of the form $\mathbb{Z}_{N}\rtimes_{\phi}\mathbb{Z}_{2}$, $N$
odd

Note that if $N=\prod\limits_{t=1}^{m}p_{t}^{\alpha_{t}}$, where $p_{i}$’s are
all distinct primes, then

$\displaystyle\mathbb{Z}_{N}$
$\displaystyle\cong\bigoplus\limits_{t=1}^{m}\mathbb{Z}_{p_{t}^{\alpha_{t}}},$
$\displaystyle\text{and }\textup{Aut}(\mathbb{Z}_{N})\cong\mathbb{Z}_{N}^{*}$
$\displaystyle\cong\prod_{t=1}^{m}\mathbb{Z}_{p_{t}^{\alpha_{t}}}^{*}$
$\displaystyle\cong\bigoplus\limits_{t=1}^{m}\mathbb{Z}_{p_{t}^{\alpha_{t}-1}(p_{t}-1)}.$

For $x\in\mathbb{Z}_{N}$ we have $x=(x_{1},x_{2},\cdots,x_{m})$ where
$x_{u}\in\mathbb{Z}_{p_{u}^{\alpha_{u}}}$. We define $p_{u}(x)=x_{u}$ for
$p_{u}\in\pi(N)$.

If $\phi:\mathbb{Z}_{2}=\\{\pm 1\\}\rightarrow\textup{Aut}(\mathbb{Z}_{N})$ is
a group homomorphism with $p_{u}(\phi(-1))=-1$ for all $p_{u}\in\pi(N)$, then
$\mathbb{Z}_{N}\rtimes_{\phi}\mathbb{Z}_{2}$ is the dihedral group of order
$2N$ and we will denote this group by $\mathfrak{D}$. When $p_{u}(-1)=1$ for
all $p_{u}\in\pi(N)$, then $\mathbb{Z}_{n}\rtimes_{\phi}\mathbb{Z}_{2}$ is the
cyclic group of order $2N$ and we will denote this group by $\mathfrak{C}$.
Now suppose $p_{u}(-1)=1$ for some $p_{u}\in\pi(n)$ and
$p_{u^{\prime}}(-1)=-1$ for some $p_{u^{\prime}}\in\pi(N)$, then the group is
isomorphic to $\mathcal{D}_{2k}\times\mathcal{C}_{l}$ for some
$k,l\in\mathbb{N}$ with $kl=N$. We denote this group by $\mathfrak{M}_{k,l}$.
We have to consider the regular embeddings for the following cases:

  1. (1)

$\mathfrak{M}_{k_{1},l_{1}}\hookrightarrow\textup{Hol}(\mathfrak{M}_{k_{2},l_{2}})$
where $k_{1}l_{1}=k_{2}l_{2}=N$ and $(k_{1},l_{1})=(k_{2},l_{2})=1$,

  2. (2)

$\mathfrak{D}\hookrightarrow\textup{Hol}(\mathfrak{M}_{k,l})$ with $k,l>1$,

  3. (3)

$\mathfrak{C}\hookrightarrow\textup{Hol}(\mathfrak{M}_{k,l})$ with $k,l>1$,

  4. (4)

$\mathfrak{D}\hookrightarrow\textup{Hol}(\mathfrak{C})$

  5. (5)

$\mathfrak{C}\hookrightarrow\textup{Hol}(\mathfrak{D})$

  6. (6)

$\mathfrak{M}_{k,l}\hookrightarrow\textup{Hol}(\mathfrak{C})$ with $k,l>1$,

  7. (7)

$\mathfrak{M}_{k,l}\hookrightarrow\textup{Hol}(\mathfrak{D})$ with $k,l>1$,

  8. (8)

$\mathfrak{D}\hookrightarrow\textup{Hol}(\mathfrak{D})$

  9. (9)

$\mathfrak{C}\hookrightarrow\textup{Hol}(\mathfrak{C})$

While counting the regular embeddings we consider the first case and all other
cases are special cases of it. We must mention here that the last two cases
have been previously discussed in [Ko20] and [B13] respectively and our
answers match with the results therein.

###  2.2. Basic results

######  Lemma 2.1.

Let $p>2$ be a prime and $\gamma\equiv 1\mod p$. Define $f_{\gamma}(0)=0$ and
for each $\delta\in\mathbb{Z}_{>0}$ define

$\displaystyle f_{\gamma}(\delta)=\sum\limits_{i=0}^{\delta-1}\gamma^{i}.$

Then

$\displaystyle f_{\gamma}(\delta_{1})\equiv f_{\gamma}(\delta_{2})\mod
p^{n}\text{ iff }\delta_{1}\equiv\delta_{2}\mod p^{n}.$

###### Proof.

See the proof of Lemma $2.17$ in [CCC20]. ∎

######  Corollary 2.2.

Let $p$ be a prime and $b\in\mathbb{Z}$ such that $b^{p^{m}}\equiv 1\mod
p^{n}$. Then

$\displaystyle p^{m}|f_{b}(p^{m})\text{and }p^{m+1}\nmid f_{b}(p^{m}).$

###### Proof.

This follows from the observation that $b\equiv 1\mod p$. ∎

##  3\. Regular embeddings and Hopf-Galois structure

We start with a presentation of the group
$\mathfrak{M}_{k,l}=\mathcal{D}_{2k}\times\mathcal{C}_{l}$. It is given by

$\displaystyle\mathfrak{M}_{k,l}=\left\langle
r,s,t:r^{k},s^{2},t^{l},srsr,sts^{-1}t^{-1},rtr^{-1}t^{-1}\right\rangle.$

We have the followings.

  1. (1)

$\textup{Hol}(\mathfrak{M}_{k,l})\cong\textup{Hol}(\mathcal{C}_{l})\times\textup{Hol}(\mathcal{D}_{2k})$,
since $(k,l)=1$,

  2. (2)

$\textup{Aut}(\mathcal{D}_{2k})\cong\left\\{\begin{pmatrix}b&a\\\
0&1\end{pmatrix}:b\in\mathbb{Z}_{k}^{*},a\in\mathbb{Z}_{k}\right\\}$ where

$\displaystyle\begin{pmatrix}b&a\\\ 0&1\end{pmatrix}\cdot
r=r^{b},\begin{pmatrix}b&a\\\ 0&1\end{pmatrix}\cdot s=r^{a}s,$

  3. (3)

$\textup{Aut}(\mathcal{C}_{l})\cong\mathbb{Z}_{l}^{*}$,

  4. (4)

$\textup{Hol}(\mathfrak{M}_{k,l})\cong\left\\{\left(\begin{pmatrix}b&a\\\
0&1\end{pmatrix},r^{i}s^{j},\begin{pmatrix}d&c\\\
0&1\end{pmatrix}\right):\displaystyle{\begin{subarray}{c}b\in\mathbb{Z}_{l}^{*},a\in\mathbb{Z}_{l},d\in\mathbb{Z}_{k}^{*},c\in\mathbb{Z}_{k},\\\
1\leq i\leq k-1,j=0,1\end{subarray}}\right\\},$

where $(r^{i}s^{j},a)$ corresponds to the element of $\mathfrak{M}_{k,l}$.

Now we want to look at the embeddings
$\Phi:\mathfrak{M}_{k_{1},l_{1}}\rightarrow\textup{Hol}(\mathfrak{M}_{k_{2},l_{2}}).$
We take

$\displaystyle\mathfrak{M}_{k_{1},l_{1}}$ $\displaystyle=\left\langle
r_{1},s_{1},t_{1}:r_{1}^{k_{1}},s_{1}^{2},t_{1}^{l_{1}},s_{1}r_{1}s_{1}r_{1},s_{1}t_{1}s_{1}^{-1}t_{1}^{-1},r_{1}t_{1}r_{1}^{-1}t_{1}^{-1}\right\rangle,$
$\displaystyle\mathfrak{M}_{k_{2},l_{2}}$ $\displaystyle=\left\langle
r_{2},s_{2},t_{2}:r_{2}^{k_{2}},s_{2}^{2},t_{2}^{l_{2}},s_{2}r_{2}s_{2}r_{2},s_{2}t_{2}s_{2}^{-1}t_{2}^{-1},r_{2}t_{2}r_{2}^{-1}t_{2}^{-1}\right\rangle.$

Let us assume that

$\displaystyle\Phi(r_{1})$ $\displaystyle=\left(\begin{pmatrix}b&a\\\
0&1\end{pmatrix},r_{2}^{i}s_{2}^{j},\begin{pmatrix}d&c\\\
0&1\end{pmatrix}\right),$ $\displaystyle\Phi(s_{1})$
$\displaystyle=\left(\begin{pmatrix}b^{\prime}&a^{\prime}\\\
0&1\end{pmatrix},r_{2}^{i^{\prime}}s_{2}^{j^{\prime}},\begin{pmatrix}d^{\prime}&c^{\prime}\\\
0&1\end{pmatrix}\right),$ $\displaystyle\Phi(t_{1})$
$\displaystyle=\left(\begin{pmatrix}b^{\prime\prime}&a^{\prime\prime}\\\
0&1\end{pmatrix},r_{2}^{i^{\prime\prime}}s_{2}^{j^{\prime\prime}},\begin{pmatrix}d^{\prime\prime}&c^{\prime\prime}\\\
0&1\end{pmatrix}\right).$

We define the set
$\mathfrak{V}=\\{a,b,i,j,c,d,a^{\prime},b^{\prime},i^{\prime},j^{\prime},c^{\prime},d^{\prime},a^{\prime\prime},b^{\prime\prime},i^{\prime\prime},j^{\prime\prime},c^{\prime\prime},d^{\prime\prime}\\}$
and refer to the elements of the set as variables.

Note that we can consider the element $a\in\mathbb{Z}_{l_{2}}$ (resp.
$b\in\mathbb{Z}_{l_{2}}^{*}$) to be an element of $\mathbb{Z}_{N}$ (resp.
$\mathbb{Z}_{N}^{*}$) by setting $p_{u}(a)=0$ (resp. $p_{u}(b)=1$) for all
$p_{u}\in\pi(N)\setminus\pi(l_{2})$. Same treatment will be applicable to all
variables in $\mathfrak{V}$ accordingly. We observe that
$N=(k_{1},l_{2})(l_{1},l_{2})(k_{1},k_{2})(l_{1},k_{2})$ and the four entities
in the right are mutually coprime. Thus it is enough to count the total number
of possibilities of the variables in each of $\mathbb{Z}_{\beta}$, where
$\beta\in\\{(k_{1},l_{2}),(l_{1},l_{2}),(k_{1},k_{2}),(l_{1},k_{2})\\}$. Now
we look at the embeddings of the groups inside the holomorph. We will
encounter several equations in this context. We put the equations in section
5.

###  3.1. Embeddings

Considering equations 5.1, 5.19 (also 5.32) we get that $p_{u}(b)=1$ for all
$p_{u}\in\pi(N)$. Similarly $p_{u}(d)=1$ for all $p_{u}\in\pi(N)$. Since
$|r_{1}|=k_{1}$ we get that $p_{u}(a)$ is a unit whenever
$p_{u}\in\pi((k_{1},l_{2}))$ and $0$ for other primes (this is equivalent to
saying $|a|=(k_{1},l_{2})$). Similarly

  1. (1)

$p_{u}(i)$ is a unit for $p_{u}\in\pi((k_{1},k_{2}))$ and $0$ otherwise,

  2. (2)

$p_{u}(c)=0$ whenever $p_{u}\in\pi(n)\setminus\pi((k_{1},k_{2}))$,

  3. (3)

$p_{u}(i^{\prime\prime})$ is a unit for $p_{u}\in\pi((l_{1},k_{2}))$ and $0$
otherwise,

  4. (4)

$p_{u}(a^{\prime\prime})$ is a unit for $p_{u}\in\pi((l_{1},l_{2}))$ and $0$
otherwise.

Point $(3)$ and $(4)$ follows from Lemma 2.1. From equations 5.1 and 5.19 we
have that $p_{u}(b)=1$ for all $p_{u}\in\pi(N)$. In each of these following
cases we only determine the coefficients of the variables for the primes
relevant to that case.

Case I: Inside $\mathbb{Z}_{(k_{1},l_{2})}$ Using equations 5.15, 5.20 and
$b=1$, we have that $a(1+b^{\prime})=0$ thus $p_{u}(b^{\prime})=-1$ for
$p_{u}\in\pi((k_{1},l_{2}))$. Referring to equation 5.11 and $p_{u}(a)$ is a
unit for $p_{u}\in\pi((k_{1},l_{2}))$ we have that $p_{u}(b^{\prime\prime})=1$
for $p_{u}\in\pi(k_{1},l_{2})$. Using equation 5.37 we have
$p_{u}(a^{\prime\prime})=0$ for all $p_{u}\in\pi((k_{1},l_{2}))$ (since
$(2,p_{u})=1$). All other variables have one possibility since $k_{2}$ is
coprime to $(k_{1},l_{2})$. Hence

  1. (1)

$a$ has $\varphi(k_{1},l_{2})$ possibilities,

  2. (2)

$a^{\prime}$ has $(k_{1},l_{2})$ possibilities.

Case II: Inside $\mathbb{Z}_{(l_{1},l_{2})}$ Here $p_{u}(a)=0$ for all
$p_{u}\in\pi((l_{1},l_{2}))$. Using equation 5.14 (equiv. 5.27) we have
$p_{u}(b^{\prime})=\pm 1$ for all $p_{u}\in\pi((l_{1},l_{2}))$. Considering
equation 5.24 (equiv. 5.37) and that $1-p_{u}(b^{\prime\prime})$ is a zero
divisor for $p_{u}\in\pi((l_{1},l_{2}))$, using equation 5.15 (equiv. 5.28) we
get $p_{u}(b^{\prime})=1$ which implies that $p_{u}(a^{\prime})=0$ in this
case. Since $\mathfrak{R}(N)$ is a Burnside number, from equation 5.6 we have
that

(3.1) $p_{u}(b^{\prime\prime})^{p_{u}^{\alpha_{u}-1}}=1\text{ for all }p_{u}.$

Hence

  1. (1)

$(a^{\prime},b^{\prime})$ has $1$ possibility,

  2. (2)

$a^{\prime\prime}$ has $\varphi(l_{1},l_{2})$ possibilities,

  3. (3)

$b^{\prime\prime}$ has $\dfrac{(l_{1},l_{2})}{\mathfrak{R}((l_{1},l_{2}))}$
possibilities.

######  Remark 3.1.

Note that the above two cases do not depend on $j^{\prime}$.

Case III: Inside $\mathbb{Z}_{(k_{1},k_{2})}$ ($j^{\prime}=0$) We have
$p_{u}((i))$ is a unit for all $p_{u}\in\pi((k_{1},k_{2}))$. Combining
equations 5.18 and 5.21 we have $p_{u}(d^{\prime})=-1$ for all
$p_{u}\in\pi((k_{1},k_{2}))$. Since $(k_{1},k_{2})$ is coprime to $l_{1}$ and
$\mathfrak{R}(N)$ is a Burnside number, using equation 5.9 we conclude that
$p_{u}(d^{\prime\prime})=1$ for all $p_{u}\in\pi((k_{1},k_{2}))$. This implies
that $p_{u}(i^{\prime\prime})=p_{u}(c^{\prime\prime})=0$ for all
$p_{u}\in\pi((k_{1},k_{2}))$, since $(l_{1},(k_{1},k_{2}))=1$. Hence

  1. (1)

$i$ has $\varphi((k_{1},k_{2}))$ possibilities

  2. (2)

each of $c,i^{\prime},c^{\prime}$ has $(k_{1},k_{2})$ possibilities.

Case IV: Inside $\mathbb{Z}_{(l_{1},k_{2})}$ ($j^{\prime}=0$) We have
$p_{u}(i)=p_{u}(c)=0$ for all $p_{u}\in\pi((l_{1},k_{2}))$. Note that
$p_{u}(d^{\prime})=\pm 1$ for all $p_{u}\in\pi((l_{1},k_{2}))$. Using equation
5.25 we have

$\displaystyle
p_{u}(i^{\prime\prime})(1-p_{u}(d^{\prime}))=p_{u}(i^{\prime})(1-p_{u}(d^{\prime\prime}))\mod
p_{u}^{\alpha_{u}}\text{ for all }p_{u}\in\pi((l_{1},k_{2})).$

Then $p_{u}(d^{\prime})=-1$ for some $p_{u}\in\pi((l_{1},k_{2}))$ implies that

$\displaystyle
2p_{u}(i^{\prime\prime})=p_{u}(i^{\prime})(1-p_{u}(d^{\prime\prime}))\mod
p_{u}^{\alpha_{u}}\text{ for all }p_{u}\in\pi((l_{1},k_{2})).$

Since $2p_{u}(i^{\prime\prime})$ is a unit and $1-p_{u}(d^{\prime\prime})$
($\neq 0$) is a zero divisor, this case does not arise. Hence we get that
$p_{u}(d^{\prime})=1$ for all $p_{u}\in\pi((l_{1},k_{2}))$. This implies that
$p_{u}(i^{\prime})=p_{u}(c^{\prime})=0$ for all $p_{u}\in\pi((l_{1},k_{2}))$.
Similar to equation 3.1 $p_{u}(d^{\prime\prime})^{p_{u}^{\alpha_{u}-1}}=1$ for
all $p_{u}\in\pi((l_{1},k_{2}))$, since $\mathfrak{R}(n)$ is Burnside. Hence

  1. (1)

$i^{\prime\prime}$ has $\varphi((l_{1},k_{2}))$ possibilities

  2. (2)

$c^{\prime\prime}$ has $(l_{1},k_{2})$ possibilities

  3. (3)

$d^{\prime\prime}$ has $\dfrac{(l_{1},k_{2})}{\mathfrak{R}((l_{1},k_{2}))}$
possibilities.

Case V: Inside $\mathbb{Z}_{(k_{1},k_{2})}$ ($j^{\prime}=1$) We have
$p_{u}(d^{\prime\prime})=1,p_{u}(c^{\prime\prime})=p_{u}(i^{\prime\prime})=0$
for all $p_{u}\in\pi((k_{1},k_{2}))$ (as in Case III). Note that
$p_{u}(d^{\prime})=\pm 1$ for all $p_{u}\in\pi((k_{1},k_{2}))$. First let us
assume that $p_{u}(d^{\prime})=1$ for some $p_{u}$. Then $p_{u}(c^{\prime})=0$
and hence $i^{\prime\prime}\in\mathbb{Z}_{p_{u}^{\alpha_{u}}}$. Using equation
5.36 we conclude that $p_{u}(c)=0$.

Next, if $p_{u}(d^{\prime})=-1$ for some $p_{u}$, using equation 5.34

$\displaystyle p_{u}(c^{\prime})=2p_{u}(i^{\prime})\mod p_{u}^{\alpha_{u}}.$

Also combining equations 5.31, 5.34 we get that $2p_{u}(i)=-p_{u}(c)$. Hence

  1. (1)

$i$ has $\varphi((k_{1},k_{2}))$ possibilities,

  2. (2)

$(d^{\prime},i^{\prime})$ has $2^{|\pi((k_{1},k_{2}))|}(k_{1},k_{2})$
possibilities.

Case VI: Inside $\mathbb{Z}_{(l_{1},k_{2})}$ ($j^{\prime}=1$) We have
$p_{u}(i)=p_{u}(c)=0$ for all $p_{u}\in\pi((l_{1},k_{2}))$. Let us assume that
$p_{u}(d^{\prime})=1$ for some $p_{u}\in\pi((l_{1},k_{2}))$. Then
$p_{u}(c^{\prime})=0$. Then using equation 5.38 we have

$\displaystyle
p_{u}(i^{\prime})(1-p_{u}(d^{\prime\prime}))=2p_{u}(i^{\prime\prime})+p_{u}(c^{\prime\prime})\mod
p_{u}^{\alpha_{u}}.$

On the other hand if $p_{u}(d^{\prime})=-1$ for some
$p_{u}\in\pi((l_{1},k_{2}))$ we get that
$p_{u}(c^{\prime})=2p_{u}(i^{\prime})$ and using equation 5.38 we get

$\displaystyle
p_{u}(i^{\prime})(1-p_{u}(d^{\prime\prime}))=p_{u}(c^{\prime\prime})\mod
p_{u}^{\alpha_{u}}.$

Hence in either of the cases $p_{u}(c^{\prime}),p_{u}(c^{\prime\prime})$ get
fixed by $p_{u}(d^{\prime}),p_{u}(i^{\prime})$. Hence

  1. (1)

$(d^{\prime},i^{\prime})$ has possibilities
$2^{|\pi((l_{1},k_{2}))|}(l_{1},k_{2})$ possibilities,

  2. (2)

$i^{\prime\prime}$ has $\varphi((l_{1},k_{2}))$ possibilities (as in Case IV)

  3. (3)

$d^{\prime\prime}$ has $\dfrac{(l_{1},k_{2})}{\mathfrak{R}((l_{1},k_{2}))}$
possibilities (as in Case IV).

###  3.2. Regularity

Now we check the regularity of these groups. Note that any element $\sigma$ in
the image of $\Phi$ is of the form

$\displaystyle\left(\begin{pmatrix}\widetilde{b}&\widetilde{a}\\\
0&1\end{pmatrix},r_{2}^{\widetilde{i}}s_{2}^{\widetilde{j}},\begin{pmatrix}\widetilde{d}&\widetilde{c}\\\
0&1\end{pmatrix}\right).$

Since $\Phi$ is a homomorphism, this element corresponds to some
$\Phi(r_{1})^{\lambda}\Phi(s_{1})^{\lambda^{\prime}}\Phi(t_{1})^{\lambda^{\prime\prime}}$,
where $0\leq\lambda\leq l_{1}-1,0\leq\lambda^{\prime}\leq
1,0\leq\lambda^{\prime\prime}\leq k_{1}-1$. First we consider the case when
$j^{\prime}=0$. Note that in this case

$\displaystyle\Phi(r_{1})^{\lambda}\Phi(s_{1})$ $\displaystyle=$
$\displaystyle\left(\begin{pmatrix}1&\lambda a\\\
0&1\end{pmatrix},r_{2}^{\lambda i},\begin{pmatrix}1&\lambda c\\\
0&1\end{pmatrix}\right)\left(\begin{pmatrix}b^{\prime}&a^{\prime}\\\
0&1\end{pmatrix},r_{2}^{i^{\prime}},\begin{pmatrix}d^{\prime}&c^{\prime}\\\
0&1\end{pmatrix}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{pmatrix}b^{\prime}&\lambda a+a^{\prime}\\\
0&1\end{pmatrix},r_{2}^{\lambda
i+i^{\prime}},\begin{pmatrix}d^{\prime}&\lambda c+c^{\prime}\\\
0&1\end{pmatrix}\right).$

By Chinese Remainder theorem there exists $0<\lambda<k_{1}$, such that
$\lambda a+a^{\prime}=0\mod(k_{1},l_{2})$ and $\lambda
i+i^{\prime}=0\mod(k_{1},k_{2})$. Also we have

$\displaystyle p_{u}(b^{\prime})$ $\displaystyle=\begin{cases}-1&\text{for all
}p_{u}\in\pi((k_{1},l_{2}))\\\ 1&\text{for all
}p_{u}\in\pi((l_{1},l_{2}))\end{cases},$ $\displaystyle p_{u}(d^{\prime})$
$\displaystyle=\begin{cases}-1&\text{for all }p_{u}\in\pi((k_{1},k_{2}))\\\
1&\text{for all }p_{u}\in\pi((l_{1},k_{2}))\end{cases}.$

Since for any such $0<\lambda<k_{1}$ the element
$\Phi(r_{1})^{\lambda}\Phi(s_{1})$ is non-trivial. Hence the group generated
in this case is not regular. Thus $j^{\prime}=0$ is not possible.

In case $j^{\prime}=1$, any term of the form
$\Phi(r_{1})^{\lambda}\Phi(s_{1})\Phi(t_{1})^{\lambda^{\prime\prime}}$ is an
element of a regular subgroup. Hence to check regularity we need to consider
the terms $\Phi(r_{1})^{\lambda}\Phi(t_{1})^{\lambda^{\prime\prime}}$ with
$\widetilde{a}=0,\widetilde{i}=0$. We have,

$\displaystyle
a^{\prime\prime}(1+b^{\prime\prime}+\cdots+(b^{\prime\prime})^{\lambda^{\prime\prime}-1})$
$\displaystyle=-\lambda a\mod l_{2}$ $\displaystyle
i^{\prime\prime}(1+d^{\prime\prime}+\cdots+(d^{\prime\prime})^{\lambda^{\prime\prime}-1})$
$\displaystyle=-\lambda i\mod k_{2}.$

Since $p_{u}(i^{\prime\prime})=0$ and $p_{u}(i)$ is a unit for all
$p_{u}\in\pi((k_{1},k_{2}))$, we get that $p_{u}(\lambda)=0$ therein. One can
also check that $p_{u}(\lambda)=0$ for all $p_{u}\in\pi((k_{1},l_{2}))$. Hence
$\lambda=0$. Similar as before, using lemma 2.1, we have
$\lambda^{\prime\prime}=0$.

######  Proposition 3.2.

If $\Gamma=\mathfrak{M}_{k_{1},l_{1}}$ and $G=\mathfrak{M}_{k_{2},l_{2}}$,
where $k_{1}l_{1}=k_{2}l_{2}=N$ is an odd number and $\mathfrak{R}(N)$ is a
Burnside number then

$\displaystyle
e^{\prime}(\Gamma,G)=\dfrac{l_{1}N}{k_{1}(l_{1},l_{2})\mathfrak{R}(l_{1})}\cdot
2^{|\pi(k_{2})|}.$

###### Proof.

From the above discussion it is evident that $j^{\prime}=1$. Thus to determine
the total number of regular embeddings we have to multiply the number of
possibilities obtained in Cases I,II,V,VI and divide it by
$\textup{Aut}(\Gamma)$. Indeed, if $\Phi_{1}(\Gamma)=\Phi_{2}(\Gamma)$ for two
different embeddings $\Phi_{1},\Phi_{2}$ then $\Phi_{1}^{-1}\Phi_{2}$ is an
automorphism of $\Gamma$. Also if $\xi$ is an automorphism of $\Phi(\Gamma)$,
then $\xi\Phi$ is also a regular embedding of $\Gamma$. Hence

$\displaystyle
e^{\prime}(\mathfrak{M}_{k_{1},l_{1}},\mathfrak{M}_{k_{2},l_{2}})$
$\displaystyle=$
$\displaystyle\dfrac{\varphi((k_{1},l_{2}))(k_{1},l_{2})\varphi((l_{1},l_{2}))(l_{1},l_{2})\varphi(k_{1},k_{2})2^{|\pi((k_{1},k_{2}))|}(k_{1},k_{2})2^{|\pi((l_{1},k_{2}))|}(l_{1},k_{2})^{2}\varphi(l_{1},k_{2})}{\mathfrak{R}((l_{1},l_{2}))|\textup{Aut}(\mathfrak{M}_{k_{1},l_{1}})|\mathfrak{R}((l_{1},k_{2}))}$
$\displaystyle=$
$\displaystyle\dfrac{\varphi(N)(l_{1},k_{2})N2^{|\pi(k_{2})|}}{\mathfrak{R}(l_{1})|\textup{Aut}(\mathfrak{M}_{k_{1},l_{1}})|}$
$\displaystyle=$
$\displaystyle\dfrac{\varphi(N)(l_{1},k_{2})N2^{|\pi(k_{2})|}}{\mathfrak{R}(l_{1})\varphi(N)k_{1}}$
$\displaystyle=$
$\displaystyle\dfrac{l_{1}(l_{1},k_{2})2^{|\pi(k_{2})|}}{\mathfrak{R}(l_{1})}$
$\displaystyle=$
$\displaystyle\dfrac{l_{1}N}{k_{1}(l_{1},l_{2})\mathfrak{R}(l_{1})}\cdot
2^{|\pi(k_{2})|}.$

Last equality is obvious and this finishes the proof of the proposition. ∎

###### Proof of Theorem 1.3.

Using Lemma 1.2, we have that

$\displaystyle e(\mathfrak{M}_{k_{1},l_{1}},\mathfrak{M}_{k_{2},l_{2}})$
$\displaystyle=$
$\displaystyle\dfrac{\textup{Aut}(\mathfrak{M}_{k_{1},l_{1}})}{\mathfrak{M}_{k_{2},l_{2}}}e^{\prime}(\mathfrak{M}_{k_{1},l_{1}},\mathfrak{M}_{k_{2},l_{2}})$
$\displaystyle=$
$\displaystyle\dfrac{l_{1}l_{2}}{(l_{1},l_{2})\mathfrak{R}(l_{1})}\cdot
2^{|\pi(k_{2})|}.$

∎

######  Remark 3.3.

The case where $l_{1}=1$ i.e. when
$\mathfrak{M}_{k_{1},l_{1}}\cong\mathcal{D}_{2N}$, does not need the
assumption that $\mathfrak{R}(N)$ is a Burnside number.

##  4\. Further results

###  4.1. Non-classical Dihedral Hopf-Galois structures

Now consider the group $\mathcal{D}_{2k}\times\mathcal{C}_{l}$ where
$kl=N,(k,l)\neq 1$. We show that
$\mathcal{D}_{2N}\not\hookrightarrow\textup{Hol}(\mathcal{D}_{2k}\times\mathcal{C}_{l})$.
We will need the following lemma

######  Lemma 4.1.

[B08, Theorem $3.2$] Let $G=H\times K$, where $H$ and $K$ have no common
direct factor. Then

$\textup{Aut}(G)=\left\\{\begin{pmatrix}A&B\\\
C&D\end{pmatrix}\Bigg{|}\begin{tabular}[]{cc}$A\in\textup{Aut}(H)$&$B\in\textup{Hom}(K,Z(H))$\\\
$C\in\textup{Hom}(H,Z(K))$&$D\in\textup{Aut}(K)$\end{tabular}\right\\}.$

######  Corollary 4.2.

The group $\textup{Hol}(\mathcal{D}_{2k}\times\mathcal{C}_{l})$ where $kl=N$
is odd ,$(k,l)\neq 1$ does not have any element of order $p^{v_{p}(N)}$ where
$p|(k,l)$.

###### Proof.

Setting $G=\mathcal{D}_{2k},H=\mathcal{C}_{l}$, we observe that

  1. (1)

$\textup{Hom}(K,Z(H))=1$. Indeed $Z(H)=\mathcal{C}_{l}$ is a group of odd
order, it has no element of order $2$.

  2. (2)

$\textup{Hom}(H,Z(K))=1$ since $Z(K)$ is trivial.

This implies that

$\displaystyle\textup{Hol}(\mathcal{D}_{2k}\times\mathcal{C}_{l})$
$\displaystyle=\mathcal{D}_{2k}\times\mathcal{C}_{l}\rtimes_{id}\textup{Aut}(\mathcal{D}_{2k}\times\mathcal{C}_{l})$
$\displaystyle\cong\textup{Hol}(\mathcal{D}_{2k})\times\textup{Hol}(\mathcal{C}_{l}).$

Since none of $\textup{Hol}(\mathcal{D}_{2k}),\textup{Hol}(\mathcal{C}_{l})$
has elements of order $p^{v_{p}(N)}$, the result follows. ∎

######  Corollary 4.3.

If $N$ is odd then
$e(\mathcal{D}_{2N},\mathcal{D}_{2k}\times\mathcal{C}_{l})=0$, whenever
$(k,l)\neq 1$.

######  Corollary 4.4.

Let $L/K$ be a finite Galois extension with Galois group isomorpic to
$\mathcal{D}_{2N}$ where $N$ is odd. Then the number of Hopf-Galois structures
on $L/K$ is at least

$\displaystyle\displaystyle{\sum\limits_{M=0}^{N}2^{M}\chi(N-M)},$

where $\chi(L)$ is the coefficient of $x^{L}$ in the polynomial
$\prod\limits_{p_{u}\in\pi(N)}(x+p_{u}^{\alpha_{u}})$.

###  4.2. Skew-braces

######  Definition 4.5.

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 $a,b,c\in\Gamma$.

Skew braces give non-degenerate set theoretic solutions of the Yang-Baxter
equation. It initially appeared in the PhD thesis of D. Bachiller and has been
studied in [BCJ16], [CJO14] et cetera. Skew braces provide group theoretic and
ring theoretic methods to understand solutions of the Yang Baxter equations.
We will need the following result which connects the skew braces and regular
subgroups.

######  Lemma 4.6.

[SV18, Proposition $A.3$] Let $\Gamma$ be a group. There exists a bijective
correspondence between isomorphism classes of skew braces with additive group
isomorphic to $\Gamma$ and classes of regular subgroups of
$\textup{Hol}(\Gamma)$ under conjugation by elements of
$\textup{Aut}(\Gamma)$.

######  Corollary 4.7.

If $(\Gamma,+)\cong\mathfrak{M}_{k_{1},l_{1}}$ and
$(\Gamma,\times)\cong\mathfrak{M}_{k_{2},l_{2}}$, where
$k_{1}l_{1}=k_{2}l_{2}=N$ is an odd number and $\mathfrak{R}(N)$ is a Burnside
number then the number of skew braces of the form $(\Gamma,+,\times)$ is given
by

$\displaystyle\dfrac{l_{1}N}{k_{1}(l_{1},l_{2})\mathfrak{R}(l_{1})}\cdot
2^{|\pi(k_{2})|}.$

###### Proof.

Follows from Proposition 3.2. ∎

##  5\. Appendix: Equations

All the notations here are adopted from section 3. Since $\Phi$ is a
homomorphism, we must have the following relations:

$\displaystyle\Phi(r_{1})^{k_{1}}$ $\displaystyle=e_{0}$
$\displaystyle\Phi(s_{1})^{2}$ $\displaystyle=e_{0}$
$\displaystyle\Phi(t_{1})^{l_{1}}$ $\displaystyle=e_{0}$
$\displaystyle\Phi(s_{1})\Phi(r_{1})\Phi(s_{1})\Phi(r_{1})$
$\displaystyle=e_{0}$ $\displaystyle\Phi(r_{1})\Phi(t_{1})$
$\displaystyle=\Phi(t_{1})\Phi(r_{1})$ $\displaystyle\Phi(s_{1})\Phi(t_{1})$
$\displaystyle=\Phi(t_{1})\Phi(s_{1}),$

where

$e_{0}=\left(\begin{pmatrix}1&0\\\
0&1\end{pmatrix},r_{2}^{0}s_{2}^{0},\begin{pmatrix}1&0\\\
0&1\end{pmatrix}\right)$

is the identity element of $\textup{Hol}(\mathfrak{M}_{k_{2},l_{2}})$. First
we observe that if $j=1$, then $\Phi(r_{1})$ has even order. Indeed

$\displaystyle\Phi(r_{1})^{2}=\left(\begin{pmatrix}b^{2}&a(1+b)\\\
0&1\end{pmatrix},r_{2}^{i(1-d)-c},\begin{pmatrix}d^{2}&c(1+d)\\\
0&1\end{pmatrix}\right)$ $\displaystyle\implies$
$\displaystyle\Phi(r_{1})^{2m+1}=\left(\begin{pmatrix}b^{2m+1}&{a}(1+b+\cdots+b^{2m})\\\
0&1\end{pmatrix},r_{2}^{\bar{i}}s,\begin{pmatrix}d^{2m+1}&c(1+d+\cdots+d^{2m})\\\
0&1\end{pmatrix}\right).$

Since $k_{1}$ is odd, this possibility does not arise. Similarly we can
conclude that $j^{\prime\prime}=0$, since $l_{1}$ is odd. Using
$\Phi(r_{1})^{k_{1}}=e_{0}$ we have

$\displaystyle\left(\begin{pmatrix}b^{k_{1}}&{a}(1+b+\cdots+b^{k_{1}-1})\\\
0&1\end{pmatrix},r_{2}^{i(1+d+\cdots+d^{k_{1}-1})}s,\begin{pmatrix}d^{k_{1}}&c(1+d+\cdots+d^{k_{1}-1})\\\
0&1\end{pmatrix}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{pmatrix}1&0\\\
0&1\end{pmatrix},r_{2}^{0}s_{2}^{0},\begin{pmatrix}1&0\\\
0&1\end{pmatrix}\right),$

which implies that

(5.1) $\displaystyle b^{k_{1}}$ $\displaystyle=1\mod{l_{2}}$ (5.2)
$\displaystyle{a}(1+b+\cdots+b^{k_{1}-1})$ $\displaystyle=0\mod{l_{2}}$ (5.3)
$\displaystyle i(1+d+\cdots+d^{k_{1}-1})$ $\displaystyle=0\mod{k_{2}}$ (5.4)
$\displaystyle d^{k_{1}}$ $\displaystyle=1\mod{k_{2}}$ (5.5) $\displaystyle
c(1+d+\cdots+d^{k_{1}-1})$ $\displaystyle=0\mod{k_{2}}.$

Using $\Phi(t_{1})^{l_{1}}=e_{0}$ we get that

(5.6) $\displaystyle(b^{\prime\prime})^{l_{1}}$ $\displaystyle=1\mod{l_{2}}$
(5.7)
$\displaystyle{a^{\prime\prime}}(1+b^{\prime\prime}+\cdots+(b^{\prime\prime})^{l_{1}-1})$
$\displaystyle=0\mod{l_{2}}$ (5.8) $\displaystyle
i(1+d^{\prime\prime}+\cdots+(d^{\prime\prime})^{l_{1}-1})$
$\displaystyle=0\mod{k_{2}}$ (5.9) $\displaystyle(d^{\prime\prime})^{l_{1}}$
$\displaystyle=1\mod{k_{2}}$ (5.10) $\displaystyle
c^{\prime\prime}(1+d^{\prime\prime}+\cdots+(d^{\prime\prime})^{l_{1}-1})$
$\displaystyle=0\mod{k_{2}}.$

Using $\Phi(r_{1})\Phi(t_{1})=\Phi(t_{1})\Phi(r_{1})$ we get that

(5.11) $\displaystyle a(1-b^{\prime\prime})$ $\displaystyle=0\mod l_{2}$
(5.12) $\displaystyle i(1-d^{\prime\prime})$ $\displaystyle=0\mod k_{2}$
(5.13) $\displaystyle c(1-d^{\prime\prime})$ $\displaystyle=0\mod k_{2}.$

Now we divide the set of equations in two parts considering $j^{\prime}=0$ and
$j^{\prime}=1$.

###  5.1. Case 1: $j^{\prime}=0$

Using $\Phi(s_{1})^{2}=e_{0}$ we have

(5.14) $\displaystyle(b^{\prime})^{2}$ $\displaystyle=1\mod l_{2}$ (5.15)
$\displaystyle a^{\prime}(1+b^{\prime})$ $\displaystyle=0\mod l_{2}$ (5.16)
$\displaystyle(d^{\prime})^{2}$ $\displaystyle=1\mod k_{2}$ (5.17)
$\displaystyle c^{\prime}(1+d^{\prime})$ $\displaystyle=0\mod k_{2}$ (5.18)
$\displaystyle i^{\prime}(1+d^{\prime})$ $\displaystyle=0\mod k_{2}.$

Using $\Phi(s_{1})\Phi(r_{1})\Phi(s_{1})\Phi(r_{1})=e_{0}$ we have

(5.19) $\displaystyle b^{2}$ $\displaystyle=1\mod l_{2}$ (5.20) $\displaystyle
a(b+b^{\prime})+a^{\prime}(1+bb^{\prime})$ $\displaystyle=0\mod l_{2}$ (5.21)
$\displaystyle(i+i^{\prime})(1+d^{\prime})$ $\displaystyle=0\mod k_{2}$ (5.22)
$\displaystyle d^{2}$ $\displaystyle=1\mod k_{2}$ (5.23) $\displaystyle
c(d+d^{\prime})+c^{\prime}(1+dd^{\prime})$ $\displaystyle=0\mod k_{2}$

Note that $b=1,d=1$ by subsection $4.1$.

Using $\Phi(s_{1})\Phi(t_{1})=\Phi(t_{1})\Phi(s_{1})$ we have

(5.24) $\displaystyle a^{\prime\prime}(1-b^{\prime})$
$\displaystyle=a^{\prime}(1-b^{\prime\prime})\mod{l_{2}}$ (5.25)
$\displaystyle i^{\prime\prime}(1-d^{\prime})$
$\displaystyle=i^{\prime}(1-d^{\prime\prime})\mod{k_{2}}$ (5.26)
$\displaystyle c^{\prime\prime}(1-d^{\prime})$
$\displaystyle=c^{\prime}(1-d^{\prime\prime})\mod{k_{2}}.$

###  5.2. Case 2: $j^{\prime}=1$

Using $\Phi(s_{1})^{2}=e_{0}$ we have

(5.27) $\displaystyle(b^{\prime})^{2}$ $\displaystyle=1\mod l_{2}$ (5.28)
$\displaystyle a^{\prime}(1+b^{\prime})$ $\displaystyle=0\mod l_{2}$ (5.29)
$\displaystyle(d^{\prime})^{2}$ $\displaystyle=1\mod k_{2}$ (5.30)
$\displaystyle c^{\prime}(1+d^{\prime})$ $\displaystyle=0\mod k_{2}$ (5.31)
$\displaystyle i^{\prime}(1-d^{\prime})$ $\displaystyle=c^{\prime}\mod k_{2}.$

Using $\Phi(s_{1})\Phi(r_{1})\Phi(s_{1})\Phi(r_{1})=e_{0}$ we have

(5.32) $\displaystyle b^{2}$ $\displaystyle=1\mod l_{2}$ (5.33) $\displaystyle
a(b+b^{\prime})+a^{\prime}(1+bb^{\prime})$ $\displaystyle=0\mod l_{2}$ (5.34)
$\displaystyle(i+i^{\prime})(1-d^{\prime})$
$\displaystyle=d^{\prime}c+c^{\prime}\mod k_{2}$ (5.35) $\displaystyle d^{2}$
$\displaystyle=1\mod l_{2}$ (5.36) $\displaystyle
c(d+d^{\prime})+c^{\prime}(1+dd^{\prime})$ $\displaystyle=0\mod l_{2}$

Note that $b=1,d=1$ by subsection $4.1$.

Using $\Phi(s_{1})\Phi(t_{1})=\Phi(t_{1})\Phi(s_{1})$ we have

(5.37) $\displaystyle a^{\prime\prime}(1-b^{\prime})$
$\displaystyle=a^{\prime}(1-b^{\prime\prime})\mod{l_{2}}$ (5.38)
$\displaystyle i^{\prime}-i^{\prime\prime}d^{\prime}$
$\displaystyle=i^{\prime\prime}+i^{\prime}d^{\prime\prime}+c^{\prime\prime}\mod{k_{2}}$
(5.39) $\displaystyle c^{\prime\prime}(1-d^{\prime})$
$\displaystyle=c^{\prime}(1-d^{\prime\prime})\mod{k_{2}}.$

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