# On the number of quaternion and dihedral braces and Hopf–Galois structures

Nigel P. Byott  Department of Mathematics & Statistics, University of Exeter,
North Park Road, Exeter EX4 4QF, United Kingdom<EMAIL_ADDRESS>https://mathematics.exeter.ac.uk/staff/NPByott and  Fabio Ferri  131 Hayward
Road, Bristol BS5 9PY, United Kingdom<EMAIL_ADDRESS>
(Date: version of )

###### Abstract.

We prove a conjecture of Guarnieri and Vendramin on the number of braces of a
given order whose multiplicative group is a generalised quaternion group. At
the same time, we give a similar result where the multiplicative group is
dihedral. We also enumerate Hopf-Galois structures of abelian type on Galois
extensions with generalised quaternion or dihedral Galois group.

###### Key words and phrases:

Braces; Hopf-Galois structures; Yang-Baxter equation; generalised quaternion
group; dihedral group

###### 2020 Mathematics Subject Classification:

20N99, 16T05, 12F10, 16T25

Byott was partially supported by the Engineering and Physical Sciences
Research Council [grant number EP/V005995/1]. Ferri was supported by the
Engineering and Physical Sciences Research Council [grant number
EP/W52265X/1].
For the purpose of open access, the authors have applied a CC BY public
copyright licence to any Author Accepted Manuscript version arising.
Data Access Statement: No data was used for the research described in this
article.

##  1\. Introduction

Braces were introduced by Rump [Rum07] as a generalisation of radical rings in
order to study set-theoretical solutions of the Yang-Baxter Equation (YBE).
Every brace can be viewed as a non-degenerate involutive solution of the YBE,
and any such solution can be embedded in a brace. Braces have since been
studied from a number of different perspectives, and generalised in a variety
of ways. Of particular relevance to this paper is the connection with Hopf-
Galois theory, first noted by Bachiller in [Bac16].

As reformulated in [CJO14], the definition of a brace may be given as follows.

######  Definition 1.1.

A left brace $(B,+,\circ)$ consists of a set $B$ and two binary operations
$+$, $\circ$ such that

  * (i)

$(B,+)$ is an abelian group;

  * (ii)

$(B,\circ)$ is a group;

  * (iii)

the brace relation $a\circ(b+c)=a\circ b-a+a\circ c$ holds for all $a$, $b$,
$c\in B$.

The left brace $(B,+,\circ)$ is trivial if the operations $+$, $\circ$
coincide.

There is an analogous notion of right brace, where (iii) is replaced by
$(a+b)\circ c=a\circ c-c+b\circ c$. In this paper, we only consider left
braces, so we will omit the adjective “left”.

Guarnieri and Vendramin [GV17] generalised braces to skew braces by removing
the requirement that the group $(B,+)$ be abelian. In the same paper, they
presented the results of computer calculations counting braces (and skew
braces) of small order. On the basis of these, they formulated a number of
conjectures, including the following:

######  Conjecture 1.2.

[GV17, Conjecture 6.6] Let $m\geq 3$ and let $q(4m)$ be the number of braces
$B$ whose multiplicative group $(B,\circ)$ is a generalised quaternion group
of order $4m$. Then

$q(4m)=\begin{cases}2&\mbox{ if }m\mbox{ is odd,}\\\ 7&\mbox{ if }m\equiv
0\pmod{8},\\\ 9&\mbox{ if }m\equiv 4\pmod{8},\\\ 6&\mbox{ if }m\equiv
2\pmod{8}\mbox{ or }m\equiv 6\pmod{8}.\end{cases}$

We note that some other enumerative conjectures from [GV17] have been proved
in [Smo18] and [Die21]. For further computational results counting braces and
skew braces, see [Ven19], [BNY20], [KSV21]. Based on their extensive
computations in [BNY20], Bardakov, Neshchadim and Yadav made conjectures on
the number of braces of order $8p$ and $12p$ for large enough primes $p$.
These have recently been proved by Crespo, Gil-Muñoz, Rio and Vela [CGMRV23a,
CGMRV23b], and we will use similar techniques in this paper; cf. Remark 12.9.

In this paper, we will prove Conjecture 1.2 (see Theorem 13.1). We will also
answer Questions 6.7 and 6.8 of [GV17] which concern the additive groups
$(B,+)$ for these braces (see Table 2). At the same time, we give
corresponding results for braces where $(B,\circ)$ is a dihedral group. We
also give enumerative results on the Hopf-Galois structures related to these
braces. More precisely, given a Galois extension of fields $L/K$ whose Galois
group is a generalised quaternion or dihedral group of a given order, we count
the number of Hopf-Galois structures on $L/K$ of each possible abelian type.

The connection between (skew) braces and Hopf-Galois theory, which is
fundamental to our approach in this paper, comes about because both can be
described in terms of regular subgroups in the holomorph of a group $N$. A
subgroup $M$ in the group $\mathrm{Perm}(X)$ of permutations of a set $X$ is
said to be regular if, given any $x$, $y\in X$, there is a unique $\pi\in M$
with $\pi(x)=y$. The conjugate of a regular subgroup by any element of
$\mathrm{Perm}(X)$ is again regular. For any group $N$, we define the
holomorph $\mathrm{Hol}(N)$ of $N$ to be $N\rtimes\mathrm{Aut}(N)$, and we
view $\mathrm{Hol}(N)$ as a group of permutations of $N$ (so that the normal
subgroup $N$ of $\mathrm{Hol}(N)$ corresponds to left translations). We next
explain how regular subgroups are related to skew braces and to Hopf-Galois
structures.

On the one hand, if $(B,+,\circ)$ is a (skew) brace, then there is a
homomorphism of groups $(B,\circ)\to\mathrm{Aut}(B,+)$ given by
$a\mapsto\lambda_{a}$, where $\lambda_{a}(b)=-a+a\circ b$. The map
$a\mapsto(a,\lambda_{a})$ embeds $(B,\circ)$ as a regular subgroup of
$\mathrm{Hol}(B,+)$. Conversely, every regular subgroup $G$ of the holomorph
of a group $N$ gives rise to a skew brace $(B,+,\circ)$ with additive group
$N$ and multiplicative group $G$ [GV17, §4]. Two regular subgroups $G_{1}$ and
$G_{2}$ of $\mathrm{Hol}(N)$ correspond to isomorphic skew braces if and only
if they are conjugate by an element of $\mathrm{Aut}(N)$ [SV18, Proposition
A.3]. By [BNY20, Lemma 2.1], $G_{1}$ and $G_{2}$ are conjugate by an element
of $\mathrm{Aut}(N)$ if and only if they are conjugate by an element of
$\mathrm{Hol}(N)$, so the isomorphism classes of skew braces with additive
group $N$ correspond to conjugacy classes of regular subgroups in
$\mathrm{Hol}(N)$.

On the other hand, let $L/K$ be a finite Galois extension of fields with
Galois group $G$. Greither and Pareigis [GP87] showed that Hopf-Galois
structures on $L/K$ correspond to regular subgroups $N$ in $\mathrm{Perm}(G)$
which are normalised by left translations by elements of $G$. We will call the
isomorphism class of $N$ the type of the corresponding Hopf-Galois structure.
Starting with an abstract group $N$, one can determine the Hopf-Galois
structures of type $N$ on $L/K$ from the regular subgroups of
$\mathrm{Hol}(N)$ which are isomorphic to $G$. More precisely, if
$e^{\prime}(G,N)$ is the number of these regular subgroups, and $e(G,N)$ is
the number of Hopf-Galois structures of type $N$ on $L/K$, then from [Byo96a]
we have the formula

(1.1) $e(G,N)=\frac{|\mathrm{Aut}(G)|}{|\mathrm{Aut}(N)|}\,e^{\prime}(G,N).$

For further background on Hopf-Galois structures, see [Chi00, Chapter 2].

In this paper we will be concerned with the case when $N$ is abelian and $G$
is either quaternion or dihedral. We will determine the abelian groups $N$ so
that $\mathrm{Hol}(N)$ contains a regular quaternion or dihedral subgroup $G$,
and we will count the number of such subgroups for each such $N$. This is the
first step both to determining the number of quaternion or dihedral braces
with additive group isomorphic to $N$, and to determining the number of Hopf-
Galois structures of type $N$ on a Galois extension of fields with quaternion
or dihedral Galois group. There are two, essentially distinct, parts in our
calculation of these numbers. Firstly, we consider the case when
$|G|=|N|=2^{n}$ for some $n\geq 2$. Our method here can be regarded as an
extension of the ideas introduced in the thesis of Featherstonhaugh [Fea03],
and further developed in [FCC12], which show that if $N$ is a finite abelian
$p$-group of rank $m<p-1$ then any regular abelian subgroup $G$ of
$\mathrm{Hol}(N)$ is isomorphic to $N$. In the setting of braces, Bachiller
[Bac16, Theorem 2.5] removed the requirement that $G$ be abelian, showing that
the number of elements of any given order is the same in $G$ and $N$. Using a
description of $\mathrm{Aut}(N)$ due to Hillar and Rhea [HR07], we obtain a
similar but weaker relation between the structures of $G$ and $N$ when the
hypothesis on $m$ is removed (Lemma 2.6). In particular, we show that if
$|N|=2^{n}$ and $\mathrm{Hol}(N)$ contains a regular subgroup $G$ which is a
dihedral or quaternion group then $N$ contains a cyclic subgroup of index at
most $8$ (Theorem 3.4). We can then find the numbers of braces and Hopf-Galois
structures in the $2$-power case by examining each of the possible groups $N$
in turn. Secondly, we show that the general case, $|N|=2^{n}s$ for $s$ odd,
can essentially be reduced to the $2$-power case. The numbers of braces and
Hopf-Galois structures are independent of $s$ when $s\geq 3$. (The number of
braces differs from that for $s=1$ in some cases.)

In some of our calculations, we use the computer algebra package Magma. All
the counts we give could in principle be obtained by purely theoretical
considerations, but to do so would have required the examination of a number
of special cases and would have significantly increased the length of the
paper.

Finally, we emphasize that (despite the occurrence of Conjecture 1.2 in the
paper [GV17] where skew braces were introduced) in this paper we are
considering braces rather than skew braces. Thus the additive group
$N\cong(B,+)$ is always assumed to be abelian.

##  2\. The holomorph of a finite abelian $p$-group

In this section, we recall from [HR07] a description of the endomorphism ring
of a finite abelian $p$-group $N$, and then use this to bound the order of
elements of $p$-power order in $\mathrm{Hol}(N)$ in terms of the rank and
exponent of $N$. Although only the case $p=2$ is needed later in the paper, in
this section we allow $p$ to be an arbitrary prime.

######  Proposition 2.1.

Let

$N=\mathbb{Z}/p^{a_{1}}\mathbb{Z}\times\mathbb{Z}/p^{a_{2}}\mathbb{Z}\times\cdots\times\mathbb{Z}/p^{a_{r}}\mathbb{Z}$

be a finite abelian $p$-group, with $1\leq a_{1}\leq\cdots\leq a_{r}$.

For $1\leq i,j\leq r$, let

$R_{ij}=p^{\max\\{0,a_{i}-a_{j}\\}}(\mathbb{Z}/p^{a_{r}}\mathbb{Z}),$

and let

$R:=(R_{ij})_{1\leq i,j\leq r},$

the set of matrices whose $(i,j)$-entry is in $R_{ij}$. Then $R$ has a ring
structure with sum and multiplication inherited from the ring
$\mathrm{Mat}_{r}(\mathbb{Z})$ of $r\times r$ matrices over $\mathbb{Z}$, and
there exists a surjective ring homomorphism

$f:R\twoheadrightarrow\mathrm{End}(N).$

Moreover,

$\ker f=(p^{a_{i}}(\mathbb{Z}/p^{a_{r}}\mathbb{Z}))_{i,j}\subseteq R.$

Finally, an element of $\mathrm{End}(N)$ represented by $(r_{ij})_{i,j}\in R$
is in $\mathrm{Aut}(N)$ if and only if the reduction of $(r_{ij})_{i,j}$ in
$\mathrm{Mat}_{r}(\mathbb{F}_{p})$ is invertible.

###### Proof.

By looking at the image of the elements of the canonical basis, one can verify
that an element of $R$ acts as an endomorphism of $N$ via its matrix
representation, and that each endomorphism has a representative in $R$. A
complete proof can be found in [HR07, Theorem 3.3]. In [HR07, Lemma 3.4] the
authors show, using adjoint matrices, that $\ker f$ consists of the matrices
whose $(i,j)$-entry is divisible by $p^{a_{i}}$. In [HR07, Theorem 3.6] they
show that the elements of $\mathrm{Aut}(N)$ are those corresponding to the
invertible matrices modulo $p$. It is worth remarking that all the proofs are
straightforward, although they involve several necessary steps to be verified.
∎

Proposition 2.1 means that we can identify $\mathrm{End}(N)$ with the ring

$\overline{R}=R/\ker f$

of $r\times r$ matrices whose $i$th row consists of elements of
$\mathbb{Z}/p^{a_{i}}\mathbb{Z}$, with the condition that the $(i,j)$-entry is
divisible by $p^{a_{i}-a_{j}}$ when $a_{i}>a_{j}$. (This condition ensures
that multiplication of matrices in $\overline{R}$ is well-defined.) We
correspondingly view elements of $N$ as column vectors with $i$th entry in
$\mathbb{Z}/p^{a_{i}}\mathbb{Z}$. Then $\mathrm{Aut}(N)$ is identified with
the group $\overline{R}^{\times}$ of units in $\overline{R}$, which consists
of elements of $\overline{R}$ whose reduction in
$\mathrm{Mat}_{r}(\mathbb{F}_{p})$ lies in $\mathrm{GL}_{r}(\mathbb{F}_{p})$.

######  Corollary 2.2.

With the hypotheses and notation of Proposition 2.1, an element of
$\mathrm{Hol}(N)$ can be represented as

$\begin{pmatrix}A&v\\\ 0&1\end{pmatrix},$

where $A$ is a matrix in $\overline{R}$ whose reduction is in
$\mathrm{GL}_{r}(\mathbb{F}_{p})$, and where $v\in N$. (We view the last row
as having entries in $\mathbb{Z}/p^{a_{r}}\mathbb{Z}$.)

We next describe a particular Sylow $p$-subgroup in $\overline{R}^{\times}$.

######  Lemma 2.3.

Let $U$ be the subgroup of unipotent upper triangular matrices in
$\mathrm{GL}_{r}(\mathbb{F}_{p})$:

$(b_{ij})\in U\Leftrightarrow b_{ii}=1\mbox{ for all }i,\mbox{ and
}b_{ij}=0\mbox{ if }i>j.$

Let $P$ be the subgroup of $\overline{R}^{\times}$ consisting of matrices
whose reduction in $\mathrm{GL}_{r}(\mathbb{F}_{p})$ lies in $U$. Then $P$ is
a Sylow $p$-subgroup of $\overline{R}^{\times}$.

###### Proof.

Let $W$ be the image of $R^{\times}$ in $\mathrm{GL}_{r}(\mathbb{F}_{p})$.
Then $W$ consists of those invertible matrices $(b_{ij})$ with $b_{ij}=0$ if
$a_{i}>a_{j}$. In particular, $U\subseteq W$. It is well-known that $U$ is a
Sylow $p$-subgroup of $\mathrm{GL}_{r}(\mathbb{F}_{p})$, so it is also a Sylow
$p$-subgroup of $W$. The kernel $K$ of the reduction map
$\overline{R}\to\mathrm{Mat}_{r}(\mathbb{F}_{p})$ clearly has $p$-power order.
By Proposition 2.1, if $A$ is an element of $R$ whose reduction is invertible,
then $A$ is invertible in $R$. Thus the kernel of the reduction map on units
$\overline{R}^{\times}\to\mathrm{GL}_{r}(\mathbb{F}_{p})$ is $1+K$. Since
$1+K$ is a $p$-group, it follows that the preimage of $U$ in
$\overline{R}^{\times}$ is a Sylow $p$-subgroup of $\overline{R}^{\times}$. ∎

We now seek to relate possible orders of elements of $\mathrm{Hol}(N)$ to the
exponent of $N$. This is done in Lemma 2.6. We first give two preliminary
results.

######  Proposition 2.4.

Let $m\geq 1$ and let $t=\lceil\log_{p}m\rceil$. Then, for any
$X\in\mathrm{GL}_{m}(\mathbb{F}_{p})$ of $p$-power order, we have
$X^{p^{t}}=I$.

###### Proof.

Replacing $X$ by a conjugate, we may assume that $X$ lies in the Sylow
$p$-subgroup $U$ of $\mathrm{GL}_{m}(\mathbb{F}_{p})$ consisting of unipotent
upper triangular matrices. By induction on $0\leq k\leq m$, we easily see that
the $(i,j)$-entry of $(X-I)^{k}$ is $0$ unless $i+j>m+k$. In particular,
$(X-I)^{m}=0$. Since $p^{t}\geq m$ we have $X^{p^{t}}-I=(X-I)^{p^{t}}=0$. ∎

######  Proposition 2.5.

Let $m\geq 1$, let $t=\lceil\log_{p}m\rceil$, and let $d\geq 1$. Then, for any
matrix $X$ of $p$-power (multiplicative) order in
$\mathrm{Mat}_{m}(\mathbb{Z}/p^{d}\mathbb{Z})$, we have $X^{p^{t+d-1}}=I$.

###### Proof.

We argue by induction on $d$. The case $d=1$ is Proposition 2.4. Now suppose
that $d>1$ and the assertion holds for $d-1$. Let
$X\in\mathrm{Mat}_{m}(\mathbb{Z}/p^{d}\mathbb{Z})$ have $p$-power order.
Reducing mod $p^{d-1}$ and applying the induction hypothesis, we have

$X^{p^{c}}-I\in p^{d-1}\mathrm{Mat}_{m}(\mathbb{Z}/p^{d}\mathbb{Z})$

where $c=t+(d-1)-1$. Now

$X^{p^{c+1}}-I=(X^{p^{c}}-I)(X^{(p-1)p^{c}}+X^{(p-2)p^{c}}+\cdots+X^{p^{c}}+I).$

Since $d-1\geq 1$, we have $X^{p^{c}}\equiv
I\pmod{p\,\mathrm{Mat}_{m}(\mathbb{Z}/p^{d}\mathbb{Z})}$, so that

$X^{(p-1)p^{c}}+X^{(p-2)p^{c}}+\cdots+X^{p^{c}}+I\equiv pI\equiv
0\pmod{p\,\mathrm{Mat}_{m}(\mathbb{Z}/p^{d}\mathbb{Z})},$

and hence

$X^{p^{t+d-1}}-I=X^{p^{c+1}}-I\in(p^{d-1}\mathrm{Mat}_{m}(\mathbb{Z}/p^{d}\mathbb{Z}))(p\,\mathrm{Mat}_{m}(\mathbb{Z}/p^{d}\mathbb{Z}))=\\{0\\}.$

This completes the induction. ∎

######  Lemma 2.6.

Let $N$ be a finite abelian $p$-group of rank $r$ and exponent $p^{d}$. If
$\mathrm{Hol}(N)$ contains an element of order $p^{k}$ then
$k<\lceil\log_{p}(r+1)\rceil+d$.

###### Proof.

We may represent $N$ as in Proposition 2.1 with $a_{r}=d$. Let $m=r+1$. Then,
by Corollary 2.2, each element of $\mathrm{Hol}(N)$ can be represented as an
invertible $m\times m$ matrix whose $i$th row consists of elements of
$\mathbb{Z}/p^{a_{i}}\mathbb{Z}$. Let $X$ be such a matrix representing an
element of order $p^{k}$, and let $\widehat{X}$ be any matrix in
$\mathrm{Mat}_{m}(\mathbb{Z}/p^{d}\mathbb{Z})$ which reduces to $X$. Then
$\widehat{X}^{p^{k-1}}\neq I$ in
$\mathrm{Mat}_{m}(\mathbb{Z}/p^{d}\mathbb{Z})$, so by Proposition 2.5 we have
$k-1<\lceil\log_{p}(r+1)\rceil+d-1$. ∎

In the setting of Lemma 2.6, if $\mathrm{Hol}(N)$ contains an element of order
$p^{d+1}$ then $r\geq p$. Thus if $r\leq p-1$ then the exponent of a
$p$-subgroup of $\mathrm{Hol}(N)$ is at most $p^{d}$. For related results on
braces and Hopf-Galois structures in the case $r\leq p-1$, see [Bac16] and
[FCC12] respectively. Our application of Lemma 2.6 will be for $p=2$ and
$r>p$.

##  3\. Preliminary results on quaternion and dihedral braces

######  Notation 3.1.

Let $n\geq 2$ be an integer and let $s$ be an odd number. We will denote a
quaternion, respectively dihedral, group $G$ by $\langle x,y\rangle_{o}$ if

$G=Q_{2^{n}s}=\langle
x,y:x^{2^{n-1}s}=1,yx=x^{-1}y,y^{2}=x^{2^{n-2}s}\rangle,$

respectively

$G=D_{2^{n}s}=\langle x,y:x^{2^{n-1}s}=1,yx=x^{-1}y,y^{2}=1\rangle$

where we mean that the order of $x$ in $G$ is exactly $2^{n-1}s$, (note that
this is stronger than simply saying that $G$ is generated by $x$ and $y$ as we
are prescribing their roles as elements of the group). We include the
degenerate case $n=2$ and $s=1$, so that $Q_{4}\cong C_{4}$ and $D_{4}\cong
C_{2}\times C_{2}$.

######  Lemma 3.2.

Let $n\geq 4$ be an integer and let $G_{1}=\langle x_{1},y_{1}\rangle_{o}$ and
$G_{2}=\langle x_{2},y_{2}\rangle_{o}$ be two groups isomorphic to either
$Q_{2^{n}}$ or $D_{2^{n}}$. Let $f:G_{1}\rightarrow G_{2}$ be an isomorphism.
Then $f(x_{1})$ is an odd power of $x_{2}$.

###### Proof.

This follows from the fact that all the elements of the type $x_{i}^{a}y_{i}$
have order $2$ or $4$, so that $f$ can only send $\langle x_{1}\rangle$ to
$\langle x_{2}\rangle$. ∎

######  Proposition 3.3.

Let $G$ be a finite group of order $2^{n}$, which is either quaternion with
$3\neq n\geq 2$ or dihedral with $n\geq 3$. Then the number of automorphisms
of $G$ is $2^{2n-3}$.

###### Proof.

See [Byo07] just before Lemma 8.1 for the complete proof. We sketch the proof
for the convenience of the reader. If $G=\langle x,y\rangle_{o}$ and
$f\in\mathrm{Aut}(G)$, then $f(x)$ can only be an odd power of $x$ and $f(y)$
can only be $x^{i}y$ for a certain $i$. We can verify that every such choice
defines an automorphism. ∎

We now apply Lemma 2.6 to find the possible types of quaternion and dihedral
braces of $2$-power order.

######  Theorem 3.4.

Let $n\geq 2$ be an integer. Let $N$ be an abelian group of order $2^{n}$ with
exponent $2^{d}$. Suppose that there is a regular quaternion or dihedral
subgroup of $\mathrm{Hol}(N)$. Then $N$ must be one of the following groups:

  * •

$C_{2^{n}}$ for $n\geq 2$;

  * •

$C_{2}\times C_{2^{n-1}}$ for $n\geq 2$;

  * •

$C_{4}\times C_{2^{n-2}}$ for $n\geq 3$;

  * •

$C_{2}\times C_{2}\times C_{2^{n-2}}$ for $n\geq 3$;

  * •

$C_{2}\times C_{2}\times C_{2}\times C_{2^{n-3}}$ for $n\geq 4$.

###### Proof.

Let $r$ be the rank of $N$. Since $\mathrm{Hol}(N)$ must contain an element of
order $2^{n-1}$, Lemma 2.6 gives $n-1<\lceil\log_{2}(r+1)\rceil+d$. Also, as
$N$ has at least one cyclic factor of order $2^{d}$, we have $r\leq n-d+1$.
Hence

$r-1\leq n-d<\lceil\log_{2}(r+1)\rceil+1.$

Thus $r\leq 4$. Moreover, if $r=2$ then $n-d=1$ or $2$; if $r=3$ then $n-d=2$;
if $r=4$ then $n-d=3$. Thus $N$ must be one of the five groups listed. ∎

##  4\. Strategy for counting braces and Hopf-Galois structures in the
$2$-power case

Let $N$ be any of the groups listed in Theorem 3.4. In this section, we
outline the strategy we will use to count the quaternion and dihedral braces
and Hopf-Galois structures corresponding to $N$. We shall then carry out this
strategy for each $N$ in turn in the subsequent sections.

Write $N$ as in Proposition 2.1:

$N=\mathbb{Z}/2^{a_{1}}\mathbb{Z}\times\mathbb{Z}/2^{a_{2}}\mathbb{Z}\times\cdots\times\mathbb{Z}/2^{a_{r}}\mathbb{Z}$

with $1\leq a_{1}\leq\cdots\leq a_{r}$, and let
$\overline{R}\cong\mathrm{End}(N)$ be as described after Proposition 2.1.
Then, by Corollary 2.2, we may view an element of $\mathrm{Hol}(N)$ as a
matrix

(4.1) $\begin{pmatrix}A&v\\\ 0&1\end{pmatrix},$

with $A$ an invertible element of $\overline{R}$ and with $v\in N$.

Let $G$ be a quaternion or dihedral regular subgroup of $\mathrm{Hol}(N)$ of
order $2^{n}=|N|$. Then $G$ is generated by elements

(4.2) $X=\begin{pmatrix}A&v\\\ 0&1\end{pmatrix},\quad Y=\begin{pmatrix}B&w\\\
0&1\end{pmatrix}$

of the above form, satisfying the relations $X^{2^{n-1}}=I$, $X^{2^{n-2}}\neq
I$, $YX=X^{-1}Y$ and either $Y^{2}=X^{2^{n-2}}$ or $Y^{2}=I$ (depending on
whether we are considering the quaternion case or dihedral case). For $k\geq
1$, we see inductively that

$X^{2^{k}}=\begin{pmatrix}A^{2^{k}}&(I+A+\cdots+A^{2^{k}-1})v\\\
0&1\end{pmatrix}.$

As $X^{2^{n-2}}\neq I$ and $G$ is regular, we therefore have

(4.3) $(I+A+\cdots+A^{2^{n-2}-1})v\neq 0.$

Since

$Y^{-1}=\begin{pmatrix}B^{-1}&-B^{-1}w\\\ 0&1\end{pmatrix},$

the remaining relations become

(4.4) $\begin{cases}A^{2^{n-1}}=I\\\ (I+A+\cdots+A^{2^{n-1}-1})v=0\\\
BA=A^{-1}B\\\ w+Bv=-A^{-1}v+A^{-1}w\\\ \begin{cases}B^{2}=A^{2^{n-2}}\text{
(quaternion)}\\\ B^{2}=I\text{ (dihedral)}\end{cases}\\\
\begin{cases}(I+B)w=(I+A+\cdots+A^{2^{n-2}-1})v\text{ (quaternion)}\\\
(I+B)w=0\text{ (dihedral)}.\end{cases}\end{cases}$

Note that any $X$ and $Y$ as in (4.2) satisfying (4.3) and (4.4) will generate
a subgroup isomorphic to $Q_{2^{n}}$ or $D_{2^{n}}$ (except in the degenerate
case $n=2$, $Y=I$).

We will determine the regular quaternion and dihedral subgroups in
$\mathrm{Hol}(N)$ by finding all solutions of (4.3) and (4.4) and imposing
further conditions to ensure regularity. We will often simplify this
calculation by working up to conjugacy in $\mathrm{Hol}(N)$. In particular,
using Lemma 2.3, we may assume that $A$ and $B$ reduce to invertible upper
triangular matrices in $\mathrm{GL}_{r}(\mathbb{F}_{2})$.

In order to avoid complications from small values of $n$, we shall carry out
the above strategy assuming that $n\geq 5$ (or $n\geq 4$ for $N=C_{2^{n}}$).
When $n\geq 5$, we shall find that only two of the five groups $N$ of order
$2^{n}$ in Theorem 3.4 give regular quaternion or dihedral subgroups. In §10
we will treat the omitted small cases $|N|=4$, $8$, $16$ using Magma.

##  5\. On the number of quaternion and dihedral braces and Hopf–Galois
structures of cyclic type

Let $N=C_{2^{n}}$, with $n\geq 4$. We apply the strategy described in §4. In
this case, the matrices $A$ and $B$ in (4.2) are single elements of
$(\mathbb{Z}/2^{n}\mathbb{Z})^{\times}$, say $\alpha$ and $\beta$
respectively, and $v,w\in\mathbb{Z}/2^{n}\mathbb{Z}$. The condition
$BA=A^{-1}B$ in (4.4) gives $\alpha^{2}=1$. Then (4.3) and the condition
$(I+A+\cdots+A^{2^{n-1}-1})v=0$ in (4.4) give $2^{n-3}(1+\alpha)v\neq 0$ and
$2^{n-2}(1+\alpha)v=0$ in $\mathbb{Z}/2^{n}\mathbb{Z}$. Hence
$(1+\alpha)v\equiv 4\pmod{8}$, and $(\ref{eq:Xreg})$ and (4.4) reduce to

$\begin{cases}(1+\alpha)v\equiv 4\pmod{8}\\\ \alpha^{2}=1\\\
(\alpha+\beta)v=(\alpha-1)w\\\ \beta^{2}=1\\\
\begin{cases}(1+\beta)w=2^{n-1}\text{ (quaternion)}\\\ (1+\beta)w=0\text{
(dihedral)}.\end{cases}\end{cases}$

Note that $\alpha^{2}\equiv 1\pmod{2^{n}}$ means that $\alpha$ can only be
$1$, $1+2^{n-1}$, $-1$ or $-1+2^{n-1}$ modulo $2^{n}$. The third and fourth
options are not compatible with the first relation, so that $\alpha$ can only
be $1$ or $1+2^{n-1}$.

If $\alpha\equiv 1+2^{n-1}\pmod{2^{n}}$, then $v\equiv 2\pmod{4}$ and by
regularity $w$ is odd. Then the last relation tells us that $\beta$ is
$-1+2^{n-1}$ in the quaternion case or $-1$ in the dihedral case. Neither of
these cases is compatible with the relation $(\alpha+\beta)v=(\alpha-1)w$.

If $\alpha\equiv 1\pmod{2^{n}}$, then again $v\equiv 2\pmod{4}$ and $w$ is
odd. The last relation tells us that $\beta$ is $-1+2^{n-1}$ in the quaternion
case or $-1$ in the dihedral case. We can see that the relation
$(\alpha+\beta)v=(\alpha-1)w$ is automatically satisfied. Hence $\alpha$ and
$\beta$ are fixed, which means we have at most $2^{n-2}$ choices for $X$. This
corresponds to at most one choice for $\langle
X\rangle\subseteq\mathrm{Hol}(N)$. In a similar fashion, we have at most
$2^{n-1}$ different choices for $Y$. Hence there is at most one possible
subset of $\mathrm{Hol}(N)$ of the form $\\{X^{i}Y\\}_{0\leq i\leq
2^{n-1}-1}$. This implies that there is at most one regular quaternion,
respectively dihedral, subgroup. Now note that the subgroups of
$\mathrm{Hol}(N)$ generated by

$X=\begin{pmatrix}1&2\\\ 0&1\end{pmatrix},Y=\begin{pmatrix}-1+2^{n-1}&1\\\
0&1\end{pmatrix}$

or

$X=\begin{pmatrix}1&2\\\ 0&1\end{pmatrix},Y=\begin{pmatrix}-1&1\\\
0&1\end{pmatrix}$

are regular.

######  Proposition 5.1.

Let $n\geq 4$ be an integer. Then there is one regular quaternion subgroup and
one regular dihedral subgroup of $\mathrm{Hol}(C_{2^{n}})$.

######  Corollary 5.2.

Let $n\geq 4$ be an integer. Then there is only one quaternion brace and one
dihedral brace of type $C_{2^{n}}$.

Using Proposition 3.3 and (1.1), we finally obtain the following.

######  Corollary 5.3.

Let $n\geq 4$ be an integer. Any Galois extension of degree $2^{n}$ with
quaternion or dihedral Galois group admits $2^{n-2}$ Hopf–Galois structures of
type $C_{2^{n}}$.

##  6\. On the number of quaternion and dihedral braces and Hopf–Galois
structures of type $C_{2}\times C_{2^{n-1}}$

We will now assume $n\geq 5$. Let $N=C_{2}\times C_{2^{n-1}}$. We look for
elements $X$, $Y$ in $\mathrm{Hol}(N)$ as in (4.2). The matrices $A$ and $B$
are in

$\begin{pmatrix}\mathbb{Z}/2\mathbb{Z}&\mathbb{Z}/2\mathbb{Z}\\\
2^{n-2}\mathbb{Z}/2^{n-1}\mathbb{Z}&\mathbb{Z}/2^{n-1}\mathbb{Z}\end{pmatrix}$

and are invertible modulo $2$, so they lie in

$\begin{pmatrix}1&\mathbb{Z}/2\mathbb{Z}\\\
2^{n-2}\mathbb{Z}/2^{n-1}\mathbb{Z}&(\mathbb{Z}/2^{n-1}\mathbb{Z})^{\times}\end{pmatrix}.$

We will write

$A=\begin{pmatrix}1&a\\\ 2^{n-2}b&\alpha\end{pmatrix},\quad
B=\begin{pmatrix}1&r\\\ 2^{n-2}s&\beta\end{pmatrix}$

with $a,b,r,s\in\mathbb{Z}/2\mathbb{Z}$ and
$\alpha,\beta\in(\mathbb{Z}/2^{n-1}\mathbb{Z})^{\times}$. We also write

$v=\begin{pmatrix}v_{1}\\\ v_{2}\end{pmatrix},\quad w=\begin{pmatrix}w_{1}\\\
w_{2}\end{pmatrix},$

with $v_{1},w_{1}\in\mathbb{Z}/2\mathbb{Z}$ and
$v_{2},w_{2}\in\mathbb{Z}/2^{n-1}\mathbb{Z}$.

We now express the conditions (4.3) and (4.4) in terms of
$a,b,\alpha,r,s,\beta$. First note that

$A^{-1}=\begin{pmatrix}1&a\\\ 2^{n-2}b&2^{n-2}ab+\alpha^{-1}\end{pmatrix}.$

so that the condition $BA=A^{-1}B$ is equivalent to

$\alpha-\alpha^{-1}=2^{n-2}(ab+as+br).$

This translates into $\alpha^{2}=1$ or $\alpha^{2}=1+2^{n-2}$, according as
$ab+as+br=0$ or $1$. In particular, note that $A^{4}=I$. Then the first
condition in (4.4) is automatically satisfied and the second condition,
together with (4.3), becomes $\alpha\equiv 1\pmod{4}$ and $v_{2}$ odd. In
particular $B^{2}$ has to be the identity in both the quaternion and dihedral
case, or equivalently we must have $2^{n-2}rs+\beta^{2}=1$. Expressing the
conditions $w+Bv=-A^{-1}v+A^{-1}w$ and those on $(I+B)w$ explicitly, we now
rewrite (4.3) and (4.4) as:

(6.1) $\begin{cases}\alpha^{2}\equiv 1+2^{n-2}(ab+as+br)\pmod{2^{n-1}}\\\
\alpha\equiv 1\pmod{4}\\\ v_{2}\equiv 1\pmod{2}\\\
aw_{2}\equiv(a+r)v_{2}\pmod{2}\\\ \beta\equiv
2^{n-2}(s+b)v_{1}v_{2}^{-1}+2^{n-2}ab-\alpha^{-1}+2^{n-2}bw_{1}v_{2}^{-1}+(2^{n-2}ab+\alpha^{-1}-1)w_{2}v_{2}^{-1}\pmod{2^{n-1}}\\\
rw_{2}\equiv 0\pmod{2}\\\ 2^{n-2}rs+\beta^{2}\equiv 1\pmod{2^{n-1}}\\\
\begin{cases}2^{n-2}sw_{1}+(1+\beta)w_{2}\equiv 2^{n-2}\pmod{2^{n-1}}\text{
(quaternion)}\\\ 2^{n-2}sw_{1}+(1+\beta)w_{2}\equiv 0\pmod{2^{n-1}}\text{
(dihedral)}.\end{cases}\end{cases}$

Recall that

$X=\begin{pmatrix}1&a&v_{1}\\\ 2^{n-2}b&\alpha&v_{2}\\\ 0&0&1\end{pmatrix}.$

We now show that we can make a convenient choice of $v_{1}$ and $v_{2}$ by
replacing our quaternion or dihedral subgroup by a conjugate. Observe that

(6.2) $\begin{pmatrix}1&1&0\\\ 0&1&0\\\
0&0&1\end{pmatrix}^{v_{1}}\begin{pmatrix}1&0&0\\\ 0&v_{2}^{-1}&0\\\
0&0&1\end{pmatrix}\begin{pmatrix}1&a&v_{1}\\\ 2^{n-2}b&\alpha&v_{2}\\\
0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\\ 0&v_{2}^{-1}&0\\\
0&0&1\end{pmatrix}^{-1}\begin{pmatrix}1&1&0\\\ 0&1&0\\\
0&0&1\end{pmatrix}^{-v_{1}}=$ $=\begin{pmatrix}1&a&0\\\
2^{n-2}b&\alpha+2^{n-2}bv_{1}&1\\\ 0&0&1\end{pmatrix}.$

This means that, up to conjugacy, we can assume $v_{1}=0$ and $v_{2}=1$.

We now impose further conditions on $a,b,r,s,\alpha,\beta$ to ensure that our
subgroup is regular. First note that $X^{4}\equiv 1\pmod{4}$, so that the
entries of $\\{X^{i}(0,0,1)^{\top}\pmod{4}\\}_{i=0,\ldots,2^{n-1}-1}$ are

$\begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix},\begin{pmatrix}0\\\ 1\\\
1\end{pmatrix},\begin{pmatrix}a\\\ 2\\\ 1\end{pmatrix},\begin{pmatrix}a\\\
3\\\ 1\end{pmatrix}.$

Since, by regularity, there exist integers $0\leq i\leq 2^{n-1}-1$ and $0\leq
j\leq 1$ such that

$X^{i}Y^{j}\begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix}=\begin{pmatrix}1\\\ 0\\\
1\end{pmatrix}$

and $j$ cannot be $0$ (else the equality could not hold modulo $4$), we may
replace $X$ by $X^{i}Y$ for a suitable $i$ and so assume that $w_{1}=1$ and
$w_{2}=0$. (This change of generators does not affect the relations satisfied
by $X$ and $Y$.)

We now rewrite the system (6.1) with the information we have on $v$ and $w$.

$\begin{cases}v_{1}=0\\\ v_{2}=1\\\ w_{1}=1\\\ w_{2}=0\\\ \alpha^{2}\equiv
1+2^{n-2}as\pmod{2^{n-1}}\\\ \alpha\equiv 1\pmod{4}\\\ r=a\\\ \beta\equiv
2^{n-2}b(1+a)-\alpha^{-1}\pmod{2^{n-1}}\\\ \begin{cases}s=1\text{
(quaternion)}\\\ s=0\text{ (dihedral)}.\end{cases}\end{cases}$

Conversely, these conditions ensure that the orbit of $0\in N$ under the
subgroup $\langle X\rangle$ of $G$ has cardinality $2^{n-1}$ (since they force
(4.3) to hold), and our choice of $w$ means that the orbit of $0$ under $G$ is
strictly larger than this. As $|G|=|N|=2^{n}$, it follows that the above
conditions are sufficient to guarantee the regularity of $G$. Moreover, they
ensure that $Y$ is completely determined by $X$ and the isomorphism class
(quaternion or dihedral) of $G$. Hence we have the following result.

######  Proposition 6.1.

Let $\langle X,Y\rangle_{o}$ be a quaternion or dihedral regular subgroup of
$\mathrm{Hol}(N)$. Let $Y^{\prime}\in\mathrm{Hol}(N)$ be such that $\langle
X,Y^{\prime}\rangle_{o}$ is also a quaternion, respectively dihedral, regular
subgroup of $\mathrm{Hol}(N)$. Then $\langle X,Y\rangle_{o}=\langle
X,Y^{\prime}\rangle_{o}$.

Once we fix $s$ (hence the isomorphism class), $a$ and $b$, there are two
possible values of $\alpha$, whose difference is $2^{n-2}$. Then everything
else is completely determined. This proves that, up to conjugation, we have
$8$ quaternion regular subgroups and $8$ dihedral regular subgroups of
$\mathrm{Hol}(N)$: we will call these subgroups the fundamental subgroups. As
an aside note the following.

######  Corollary 6.2.

Let

$X=\begin{pmatrix}1&a&v_{1}\\\ 2^{n-2}b&\alpha&v_{2}\\\ 0&0&1\end{pmatrix},$

be a matrix such that $a,b,v_{1}\in\mathbb{Z}/2\mathbb{Z}$,
$v_{2}\in(\mathbb{Z}/2^{n-1}\mathbb{Z})^{\times}$, $\alpha\equiv 1\pmod{4}$
and either $\alpha^{2}\equiv 1\pmod{2^{n-1}}$ or $\alpha^{2}\equiv
1+2^{n-2}\pmod{2^{n-1}}$. Then there exists $Y\in\mathrm{Hol}(N)$ such that
$\langle X,Y\rangle_{o}$ is a quaternion, respectively dihedral, regular
subgroup of $\mathrm{Hol}(N)$.

###### Proof.

By (6.2), $X$ can be conjugated via $\mathrm{Aut}(N)$ to a matrix $X^{\prime}$
such that $\langle X^{\prime},Y^{\prime}\rangle_{o}$ is a fundamental subgroup
for some $Y^{\prime}\in\mathrm{Hol}(N)$, where $a=a^{\prime}$, $b=b^{\prime}$
and $\alpha^{\prime}$ is either $\alpha$ or $\alpha+2^{n-2}$. If
$g\in\mathrm{Hol}(N)$ is such that $gXg^{-1}=X^{\prime}$, then $\langle
X,g^{-1}Y^{\prime}g\rangle_{o}$ is a regular subgroup of $\mathrm{Hol}(N)$. ∎

We now show that the fundamental subgroups represent six conjugacy classes for
either possibility for $s$. As noted in the introduction, if two regular
subgroups are conjugate in $\mathrm{Hol}(N)$ then they are conjugate by an
element of $\mathrm{Aut}(N)$. Thus we only need to check for conjugation by
elements of the form

$M=\begin{pmatrix}1&p&0\\\ 2^{n-2}q&\gamma&0\\\ 0&0&1\end{pmatrix}$

with $p,q\in\mathbb{Z}/2\mathbb{Z}$ and
$\gamma\in(\mathbb{Z}/2^{n-1}\mathbb{Z})^{\times}$. By Lemma 3.2, a necessary
condition for $\langle X,Y\rangle_{o}$ to be conjugate to $\langle
X_{*},Y_{*}\rangle_{o}$ (with parameters denoted by $a_{*}$, $b_{*}$ etc) is
that, for a certain odd exponent $d$, we have $MXM^{-1}=X_{*}^{d}$. Moreover,
by (6.2) we may assume $v_{1*}=0$, $v_{2*}=1$. We calculate

$MXM^{-1}=\begin{pmatrix}1&p&0\\\ 2^{n-2}q&\gamma&0\\\
0&0&1\end{pmatrix}\begin{pmatrix}1&a&0\\\ 2^{n-2}b&\alpha&1\\\
0&0&1\end{pmatrix}\begin{pmatrix}1&p&0\\\ 2^{n-2}q&\gamma^{-1}+2^{n-2}pq&0\\\
0&0&1\end{pmatrix}=$ $=\begin{pmatrix}1&a&p\\\
2^{n-2}b&\alpha+2^{n-2}(aq+bp)&\gamma\\\ 0&0&1\end{pmatrix}.$

Note that

$X_{*}^{-1}=\begin{pmatrix}A_{*}^{-1}&-A_{*}^{-1}v_{*}\\\
0&1\end{pmatrix}=\begin{pmatrix}1&a_{*}&a_{*}\\\
2^{n-2}b_{*}&2^{n-2}a_{*}b_{*}+\alpha_{*}^{-1}&2^{n-2}a_{*}b_{*}-\alpha_{*}^{-1}\\\
0&0&1\end{pmatrix}.$

Since $A_{*}^{4}=I$, for every positive integer $k$ we have

$X_{*}^{4k}=\begin{pmatrix}1&0&0\\\ 0&1&k(1+\alpha_{*})(1+\alpha_{*}^{2})\\\
0&0&1\end{pmatrix}.$

(Recall that $(1+\alpha_{*})(1+\alpha_{*}^{2})\equiv 4\pmod{8}$.) Thus,
depending on $d$ modulo $4$, we find that $X_{*}^{d}$ is either of the form

(6.3) $X_{*}^{d}=X_{*}^{4k+1}=\begin{pmatrix}1&a_{*}&0\\\
2^{n-2}b_{*}&\alpha_{*}&k\alpha_{*}(1+\alpha_{*})(1+\alpha_{*}^{2})+1\\\
0&0&1\end{pmatrix}$

or of the form

(6.4) $X_{*}^{d}=X_{*}^{4k-1}=\begin{pmatrix}1&a_{*}&a_{*}\\\
2^{n-2}b_{*}&2^{n-2}a_{*}b_{*}+\alpha_{*}^{-1}&k(1+\alpha_{*})(1+\alpha_{*}^{2})-\alpha_{*}^{-1}+2^{n-2}a_{*}b_{*}\\\
0&0&1\end{pmatrix}.$

Since $MXM^{-1}=X_{*}^{d}$, it follows that $a=a_{*}$ and $b=b_{*}$ (therefore
also $r=r_{*}$ and, for isomorphism class reasons, $s=s_{*}$) and either $p=0$
or $p=a$, depending on $d\pmod{4}$. Looking at the central entry, this means
that either

$\alpha+2^{n-2}aq=\alpha_{*},\ \ \text{ or }\ \
\alpha+2^{n-2}aq=\alpha_{*}^{-1}.$

If $a=0$ then the above, together with the fact that in this case
$\alpha^{2}=1$, implies that $\alpha=\alpha_{*}$. If $a=1$, then we always
find a choice for $q$ (and possibly $\gamma$) such that
$\alpha_{*}\neq\alpha$. By Proposition 6.1 and Corollary 6.2, there is only
one possible choice for $Y$ such that $\langle X,Y\rangle_{o}$ is a
quaternion, respectively dihedral, regular subgroup. This means that the
conjugations we found when $a=1$ actually pair fundamental groups. We have
therefore shown the following result:

######  Proposition 6.3.

Let $n\geq 5$ be an integer. Then there are $6$ quaternion braces and $6$
dihedral braces of type $C_{2}\times C_{2^{n-1}}$.

We now count the number of Hopf–Galois structures. We first note that the
eight fundamental subgroups are distinct from each other. Suppose that
$\langle X_{1},Y_{1}\rangle_{o}=\langle X_{2},Y_{2}\rangle_{o}$. We will
denote by $a_{i}$, $b_{i}$ etc. their respective parameters for $i=1,2$. Then,
by Lemma 3.2, $\langle X_{1}\rangle=\langle X_{2}\rangle$. By (6.3) and (6.4),
we must have that $a_{1}=a_{2}$ and $b_{1}=b_{2}$. Note that in $X_{1}^{4k-1}$
the $(2,3)$-entry is $3$ modulo $4$, so that we cannot have
$X_{1}^{4k-1}=X_{2}$. If $X_{1}^{4k+1}=X_{2}$, then also
$\alpha_{1}=\alpha_{2}$.

######  Proposition 6.4.

Let $n\geq 5$ be an integer. Then any Galois extension of degree $2^{n}$ with
quaternion or dihedral Galois group admits $2^{n+1}$ Hopf–Galois structures of
type $C_{2}\times C_{2^{n-1}}$.

###### Proof.

We first count the number of regular subgroups of $\mathrm{Hol}(N)$. Let
$\langle X,Y\rangle_{o}$ be a quaternion or dihedral regular subgroup of
$\mathrm{Hol}(N)$. Then $X$ is of the form

$\begin{pmatrix}1&a&v_{1}\\\ 2^{n-2}b&\alpha&v_{2}\\\ 0&0&1\end{pmatrix},$

where $a,b\in\mathbb{Z}/2\mathbb{Z}$,
$v_{2}\in(\mathbb{Z}/2^{n-1}\mathbb{Z})^{\times}$, $\alpha\equiv 1\pmod{4}$
and either $\alpha^{2}\equiv 1\pmod{2^{n-1}}$ or $\alpha^{2}\equiv
1+2^{n-2}\pmod{2^{n-1}}$, $v_{1}=0,1$. Note that we have two choices for $a$,
two choices for $b$, two choices for $\alpha$, two for $v_{1}$ and $2^{n-2}$
for $v_{2}$ (as we proved $v_{2}$ has to be odd), for a total of $2^{n+2}$. By
Proposition 6.1 and Corollary 6.2, each such matrix belongs to a unique
regular subgroup of $\mathrm{Hol}(N)$. As the number of generators in $\langle
X\rangle$ is $2^{n-2}$, this implies that there are $16$ regular subgroups in
each isomorphism case.

By Proposition 3.3, $G$ has $2^{2n-3}$ automorphisms in both cases. From the
matrix description, we also easily see that $N$ has $2^{n}$ automorphisms.
Hence in both cases, by (1.1) the number of Hopf–Galois structures is

$\frac{2^{2n-3}}{2^{n}}\cdot 16=2^{n+1}.$

∎

##  7\. On the number of quaternion and dihedral braces and Hopf–Galois
structures of type $C_{4}\times C_{2^{n-2}}$

Using Magma (or by a direct calculation) we easily verify the following:

######  Lemma 7.1.

$\mathrm{Hol}(C_{4}\times C_{8})$ has no elements of order $16$.

Lemma 7.1 tells us that there are no quaternion or dihedral regular subgroups
of $\mathrm{Hol}(C_{4}\times C_{8})$. This implies that we cannot find any
brace or Hopf–Galois structure with additive group $C_{4}\times C_{8}$ and
quaternion or dihedral multiplicative group.

Now let $n\geq 6$. Let $N=C_{4}\times C_{2^{n-2}}$. Let $G$ be a quaternion or
dihedral regular subgroup of $\mathrm{Hol}(N)$. Then $G$ acts transitively on
$N$. Note that the $2^{n-5}$-th powers generate a characteristic subgroup $M$
of $N$ such that $N/M\cong C_{4}\times C_{8}$, so that we find a transitive
action of $G$ on $C_{4}\times C_{8}$. The image of $G$ in
$\mathrm{Perm}(C_{4}\times C_{8})$ lies in $\mathrm{Hol}(C_{4}\times C_{8})$
as $G$ is a subgroup of $\mathrm{Hol}(N)$. The kernel of the map
$f:G\rightarrow\mathrm{Hol}(C_{4}\times C_{8})$ we just obtained has index at
least $32$, as the action of $G$ on $C_{4}\times C_{8}$ is transitive.
Therefore $G/\ker f$, and hence also $\mathrm{Hol}(C_{4}\times C_{8})$, must
have an element of order $16$. This contradicts Lemma 7.1.

######  Proposition 7.2.

Let $n\geq 5$ be an integer. Then there are no quaternion or dihedral braces
type $C_{4}\times C_{2^{n-2}}$. A Galois extension of degree $2^{n}$ with
quaternion or dihedral Galois group does not admit any Hopf-Galois structures
of type $C_{4}\times C_{2^{n-2}}$.

##  8\. On the number of quaternion and dihedral braces and Hopf–Galois
structures of type $C_{2}\times C_{2}\times C_{2^{n-2}}$

Let $n\geq 5$ be an integer, let $N=C_{2}\times C_{2}\times C_{2^{n-2}}$, and
let $G=\langle X,Y\rangle_{o}$ be a quaternion or dihedral regular subgroup of
$\mathrm{Hol}(N)$. Then, using Corollary 2.2, and conjugating to ensure that
$G$ lies in the Sylow $2$-subgroup of $\mathrm{Hol}(N)$ described in Lemma
2.3, we can write

$X=\begin{pmatrix}1&a&b&v_{1}\\\ 0&1&c&v_{2}\\\
2^{n-3}d&2^{n-3}e&\alpha&v_{3}\\\ 0&0&0&1\end{pmatrix}$

where $\alpha\in(\mathbb{Z}/2^{n-2}\mathbb{Z})^{\times}$,
$a,b,c,d,e,v_{1},v_{2}\in\mathbb{Z}/2\mathbb{Z}$,
$v_{3}\in\mathbb{Z}/2^{n-2}$. We now compute $X^{2}$ and $X^{4}$:

$X^{2}=\begin{pmatrix}1&0&ac&av_{2}+bv_{3}\\\ 0&1&0&cv_{3}\\\
0&2^{n-3}ad&2^{n-3}(bd+ce)+\alpha^{2}&2^{n-3}dv_{1}+2^{n-3}ev_{2}+(1+\alpha)v_{3}\\\
0&0&0&1\end{pmatrix},$ $X^{4}=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\
0&0&\alpha^{4}&2^{n-3}acdv_{3}+(1+\alpha)(1+\alpha^{2})v_{3}\\\
0&0&0&1\end{pmatrix}.$

Note that $2^{n-3}acdv_{3}+(1+\alpha)(1+\alpha^{2})v_{3}$ is divisible by $4$
since $n\geq 5$, and that $\alpha^{4}\equiv 1\pmod{4}$. So $X^{4}\equiv
I\pmod{4}$, and an easy induction shows that $X^{2^{k}}\equiv I\pmod{2^{k}}$
for $k\geq 2$. Thus $X^{2^{n-2}}=I$ in $\mathrm{Hol}(N)$, contradicting the
relations defining $G$.

######  Proposition 8.1.

Let $n\geq 5$ be an integer. Then there are no quaternion or dihedral braces
of type $C_{2}\times C_{2}\times C_{2^{n-2}}$. A Galois extension of degree
$2^{n}$ with quaternion or dihedral Galois group does not admit any Hopf-
Galois structures of type $C_{2}\times C_{2}\times C_{2^{n-2}}$.

##  9\. On the number of quaternion and dihedral braces and Hopf–Galois
structures of type $C_{2}\times C_{2}\times C_{2}\times C_{2^{n-3}}$

Let $n\geq 5$ be an integer, let $N=C_{2}\times C_{2}\times C_{2}\times
C_{2^{n-3}}$ and let $\langle X,Y\rangle_{o}$ be a quaternion or dihedral
regular subgroup of $\mathrm{Hol}(N)$. Then, using Corollary 2.2 and Lemma 2.3
as before, we can write

$X=\begin{pmatrix}1&a&b&c&v_{1}\\\ 0&1&d&e&v_{2}\\\ 0&0&1&f&v_{3}\\\
2^{n-4}g&2^{n-4}h&2^{n-4}i&\alpha&v_{4}\\\ 0&0&0&0&1\end{pmatrix}$

where $\alpha\in(\mathbb{Z}/2^{n-3}\mathbb{Z})^{\times}$,
$a,b,c,d,e,f,g,h,i,v_{1},v_{2},v_{3}\in\mathbb{Z}/2\mathbb{Z}$,
$v_{4}\in\mathbb{Z}/2^{n-3}$. Now note that

$X^{2}=\begin{pmatrix}1&0&ad&ae+bf&av_{2}+bv_{3}+cv_{4}\\\
0&1&0&df&dv_{3}+ev_{4}\\\ 0&0&1&0&fv_{4}\\\
0&2^{n-4}ga&2^{n-4}(gb+hd)&\alpha^{2}+2^{n-4}(gc+he+if)&(1+\alpha)v_{4}+2^{n-4}(gv_{1}+hv_{2}+iv_{3})\\\
0&0&0&0&1\end{pmatrix}.$

and

$X^{4}=\begin{pmatrix}1&0&0&0&adfv_{4}\\\ 0&1&0&0&0\\\ 0&0&1&0&0\\\
0&0&0&\alpha^{4}+2^{n-4}gadf&(1+\alpha)(1+\alpha^{2})v_{4}+2^{n-4}k\\\
0&0&0&0&1\end{pmatrix}$

for a certain integer $k$. If $n\geq 6$ then $X^{8}\equiv I\pmod{8}$ and,
analogously to §8, we obtain $X^{2^{n-2}}=I$ in $\mathrm{Hol}(N)$. If $n=5$
then $2^{n-3}=4$ and we already have $X^{4}=I$ in $\mathrm{Hol}(N)$. In both
cases, we have a contradiction.

######  Proposition 9.1.

Let $n\geq 5$ be an integer. Then there are no quaternion or dihedral braces
or Hopf–Galois structures of type $C_{2}\times C_{2}\times C_{2}\times
C_{2^{n-3}}$. A Galois extension of degree $2^{n}$ with quaternion or dihedral
Galois group does not admit any Hopf-Galois structures of type $C_{2}\times
C_{2}\times C_{2}\times C_{2^{n-3}}$.

##  10\. On quaternion and dihedral braces and Hopf–Galois structures of order
$4$, $8$ and $16$

For each quaternion or dihedral group $G$ of order $m=4$, $8$, $16$, we will
use Theorem 3.4 to read off all the possible isomorphism classes of $N$.
Suppose that in the standard labelling notation we write $N$ as
SmallGroup($m$,$j$) and $G$ as SmallGroup($m$,$k$). We will implement the
following Magma code.

     N:=SmallGroup(m,j);
     G:=SmallGroup(m,k);
     H:=Holomorph(N);
     R:=RegularSubgroups(H);
     R;

The last command will list all the conjugacy classes of the regular subgroups
of $\mathrm{Hol}(N)$, labelling them and displaying their length. The last
label will be the total number of conjugacy classes, which we denote by $l$.
Then we run the following.

     for i in [1..l] do
        if IsIsomorphic(R[i]‘subgroup,G) eq true then
            print i;
        end if;
     end for;

The number of lines in the output will tell us the number of conjugacy classes
of regular subgroups of $\mathrm{Hol}(N)$, hence the number $c(j,k)$ of
isomorphism classes of braces with additive group $N$ and multiplicative group
$G$. The sum of the lengths will count the number $r(j,k)$ of regular
subgroups of $\mathrm{Hol}(N)$, which multiplied by
$\frac{|\mathrm{Aut}(G)|}{|\mathrm{Aut}(N)|}$ will give the number $h(j,k)$ of
Hopf–Galois structures with Galois group $G$ and type $N$. The results are
shown in Table 1. The reader can refer to [Dok18] for the ‘Small Group’
labelling as well the automorphism groups of specific groups and other general
group properties.

Table 1. Numbers of braces $c(j,k)$, regular subgroups $r(j,k)$ and Hopf-Galois structures $h(j,k)$ for $m=4$, $8$, $16$. $N$ | $G$ | $j$ | $k$ | $l$ | $c(j,k)$ | $r(j,k)$ | $h(j,k)$
---|---|---|---|---|---|---|---
$C_{4}$ | $C_{4}$ | 1 | 1 | 2 | 1 | 1 | 1
$C_{2}\times C_{2}$ | $C_{4}$ | 2 | 1 | 2 | 1 | 3 | 1
$C_{4}$ | $C_{2}\times C_{2}$ | 1 | 2 | 2 | 1 | 1 | 3
$C_{2}\times C_{2}$ | $C_{2}\times C_{2}$ | 2 | 2 | 2 | 1 | 1 | 1
$C_{8}$ | $Q_{8}$ | 1 | 4 | 5 | 1 | 1 | 6
$C_{2}\times C_{4}$ | $Q_{8}$ | 2 | 4 | 14 | 1 | 2 | 6
$C_{2}\times C_{2}\times C_{2}$ | $Q_{8}$ | 5 | 4 | 8 | 1 | 14 | 2
$C_{8}$ | $D_{8}$ | 1 | 3 | 5 | 1 | 1 | 2
$C_{2}\times C_{4}$ | $D_{8}$ | 2 | 3 | 14 | 5 | 14 | 14
$C_{2}\times C_{2}\times C_{2}$ | $D_{8}$ | 5 | 3 | 8 | 2 | 126 | 6
$C_{16}$ | $Q_{16}$ | 1 | 9 | 8 | 1 | 1 | 4
$C_{2}\times C_{8}$ | $Q_{16}$ | 5 | 9 | 66 | 4 | 8 | 16
$C_{4}\times C_{4}$ | $Q_{16}$ | 2 | 9 | 83 | 2 | 48 | 16
$C_{2}\times C_{2}\times C_{4}$ | $Q_{16}$ | 10 | 9 | 161 | 1 | 48 | 8
$C_{2}\times C_{2}\times C_{2}\times C_{2}$ | $Q_{16}$ | 14 | 9 | 39 | 1 | 5040 | 8
$C_{16}$ | $D_{16}$ | 1 | 7 | 8 | 1 | 1 | 4
$C_{2}\times C_{8}$ | $D_{16}$ | 5 | 7 | 66 | 6 | 16 | 32
$C_{4}\times C_{4}$ | $D_{16}$ | 2 | 7 | 83 | 0 | 0 | 0
$C_{2}\times C_{2}\times C_{4}$ | $D_{16}$ | 10 | 7 | 161 | 0 | 0 | 0
$C_{2}\times C_{2}\times C_{2}\times C_{2}$ | $D_{16}$ | 14 | 7 | 39 | 0 | 0 | 0

For $m=4$ and $m=8$, the number of braces $c(j,k)$ and the number of Hopf-
Galois structures $h(j,k)$ are already known, but we include them here for
completeness. For $m=4$, see [Bac15, Proposition 2.4] for braces and [Byo96a,
Byo96b] for Hopf-Galois structures. For $m=8$, see [NZ18, §5.1] for both.

##  11\. Summary of the $2$-power case

Recall that $q(2^{n})$ denotes the number of isomorphism classes of quaternion
braces of order $2^{n}$. Let $d(2^{n})$ denote the number of isomorphism
classes of dihedral braces of order $2^{n}$. Summarising the information
contained in Corollaries 5.2, 5.3, Propositions 6.3, 6.4, 7.2, 8.1, 9.1 and
Table 1, we obtain the following values for $q(n)$ and $d(n)$, as well as
corresponding totals for the number of Hopf-Galois structures of abelian type
on a quaternion or dihedral Galois extension of degree $2^{n}$.

######  Theorem 11.1.

For $n\geq 5$, we have $q(2^{n})=d(2^{n})=7$; also $q(16)=9$, $q(8)=3$,
$q(4)=2$ and $d(16)=7$, $d(8)=8$, $d(4)=2$.

If $n\geq 5$, then there are $9\cdot 2^{n-2}$ Hopf-Galois structures of
abelian type on any Galois extension of degree $2^{n}$ whose Galois group is
quaternion (respectively dihedral). On a quaternion (respectively, dihedral)
extension of degree $16$ there are $52$ (respectively $36$) Hopf-Galois
structures of abelian type. On a quaternion (respectively dihedral) extension
of degree $8$ there are $14$ (respectively, $22$) Hopf-Galois structures of
abelian type. On a cyclic (respectively, elementary abelian) extension of
degree $4$ there are $2$ (respectively $4$) Hopf-Galois structures.

When $n\geq 5$, the only possible additive groups for a quaternion or dihedral
brace of order $2^{n}$ (or abelian types for a Hopf-Galois structure on a
quaternion or dihedral field extension of degree $2^{n}$) are $C_{2^{n}}$ and
$C_{2}\times C_{2^{n-1}}$. The same is true for $n=4$ in the dihedral case,
but all $5$ abelian groups of order $16$ may occur in the quaternion case. For
$n=3$, all $3$ abelian groups of order $8$ may occur in either case.

##  12\. The non $2$-power case

Let $G$ be a quaternion or dihedral group of order $2^{n}s$, where $n\geq 2$,
$s$ is odd and $s\geq 3$. We recall that this means that either

$G=\langle x,y:x^{2^{n-1}s}=1,yx=x^{-1}y,y^{2}=x^{2^{n-2}s}\rangle\cong
Q_{2^{n}s}$

or

$G=\langle x,y:x^{2^{n-1}s}=1,yx=x^{-1}y,y^{2}=1\rangle\cong D_{2^{n}s}.$

The following result is immediate.

######  Lemma 12.1.

The subgroup $C=\langle x^{2^{n-1}}\rangle$ is the unique subgroup of $G$ of
order $s$.

######  Corollary 12.2.

If $G_{2}$ is any Sylow $2$-subgroup of $G$, we can write $G=C\rtimes G_{2}$.
Moreover $G_{2}$ is quaternion or dihedral, respectively.

###### Proof.

The first statement is clear as $C$ is normal in $G$. For the second, note
that one Sylow $2$-subgroup is $\langle x^{s},y\rangle$. (Recall from Notation
3.1 that we allow the groups of order $4$ as quaternion or dihedral.) ∎

######  Lemma 12.3.

The commutator $G^{\prime}$ of $G$ is generated by $x^{2}$.

###### Proof.

We note that $x^{2}=xyx^{-1}y^{-1}$, so that $x^{2}$ lies in $G^{\prime}$.
Also $G/\langle x^{2}\rangle$ is abelian: it is isomorphic to $C_{4}$ if $n=2$
and $G$ is quaternion, and to $C_{2}\times C_{2}$ in all other cases. This
shows that $G^{\prime}=\langle x^{2}\rangle$. ∎

Now let $N$ be a finite abelian group of order $2^{n}s$ and let $G$ be a
quaternion or dihedral regular subgroup of $\mathrm{Hol}(N)$. This corresponds
to giving the set $N$ an operation $\circ$ such that $(N,+,\circ)$ is a brace
with $(N,\circ)\cong G$. Thus we have $G=\\{g_{a}:a\in N\\}$ where
$g_{a}=(a,\lambda_{a})$ with $\lambda_{a}\in\mathrm{Aut}(N)$ given by
$\lambda_{a}(b)=-a+a\circ b$. We have a canonical isomorphism $G\to(N,\circ)$
given by $g_{a}\mapsto a$, so that $g_{a}g_{b}=g_{a\circ b}$ and the action of
$G$ on $N$ is given by $g_{a}\cdot b=a\circ b$.

For any characteristic subgroup $M$ of $(N,+)$, we have $\lambda_{a}(M)=M$ for
all $a\in N$, so that $M$ is also a subgroup of $(M,\circ)$ and the brace
structure of $N$ restricts to make $(M,+,\circ)$ into a subbrace. In
particular, for each prime $p$ dividing $2^{n}s$, this applies to the (unique)
Sylow $p$-subgroup $N_{p}$ of $N$. It also applies to the unique subgroup
$N_{s}$ of order $s$ (this being the direct product of the $N_{p}$ over the
primes $p$ dividing $s$).

Let $G_{s}=\\{(a,\lambda_{a}):a\in N_{s}\\}$ and let
$G_{2}=\\{(b,\lambda_{b}):b\in N_{2}\\}$. Thus $G_{s}$ is the unique subgroup
$C$ of order $s$ in $G$, as in Lemma 12.1, and $G_{2}$ is a specific Sylow
$2$-subgroup of $G$ distinguished by the inclusion
$G\subseteq\mathrm{Hol}(N)$. The regular action of $G$ on $N$ restricts to a
regular action of $G_{s}$ on $N_{s}$ (respectively, of $G_{2}$ on $N_{2}$),
via which we obtain a group operation $\circ_{s}$ on $N_{s}$ (respectively
$\circ_{2}$ on $N_{2}$), which is simply the restriction of $\circ$. We then
have canonical isomorphisms $G_{s}\to(N_{s},\circ_{s})$ and
$G_{2}\to(N_{2},\circ_{2})$.

Our goal is now to reconstruct the brace structure on $N$ from more
fundamental components. For the additive group, we have the canonical
decomposition $N=N_{s}\times N_{2}$, and from now on we will write elements of
$N$ in the form $(a,b)$ with $a\in N_{s}$ and $b\in N_{2}$. Since $N_{s}$ and
$N_{2}$ are both characteristic in $N$, we have canonical isomorphisms
$\mathrm{Aut}(N)\to\mathrm{Aut}(N_{s})\times\mathrm{Aut}(N_{2})$ and
$\mathrm{Hol}(N)\to\mathrm{Hol}(N_{s})\times\mathrm{Hol}(N_{2})$. We seek to
determine the operation $\circ$ on $N$, or equivalently, the function
$\lambda:N\to\mathrm{Aut}(N,+)$ such that

$(a,b)\circ(a^{\prime},b^{\prime})=(a,b)+\lambda_{(a,b)}(a^{\prime},b^{\prime})\mbox{
for }a,a^{\prime}\in N_{s}\mbox{ and }b,b^{\prime}\in N_{2}.$

We first evaluate $\lambda$ on the summands $N_{s}$, $N_{2}$ separately.

######  Proposition 12.4.

For $a$, $a^{\prime}\in N_{s}$ we have
$\lambda_{(a,0)}(a^{\prime},0)=(a^{\prime},0)$. Thus $(N_{s},+,\circ_{s})$ is
a trivial brace.

###### Proof.

For each prime $p$ dividing $s$, let $G_{p}=\\{(b,\lambda_{b}):b\in N_{p}\\}$.
Then $G_{p}$ is a Sylow $p$-subgroup of $G$. As $p$ is odd, it follows that
$G_{p}$ is cyclic. Moreover, the image of $G_{p}$ under the projection
$\mathrm{Hol}(N_{s})\to\mathrm{Hol}(N_{p})$ is a regular subgroup of
$\mathrm{Hol}(N_{p})$. By [Koh98, Theorem 4.4], $N_{p}$ must then also be
cyclic. Hence $N_{s}$ is cyclic, and therefore $\mathrm{Aut}(N_{s})$ is
abelian. By Lemma 12.3, the canonical homomorphism
$G\to\mathrm{Aut}(N)\twoheadrightarrow\mathrm{Aut}(N_{s})$ is then trivial on
$G_{s}$. Thus $\lambda_{(a,0)}(a^{\prime},0)=(a^{\prime},0)$, and
$(a,0)\circ(a^{\prime},0)=(a+a^{\prime},0)$. ∎

######  Proposition 12.5.

For $a\in N_{s}$ and $b^{\prime}\in N_{2}$ we have
$\lambda_{(a,0)}(0,b^{\prime})=(0,b^{\prime})$.

###### Proof.

The action of $G$ on $N$ induces a transitive action of $G$ on $N_{2}$ via the
projection $\mathrm{Hol}(N)\to\mathrm{Hol}(N_{2})$. The stabiliser of any
element $b^{\prime}$ of $N_{2}$ is a subgroup of $G$ of order $s$, and by
Lemma 12.1 this can only be $G_{s}$. Hence for any $a\in N_{s}$ and
$b^{\prime}\in N_{2}$, the image of $g_{a}\cdot
b^{\prime}=(a,0)\circ(0,b^{\prime})$ under the projection $N\to N_{2}$ must
coincide with $b^{\prime}$. But
$(a,0)\circ(0,b^{\prime})=(a,0)+\lambda_{(a,0)}(0,b^{\prime})$, and
$\lambda_{(a,0)}(N_{2})=N_{2}$ as $N_{2}$ is characteristic in $N$. Hence
$\lambda_{(a,0)}(0,b^{\prime})=(0,b^{\prime})$. ∎

By Lemma 12.1, the group $G_{s}$ is normal in $G$. The subgroup $G_{2}$ is a
complement to $G_{s}$. Hence $(N_{s},\circ)$ is normal in $(N,\circ)$ with
complement $(N_{2},\circ)$, so conjugation in $(N,\circ)$ gives an action
$\tau:(N_{2},\circ_{2})\to\mathrm{Aut}(N_{s},\circ_{s})$ making $(N,\circ)$
into the semidirect product
$(N_{s},\circ_{s})\rtimes_{\tau}(N_{2},\circ_{2})$. We write $\tau_{b}$ for
the image of $b\in N_{2}$ under $\tau$, so in $(N,\circ)$ we have the relation

(12.1) $b\circ a=\tau_{b}(a)\circ b\mbox{ for }a\in N_{s}\mbox{ and }b\in
N_{2}.$

######  Proposition 12.6.

For $b\in N_{2}$ and $a^{\prime}\in N_{s}$ we have
$\lambda_{(0,b)}(a^{\prime},0)=(\tau_{b}(a^{\prime}),0)$.

###### Proof.

$\displaystyle(0,b)+\lambda_{(0,b)}(a^{\prime},0)$ $\displaystyle=$
$\displaystyle(0,b)\circ(a^{\prime},0)$ $\displaystyle=$
$\displaystyle(\tau_{b}(a^{\prime}),0)\circ(0,b)$ $\displaystyle=$
$\displaystyle(\tau_{b}(a^{\prime}),b),$

where the second equality is a restatement of (12.1) and the third follows
from Proposition 12.5. ∎

The brace $(N_{2},+,\circ_{2})$ of order $2^{n}$ is one of those listed in
Theorem 11.1, so is not a trivial brace except possibly when $n=2$. Let
$\lambda^{(2)}:(N_{2},\circ_{2})\to\mathrm{Aut}(N_{2},+)$ be given by
$\lambda^{(2)}_{b}(b^{\prime})=-b+b\circ b^{\prime}$ for $b$, $b^{\prime}\in
N_{2}$. Then we have

(12.2) $\lambda_{(0,b)}(0,b^{\prime})=(0,\lambda^{(2)}_{b}(b^{\prime}))\mbox{
for }b,b^{\prime}\in N_{2}.$

Putting these pieces together, we can reconstruct the brace $N$ from the
(trivial) brace $N_{s}$, the brace $N_{2}$ and the homomorphism
$\tau:(N_{2},\circ)\to\mathrm{Aut}(N_{s},\circ)$. In fact, the brace $N$ is
the semidirect product of the braces $N_{s}$ and $N_{2}$, cf. [SV18, Corollary
2.36].

######  Lemma 12.7.

The operations in the brace $(N,+,\circ)$ are given by

$(a,b)+(a^{\prime},b^{\prime})=(a+a^{\prime},b+b^{\prime}),$
$(a,b)\circ(a^{\prime},b^{\prime})=(a\circ\tau(a^{\prime}),b\circ_{2}b^{\prime})=(a+\tau(a^{\prime}),b\circ_{2}b^{\prime})$

for all $a$, $a^{\prime}\in N_{s}$ and $b$, $b^{\prime}\in N_{2}$.

###### Proof.

The first equation is immediate from the direct product decomposition
$N=N_{s}\times N_{2}$ of the additive group.

As for the multiplicative group, we have from Proposition 12.5 that
$(a,b)=(a,0)+(0,b)=(a,0)\circ(0,b)$. Since
$\lambda:(N,\circ)\to\mathrm{Aut}(N,+)$ is a group homomorphism, it follows
that

$\lambda_{(a,b)}(a^{\prime},b^{\prime})=\lambda_{(a,0)}\lambda_{(0,b)}(a^{\prime},b^{\prime})=\lambda_{(a,0)}\lambda_{(0,b)}(a^{\prime},0)+\lambda_{(a,0)}\lambda_{(0,b)}(0,b^{\prime})$

where

$\lambda_{(a,0)}\lambda_{(0,b)}(a^{\prime},0)=(\tau_{b}(a^{\prime}),0)$

by Propositions 12.6 and 12.4, and

$\lambda_{(a,0)}\lambda_{(0,b)}(0,b^{\prime})=(0,\lambda^{(2)}_{b}(b^{\prime}))$

by (12.2) and Proposition 12.5. Thus

$\displaystyle(a,b)\circ(a^{\prime},b^{\prime})$ $\displaystyle=$
$\displaystyle(a,b)+(\tau_{b}(a^{\prime}),0)+(0,\lambda^{(2)}_{b}(b^{\prime}))$
$\displaystyle=$
$\displaystyle(a+\tau_{b}(a^{\prime}),b+\lambda^{(2)}(b^{\prime}))$
$\displaystyle=$
$\displaystyle(a\circ\tau_{b}(a^{\prime}),b\circ_{2}b^{\prime}),$

as $a\circ\tau_{b}(a^{\prime})=a+\tau_{b}(a^{\prime})$ by Proposition 12.4.
This completes the proof. ∎

Keeping the abelian group $N=N_{s}\times N_{2}$ fixed, we now allow $G$ to
vary amongst the regular quaternion (respectively dihedral) subgroups of
$\mathrm{Hol}(N)$. We will describe these groups $G$ in terms of the regular
quaternion (respectively dihedral) subgroups $H$ of $\mathrm{Hol}(N_{2})$. Any
such $H$ determines a group operation $\circ_{H}$ on $N_{2}$ making
$(N_{2},+,\circ_{H})$ into a quaternion (respectively dihedral) brace. Let
$\lambda^{H}:(N_{2},\circ_{H})\to\mathrm{Aut}(N_{s})$ be the corresponding
homomorphism, so $\lambda^{H}_{b}(b^{\prime})=-b+b\circ_{H}b^{\prime}$ for
$b$, $b^{\prime}\in N_{2}$. We denote by $T_{H}$ the set of homomorphisms
$\tau:(N_{2},\circ_{H})\to\mathrm{Aut}(N_{s})$ for which
$N_{s}\rtimes_{\tau}(N_{2},\circ_{H})$ is a quaternion (respectively dihedral)
group.

There is an action of the group $\mathrm{Aut}(N_{2})$ on the set of all
functions $\sigma:N_{2}\to\mathrm{Aut}(N_{s})$ by
$(\beta\cdot\sigma)_{b}=\sigma_{\beta^{-1}(b)}$ for
$\beta\in\mathrm{Aut}(N_{2})$ and $b\in N_{2}$.

######  Theorem 12.8.

Let $n\geq 2$, let $s\geq 3$ be odd, and let $N=N_{s}\times N_{2}$ be a finite
abelian group of order $2^{n}s$, where $N_{s}$ is cyclic of order $s$. Then
there is a bijective correspondence between regular quaternion (respectively,
dihedral) subgroups $G$ in $\mathrm{Hol}(N)$ and pairs $(H,\tau)$ where $H$ is
a regular quaternion (respectively, dihedral) subgroup of
$\mathrm{Hol}(N_{2})$ and $\tau\in T_{H}$. Moreover, if $G$ corresponds to the
pair $(H,\tau)$, then, for $\alpha\in\mathrm{Aut}(N_{s})$,
$\beta\in\mathrm{Aut}(N_{2})$, the group $(\alpha,\beta)G(\alpha,\beta)^{-1}$
corresponds to the pair $(\beta H\beta^{-1},\beta\cdot\tau)$.

###### Proof.

Let $G$ be a regular quaternion (respectively, dihedral) subgroup of
$\mathrm{Hol}(N)$. As above, $G$ determines a group operation $\circ$ on $N$
and also determines subgroups $G_{s}$, $G_{2}$ of $\mathrm{Hol}(N)$. Let $H$
be the image of $G_{2}$ under the projection
$\mathrm{Hol}(N)=\mathrm{Hol}(N_{s})\times\mathrm{Hol}(N_{2})\to\mathrm{Hol}(N_{2})$.
Then $H$ is a regular quaternion (respectively, dihedral) subgroup of
$\mathrm{Hol}(N_{2})$, and the group operation $\circ_{H}$ on $N_{2}$
determined by $H$ is just the restriction $\circ_{2}$ of $\circ$ to $N_{2}$.
Moreover, by Proposition 12.4, the restriction of $\circ$ to $N_{s}$ makes
$N_{s}$ into a trivial brace, and, as explained just before Proposition 12.6,
there is a homomorphism
$\tau:(N_{2},\circ_{2})\to\mathrm{Aut}(N_{s},\circ)=\mathrm{Aut}(N_{s},+)$ so
that $N_{s}\rtimes_{\tau}(N_{2},\circ_{2})=(N,\circ)\cong G$. Since
$\circ_{2}$ coincides with $\circ_{H}$, we have $\tau\in T_{H}$. We associate
the pair $(H,\tau)$ to $G$. Conversely, let $(H,\tau)$ be a pair as in the
statement of the theorem. Then $H$ gives rise to a group operation $\circ_{H}$
on $N_{2}$ and a canonical isomorphism $(N_{2},\circ_{H})\cong H$. Using
$\circ_{H}$ and $\tau$, we define an operation $\circ$ on $N$ by

$(a,b)\circ(a^{\prime},b^{\prime})=(a+\tau(a^{\prime}),b\circ_{H}b^{\prime})$

for all $a$, $a^{\prime}\in N_{s}$ and $b$, $b^{\prime}\in N_{2}$. It is
straightforward to check that this makes $N$ into a quaternion (respectively,
dihedral) brace which therefore corresponds to a regular quaternion
(respectively, dihedral) subgroup $G$ of $\mathrm{Hol}(N)$. It is clear that
these two constructions are mutually inverse. This establishes the bijective
correspondence in the theorem.

Now let $(H,\tau)$ correspond to $G$. We will write elements of
$\mathrm{Hol}(N)=(N_{s}\rtimes\mathrm{Aut}(N_{s}))\times\mathrm{Hol}(N_{2})$
as triples $(a,\alpha,h)$ with $a\in N_{s}$, $\alpha\in\mathrm{Aut}(N_{s})$,
$h\in\mathrm{Hol}(N_{2})$. Then

$G=\\{(a,\tau_{b},h_{b}):a\in N_{s},b\in N_{2}\\}$

where $h_{b}=(b,\lambda^{H}_{b})\in H$. For $\alpha\in\mathrm{Aut}(N_{s})$ and
$\beta\in\mathrm{Aut}(N_{2})\subset\mathrm{Hol}(N_{2})$, we have

$(0,\alpha,\beta)(a,\tau_{b},h_{b})(0,\alpha,\beta)^{-1}=(\alpha(a),\tau_{b},\beta
h_{b}\beta^{-1}),$

where we have again used the fact that $\mathrm{Aut}(N_{s})$ is abelian.
Moreover in $\mathrm{Hol}(N_{2})$ we have

$\beta
h_{b}\beta^{-1}=(0,\beta)(b,\lambda^{H}_{b})(0,\beta^{-1})=(\beta(b),\beta\lambda^{H}_{b}\beta^{-1}).$

Writing $H^{\prime}=\beta H\beta^{-1}=\\{h^{\prime}_{b}:b\in N_{2}\\}$ with
$h^{\prime}_{b}=(b,\lambda^{H^{\prime}}_{b})$, we therefore have $\beta
h_{b}\beta^{-1}=h^{\prime}_{\beta(b)}$. Hence

$\displaystyle(\alpha,\beta)G(\alpha,\beta)^{-1}$ $\displaystyle=$
$\displaystyle\\{(\alpha(a),\tau_{b},h^{\prime}_{\beta(b)}):a\in N_{s},b\in
N_{2}\\}$ $\displaystyle=$
$\displaystyle\\{(a,(\beta\cdot\tau)_{\beta(b)},h^{\prime}_{\beta(b)}):a\in
N_{s},b\in N_{2}\\}$ $\displaystyle=$
$\displaystyle\\{(a,(\beta\cdot\tau)_{b},h^{\prime}_{b}):a\in N_{s},b\in
N_{2}\\}$

which is the regular subgroup of $\mathrm{Hol}(N)$ corresponding to the pair
$(\beta H\beta^{-1},\beta\cdot\tau)$. ∎

######  Remark 12.9.

The proof of Theorem 12.8 is inspired by [CGMRV23a, Proposition 4]. Note in
fact that their proposition holds under the general assumptions (satisfied in
both cases) that the braces we are studying are semidirect products of braces
such that the normal component is a trivial brace.

We next investigate the sets $T_{H}$.

######  Proposition 12.10.

Let $H$ be a regular quaternion (respectively dihedral) subgroup of
$\mathrm{Hol}(N_{2})$. If $H\cong Q_{2^{n}}$ with $n\neq 3$, or $H\cong
D_{2^{n}}$ with $n\neq 2$, then $|T_{H}|=1$. If $H\cong Q_{8}$ or $D_{4}$ then
$|T_{H}|=3$. In particular, $T_{H}$ does not depend on $s$.

###### Proof.

A homomorphism $\tau:(N_{2},\circ_{H})\to\mathrm{Aut}(N_{s})$ belongs to
$T_{H}$ if and only if $N_{s}\rtimes_{\tau}(N_{2},\circ_{H})$ is a quaternion
(respectively dihedral) group. This occurs precisely when the image of $\tau$
is the subgroup of order $2$ in $\mathrm{Aut}(N_{s})$ generated by inversion.
Thus $\tau$ is completely determined by its kernel, which may be any subgroup
of index $2$ in $(N_{2},\circ_{H})\cong H$. If $H\cong Q_{2^{n}}$ with $n\neq
3$, or $H\cong D_{2^{n}}$ with $n\neq 2$, then $H$ has a unique subgroup of
index $2$. Thus $|T_{H}|=1$. In the remaining cases, $H$ has $3$ subgroups of
index $2$, so $|T_{H}|=3$. ∎

We now modify the notation used in Table 1 to remove the dependence on the
‘Small Group’ labelling. For an abelian group $N$ and a group $J$ with
$|N|=|J|=2^{n}s$, let $c(N,J)$ denote the number of isomorphism classes of
braces with additive group isomorphic to $N$ and with multiplicative group
isomorphic to $J$. This is just the number of orbits of regular subgroups in
$\mathrm{Hol}(N)$ isomorphic to $J$ under conjugation by $\mathrm{Aut}(N)$.
Similarly, let $r(N,J)$ be the total number of regular subgroups in
$\mathrm{Hol}(N)$ isomorphic to $J$, and let $h(N,J)$ denote the number of
Hopf-Galois structures of type $N$ on a Galois extension whose Galois group is
isomorphic to $J$.

The next result shows that, when $J$ is a quaternion or dihedral group, the
quantities $r(N,J)$ and $c(N,J)$ almost always coincide with the corresponding
quantities for the Sylow $2$-subgroups of $N$ and $J$.

######  Theorem 12.11.

Let $N=N_{s}\times N_{2}$ be as in Theorem 12.8, let $J\cong Q_{2^{n}s}$ or
$D_{2^{n}s}$, and let $J_{2}$ be a Sylow $2$-subgroup of $J$.

If $J\cong Q_{2^{n}s}$ with $n\neq 3$, or $J\cong D_{2^{n}s}$ with $n\neq 2$,
then $r(N,J)=r(N_{2},J_{2})$ and $c(N,J)=c(N_{2},J_{2})$.

If $J\cong Q_{8s}$ or $D_{4s}$ then $r(N,J)=3r(N_{2},J_{2})$. Also for $n=3$
we have $c(N,Q_{8s})=c(C_{3}\times N_{2},Q_{24})$, and for $n=2$ we have
$c(N,D_{4s})=c(C_{3}\times N_{2},D_{12})$.

###### Proof.

If $J\cong Q_{2^{n}s}$ with $n\neq 3$, or $J\cong D_{2^{n}s}$ with $n\neq 2$,
then $|T_{H}|=1$ by Proposition 12.10. Thus Theorem 12.8 gives a bijection
between regular quaternion (respectively dihedral) subgroups $G$ of
$\mathrm{Hol}(N)$ and regular quaternion (respectively dihedral) subgroups $H$
of $\mathrm{Hol}(N_{2})$. Hence $r(N,J)=r(N_{2},J_{2})$. Writing
$S_{H}=\\{\beta\in\mathrm{Aut}(N_{2}):\beta H\beta^{-1}=H\\}$ for the
stabiliser of $H$ in $\mathrm{Aut}(N_{2})$, we see that the stabiliser of $G$
in $\mathrm{Aut}(N)=\mathrm{Aut}(N_{s})\times\mathrm{Aut}(N_{2})$ is
$\mathrm{Aut}(N_{s})\times S_{H}$. Thus the length of the conjugacy class of
$G$ in $\mathrm{Aut}(N_{2})$ is

$\frac{|\mathrm{Aut}(N_{s})\times\mathrm{Aut}(N_{2})|}{|\mathrm{Aut}(N_{s})\times
S_{H}|}=\frac{|\mathrm{Aut}(N_{2})|}{|S_{H}|},$

which coincides with the length of the conjugacy class of $H$ in
$\mathrm{Aut}(N_{2})$. Thus we also have $c(N,J)=c(N_{2},J_{2})$.

In the two remaining cases, we have $|T_{H}|=3$ for each $H$. Thus each
regular quaternion (respectively dihedral) subgroup $H$ in $\mathrm{Hol}(N)$
occurs in $3$ pairs $(H,\tau)$ in Theorem 12.8, so $r(N,J)=3r(N_{2},J_{2})$.
We consider the conjugacy class of the regular subgroup $G$ of
$\mathrm{Hol}(N)$ corresponding to the pair $(H,\tau)$. Now
$(\alpha,\beta)\in\mathrm{Aut}(N)$ is in the stabiliser of $G$ if and only if
$\beta H\beta^{-1}$ and $\beta\cdot\tau=\tau$ (with no condition on $\alpha$).
Then $\beta$ must belong firstly to $S_{H}$, and further to the stabiliser
$S_{H,\tau}$ of $\tau$ in $S_{H}$. Here the action of $S_{H}$ on $T_{H}$
corresponds to its action on the subgroups of index $2$ in $H$. Thus
$S_{H,\tau}$ may depend on the subgroup $H$, but is independent of $s$. The
lengths of the conjugacy classes of the groups $G$ are therefore the same for
arbitrary odd $s\geq 3$ as for $s=3$. Hence $c(N,J)$ is as stated. ∎

To complete the evaluation of $c(N,J)$ in all cases, it only remains to handle
$J\cong Q_{24}$ and $J\cong D_{12}$. We do this using Magma. We start with the
case $N=C_{24}$ and $J=Q_{24}$. As in §10, we run the following code. We omit
the output of ‘R;’ which we use to count the number of conjugacy classes of
regular subgroups. In this case, there are $14$ conjugacy classes.

     N:=SmallGroup(24,2);
     G:=SmallGroup(24,4);
     H:=Holomorph(N);
     R:=RegularSubgroups(H);
     R;
     for i in [1..14] do
        if IsIsomorphic(R[i]‘subgroup,G) eq true then
            print i;
        end if;
     end for;
     6
     14

There are therefore two conjugacy classes of regular subgroups isomorphic to
$Q_{24}$. Hence there are two isomorphism classes of braces of order $24$ with
cyclic additive group and quaternion multiplicative group. Repeating the
calculation with $N=C_{3}\times C_{2}\times C_{4}$ (SmallGroup(24,9)) and
$N=C_{3}\times C_{2}^{3}$ (SmallGroup(24,15)), we find three conjugacy classes
and one conjugacy class of regular subgroups isomorphic to $Q_{24}$,
respectively.

######  Corollary 12.12.

Let $s\geq 3$ be an odd number. Then there are $6$ isomorphism classes of
braces with multiplicative group isomorphic to $Q_{8s}$.

Analogously, we can analyse the $D_{12}$ case, starting from $N=C_{12}$.

     N:=SmallGroup(12,2);
     G:=SmallGroup(12,4);
     H:=Holomorph(N);
     R:=RegularSubgroups(H);
     R;
     for i in [1..5] do
        if IsIsomorphic(R[i]‘subgroup,G) eq true then
            print i;
        end if;
     end for;
     2
     5

In this case there are two conjugacy classes. Hence there are two isomorphism
classes of braces with cyclic additive group and quaternion multiplicative
group. Repeating the calculation with $N=C_{3}\times C_{2}\times C_{2}$
(SmallGroup(12,5)), we find one conjugacy class of regular subgroups of
$\mathrm{Hol}(N)$ isomorphic to $D_{12}$.

######  Corollary 12.13.

Let $s\geq 3$ be an odd number. Then there are $3$ isomorphism classes of
braces with multiplicative group isomorphic to $D_{4s}$.

##  13\. Final statement about braces

Putting together Theorem 11.1 and the conclusions in §12, we have now proved
Conjecture 6.6 of [GV17] which we stated as Conjecture 1.2.

######  Theorem 13.1.

Let $m\geq 3$ be an integer and let $q(4m)$ be the number of isomorphism
classes of braces with multiplicative group isomorphic to $Q_{4m}$. Then

$q(4m)=\begin{cases}2\ \ \text{ if $m$ is odd;}\\\ 6\ \ \text{ if }m\equiv
2\pmod{4};\\\ 9\ \ \text{ if }m\equiv 4\pmod{8};\\\ 7\ \ \text{ if }m\equiv
0\pmod{8}.\end{cases}$

We also have the analogue of Theorem 13.1 for dihedral braces.

######  Theorem 13.2.

Let $m\geq 3$ be an integer and let $d(4m)$ be the number of isomorphism
classes of braces with multiplicative group isomorphic to $D_{4m}$. Then

$d(4m)=\begin{cases}3\ \ \text{ if $m$ is odd;}\\\ 8\ \ \text{ if }m\equiv
2\pmod{4};\\\ 7\ \ \text{ if }m\equiv 4\pmod{8};\\\ 7\ \ \text{ if }m\equiv
0\pmod{8}.\end{cases}$

Table 2. Number of isomorphism classes of quaternion or dihedral braces for each possible additive group $N$ $N$ | Conditions | Quaternion braces | Dihedral braces
---|---|---|---
$C_{s}\times C_{2^{n}}$ |  $n\geq 5$, $s$ odd | 1 | 1
$C_{s}\times C_{2}\times C_{2^{n-1}}$ |  $n\geq 5$, $s$ odd | 6 | 6
$C_{s}\times C_{16}$ |  $s$ odd | 1 | 1
$C_{s}\times C_{2}\times C_{8}$ |  $s$ odd | 4 | 6
$C_{s}\times C_{4}\times C_{4}$ |  $s$ odd | 2 | 0
$C_{s}\times C_{2}\times C_{2}\times C_{4}$ |  $s$ odd | 1 | 0
$C_{s}\times C_{2}\times C_{2}\times C_{2}\times C_{2}$ |  $s$ odd | 1 | 0
$C_{s}\times C_{8}$ |  $s\geq 3$ odd | 2 | 1
$C_{s}\times C_{2}\times C_{4}$ |  $s\geq 3$ odd | 3 | 5
$C_{s}\times C_{2}\times C_{2}\times C_{2}$ |  $s\geq 3$ odd | 1 | 2
$C_{8}$ |  | 1 | 1
$C_{4}\times C_{2}$ |  | 1 | 5
$C_{2}\times C_{2}\times C_{2}$ |  | 1 | 2
$C_{s}\times C_{4}$ |  $s\geq 3$ odd | 1 | 2
$C_{s}\times C_{2}\times C_{2}$ |  $s\geq 3$ odd | 1 | 1
$C_{4}$ |  | 1 | 1
$C_{2}\times C_{2}$ |  | 1 | 1

Question 6.7 of [GV17] asks which finite abelian groups $A$ appear as the
additive group of a quaternion brace, and Question 6.8 asks for the number of
isomorphism classes of quaternion braces for each such $A$. The answers to
these questions, and the corresponding answers for dihedral braces, are given
in Table 2.

##  14\. Final statement about Hopf–Galois structures

We start by generalising Proposition 3.3.

######  Proposition 14.1.

Let $s\geq 3$ be an odd integer and let $n\geq 2$ be an integer. Then
$|\mathrm{Aut}(Q_{2^{n}s})|=|\mathrm{Aut}(D_{2^{n}s})|=2^{2n-3}s\varphi(s)$.

###### Proof.

Let $G=\langle x,y\rangle_{o}$ be $Q_{2^{n}s}$ or $D_{2^{n}s}$. The cyclic
group $\langle x\rangle$ of order $2^{n-1}s$ is characteristic in $G$, so that
every automorphism can only send $x$ to $\varphi(2^{n-1}s)$ other possible
elements. Moreover, $y$ can only go to a power of $x$ times $y$. We can verify
that each of these possibilities defines an automorphism. ∎

Using the notation $h(N,J)$ introduced before Theorem 12.11, we can now reduce
the enumeration of Hopf-Galois structures on quaternion or dihedral extensions
to the $2$-power case.

######  Proposition 14.2.

Let $s\geq 3$ be odd and let $n\geq 2$. Let $J\cong Q_{2^{n}s}$ or $J\cong
D_{2^{n}s}$, let $J_{2}$ be a Sylow $2$-subgroup of $J$, and let
$N=N_{s}\times N_{2}$ be an abelian group of order $2^{n}s$ with $N_{s}$
cyclic of order $s$. Then $h(N,J)=h(N_{2},J_{2})s$ in all cases.

###### Proof.

By (1.1) we have

$h(N,J)=\frac{|\mathrm{Aut}(J)|}{|\mathrm{Aut}(N)|}\;r(N,J)\mbox{ and
}h(N_{2},J_{2})=\frac{|\mathrm{Aut}(J_{2})|}{|\mathrm{Aut}(N_{2})|}\;r(N_{2},J_{2}).$

As $\mathrm{Aut}(N)=\mathrm{Aut}(N_{s})\times\mathrm{Aut}(N_{2})$ and
$|\mathrm{Aut}(N_{s})|=\varphi(s)$, we deduce that

(14.1)
$h(N,J)=\frac{|\mathrm{Aut}(J)|}{|\mathrm{Aut}(J_{2})|}\cdot\frac{1}{{\varphi(s)}}\cdot\frac{r(N,J)}{r(N_{2},J_{2})}\cdot
h(N_{2},J_{2}).$

If $J\cong Q_{2^{n}s}$ with $n\neq 3$, or $J\cong D_{2^{n}s}$ with $n\neq 2$,
then from Propositions 3.3 and 14.1 we have
$|\mathrm{Aut}(J)|=s\varphi(s)|\mathrm{Aut}(J_{2})|$. Also
$r(N,J)=r(N_{2},J_{2})$ by Theorem 12.11. Then (14.1) gives
$h(N,J)=h(N_{2},J_{2})s$.

In the remaining cases $J\cong Q_{8s}$ and $J\cong D_{4s}$, we have
$|\mathrm{Aut}(J_{2})|=3\cdot 2^{2n-3}$. (The extra factor $3$ arises since
there are $3$ subgroups of index $2$ in $Q_{8}$ and $D_{4}$.) Thus
$|\mathrm{Aut}(J)|=\frac{1}{3}s\varphi(s)|\mathrm{Aut}(J_{2})|$. But by
Theorem 12.11, $r(N,J)=3r(N_{2},J_{2})$ in these cases, so again (14.1) yields
$h(N,J)=h(N_{2},J_{2})s$. ∎

Combining Proposition 14.2 and Table 1, we obtain our final result.

######  Theorem 14.3.

Let $n\geq 2$ be an integer and let $s\geq 3$ be an odd integer. If $n\geq 5$,
then there are $2^{n-2}\cdot 9s$ Hopf-Galois structures of abelian type on a
Galois extension of degree $2^{n}s$ with quaternion or dihedral Galois group.
There are $2s$, respectively $14s$, respectively $52s$, Hopf-Galois structures
of abelian type on a quaternion Galois extension of degree $4s$, respectively
$8s$, respectively $16s$. There are $4s$, respectively $22s$, respectively
$36s$, Hopf-Galois structures of abelian type on a dihedral Galois extension
of degree $4s$, respectively $8s$, respectively $16s$. The numbers of Hopf-
Galois structures of each possible abelian type are as shown in Table 3.

Table 3. Number of Hopf-Galois structures of each possible abelian type $N$ on a Galois extension of degree $2^{n}s$ ($s\geq 1$, odd) with quaterrnionic or dihedral Galois group $G$ $N$ | Conditions |  $G$ quaternion |  $G$ dihedral
---|---|---|---
$C_{s}\times C_{2^{n}}$ | $n\geq 5$ | $2^{n-2}s$ | $2^{n-2}s$
$C_{s}\times C_{2}\times C_{2^{n-1}}$ | $n\geq 5$ | $2^{n+1}s$ | $2^{n+1}s$
$C_{s}\times C_{16}$ |  | $4s$ | $4s$
$C_{s}\times C_{2}\times C_{8}$ |  | $16s$ | $32s$
$C_{s}\times C_{4}\times C_{4}$ |  | $16s$ | 0
$C_{s}\times C_{2}\times C_{2}\times C_{4}$ |  | $8s$ | 0
$C_{s}\times C_{2}\times C_{2}\times C_{2}\times C_{2}$ |  | $8s$ | $0$
$C_{s}\times C_{8}$ |  | $6s$ | $2s$
$C_{s}\times C_{2}\times C_{4}$ |  | $6s$ | $14s$
$C_{s}\times C_{2}\times C_{2}\times C_{2}$ |  | $2s$ | $6s$
$C_{s}\times C_{4}$ |  | $s$ | $3s$
$C_{s}\times C_{2}\times C_{2}$ |  | $s$ | $s$

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