On the Galois correspondence for Hopf Galois structures arising from finite radical algebras and Zappa-Sz\'ep products

TL;DR

This paper investigates the Galois correspondence ratio in Hopf Galois structures derived from finite radical algebras and Zappa-Szép products, revealing how subgroup counts influence the correspondence in specific algebraic contexts.

Contribution

It introduces a method to compute the Galois correspondence ratio using skew braces from radical algebras and Zappa-Szép products, especially for bi-skew braces of squarefree order.

Findings

Galois correspondence ratio approaches 1/2 for certain bi-skew braces.

The ratio is near 0 in some configurations involving dihedral groups.

Explicit examples demonstrate the impact of algebraic structure on the Galois correspondence.

Abstract

Let $L/K$ be a $G$-Galois extension of fields with an $H$-Hopf Galois structure of type $N$. We study the ratio $GC(G, N)$, which is the number of intermediate fields $E$ with $K \subseteq E \subseteq L$ that are in the image of the Galois correspondence for the $H$-Hopf Galois structure on $L/K$, divided by the number of intermediate fields. By Galois descent, $L \otimes_K H = LN$ where $N$ is a $G$-invariant regular subgroup of $\mathrm{Perm}(G)$, and then $GC(G, N)$ is the number of $G$-invariant subgroups of $N$, divided by the number of subgroups of $G$. We look at the Galois correspondence ratio for a Hopf Galois structure by translating the problem into counting certain subgroups of the corresponding skew brace. We look at skew braces arising from finite radical algebras $A$ and from Zappa-Sz\'ep products of finite groups, and in particular when $A^3 = 0$ or the Zappa-Sz\'ep product is a semidirect product, in which cases the corresponding skew brace is a bi-skew brace, that is, a set $G$ with two group operations $\circ$ and $\star$ in such a way that $G$ is a skew brace with either group structure acting as the additive group of the skew brace. We obtain the Galois correspondence ratio for several examples. In particular, if $(G, \circ, \star)$ is a bi-skew brace of squarefree order $2m$ where $(G, \circ) \cong Z_{2m}$ is cyclic and $(G, \star) = D_m$ is dihedral, then for large $m$, $GC(Z_{2m},D_m), $ is close to 1/2 while $GC(D_m, Z_{2m})$ is near 0.

Equations

$$H^{\prime }\mapsto L^{H^{\prime }}=\{x\in L|h^{\prime }x=ϵ(h^{\prime })x\text{ for all h′ in }{H}^{\prime }\},$$

$$g_1\star g_2=b(n_1{n}_2).$$

$$n_1(g_2)$$

$$=b(n_1{n}_2)=g_1\star g_2={\lambda }_{\star }(g_1)(g_2).$$

$$a\circ (b\star c)=(a\circ b)\star a^{-1}\star (a\circ c)$$

$${\lambda }_{\star }:$$

$${\lambda }_{\circ }:$$

$$\mathrm{Hol}(B,\star )={\lambda }_{\star }(B)\mathrm{Aut}(B,\star )\subset \mathrm{Perm}(B),$$

$$\mathrm{Aut}(B,\circ )/{\mathrm{Aut}}_{sb}(B,\circ ,\star ).$$

$$GC((G,\circ ),(G,\star ))=\frac{|\{\text{E in the image of the Galois correspondence for H}\}|}{|\{E:K\subset E\subset L\}|}.$$

$$\text{For all }g\in G,g^{\prime }\in G^{\prime },(g\circ g^{\prime })\star g^{-1}=h^{\prime }\text{ is in }{G}^{\prime }.$$

$${\rho }_g(g^{\prime })=(g\circ g^{\prime })\star g^{-1}.$$

$$GC((G,\circ ),(G,\star ))=\frac{|\{\text{∘-stable subgroups of (G,⋆)}\}|}{|\{\text{subgroups of (G,∘)}\}|}.$$

$$a\circ b=a+b+a\cdot b$$

$$a\circ b=a+b+ab,$$

$$\overline{a}=-a+a^2-a^3+\dots ,$$

$$g+g^{\prime }+g{g}^{\prime }$$

$$g{g}^{\prime }$$

$$g\circ g^{\prime }=g+g^{\prime }+g{g}^{\prime }=h^{\prime }+g,$$

$$GC((A,\circ ),(A,\star ))=\frac{|\{\text{left ideals of A}\}|}{|\{\text{subgroups of (A,∘)}|}.$$

$$\left(\begin{array}{cccc}1& r& 0& s\\ 0& 0& 1& r\end{array}\right).$$

$$p+1+1+p+p+1+1+p^2=p^2+3p+4$$

$$GC((G,\circ ),(G,+))=\frac{|\{\text{ left ideals of A}\}|}{|\{\text{subgroups of (A,∘)}\}|}.$$

$$(ra+sb+tc+ud{)}^{\circ m}=mra+msb+(m+\left(\genfrac{}{}{0ex}{}{m}{2}\right))r^{2}c+(m+\left(\genfrac{}{}{0ex}{}{m}{2}\right))rsd$$

$${\alpha }_1^{-1}=-a-sb-w_1+c+sd,$$

$${\alpha }_1=a+sb+w_1\text{ and }{\alpha }_2=a+s^{\prime }b+w_2.$$

$${\alpha }_1^{-1}\circ {\alpha }_2=(s^{\prime }-s)b+w_3.$$

$$(-s^{\prime }b-w_5)\circ {\alpha }_2=(-s^{\prime }b-w_5)+a+s^{\prime }b+w_2=a+w_6,$$

$$(-b-w_4)\circ (a+w_6)\circ (b+w_4)\circ (-a-w_6+c)$$

$$=(a-b-w_4+w_6)\circ (-a+b+w_4-w_6+c$$

$$=c-c+d=d.$$

$$G^{\prime }=⟨a+sc,b+tc,d⟩$$

$$GC((G,\circ ),(G,+))=\frac{p^2+3p+5}{2p^3+4p^2+3p+5}.$$

Full text

1. Introduction

In 1969 S. Chase and M. Sweedler [CS69] defined the notion of an $H$-Hopf Galois structure on a finite extension of commutative rings. The notion in particular applies to the case of a separable field extension $L/K$, where it extends the classical notion of a Galois extension, the case where $L/K$ is normal, $G$ is the Galois group of $L/K$, and $H$ is the group ring $KG$.

Let $L/K$ be a $G$-Galois extension of fields. Let $H$ be a cocommutative $K$-Hopf algebra that acts on $L$ as an $H$-module algebra, and suppose $L$ is an $H$-Hopf Galois extension of $K$. Then, as shown in [GP87], $L\otimes_{K}H=LN$ for some regular subgroup $N$ of $\mathrm{Perm}(G)$, where $N$ is normalized by the image $\lambda(G)$ of the left regular representation $\lambda:G\to\mathrm{Perm}(G)$. In turn, if $N$ is a regular subgroup of $\mathrm{Perm}(G)$ normalized by $\lambda(G)$, then $N$ yields by Galois descent a $K$-Hopf algebra $H=L[N]^{G}$ which acts on $L/K$ making $L/K$ a $H$-Hopf Galois extension. In this way there is a bijection between Hopf Galois structures on the $G$-Galois extension $L/K$ and regular subgroups of $\mathrm{Perm}(G)$ normalized by $\lambda(G)$.

Given an $H$-Hopf Galois structure on $L/K$, there is a Galois correspondence, originally described by Chase and Sweedler [CS69], namely, an injective correspondence from $K$-subHopf algebras of $H$ to fields $E$ with $K\subset E\subset L$, by

[TABLE]

where $\epsilon:H\to K$ is the counit map. But in contrast to classical Galois theory, the Galois correspondence for a Hopf Galois extension can fail to be surjective. The first set of examples where surjectivity fails is in [GP87]: let $N$ be the regular subgroup $N=\lambda(G)$ of $\mathrm{Perm}(G)$. Then $N$ is normalized by itself, and the subgroups $N^{\prime}$ of $N=\lambda(G)$ that are normalized by $\lambda(G)$ are the subgroups $\lambda(G^{\prime})$ where $G^{\prime}$ is a normal subgroup of $G$. So if $G$ is non-abelian and has non-normal subgroups, then surjectivity fails. (The classical Galois structure on $L/K$ given by the Galois group corresponds to $\rho(G)$, the image in $\mathrm{Perm}(G)$ of the right regular representation of $G$ in $\mathrm{Perm}(G)$, $\rho(g)(h)=hg^{-1}$ for $g,h$ in $G$: $\rho(G)$, hence every subgroup of $\rho(G)$, is centralized, hence normalized by $\lambda(G)$.)

But except for the Greither-Pareigis examples, very little was known about the image of the Galois correspondence for an $H$-Hopf Galois structure on a $G$-Galois extension $L/K$ of fields until [CRV16] observed that the $K$-subHopf algebras of $H$ correspond bijectively to the subgroups $N^{\prime}$ of $N$ that are normalized by $\lambda(G)$. Using that result, [Ch17] examined $G$-Galois extensions $L/K$ where $G$ is an abelian $p$-group of order $p^{n}$ and $L/K$ has an $H$-Hopf Galois structure of type $N$, also an abelian $p$-group of order $p^{n}$. If $H$ has type $N$, then there is a regular embedding of $G$ into $\mathrm{Hol}(N)$: call the image $T$. Then ([CDVS06], [FCC12]), that regular subgroup $T$ defines a commutative nilpotent ring structure $A=(N,+,\cdot)$ on the additive abelian group $N$ so that the regular subgroup $T$, and hence the Galois group $G$, is isomorphic to the adjoint group $(A,\circ)$ of the nilpotent ring $A$. In that setting, for the Hopf Galois structure given by $H$, the image of the Galois correspondence is in bijective correspondence with the ideals of $A$.

If $A^{e}=0$ and $e<p$, hence $G$ and $N$ are elementary abelian $p$-groups, [CG18] computed upper and lower bounds on the proportion of subgroups of $A$ that are ideals, and showed, for example, that if $4\leq e<p$, then the proportion of subgroups that are ideals is $<.01$.

We note that the only known cases of a non-classical $H$-Hopf Galois structure on a $G$-Galois field extension where the Galois correspondence for $H$ is surjective are for $G$ a non-abelian Hamiltonian group (every subgroup is normal) where $H$ corresponds to $\lambda(G)$, or for $G$ cyclic of odd prime power order and $H$ any Hopf Galois structure ([Ch17], Proposition 4.3).

The remainder of the paper is organized as follows. Section 2 describes the relationship between Hopf Galois structures and skew braces, and section 3 describes the Galois correspondence ratio in the skew brace setting. Section 4 looks at that ratio for skew braces arising from radical algebras, illustrated by a four-dimensional example $A=A^{0}_{4,25}$. Section 5 introduced bi-skew braces and finds the Galois correspondence ratio for the other skew brace structure for $A$, which exists because $A^{3}=0$. Section 6 looks at the Galois correspondence ratios arising from fixed point free pairs of homomorphisms to a Zappa-Szép product of two finite groups from the corresponding direct product, with an example. Section 7 specializes to semi-direct products of groups, which yield bi-skew braces, hence two Galois correspondence ratios. Section 8 looks at examples where the order of $G$ is square-free, and in particular determines the two highly divergent ratios for $G$ a class of generalized dihedral groups.

Many thanks to the referees for their numerous helpful comments and corrections on a previous version of this paper.

2. Hopf Galois structures and skew braces

The concept of skew brace was first defined in [GV17] and the connection of skew braces with Hopf Galois structures was described in [SV18]. Following [SV18], in what follows, “skew brace” and “brace” will always mean “skew left brace” and “left brace”, respectively.

Finite radical rings are skew braces, so it was natural to generalize the description of the Galois correspondence ratio for radical rings in [Ch17] to the setting of skew braces. This was done in [Ch18].

To see how this works, we first reexamine the description of a Hopf Galois structure.

Let $G$ be a finite group, and denote the operation on $G$ by $\circ$. Let $L/K$ be a $(G,\circ)$-Galois extension of fields. If $L$ is also an $H$-Hopf Galois extension of $K$, then $L\otimes_{K}H=LN$ for some regular subgroup $N$ of $\mathrm{Perm}(G)$, where $N$ is normalized by the image $\lambda_{\circ}(G)$ of the left regular representation $\lambda_{\circ}:G\to\mathrm{Perm}(G)$.

Since $N$ is a regular subgroup of $\mathrm{Perm}(G)$, the map $b:N\to G$ given by $n\mapsto n(e)$ is a bijection. Then $b$ defines an operation $\star$ on $G$, as follows: if $b(n_{1})=g_{1},b(n_{2})=g_{2}$, we define

[TABLE]

Then $N=\lambda_{\star}(G)$, the image of the left regular representation map $\lambda_{\star}$ corresponding to the operation $\star$. For setting $b(n_{1})=g_{1},b(n_{2})=g_{2}$ and $g_{1}\star g_{2}=b(n_{1}n_{2})$, then the action of $N\subset\mathrm{Perm}(G)$ on $G$ is

[TABLE]

If $N=\lambda_{\star}(G)$ in $\mathrm{Perm}(G)$ is normalized by $\lambda_{\circ}(G)$, then $\lambda_{\circ}(G)$ is contained in $\mathrm{Hol}(G,\star)$, the normalizer of $\lambda_{\star}(G)$ in $\mathrm{Perm}(G)$. This observation connects Hopf Galois structures with skew braces.

Definition 1**.**

A skew brace is a finite set $B$ with two operations, $\star$ and $\circ$, so that $(B,\star)$ is a group (the “additive group”), $(B,\circ)$ is a group, and the compatibility condition

[TABLE]

holds for all $a,b,c$ in $B$. Here $a^{-1}$ is the inverse of $a$ in $(B,\star)$. Denote the inverse of $a$ in $(B,\circ)$ by $\overline{a}$.

If $B$ has two operations $\star$ and $\circ$ and is a skew brace with $(B,\star)$ the additive group, then we write $B=B(\circ,\star)$ (i. e. the additive group operation is on the right).

A brace is a skew brace with abelian additive group. Every brace $(A,\circ,+)$ with abelian circle group is a radical algebra $(A,+,\cdot)$ [Ru07], where the algebra multiplication on $A$ is defined by $a\cdot b=a\circ b-a-b$.

A set $B$ with two group operations $\circ$ and $\star$ has two left regular representation maps:

[TABLE]

Then Guarnieri and Vendramin proved ([GV17], Proposition 1.9):

Theorem 2.1**.**

A set $(B,\circ,\star)$ with two group operations is a skew brace if and only if the group homomorphism $\lambda_{\circ}:(B,\circ)\to\mathrm{Perm}(B)$ has image in

[TABLE]

the normalizer in $\mathrm{Perm}(B)$ of $\lambda_{\star}(G)$.

Let $L/K$ be a Galois extension with group $G=(G,\circ)$. Let $H$ be a $K$-Hopf algebra giving a Hopf Galois structure of type $N$ on $L/K$. Using $N$ to define the group structure $(G,\star)$ on $G$ and then identifying $N$ with $(G,\star)$ as above, then $\lambda_{\circ}(G)$ normalizes $N=(G,\star)$, so $\lambda_{\circ}$ is contained in $\mathrm{Hol}(G,\star)$. Thus $(G,\circ,\star)$ is a skew brace.

Conversely, let $(G,\circ,\star)$ be a skew brace. Let $L/K$ be a Galois extension with Galois group $(G,\circ)$. Then $L/K$ has a Hopf Galois structure of type $(G,\star)$. For given the skew brace structure $(G,\circ,\star)$ on the Galois group $(G,\circ)$ of $L/K$, then $\lambda_{\circ}(G)$ is contained in $\mathrm{Hol}(G,\star)$, and so the subgroup $N=\lambda_{\star}(G)\subset\mathrm{Perm}(G)$ is normalized by $\lambda_{\circ}(G)$. So $N$ corresponds by Galois descent to a Hopf Galois structure on $L/K$ of type $(G,\star)$.

Remark 2.2**.**

We note that the correspondence that connects regular subgroups $N$ of $\mathrm{Perm}(\Gamma)$ normalized by $\lambda(\Gamma)$ and isomorphic to $(G,\star)$ to isomorphism types of skew braces $(G,\circ,\star)$ with $(G,\circ)\cong\Gamma$ and $(G,\star)\cong N$ is not bijective. We have (c. f. [NZ19], Corollary 2.4):

Proposition 2.3** (Byott, Nejabati Zenouz).**

Given an isomorphism type $(B,\circ,\star)$ of skew brace, the number of Hopf Galois structures on a Galois extension $L/K$ with Galois group isomorphic to $(B,\circ)$ and skew brace isomorphic to $(B,\circ,\star)$ is

[TABLE]

Here $\mathrm{Aut}_{sb}(B,\circ,\star)$ is the group of skew brace automorphisms of $(B,\circ,\star)$, that is, maps from $B$ to $B$ that are simultaneously group automorphisms of $(B,\star)$ and of $(B,\circ)$.

3. The Galois correspondence for skew braces

Given that an $H$-Hopf Galois structure on a $G$-Galois extension corresponds to a skew brace $(G,\circ,\star)$ so that $G\cong(G,\circ)$ and $H$ has type $(G,\star)$, we can rephrase the question of identifying the $K$-sub-Hopf algebras of $H$, and hence the question of counting the size of the image of the Galois correspondence for $H$, into a question of identifying and counting certain subgroups of the skew brace $(G,\circ,\star)$. This was done in [Ch18].

Let $L/K$ be Galois with group $(G,\circ)$ and $H$- Hopf Galois where $H$ has type $(G,\star)$, so $(G,\circ,\star)$ is a skew brace. We’re interested in the Galois correspondence ratio,

[TABLE]

The numerator counts the $\lambda_{\circ}(G)$-invariant subgroups of $\lambda_{\star}(G)$. Looking at them in the skew brace setting, we have

Definition 2**.**

Let $(G,\circ,\star)$ be a skew brace. A subgroup $(G^{\prime},\star)$ of $(G,\star)$ is $\circ$-stable if $\lambda_{\star}(G^{\prime})$ is closed under conjugation in $\mathrm{Perm}(G)$ by $\lambda_{\circ}(G)$.

This condition is equivalent to

[TABLE]

Remark 3.1**.**

Lemma 2.2 of [DeC19] introduces the automorphism $\rho_{g}$ of $(G,\star)$ defined by

[TABLE]

So a $\circ$-stable subgroup $G^{\prime}$ of $G$ is a subgroup invariant under the action of $\rho_{g}$ for all $g$.

Section 5 of [KT19] recasts the Galois correspondence for a Hopf Galois structure of type $(G,\star)$ on a $(G,\circ)$-Galois extension by replacing the skew brace $(G,\circ,\star)$ by its opposite skew brace $(G,\circ,\star^{\prime})$. Then a $\circ$-stable subgroup $G^{\prime}$ of $G$ is what they call a quasi-ideal of the opposite brace. A quasi-ideal $G^{\prime}$ is an ideal of $G$ ([GV17], Definition 2.1) if and only if $G^{\prime}$ is a normal subgroup of $(G,\circ)$.

We observed in Proposition 4.1 of [Ch18] that a $\circ$-stable subgroup of $(G,\circ,\star)$ is a subgroup of both $(G,\circ)$ and $(G,\star)$.

Thus, if $L/K$ is a $(G,\circ)$-Galois extension and an $H$-Galois extension where the $H$ structure corresponds to a skew brace structure $(G,\circ,\star)$ on $G$, so that $H$ has type $(G,\star)$, then the Galois correspondence ratio becomes

[TABLE]

4. Radical algebras and the Galois correspondence for corresponding Hopf Galois structures

A ring $A=(A,+,\cdot)$ (without unit) is a radical ring if the operation

[TABLE]

for all $a$, $b$ in $A$ makes $(A,\circ)$ into a group with identity 0. If $A$ is a finite ring, then $A$ is Artinian (has descending chain condition on left (or right) ideals), so $A$ is nilpotent (every element of $A$ is nilpotent), c. f. Section 1.2 of [He61].

Conversely, a nilpotent ring $A$ is a radical ring. That means, if we define the operation $\circ$ on $A$ by

[TABLE]

then $(A,\circ)$ is a group. Associativity is clear, the identity is 0, and the inverse of $a$ in $(A,\circ)$, denoted by $\overline{a}$, is

[TABLE]

a finite sum because $A$ is nilpotent. (The circle group of $A$ is isomorphic (by $a\mapsto-a$) to what is sometimes called the adjoint group of $A$.)

A radical ring $A$ is a brace $(A,\circ,+)$, as was first observed in [Ru07], and hence is a skew brace.

Let $(A,+,\cdot)$ be a finite radical ring, and let $a\circ b=a+b+a\cdot b$. In [Ch18] we observed:

Proposition 4.1**.**

If $(A,\circ,+)$ is the skew brace arising from a radical ring $A$, then the $\circ$-stable subgroups of $(A,+)$ are the left ideals of the ring $A$.

For the reader’s convenience, here is a proof.

Proof.

Suppose $(G^{\prime},+)$ is a $\circ$-stable subgroup of $(A,+)$. Then for all $g$ in $A$, $g^{\prime}$ in $G^{\prime}$, there is some $h^{\prime}$ in $G^{\prime}$ so that $g\circ g^{\prime}=h^{\prime}+g$, or equivalently:

[TABLE]

So $G^{\prime}$ is closed under left multiplication by elements of $A$, hence is a left ideal of $A$.

Conversely, if $G^{\prime}$ is a left ideal of $A$, then for all $g^{\prime}$ in $G^{\prime}$, $g$ in $A$, $gg^{\prime}$ is in $G^{\prime}$, so $gg^{\prime}+g^{\prime}=h^{\prime}$ is in $G^{\prime}$. So

[TABLE]

the condition that $(G^{\prime},+)$ is a $\circ$-stable subgroup of $(A,+)$. ∎

Thus if $A$ is a finite radical ring and $L/K$ is a Galois extension with Galois group $(A,\circ)$ with a Hopf Galois structure with $H$ of type $(A,+)$, then the Galois correspondence ratio for $H$ acting on $L/K$ is

[TABLE]

In [Ch18] we illustrated this result by looking at two non-commutative nilpotent $\mathbb{F}_{p}$-algebras of dimension 3 of [DeG17]. Here is a four-dimensional example.

Example 4.2**.**

We look at de Graaf’s [DeG17] $\mathbb{F}_{p}$-algebra $A^{0}_{4,21}$ with an $\mathbb{F}_{p}$-basis consisting of elements $a,b,c,d$ with multiplication given by $a^{2}=c,ab=d$ and all other products of basis elements $=0$. Assume $p>2$. Then a subset $J$ of $A$ is a left ideal of $A$ if and only if $aJ\subset J$. So if $J$ is a left ideal of $A$ and $r=r_{1}a+r_{2}b+r_{3}c+r_{4}d$ is in $J$, then $ar=r_{1}c+r_{2}d$ is also in $J$.

A 4 $\times$ 4 row-reduced echelon matrix in $M_{4}(\mathbb{F}_{p})$ with pivots (leading ones) in columns $c_{1},\ldots,c_{r}$ will be called a matrix of form $(c_{1}\ldots c_{r})$. Every non-zero subspace of $\mathbb{F}_{p}^{4}$ is generated by the non-zero rows of a unique row-reduced echelon matrix.

From that viewpoint, we find that the non-zero left ideals of $A$ correspond to all echelon matrices of forms (3), (4), (34), (24), (134), (234), (1234) $=I$, and also matrices of form (13) with two parameters $r$ and $s$: the non-zero rows of such a echelon matrix form the matrix

[TABLE]

Counting the number of parameters for each form gives

[TABLE]

non-zero left ideals of $A$.

Suppose $L/K$ is a Galois extension with Galois group $(A,\circ)$ and has a $H$-Hopf Galois structure of type $(A,+)$ corresponding to the skew brace $(A,\circ,+)$. Then the Galois correspondence ratio for the $H$-Hopf Galois structure is

[TABLE]

Including the zero ideal, the numerator is $p^{2}+3p+5$.

To find the denominator, the number of subgroups of $(A,\circ)$, is a bit more challenging.

Now $A$ is the $\mathbb{F}_{p}$-algebra with $\mathbb{F}_{p}$-basis $\{a,b,c,d\}$, where $a^{2}=c,ab=d$ and all other products $=0$. For $x,y$ in $A$, define $x\circ y=x+y+xy$. Since $p>2$, $(A,\circ)$ is a group of exponent $p$, for given any $r,s,t,u$ in $\mathbb{F}_{p}$, $ra\cdot ra=r^{2}c,ra\cdot sb=rsd$, so one sees easily by induction that

[TABLE]

for all $m\geq 1$.

Since $(\langle b,c,d\rangle,\circ)=(\langle b,c,d\rangle,+)$, there are $2p^{2}+2p+4$ subgroups of $\langle b,c,d\rangle$.

Then we must count the subgroups of $(A,\circ)$ that contain an element of the form $\alpha=ra+sb+tc+wd$ with $r\neq 0$. By the formula above, we can assume $r=1$. Thus we have $p^{3}$ cyclic subgroups of order $p$. We need also to count non-cyclic subgroups $G^{\prime}$ that are not contained in $\langle b,c,d\rangle$.

Let $w_{1},w_{2},\ldots,$ denote elements of $\langle c,d\rangle$. Any subgroup of $(A,\circ)$ that contains $\alpha_{1}=a+sb+w_{1}$ also contains

[TABLE]

where $\alpha_{1}\circ\alpha_{1}^{-1}=0$.

Let $G^{\prime}=\langle\alpha_{1},\alpha_{2}\rangle$, where

[TABLE]

First suppose $s^{\prime}\neq s$. Then $G^{\prime}$ contains

[TABLE]

Now the $\circ$-subgroup $\langle b,c,d\rangle$ of $(A,\circ)$ is isomorphic to $\mathbb{F}_{p}^{3}$, so if $(s^{\prime}-s)b+w_{3}$ is in $G^{\prime}$ and $s-s^{\prime}\neq 0$, then $G^{\prime}$ contains $b+w_{4}$, with inverse $-b-w_{4}$, so contains

[TABLE]

with $\circ$-inverse $-a-w_{6}+c$. So $G^{\prime}$ contains $a+w_{6}$ and $b+w_{4}$. But then $G^{\prime}$ contains

[TABLE]

So if $G^{\prime}=\langle\alpha_{1},\alpha_{2}\rangle$, then

[TABLE]

for $s,t$ in $\mathbb{F}_{p}$. There are $p^{2}$ such groups.

If $s-s^{\prime}=0$, then $G^{\prime}=\langle a+sb+tc+ud,t^{\prime}c+u^{\prime}d\rangle$ for some $t,t^{\prime},u,u^{\prime}$ in $\mathbb{F}_{p}$. If $t^{\prime}\neq 0$, then $G^{\prime}=\langle a+sb+u^{\prime\prime}d,c+u^{\prime\prime\prime}d\rangle$ for $s,u^{\prime\prime},u^{\prime\prime\prime}$ in $\mathbb{F}_{p}$, so there are $p^{3}$ groups of that form. If $t^{\prime}=0$, then $G^{\prime}=\langle a+sb+tc,d\rangle$, so there are $p^{2}$ groups of that form.

Adjoining $c$ to any of the groups of the last two forms yields $p$ groups of the form $G^{\prime}=\langle a+sb,c,d\rangle$.

Including $G^{\prime}=G$, we have $2p^{3}+4p^{2}+3p+5$ subgroups of $(G,\circ)$.

Thus the Galois correspondence ratio is

[TABLE]

For large $p$ this is near $1/2p$.

5. Bi-skew braces

Definition 3**.**

A bi-skew brace is a finite set $B$ with two operations, $\star$ and $\circ$ so that $(B,\star)$ is a group, $(B,\circ)$ is a group, and $B$ is a skew brace with either group acting as the additive group: that is, the two compatibility conditions

[TABLE]

and

[TABLE]

hold for all $a,b,c$ in $B$.

In Proposition 4.1 of [Ch19] we showed that radical algebras $A$ with $A^{3}=0$ yield bi-skew braces. That means that with $\circ$ defined as before, $(A,+,\circ)$ is a skew brace. Thus if $L^{\prime}/K^{\prime}$ is a Galois extension with Galois group $(A,+)$, then $L^{\prime}/K^{\prime}$ has a Hopf Galois structure of type $(A,\circ)$ coming from the skew brace structure $(A,+,\circ)$ on $A$. The image of the Galois correspondence for that Hopf Galois structure is then in bijective correspondence with the set of $+$-stable subgroups of $(A,\circ)$.

A subgroup $J$ of $(A,\circ)$ is a $+$-stable subgroup if for all $g$ in $A$ and $h$ in $J$, there is some $h^{\prime}$ in $J$ so that

[TABLE]

Let $g,h$ be arbitrary elements of $J$ and note that $J$ is closed under $\circ$. Then $J$ is also closed under $+$, hence is a subgroup of $(A,+)$. Since $h^{\prime}\circ g=h^{\prime}g+h^{\prime}+g$, an alternative version of $+$-stability is that for all $g$ in $A$, $h$ in $J$, there is $h^{\prime}$ in $J$ so that

[TABLE]

or $h=h^{\prime}+h^{\prime}g$.

We show:

Proposition 5.1**.**

Let $(A,+,\circ)$ be the skew brace arising from the finite radical ring $(A,+,\cdot)$ with $A^{3}=0$. Then the $+$-stable subgroups of $(A,\circ)$ are the right ideals of the ring $A$.

Proof.

Let $J$ be a right ideal of the radical ring $A$: then $J$ is closed under addition and scalar multiplication on the right. We show that $J$ is $+$-stable. Given $h$ in $J$, $g$ in $G$, let $h^{\prime}=h-hg$. Then $h^{\prime}$ is in $J$ since $J$ is a right ideal, and since $A^{3}=0$,

[TABLE]

So $J$ is $+$-stable.

Conversely, if $J$ is $+$-stable, then $J$ is an additive subgroup of $A$, and for all $h$ in $J$, $g$ in $A$, there is some $h^{\prime}$ in $J$ so that $h^{\prime}g=h-h^{\prime}$. Since $A^{3}=0$, $0=h^{\prime}gg=hg-h^{\prime}g$, so $hg=h^{\prime}g=h-h^{\prime}$. Thus for all $h$ in $J$, $g$ in $G$, $hg$ is in $J$, and so $J$ is a right ideal of the radical ring $A$.∎

Thus for a $G=(A,+)$-Galois extension $L^{\prime}/K^{\prime}$ with an $H^{\prime}$-Hopf Galois structure of type $(A,\circ)$, the Galois correspondence ratio is

[TABLE]

Example 5.2**.**

One motivation for this paper was to see if there might be any relationship between the Galois correspondence ratios for the two skew braces associated to a bi-skew brace.

So we look again at de Graaf’s [DeG17] example $A^{0}_{4,21}$ with an $\mathbb{F}_{p}$-basis consisting of elements $a,b,c,d$ with multiplication given by $a^{2}=c,ab=d$ and all other products of basis elements $=0$. Then a subset $J$ of $A$ is a right ideal of $A$ if and only if $Ja\subset J$ and $Jb\subset J$, hence if $J$ is a right ideal of $A$ and $r=r_{1}a+r_{2}b+r_{3}c+r_{4}d$ is in $A$, then $ra^{2}=r_{1}c$ and $rab=r_{1}d$ are in $J$. Thus the non-zero right ideals of $A$ correspond to all echelon matrices of forms (2), (3), (4), (23), (24), (34), (134), (234), (1234). Counting the number of parameters for each form gives

[TABLE]

non-zero right ideals of $A$.

To determine the proportions of intermediate fields that are in the image of the Galois correspondence for the Hopf Galois structures corresponding to the skew brace $(A,+,\circ)$, we also need the numbers of subgroups of $(A,+)$. But since $(A,+)\cong\mathbb{F}_{p}^{4}$, the number of subgroups of $(A,+)$ is equal to the number of subspaces of $\mathbb{F}_{p}^{4}$, namely $p^{4}+3p^{3}+4p^{2}+3p+5$.

Thus if $L^{\prime}/K^{\prime}$ is a $(A,+)$-Galois extension with an $H^{\prime}$-Hopf Galois structure of type $(A,\circ)$, then the proportion of subgroups of $(A,+)$ that are in the image of the Galois correspondence for $H$ is

[TABLE]

For large $p$ this is near $2/p^{2}$.

The two Hopf Galois extensions, related by the bi-skew brace arising from the radical algebra $A$ with $A^{3}=0$, have Galois correspondence ratios that behave like $1/2p$ and $2/p^{2}$ for large primes $p$.

6. Zappa-Szép products and the Galois correspondence for corresponding Hopf Galois structures

A finite group $G$ with identity $e$ and subgroups $G_{L}$ and $G_{R}$ is an internal Zappa-Szép product if $G=G_{L}G_{R}$ and $G_{L}\cap G_{R}={e}$. Other terminology: two subgroups $G_{L}$ and $G_{R}$ of a finite group are complementary if $|G_{L}|\cdot|G_{R}|=|G|$ and $G_{L}\cap G_{R}={e}$ ([By15], Section 7), or $G$ admits an exact factorization through the subgroups $G_{L}$ and $G_{R}$ ([SV18], Example 3.6). Thus every element $g$ of $G$ can be uniquely written as $g=g_{L}g_{R}$ for $g_{L}$ in $G_{L}$, $g_{R}$ in $G_{R}$.

Denote the group operation on $G$ by $\cdot$, usually omitted.

In general, for groups $\Gamma,G$ of the same finite order, a fixed point free pair of homomorphisms from $\Gamma$ to $G$ yields a Hopf Galois structure of type $G$ on a $\Gamma$-Galois extension of fields, or equivalently, a skew brace structure $(G,\circ,\cdot)$ on the additive group $G=(G,\cdot)$, where $(G,\circ)\cong\Gamma$. In the Hopf Galois setting the method was first used for $\Gamma=G$ in [CCo07] and in general in [BC12], c. f. Remark 7.2 of [By15]. In the skew brace setting the construction is noted without details in Example 3.6 of [SV18].

Here is how it works for Zappa-Szép products.

Given a Zappa-Szép product $G=G_{L}\cdot G_{R}$ there is a pair of homomorphisms $\beta_{L}$ and $\beta_{R}:G_{L}\times G_{R}\to G$ given by $\beta_{L}(g_{L},g_{R})=g_{L}$, $\beta_{R}(g_{L},g_{R})=g_{R}$. Since $G_{L}\cap G_{R}=\{e\}$, $(\beta_{L},\beta_{R})$ is a fixed point free pair of homomorphisms from $G_{L}\times G_{R}$ to $G$: $\beta_{L}(g_{L},g_{R})=\beta_{R}(g_{L},g_{R})$ if and only if $g_{L}=g_{R}=e$.

Using the left and right regular representations: $\lambda,\rho:G\to\mathrm{Perm}(G)$ given by $\lambda(g)(x)=gx,\rho(g)(x)=xg^{-1}$ for $g,x$ in $G$, the fixed point free pair $(\beta_{L},\beta_{R})$ yields a regular embedding

[TABLE]

defined for $x$ in $G$ by

[TABLE]

The regular embedding $\beta:G_{L}\times G_{R}\to\mathrm{Hol}(G,\cdot)$ yields a bijection $b:G_{L}\times G_{R}\to G$ by

[TABLE]

Then $b$ defines a new group operation $\circ$ on $G$ from the direct product operation on $G_{L}\times G_{R}$ by

[TABLE]

that is,

[TABLE]

or more concisely, $g_{L}g_{R}^{-1}\circ h=g_{L}hg_{R}^{-1}$. Thus $b:G_{L}\times G_{R}\to(G,\circ)$ is an isomorphism, $\beta:G_{L}\times G_{R}\to\mathrm{Hol}(G,\cdot)$ becomes the left regular embedding $\lambda_{\circ}:(G,\circ)\to\mathrm{Hol}(G,\cdot)$, and the circle operation on $G=G_{L}G_{R}$ makes $G$ into a skew brace by Theorem 2.1.

Suppose $L/K$ is a Galois extension with Galois group $(G,\circ)\cong G_{L}\times G_{R}$. Since $(G,\circ,\cdot)$ is a skew brace, $L/K$ has a Hopf Galois structure by a $K$-Hopf algebra $H$ of type $(G,\cdot)=G_{L}G_{R}$. Then the image of the Galois correspondence for $H$ corresponds to the $\lambda_{\circ}(G)$-invariant subgroups of $(G,\cdot)$, and from Section 3 above, these are the $\circ$-stable subgroups of $(G,\cdot)$, namely, the subgroups $G^{\prime}$ of $(G,\cdot)$ with the property that for all $h$ in $G^{\prime}$, $g$ in $G$, the element $(g\circ h)\cdot g^{-1}$ is in $G^{\prime}$.

Proposition 6.1**.**

Let $(G,\circ,\cdot)$ be the skew brace where $(G,\cdot)\cong G_{L}G_{R}$, a Zappa-Szép product and let $\circ$ be the direct product operation on $G$ given by $g\circ x=g_{L}xg_{R}^{-1}$ for $g=g_{L}g_{R}^{-1}$ in $G$. Then a subgroup $G^{\prime}$ of $(G,\cdot)$ is $\circ$-stable if and only if $G^{\prime}$ is normalized by $G_{L}$.

Proof.

$G^{\prime}$ is a $\circ$-stable subgroup of $(G,\cdot)\cong G_{L}G_{R}$ iff for all $x$ in $G^{\prime}$, $g$ in $G$, $(g\circ x)\cdot g^{-1}$ is in $G^{\prime}$, and

[TABLE]

∎

Example 6.2**.**

Let $(G,\cdot)=A_{5}=C_{5}\cdot A_{4}$ where $C_{5}=\langle(1,2,3,4,5)\rangle$ and $A_{4}=\mathrm{Perm}(\{1,2,3,4\})$ is the stabilizer of 5 (c.f. [By15], Example 7.4). Let $\sigma=(1,2,3,4,5)$. A subgroup $G^{\prime}$ of $A_{5}$ is $\circ$-stable iff $G^{\prime}$ is normalized by $G_{L}$. Three obvious $\circ$-stable subgroups are $\{(1)\}$, $C_{5}$ and $A_{5}$. Claim: there is exactly one other $\circ$-stable subgroup.

Since $\sigma^{-1}(a_{1},\ldots,a_{r})\sigma=(a_{1}+1,\ldots,a_{r}+1)$, any non-trivial subgroup $G^{\prime}$ of $A_{5}$ that is normalized by $\sigma$ is transitive, so has order a multiple of 5. Since $A_{5}$ has no non-trivial subgroups of order $>12$, $G^{\prime}$ must have order 5 or 10, and has a characteristic subgroup $H$ of order 5, normalized by $\sigma$. Thus $H=\langle\sigma\rangle$ and $G^{\prime}=\langle\sigma,\rho\rangle$ where $\rho$ has order 2 and $\rho\sigma\rho^{-1}=\sigma^{-1}$. (We can choose $\rho=(1,2)(3,5)$.)

So there are four $\circ$-stable subgroups of $A_{5}$. If $L/K$ is a Galois extension with Galois group $G=C_{5}\times A_{4}$ with a Hopf Galois structure by $H$ of type $A_{5}$ corresponding to the skew brace defined by the fixed point free pair of homomorphisms from $G$ to $A_{5}$ as above, then of the intermediate fields $E$ with $K\subseteq E\subseteq L$, exactly four are in the image of the Galois correspondence for $H$.

How many intermediate subfields are there? How many subgroups are there of $C_{5}\times A_{4}$? Since $(|C_{5}|,|A_{4}|)=(12,5)=1$, the answer is: twice the number of subgroups of $A_{4}$, hence $2\cdot 10=20$ subgroups of $C_{5}\times A_{4}$. (See [DF99], page 112, for the lattice diagram of subgroups of $A_{4}$.) So the proportion of intermediate subfields of $L/K$ that are in the image of the Galois correspondence for $H$ is

[TABLE]

7. Semi-direct products

One set of examples of Zappa-Szép products are semidirect products of groups.

Let $G=G_{L}\rtimes G_{R}$ be a semidirect product of two finite groups $G_{L}$ and $G_{R}$, where $G_{L}$ is normal in $G$ and the action of $G_{R}$ on $G_{L}$ is by conjugation.

Denote the group operation in $G$ by $\cdot$, which we will often omit. Thus for $x,y$ in $G$, $xy=x\cdot y$.

An element of $G$ has a unique decomposition as $x=x_{L}x_{R}^{-1}$ for $x_{L}$ in $G_{L}$, $x_{R}$ in $G_{R}$. In the semidirect product, an element $y_{R}$ of $G_{R}$ acts on $x_{L}$ in $G_{L}$ by conjugation:

[TABLE]

Then

[TABLE]

Along with the given group operation on $G$ we also have the direct product operation $\circ$ on $G$, as follows:

[TABLE]

Proposition 7.1**.**

$(G,\circ,\cdot)$* is a bi-skew brace.*

This was proved in [Ch19].

So suppose $L^{\prime}/K^{\prime}$ is a Galois extension with Galois group $(G,\cdot)=G_{L}\rtimes G_{R}$. Since $(G,\cdot,\circ)$ is a skew brace, $L^{\prime}/K^{\prime}$ has a Hopf Galois structure by a $K^{\prime}$-Hopf algebra $H^{\prime}$ of type $(G,\circ)\cong G_{L}\times G_{R}$. Then the image of the Galois correspondence for $H^{\prime}$ corresponds to the $\lambda_{\cdot}(G)$-invariant subgroups of $(G,\circ)$, and these are the $\cdot$-stable subgroups of $(G,\circ)$, namely, the subgroups $G^{\prime}$ of $(G,\circ)$ with the property that for all $x$ in $G^{\prime}$, $g$ in $G$, the element $(g\cdot x)\circ\overline{g}$ is in $G^{\prime}$.

Proposition 7.2**.**

Let $(G,\circ,\cdot)$ be the bi-skew brace where $(G,\cdot)\cong G_{L}\rtimes G_{R}$, a semidirect product where $G_{R}$ acts on $G_{L}$ by conjugation, and let $\circ$ be the direct product operation on $G$ given by $g\circ x=g_{L}xg_{R}^{-1}$ for $g=g_{L}g_{R}^{-1}$ in $G$. Then a subgroup $G^{\prime}$ of $(G,\circ)$ is $\cdot$-stable if and only if for every $x=x_{L}x_{R}^{-1}$ in $G^{\prime}$ and all $g$ in $G$, $gx_{L}g^{-1}x_{R}^{-1}$ is in $G^{\prime}$.

Proof.

. A subgroup $G^{\prime}$ of $(G,\circ)$ is $\cdot$-stable iff for all $x$ in $G^{\prime}$, $g$ in $G$, the element $(g\cdot x)\circ\overline{g}$ is in $G^{\prime}$. Note that if $g=g_{L}g_{R}^{-1}$, then $\overline{g}=g_{L}^{-1}g_{R}$. So for $g,x,y$ in $G$,

[TABLE]

∎

The remainder of the paper is devoted to examples. In the examples, the circle operation $\circ$ is $+$, the usual addition of modular arithmetic.

Example 7.3**.**

Let $(G,+)=G_{L}\times G_{R}=\mathbb{Z}_{9}\times\mathbb{Z}_{6}$, the direct product with the usual operation, $(r,s)+(r^{\prime},s^{\prime})=(r+r^{\prime},s+s^{\prime})$, and identify $(r,s)$ with $r\cdot 2^{s}$ in $(G,\cdot)=\mathbb{Z}_{9}\rtimes_{2}U_{9}\cong\mathrm{Hol}(C_{9})$. So $(r,s)\cdot(r^{\prime},s^{\prime})=(r+2^{s}r^{\prime},s+s^{\prime})$.

We wish to find the $+$-stable subgroups of $(G,\cdot)$ and the $\cdot$-stable subgroups of $(G,+)$. Since a $+$-stable subgroup of $(G,\cdot)$ is a subgroup of $(G,+)$, and a $\cdot$-stable subgroup of $(G,+)$ is a subgroup of $(G,\cdot)$, all subgroups of interest are subgroups of the abelian group $(G,+)$. So we begin by finding the subgroups of the direct product $(G,+)$, then see which are $+$-stable and which are $\cdot$-stable.

We find that there are 20 subgroups of $(G,+)=\mathbb{Z}_{9}\times\mathbb{Z}_{6}$: sixteen are cyclic groups, with generators:

[TABLE]

The non-cyclic subgroups of $(G,+)$ are

[TABLE]

All are subgroups of $(G,\cdot)$.

A subgroup $G^{\prime}$ of $(G,+)$ is $\cdot$-stable if for all $(r,s)$ in $G^{\prime}$ and $(t,-u)$ in $G$,

[TABLE]

is in $G^{\prime}$. Since $(-2^{u}t,u)(t,-u)=(-2^{u}t+2^{u}t,u+(-u))=(0,0)$, we have that

[TABLE]

is in $G^{\prime}$. Setting $u=1$ implies that $(2r,s)$ is in $G^{\prime}$. Since $G^{\prime}$ is a subgroup of $(G,+)$, therefore

[TABLE]

is in $G^{\prime}$. Thus:

A subgroup $G^{\prime}$ of $(G,\cdot)$ is $\cdot$-stable iff for all $(r,s)$ in $G^{\prime}$, $(r,0)$ is in $G^{\prime}$.

Using this criterion, we find that the four non-cyclic subgroups of $(G,+)$ are $\cdot$-stable, and also the cyclic groups with generators

[TABLE]

the last two because $10(1,3)=(1,0)$ and $4(3,3)=(3,0)$. Thus there are 12 $\cdot$-stable subgroups of $(G,+)$.

There are 32 subgroups of $(G,\cdot)=\mathbb{Z}_{9}\rtimes\mathbb{Z}_{6}$, as follows: There are 26 cyclic subgroups: $\langle 0,0\rangle$ of order 1; $\langle r,3\rangle$ of order 3 for $0\leq r<9$; $\langle 0,2\rangle,\langle 3,0\rangle,\langle 3,2\rangle,\langle 3,4\rangle$ of order 3; $\langle r,1\rangle$ of order 6 for $0\leq r<9$; and $\langle 1,2s\rangle$ of order 3 for $0\leq s<3$. There are six non-cyclic subgroups, $\langle(a,0),(0,b)\rangle$ for $a=1,3$ and $b=1,2,3$.

So if $L/K$ is $G$-Galois with $G\cong\mathbb{Z}_{9}\rtimes\mathbb{Z}_{6}\cong(G,\cdot)$, then for the Hopf Galois structure on $L/K$ corresponding to $(G,+)\cong\mathbb{Z}_{9}\times\mathbb{Z}_{6}$, the ratio

[TABLE]

Now we look at $+$-stable subgroups of $(G,\cdot)$.

A subgroup $G^{\prime}$ of $(G,\cdot)$ is $+$-stable if $G^{\prime}$ is a subgroup of both $(G,+)$ and $(G,\cdot)$ and is normalized by $G_{L}$ in $(G,\cdot)$, that is, $(-t,0)(r,s)(t,0)$ is in $G^{\prime}$ for all $t$ and all $(r,s)$ in $G^{\prime}$. Now

[TABLE]

Since $G^{\prime}$ is a group under $+$, $G^{\prime}$ contains

[TABLE]

Setting $t=1$, we have:

[TABLE]

Thus, if $s=1$, then $(1,0)$ is in $G^{\prime}$; if $s=2$ then $(3,0)$ is in $G^{\prime}$; if $s=3$ then $(7,0)$, hence $(1,0)$ is in $G^{\prime}$; if $s=4$ then $(15,0)$, hence $(3,0)$ is in $G^{\prime}$; and if $s=5$ then $(31,0)$, hence $(1,0)$ is in $G^{\prime}$ .

Among the cyclic subgroups of $(G,\cdot)$, this condition holds trivially for those with generators $(0,0),(1,0)$ and $(3,0)$, and also for $(1,2)$ and $(1,4)$ since $(3,0)=(1,2)\circ(1,2)\circ(1,2)=(1,4)(1,4)(1,4)$. But the condition fails for all other cyclic subgroups of $(G,\cdot)$.

The condition also holds for the non-cyclic subgroups

[TABLE]

but not for $H=\langle(3,0),(0,1)\rangle$ or $\langle(3,0),(0,3)\rangle$ because $(1,0)$ is not in $H$.

Thus there are nine $+$-stable subgroups of $(G,\cdot)$. There are 20 subgroups of $(G,+)$. So if $L^{\prime}/K^{\prime}$ is $G$-Galois with $G\cong\mathbb{Z}_{9}\times\mathbb{Z}_{6}\cong(G,+)$, then for the $H^{\prime}$-Hopf Galois structure corresponding to $(G,\cdot)\cong\mathbb{Z}_{9}\rtimes\mathbb{Z}_{6}$, the ratio

[TABLE]

8. Groups of squarefree order

Hopf Galois structures on groups of squarefree order were studied by [AB18], whose results included showing that if the field extension $L/K$ has a Galois group $G$ cyclic of squarefree order $mn$, then $L/K$ has a Hopf Galois structure of type $N$ for every group $N$ of order $mn$: each such group $N$ must be a semidirect product of cyclic groups.

We look at some of those examples.

Let $(G,+)=\mathbb{Z}_{m}\times\mathbb{Z}_{n}$ under componentwise addition, where $m$ and $n>1$ are coprime and squarefree and $n$ divides $\phi(m)$. Then $(G,+)$ is cyclic of order $mn$, and every element of $G$ may be written as $(r,s)=(r,0)+(0,s)$ for $r$ modulo $m$, $s$ modulo $n$. Also, $\langle(r,s)\rangle$ contains $(r,0)$ and $(0,s)$ because $m$ and $n$ are coprime.

The subgroups of $(G,+)$ are generated by $(r,s)$ where $r$ divides $m$ and $s$ divides $n$, so there are $d(m)d(n)$ subgroups of $(G,+)$, where $d(m)$ is the number of divisors of $m$. If $m$ is a product of $g$ distinct primes, and $n$ is a product of $h$ distinct primes, then $d(m)=2^{g}$, $d(n)=2^{h}$. Hence the number of subgroups of $(G,+)$ is $2^{g+h}$.

Let $b$ have order $n$ in $U_{m}$, the group of units modulo $m$. Form the semidirect product $(G,\cdot)=\mathbb{Z}_{m}\rtimes_{b}\mathbb{Z}_{n}$ with the operation

[TABLE]

Then $(G,+,\cdot)$ is a bi-skew brace.

We observe that every subgroup of $(G,+)$ is also a subgroup of $(G,\cdot)$. For let $G^{\prime}=\langle(r,s)\rangle=\langle(r,0),(0,s)\rangle$ be a subgroup of $(G,+)$. For elements $(cr,ds),(er,fs)$ of $G^{\prime}$,

[TABLE]

which is in $\langle(r,0),(0,s)\rangle$.

Since the $+$-stable subgroups and the $\cdot$-stable subgroups of $(G,+,\cdot)$ are subgroups of both $(G,+)$ and $(G,\cdot)$, we may search for each from among the subgroups $\langle(r,s)\rangle$ of the cyclic group $(G,+)$, where $r$ divides $m$ and $s$ divides $n$.

We first find the $\cdot$-stable subgroups of $(G,+)$.

Proposition 8.1**.**

Let $(G,+)=\mathbb{Z}_{m}\times\mathbb{Z}_{n}\cong\mathbb{Z}_{mn}$ with $(m,n)=1$, and $(G,\cdot)=\mathbb{Z}_{m}\rtimes_{b}\mathbb{Z}_{n}$ with $b$ of order $n$ modulo $m$. Then every subgroup of $(G,+)$ is $\cdot$-stable.

Proof.

A subgroup $G^{\prime}$ of $(G,+)$ is $\cdot$-stable if and only if for all $(x,y)$ in $G^{\prime}$ and all $g$ in $G$, $(g\cdot x\cdot g^{-1},y)$ is in $G^{\prime}$ (where $g^{-1}$ is the $\cdot$ inverse of $G$).

Let $(a,-h)$ be in $G$, then $(a,-h)^{-1}=(-b^{h}a,h)$. So for $(x,y)=(x,0)+(0,y)$ in $G^{\prime}$,

[TABLE]

This is in $G^{\prime}$ because $(b^{h}x,0)$ and $(0,y)$ are in $G^{\prime}$.

Thus every subgroup of $(G,+)$ is $\cdot$-stable. ∎

It follows that for a Galois extension $L/K$ with Galois group isomorphic to $(G,\cdot)\cong\mathbb{Z}_{m}\rtimes\mathbb{Z}_{n}$ with a Hopf Galois extension of type $(G,+)\cong\mathbb{Z}_{mn}$ corresponding to the skew brace $(G,\cdot,+)$, the proportion of intermediate subfields of $L/K$ that are in the image of the Galois correspondence for the Hopf Galois structure on $L/K$ is

[TABLE]

Now we look at the $+$-stable subgroups of $(G,\cdot)$.

A subgroup $\langle(r,0),(0,s)\rangle$ of $(G,\cdot)=\mathbb{Z}_{m}\rtimes Z_{n}$ is $+$-stable iff $(b^{s}-1,0)$ is in $\langle(r,0)\rangle$, iff $r$ divides $b^{s}-1$.

Now $m=p_{1}\cdots p_{g}$ and $n=q_{1}\cdots q_{h}$, products of distinct primes. So if $r=p_{i_{1}}\cdots p_{i_{k}}$ and $b$ has order $n_{i}$ modulo $p_{i}$, then $(b^{s}-1,0)$ is in $\langle(r,0)\rangle$ if and only if the least common multiple $[n_{i_{1}},\ldots,n_{i_{k}}]$ divides $s$.

Example 8.2**.**

Consider $(G,+)=\mathbb{Z}_{pq}\cong\mathbb{Z}_{p}\times\mathbb{Z}_{q}$ where $q$ is a prime divisor of $p-1$. Let $b$ have order $q$ modulo $p-1$ and let $(G,\cdot)=\mathbb{Z}_{p}\rtimes_{b}\mathbb{Z}_{q}$ with operation

[TABLE]

Then $(G,+)$ has four subgroups, generated by $(1,0),(0,1),(1,1)$ and $(0,0)$, of orders $p,q,pq$ and 1, respectively. A subgroup $G^{\prime}=\langle(r,s)\rangle$ of $(G,\cdot)$ is $+$-stable if and only if it is a subgroup of $(G,+)$ and $b^{s}\equiv 1\pmod{r}$. The only subgroup of $(G,+)$ that is not $+$-stable is $(0,1)$, because $b-1$ is not in $\langle 0\rangle\subset\mathbb{Z}_{p}$. So for a $(G,+)$-Galois extension $L/K$ with an $H$-Hopf Galois extension of type $(G,\cdot)$, the ratio,

[TABLE]

Every subgroup of $(G,+)$ is $\cdot$-stable. A computation shows that the number of subgroups of $(G,\cdot)$ is $p+3$: the $p$ cyclic subgroups of order $q$ generated by $(r,1)$ for $r=0,\ldots,p-1$, together with the group generated by $(1,0)$, of order $p$, and the two trivial groups, $\{0\}$ and $G$. So for a $(G,\cdot)$-Galois extension $L^{\prime}/K^{\prime}$ with an $H^{\prime}$-Hopf Galois extension of type $(G,+)$, the ratio

[TABLE]

Example 8.3**.**

Now let $(G,+)=\mathbb{Z}_{mn}$ where $m=p_{1}\cdots p_{g}$, and $n=q_{1}\cdots q_{g}$, all pairwise distinct primes, where for $i=1,\ldots g$, $q_{i}$ divides $p_{i}-1$. Let $b$ have order $q_{i}$ modulo $p_{i}$ for all $i$, so $b$ has order $n$ modulo $m$. Then

Proposition 8.4**.**

With $m,n,b$ chosen as above,

[TABLE]

and every subgroup of $(G,+)$, resp. $(G,\cdot)$ is a direct product of its projections onto the corresponding subgroups.

The decomposition of $(G,+)$ is obvious; that of $(G,\cdot)$ is a routine induction argument from the case $g=2$, which in turn follows because the order of $b$ modulo $p_{1}p_{2}$ is the product of the coprime orders of $b$ modulo $p_{1}$ and modulo $p_{2}$. The statements about subgroups follow from Goursat’s Lemma and the fact that the direct factors have coprime order (or by a Chinese Remainder Theorem argument).

Thus the ratios of Proposition 8.2 multiply to yield

Corollary 8.5**.**

With $(G,+)$ and $(G,\cdot)$ as above,

[TABLE]

and

[TABLE]

Both ratios go to zero for large $g$.

For a final example, we look at a generalization of the dihedral group $D_{m}$ where $m$ is odd and squarefree.

Example 8.6**.**

Let $m=p_{1}\cdots p_{g}$, a product of distinct primes, and let $n=q_{1}\cdots q_{h}$ where $q_{1},\ldots q_{h}$ are distinct primes that divide $p-1$ for every prime $p$ dividing $m$. (The dihedral case is $h=1,n=2$). Let $b$ have order $n$ modulo $p_{i}$ for every $i$. Let $(G,+)=\mathbb{Z}_{m}\times\mathbb{Z}_{n}\cong\mathbb{Z}_{mn}$, $(G,\cdot)=\mathbb{Z}_{m}\rtimes_{b}\mathbb{Z}_{n}$. Then the subgroups of $(G,+)$ all have the form $\langle(r,s)\rangle=\langle(r,0),(0,s)\rangle$ where $r$ divides $m$, $s$ divides $n$.

Since every $+$-stable subgroup $G^{\prime}$ of $(G,\cdot)$ is a subgroup of $(G,+)$, we can assume $G^{\prime}=\langle(r,s)\rangle$ where $r$ divides $m$, $s$ divides $n$. Now $G^{\prime}$ is $+$-stable iff $b^{s}-1$ is in $\langle r\rangle$. We have two cases:

Case 1: $r=1$ and $b^{s}-1$ is in $\langle 1\rangle$ for all $s$.

Case 2: $1<r$ and $r$ divides $m$. Then some $p_{i}$ divides $r$, so $b$ has order $n$ modulo $p_{i}$. Then $b^{s}\equiv 1\pmod{r}$ if and only if $s\equiv 0\pmod{n}$.

Thus $\langle(r,s)\rangle$ is $+$-stable for $r=1$ and all $s$, or for $r\neq 1$ and $s=n$. Since we may assume that $r$ divides $m$, $s$ divides $n$, then the number of $(r,s)$ in Case 1 is $2^{h}$, and the number of $(r,s)$ in Case 2 is $2^{g}-1$. So the number of $+$-stable subgroups of $G$ is $2^{h}+2^{g}-1$.

The number of subgroups of $(G,+)$ is $2^{h}\cdot 2^{g}$. So the ratio

[TABLE]

On the other hand, by Proposition 8.1, every subgroup of $(G,+)$ is $\cdot$-stable, so the ratio

[TABLE]

Since $\mathbb{Z}_{m}\rtimes\mathbb{Z}_{n}$ is metabelian, every subgroup of $(G,\cdot)=\mathbb{Z}_{m}\rtimes\mathbb{Z}_{n}$ has the form $H\rtimes K$ where $H<\mathbb{Z}_{m}$, $K<\mathbb{Z}_{n}$. If $K=(0)$ then we’re counting the $2^{g}$ subgroups of $\mathbb{Z}_{m}$. For each $s$ dividing $n$ and $r$ dividing $m$ there are $r$ subgroups of order $(m/r)(n/s)$ of the form $\langle(r,1),(t,s)\rangle$ for $0\leq t<r$. The total number of subgroups of $(G,\cdot)$ is then

[TABLE]

where $\sigma(m)$, the sum of the divisors of $m=p_{1}\cdot\ldots\cdot p^{g}$, is

[TABLE]

Thus

[TABLE]

Since $\sigma(m)\geq 3^{g}$,

[TABLE]

which is close to 0 for large $g$.

In particular, for $\mathbb{Z}_{n}=\mathbb{Z}_{2}$, the dihedral case with $m$ odd and squarefree, the ratio

[TABLE]

goes to 0 with $g$, while

[TABLE]

for all $g$.

Responding to the question raised in Example 5.2, this example shows that given a bi-skew brace $(G,\cdot,+)$ with $(G,\cdot)\cong D_{m}$ and $(G,+)\cong\mathbb{Z}_{2m}$ for highly composite square-free odd $m$, the ratios describing the images of the Galois correspondences for the Hopf Galois structures of type $\mathbb{Z}_{2m}$, resp. $D_{m}$, on Galois extensions with Galois group $D_{m}$, resp. $\mathbb{Z}_{2m}$, corresponding to the bi-skew brace could hardly be more dissimilar.