On Hopf Galois structures and complete groups

1.  Regular embeddings

Given a Galois extension $L/K$ with Galois group $\Gamma$, and a group $G$ of cardinality that of $\Gamma$, the number $e\left(\Gamma ,G\right)$ of Hopf Galois structures on $L/K$ whose Hopf algebra $H$ has assocated group $G$ is equal to the number of equivalence classes of regular embeddings of $\Gamma$ into $\mathrm{Hol}\left(G\right)$, the semidirect product of $G$ and $\mathrm{Aut}\left(G\right)$. Here an embedding $\beta :\Gamma \to \mathrm{Hol}\left(G\right)$ is regular if, when viewing $\mathrm{Hol}\left(G\right)$ inside $\mathrm{Perm}\left(G\right)$ via $\left(g,\alpha \right)\left(x\right)=\alpha \left(x\right)g^{-1}$, the orbit of the identity element $1$ of $G$ under $\beta \left(\Gamma \right)$ is all of $G$.

Obtaining $e\left(\Gamma ,G\right)$ in any particular case involves a number of steps:

(a) determining $\mathrm{Aut}\left(G\right)$, hence $\mathrm{Hol}\left(G\right)$;

(b) finding monomorphisms $\beta$ from $\Gamma$ to $\mathrm{Hol}\left(G\right)$ by defining $\beta$ on generators of $\Gamma$ and checking $\beta$ on the relations among those generators: in particular, it is helpful to know the orders of elements of $\mathrm{Hol}\left(G\right)$ in order to choose where to send the generators of $\Gamma$;

(c) checking for regularity of $\beta$;

(d) simplifying $\beta$ under the equivalence relation of conjugation by elements of $\mathrm{Aut}\left(G\right)$ inside $\mathrm{Hol}\left(G\right)$ -- that is, finding a complete set of representatives for the equivalence classes of regular embeddings $\beta$;

(e) counting the representatives.

Of these tasks, checking regularity is the least natural. If we view $\mathrm{Hol}\left(G\right)$, the semidirect product of $G$ and $\mathrm{Aut}\left(G\right)$, as $G\times \mathrm{Aut}\left(G\right)$ as sets, the operation is $$\left(g,\alpha \right)·\left(g^{\prime },{\alpha }^{\prime }\right)=\left(g\alpha \left(g^{\prime }\right),\alpha {\alpha }^{\prime }\right)$$ for $g,g^{\prime }\in G,\alpha ,{\alpha }^{\prime }\in \mathrm{Aut}\left(G\right)$. Then $G\cong \left\{\left(g,1\right)\right\}$ is a normal subgroup of $\mathrm{Hol}\left(G\right)$, and the projection ${\pi }_2$ onto $\mathrm{Aut}\left(G\right)$ by ${\pi }_2\left(g,\alpha \right)=\alpha$ is a homomorphism; however the projection ${\pi }_1$ onto $G$, ${\pi }_1\left(g,\alpha \right)=g$, is not. But since an element $\left(g,\alpha \right)$ viewed in $\mathrm{Perm}\left(G\right)$ acts on the identity element $1$ of the set $G$ by $\left(g,\alpha \right)\left(1\right)=\alpha \left(1\right)g^{-1}=g^{-1}$, checking regularity of a 1-1 homomorphism $\beta :\Gamma \to \mathrm{Hol}\left(G\right)$ is the same as determining whether the function (non-homomorphism) ${\pi }_1\beta :\Gamma \to G$ is bijective.

Let $\mathrm{Inn}\left(G\right)$ be the group of inner automorphisms of $G$, then $\mathrm{Inn}\left(G\right)$ is normal in $\mathrm{Aut}\left(G\right)$ and $\mathrm{Aut}\left(G\right)/\mathrm{Inn}\left(G\right)=O\left(G\right)$, the outer automorphism group, fits in the exact sequence $$1\to Z\left(G\right)\to G\to \mathrm{Aut}\left(G\right)\to O\left(G\right)\to 1$$ where the map $C:G\to \mathrm{Aut}\left(G\right)$ is conjugation: $C\left(g\right)\left(x\right)=gx{g}^{-1}$ for $g,x\in G$, and $Z\left(G\right)$ is the center of $G$.

Suppose $Z\left(G\right)=\left(1\right)$ and the composition of the 1-1 homomorphism $\beta :\Gamma \to \mathrm{Hol}\left(G\right)$ with the homomorphism $\mathrm{Hol}\left(G\right)\to O\left(G\right)$ yields a trivial homomorphism from $\Gamma$ to $O\left(G\right)$. Then $\beta$ maps into $G⋊\mathrm{Inn}\left(G\right)$, and, following [CC99], we may decompose $\beta$ as follows: we have a homomorphism $$j:G⋊\mathrm{Inn}\left(G\right)\to G\times G$$ by $j\left(g,C\left(h\right)\right)=\left(gh,h\right)$ for $g,h\in G$, with inverse sending $\left(g,h\right)$ to $\left(g{h}^{-1},C\left(h\right)\right)$. Letting $p_i:G\times G\to G$ be projection onto the $i$th factor, $i=1,2$, the homomorphism $\beta$ yields homomorphisms ${\beta }_1=p_{1}j\beta$ and ${\beta }_2=p_{2}j\beta :\Gamma \to G$ such that $$\beta \left(\gamma \right)=\left({\beta }_1\left(\gamma \right){\beta }_2{\left(\gamma \right)}^{-1},C\left({\beta }_2\right)\right).$$ Then $\beta$ is regular iff $$\left\{{\pi }_1\beta \left(\gamma \right)\right|\gamma \in \Gamma \}=\{{\beta }_1\left(\gamma \right){\beta }_2{\left(\gamma \right)}^{-1}|\gamma \in \Gamma \}=G.$$ Thus whenever $Z\left(G\right)=\left(1\right)$ and $\beta \left(\Gamma \right)\subset G⋊\mathrm{Inn}\left(G\right)$, we may describe $\beta$, and in particular the function ${\pi }_1\beta :\Gamma \to G$, in terms of the homomorphisms ${\beta }_1,{\beta }_2:\Gamma \to G$, namely, $${\pi }_1\beta \left(\gamma \right)={\beta }_1\left(\gamma \right){\beta }_2{\left(\gamma \right)}^{-1}.$$

A class of groups $G$ where $Z\left(G\right)=\left(1\right)$ and $\beta \left(\Gamma \right)\subset G⋊\mathrm{Inn}\left(G\right)$ is the class of complete groups, that is, finite groups $G$ with $Z\left(G\right)=\left(1\right)$ and $O\left(G\right)=\left(1\right)$. The best-known examples of finite complete groups are $\mathrm{Aut}\left(A\right)$ where $A$ is simple and non-abelian, $S_n$ for $n\ge 3,n\ne 6$, and $\mathrm{Hol}\left(Z_m\right)$ where $m$ is odd. (See [Sch65, III.4.u-w] .) Another class is the class of simple groups, for if $G$ is simple, then $O\left(G\right)$ is solvable, so any $\beta :G\to \mathrm{Hol}\left(G\right)$ has image in $G⋊\mathrm{Inn}\left(G\right)$.

In [CC99] we let $\Gamma =G$ and determined $e\left(G,G\right)$, the number of regular embeddings of $G$ to $\mathrm{Hol}\left(G\right)$, when $G$ is simple non-abelian or $S_n,n\ge 4$. In the next section we examine the case where $\Gamma =G=\mathrm{Hol}\left(Z_q\right)$ where $q=p^e$ and $p$ is an odd prime.

2.  $\mathrm{Hol}\left(Z_{p^e}\right)$

Let $G=Z_{p^e}⋊{Z_{p^e}}^{*}$, the holomorph of $Z_{p^e}$. Let $b$ be a number $<p^e$ that generates ${Z_{p^e}}^{*}$. Let $\Gamma =G$. In this section we prove:

Theorem 2.1.   If $G=\mathrm{Hol}\left(Z_{p^e}\right)$, $p$ odd, then up to equivalence there are $$e\left(G,G\right)=2p^{e-1}\varphi \left(p^{e-1}\right)+2p^e\varphi \left(p^{e-1}\right)\left(\varphi \left(p-1\right)-1\right)$$ regular embeddings of $G$ into $\mathrm{Hol}\left(G\right)$. Thus $e\left(G,G\right)$ is the number of $H$-Hopf Galois structures on a Galois extension $L/K$ with Galois group $G$ where the associated group of $H$ is $G$.

Proof. We wish to find equivalence classes of regular embeddings of $G$ in $\mathrm{Hol}\left(G\right)$. Since $G$ is complete, we know from the last section that any homomorphism $\beta :G\to \mathrm{Hol}\left(G\right)$ may be decomposed as $$\beta \left(g\right)={\beta }_1\left(g\right){\beta }_2{\left(g\right)}^{-1}C\left({\beta }_2\left(g\right)\right)$$ for homomorphisms ${\beta }_1,{\beta }_2:G\to G$. So we begin by describing the homomorphisms from $G$ to $G$.

Let $\alpha :G\to G$ be a homomorphism. Then $\alpha$ is determined by $$\begin{array}{cccc}& \hfill \alpha \left(1,1\right)& \hfill =\hfill & \left(m,c\right)\text{of order dividing}{p}^e,\text{and}\hfill \\ & \hfill \alpha \left(0,b\right)& \hfill =\hfill & \left(n,d\right)\text{of order dividing}{p}^{e-1}\left(p-1\right)\text{.}\hfill \end{array}$$ If $\alpha \left(1,1\right)$ has order dividing $p^e$, then $c=1$. To see this, first note that for any $s>0$, $${\left(m,c\right)}^s=\left(m\left(1+c+c^2+\dots c^{s-1}\right),c^s\right)$$ so $c$ must have order dividing $p^e$, which implies that $c\equiv 1\left(modp\right)$. Also, for $\alpha \left(0,b\right)$ to have order dividing $p^{e-1}\left(p-1\right)$ we require that $d\not\equiv 1\left(modp\right)$ or $p$ divides $n$. For if $d\equiv 1\left(modp\right)$ one sees by induction that ${\left(n,d\right)}^{p^{e-1}}=\left(p^{e-1}{n}^{\prime },1\right)$ where $p$ divides $n^{\prime }$ iff $p$ divides $n$. Thus if $p$ does not divide $n$, then $\left(n,d\right)$ has order $p^e$. On the other hand, if $d\not\equiv 1\left(modp\right)$, then $${\left(n,d\right)}^{p-1}=\left(n\left(\frac{d^{p-1}-1}{d-1}\right),d^{p-1}\right)$$ and $p$ divides $d^{p-1}-1$ but not $d-1$, so the order of $\left(n,d\right)$ divides $p^{e-1}\left(p-1\right)$.

Now we check the relation $$\left(b,1\right)\left(0,b\right)=\left(b,b\right)=\left(0,b\right)\left(1,1\right).$$ Applying $\alpha$ yields: $$\left(m\left(1+c+\dots c^{b-1}\right),c^b\right)\left(n,d\right)=\left(n,d\right)\left(m,c\right),$$ hence $$\left(m\left(1+c+\dots c^{b-1}\right)+c^{b}n,c^{b}d\right)=\left(n+dm,dc\right).$$ Thus $c^{b}d=cd$. Since $d$ is a unit modulo $p^e$ it follows that $c^b=c$, hence $c^{b-1}=1$. But since $c=1+pf$ for some $f$, $$1={\left(1+pf\right)}^{b-1}.$$ Since $b-1$ is relatively prime to $p$ and ${\left(1+pf\right)}^{p^{e-1}}=1$, it follows that $1+pf=1$, hence $f=0$ and $c=1$.

Thus for any homomorphism $\alpha :G\to G$, $$\alpha \left(1,1\right)=\left(m,1\right)$$ for some $m$.

Since $c=1$, the requirement $$\left(m\left(1+c+\dots c^{b-1}\right)+c^{b}n,c^{b}d\right)=\left(n+dm,dc\right)$$ becomes $$\left(mb+n,d\right)=\left(n+dm,d\right)$$ which implies that $bm=dm$. Thus if $m\ne 0$, then $b\equiv d\left(modp\right)$, and if $m$ is a unit (i.e. relatively prime to $p$), then $b=d$. In the latter case, $\alpha \left(1,1\right)=\left(m,1\right)$ has order $p^e$ and $\alpha \left(0,b\right)=\left(n,b\right)$ has order $p^{e-1}\left(p-1\right)$.

Clearly if $\alpha$ is an automorphism then $m$ is relatively prime to $p$ and so $b=d$. Conversely, if $\alpha \left(1,1\right)=\left(m,1\right)$ with $m$ relatively prime to $p$, then $\alpha \left(0,b\right)=\left(n,b\right)$ for some $n$. Conjugating $\alpha$ by $\left(h,c\right)$ in $G$ yields $$\begin{array}{cccc}& \hfill \left(h,c\right)\alpha \left(1,1\right){\left(h,c\right)}^{-1}& \hfill =\hfill & \left(h,c\right)\left(m,1\right)\left(-c^{-1}h,c^{-1}\right)\hfill \\ & \hfill & \hfill =\hfill & \left(h+cm-h,1\right)\hfill \\ & \hfill & \hfill =\hfill & \left(cm,1\right)\text{.}\hfill \end{array}$$ We choose $c$ so that $cm=1$. Then we choose $h$ so that $$\begin{array}{cccc}& \hfill \left(0,b\right)=\left(h,c\right)\alpha \left(0,b\right){\left(h,c\right)}^{-1}& \hfill =\hfill & \left(h,c\right)\left(n,b\right)\left(-c^{-1}h,c^{-1}\right)\hfill \\ & \hfill & \hfill =\hfill & \left(h+cn,bc\right)\left(-c^{-1}h,c^{-1}\right)\hfill \\ & \hfill & \hfill =\hfill & \left(h+cn-bh,b\right)\hfill \\ & \hfill & \hfill =\hfill & \left(\left(1-b\right)h+cn,b\right):\hfill \end{array}$$ we set $h=-{\left(1-b\right)}^{-1}cn$, possible because $b$ has order $p^{e-1}\left(p-1\right)\left(mod{p}^e\right)$ and hence $b\not\equiv 1\left(modp\right)$. With these choices of $h$ and $c$, the homomorphism $C\left(h,c\right)\alpha :G\to G$ is the identity on the generators $\left(1,1\right)$ and $\left(0,b\right)$ of $G$, and so $\alpha =C{\left(h,c\right)}^{-1}$ is an automorphism of $G$.

Now we ask about regularity: for which pairs of endomorphisms $\left({\beta }_1,{\beta }_2\right)$ is $$\left\{{\beta }_1\left(g\right){\beta }_2{\left(g\right)}^{-1}\right|g\in G\}=G,$$ or equivalently, ${\pi }_1\beta ={\beta }_1·{{\beta }_2}^{-1}$ is a 1-1 function from $G$ to $G$? Let ${\beta }_i\left(1,1\right)=\left(m_i,1\right),{\beta }_i\left(0,b\right)=\left(n_i,d_i\right)$ for $i=1,2$.

If neither ${\beta }_1$ nor ${\beta }_2$ is an automorphism, then $p$ divides $m_1$ and $m_2$, so $$\begin{array}{cccc}& \hfill {\beta }_1\left(p^{e-1}l,1\right){\beta }_2{\left(p^{e-1}l,1\right)}^{-1}& \hfill =\hfill & \left(p^{e-1}l{m}_1,1\right)\left(-p^{e-1}l{m}_2,1\right)\hfill \\ & \hfill & \hfill =\hfill & \left(0,1\right)\left(0,1\right)=\left(0,1\right)\hfill \end{array}$$ for all $l$, and so ${\beta }_1·{{\beta }_2}^{-1}$ is not 1-1.

Suppose both ${\beta }_1$ and ${\beta }_2$ are automorphisms. If $m_1\equiv m_2\left(modp\right)$ then $$\begin{array}{cccc}& \hfill {\beta }_1·{{\beta }_2}^{-1}\left(p^{e-1},1\right)& \hfill =\hfill & \left(p^{e-1}{m}_1-p^{e-1}{m}_2,1\right)\hfill \\ & \hfill & \hfill =\hfill & \left(0,1\right)\hfill \\ & \hfill & \hfill =\hfill & {\beta }_1·{{\beta }_2}^{-1}\left(0,1\right)\hfill \end{array}$$ so ${\beta }_1·{{\beta }_2}^{-1}$ is not 1-1. If $m_1\not\equiv m_2\left(modp\right)$, then let $s\left(m_1-m_2\right)=n_1-n_2$. Then $$\begin{array}{cccc}& \hfill {\beta }_1·{{\beta }_2}^{-1}\left(0,b\right)& \hfill =\hfill & \left(n_1-n_2,1\right)\hfill \\ & \hfill & \hfill =\hfill & \left(\left(m_1-m_2\right)s,1\right)\hfill \\ & \hfill & \hfill =\hfill & {\beta }_1·{{\beta }_2}^{-1}\left(s,1\right),\hfill \end{array}$$ so again, ${\beta }_1·{{\beta }_2}^{-1}$ is not 1-1.

Thus for $\beta$ to be regular, exactly one of ${\beta }_1$ and ${\beta }_2$ is an automorphism.

We return to looking at regularity after we look at equivalence by $\mathrm{Aut}\left(G\right)=\mathrm{Inn}\left(G\right)$ inside $\mathrm{Hol}\left(G\right)$.

For $g,h,k\in G$ we have $$\begin{array}{cccc}& \hfill \left(1,C\left(g\right)\right)\left(h,C\left(k\right)\right)\left(1,C{\left(g\right)}^{-1}\right)& \hfill =\hfill & \left(C\left(g\right)\left(h\right),C\left(g\right)C\left(k\right)C{\left(g\right)}^{-1}\right)\hfill \\ & \hfill & \hfill =\hfill & \left(gh{g}^{-1},C\left(gk{g}^{-1}\right)\right)\text{.}\hfill \end{array}$$ Thus if $\beta \left(x\right)=\left({\beta }_1\left(x\right){\beta }_2{\left(x\right)}^{-1},C\left({\beta }_2\left(x\right)\right)\right)$ for $x\in G$, then $\beta \sim {\beta }^{\prime }$ with $$\begin{array}{cccc}& \hfill {\beta }^{\prime }\left(x\right)& \hfill =\hfill & \left(g{\beta }_1\left(x\right){\beta }_2{\left(x\right)}^{-1}{g}^{-1},C\left(g{\beta }_2\left(x\right)g^{-1}\right)\right)\hfill \\ & \hfill & \hfill =\hfill & \left(g{\beta }_1\left(x\right)g^{-1}·{\left(g{\beta }_2\left(x\right)g^{-1}\right)}^{-1},C\left(g{\beta }_2\left(x\right)g^{-1}\right)\right):\hfill \end{array}$$ we get from $\beta$ to ${\beta }^{\prime }$ by simultaneously conjugating ${\beta }_1$ and ${\beta }_2$ by $g\in G$.

Assume ${\beta }_2$ is an automorphism, then, up to equivalence, we may assume that ${\beta }_2$ is the identity automorphism on $G$ and ${\beta }_1$ is not an automorphism. Then $e\left(G,G\right)$ will be twice the number of possible ${\beta }_1$, since the case where ${\beta }_1$ is an automorphism and ${\beta }_2$ is not is the same.

Returning to the regularity question, we ask, for which ${\beta }_1$ is $\left\{{\beta }_1\left(g\right)g^{-1}\right\}=G$?

Assume $$\begin{array}{cccc}& \hfill {\beta }_1\left(1,1\right)& \hfill =\hfill & \left(m,1\right)\hfill \\ & \hfill {\beta }_1\left(0,b\right)& \hfill =\hfill & \left(n,d\right)\hfill \end{array}$$ for some $m$ divisible by $p$, some $d\ne 1$ and some $n$. Then $$\begin{array}{cccc}& \hfill {\beta }_1\left(l,b^k\right)& \hfill =\hfill & {\beta }_1\left(l,1\right){\beta }_1\left(0,b^k\right)\hfill \\ & \hfill & \hfill =\hfill & \left(lm,1\right)\left(n\left(\frac{d^k-1}{d-1}\right),d^k\right)\hfill \\ & \hfill & \hfill =\hfill & \left(lm+n\left(\frac{d^k-1}{d-1}\right),d^k\right)\text{.}\hfill \end{array}$$ For any $h,r$ we want to find $l,k$ so that $$\begin{array}{cccc}\text{(**)}& \hfill {\beta }_1\left(l,b^k\right){\left(l,b^k\right)}^{-1}& \hfill =\hfill & \left(lm+n\left(\frac{d^k-1}{d-1}\right),d^k\right)\left(-b^{-k}l,b^{-k}\right)\hfill \\ & \hfill & \hfill =\hfill & \left(lm+n\left(\frac{d^k-1}{d-1}\right)-d^k{b}^{-k}l,d^k{b}^{-k}\right)\hfill \\ & \hfill & \hfill =\hfill & \left(h,b^r\right)\text{.}\hfill \end{array}$$ In order that for any $r$, there is a $k$ so that $b^r={\left(d{b}^{-1}\right)}^k$, we require that $d{b}^{-1}$ generates ${Z_{p^e}}^{*}$. Now for ${\beta }_1$ to be a homomorphism, we need $bm=dm$, and this is possible only if $m=0$ or $b\equiv d\left(modp\right)$. But in the latter case, $d{b}^{-1}\equiv 1\left(modp\right)$, so cannot generate ${Z_{p^e}}^{*}$. Thus if $\left\{{\beta }_1\left(g\right)g^{-1}\right\}=G$, then $m=0$, and $d=b^{f+1}$ such that $b^f$ generates ${Z_{p^e}}^{*}$.

Thus ${\beta }_1$ is defined by $$\begin{array}{cccc}& \hfill {\beta }_1\left(1,1\right)& \hfill =\hfill & \left(0,1\right)\hfill \\ & \hfill {\beta }_1\left(0,b\right)& \hfill =\hfill & \left(n,d\right),\hfill \end{array}$$ and (**) becomes $$\left(n\left(1+d+\dots +d^{k-1}\right)-b^{fk}l,b^{fk}\right)=\left(h,b^r\right),$$ which is solvable for all $n$ and for all $f$ such that $b^f$ generates ${Z_{p^e}}^{*}$ . We have two cases:

If $d=b^{f+1}\equiv 1\left(modp\right)$, then $p-1$ divides $f+1$ and $p$ divides $n$ (or else $\beta$ is not a homomorphism).

If $d=b^{f+1}\not\equiv 1\left(modp\right)$, then $n$ is arbitrary.

The first case gives $p^{e-1}$ choices for $n$ and $\varphi \left(p^{e-1}\right)$ choices for $f$.

The second case gives $p^e$ choices for $n$ and $$\varphi \left(p^{e-1}\left(p-1\right)\right)-\varphi \left(p^{e-1}\right)=\varphi \left(p^{e-1}\right)\left(\varphi \left(p-1\right)-1\right)$$ choices for $f$. The theorem follows.

Corollary 2.2.   If $G=\mathrm{Hol}\left(Z_p\right)$, then $e\left(G,G\right)=2\left(1+p\left(\varphi \left(p-1\right)-1\right)\right)$.

Example 2.3.   For $p=5$ there are, up to equivalence, $2\left(1+5\left(2-1\right)\right)=12$ regular embeddings of $G=Z_5⋊{Z_5}^{*}$ into $\mathrm{Hol}\left(G\right)$. If we choose $b=2$ then $b$ and $b^3$ generate ${Z_5}^{*}$, so $d=b^0=1$ or $d=b^2=4$. Thus if we let ${\beta }_2$ be the identity automorphism, then ${\beta }_1\left(1,1\right)=\left(0,1\right)$ and ${\beta }_1\left(0,2\right)=\left(n,4\right)$, $n=0,1,2,3,4$, or $\left(0,1\right)$.

3.  Groups of order $p\left(p-1\right)$

Let $\Gamma =\mathrm{Hol}\left(Z_p\right)$ with $p$ an odd prime, as above, and let $p-1=2q$. Then there are at least five non-isomorphic groups $G$ of order $2qp$ other than $\Gamma$, namely, $Z_{2qp},D_{qp},D_p\times Z_q,D_q\times Z_p$, and $\left(Z_p⋊Z_q\right)\times Z_2$ (where $Z_q$ is identified as the subgroup of $\mathrm{Aut}\left(Z_p\right)$ of index 2, and $D_n$ is the dihedral group of order $2n$). The five groups are non-isomorphic because their centers have orders $2pq,1,q,p$ and $2$ respectively.

If $q$ is also an odd prime, then $q$ is called a Sophie Germain prime, resp. $p$ is called a safeprime, and in that case, these five groups, together with $\Gamma =\mathrm{Hol}\left(Z_p\right)$, are the only groups of order $2pq$, up to isomorphism. To see this, we first obtain the following lemma, needed also in the next section:

Lemma 3.1.   Let $p$ be prime, $p=2q+1$. Then $$\mathrm{Aut}\left(D_p\right)=\mathrm{Aut}\left(Z_p⋊Z_q\right)=\mathrm{Aut}\left(\mathrm{Hol}\left(Z_p\right)\right)=\mathrm{Inn}\left(\mathrm{Hol}\left(Z_p\right)\right).$$

Proof. Let $G=Z_p⋊Z_a$ with $a=2,q$ or $2q=p-1$. Write $Z_p$ additively and view $Z_a$ as the cyclic subgroup of order $a$ inside ${Z_p}^{*}=\mathrm{Aut}\left(Z_p\right)$ and $G\subset \mathrm{Hol}\left(Z_p\right)$. Let $b$ generate $Z_a$. If $\alpha$ is an automorphism of $G$, then:

1. $\alpha \left(1,1\right)=\left(m,1\right)$ for $p$ not dividing $m$, and

2. $\alpha \left(0,b\right)=\left(n,b\right)$ for any $n$ (as one sees by applying $\alpha$ to the relation $\left(b,1\right)\left(0,b\right)=\left(b,b\right)=\left(0,b\right)\left(1,1\right)$.)

So there are $p\left(p-1\right)=\left|\mathrm{Hol}\left(Z_p\right)\right|$ automorphisms of $G$. Now if $\left(l,d\right)$ is any element of $\mathrm{Hol}\left(Z_p\right)$, then conjugation by $\left(l,d\right)$ is an automorphism of $G$, since $$\left(l,d\right)\left(m,b^r\right){\left(l,d\right)}^{-1}=\left(l+dm-b^{r}l,b^r\right)\in G.$$ Hence the conjugation map $C:\mathrm{Hol}\left(Z_p\right)\to \mathrm{Aut}\left(G\right)$, $$C\left(l,d\right)\left(m,b^r\right)=\left(l,d\right)\left(m,b^r\right){\left(l,d\right)}^{-1}=\left(l+dm-b^{r}l,b^r\right),$$ is defined. Then $C$ is 1-1. For if $C\left(l_1,d_1\right)=C\left(l_2,d_2\right)$ on $G$, then $$l_1+d_{1}m-b^r{l}_1=l_2+d_{2}m-b^r{l}_2$$ for all $m,r$: in particular for $m=1,r=0$ we get $d_1=d_2$ and for $m=0,r=1$ we get $l_1=l_2$. Thus $C$ is an isomorphism.

Proposition 3.2.   If $q$ and $p=2q+1$ are odd primes, then, up to isomorphism, there are exactly six groups of order $p\left(p-1\right)$.

This is probably well-known, but we sketch a proof for the reader's convenience.

Proof. If $G$ has order $p\left(p-1\right)$ then by the first Sylow Theorem, $G$ has a unique normal subgroup $G_p$ of order $p$. By Schur's Theorem, $G_p$ has a complementary subgroup of order $2q$, and hence a subgroup $K$ of order $q$. Then $G_{p}K=J$ is a subgroup of $G$ of order $pq$, hence normal in $G$. By Schur's Theorem again, $J$ has a complementary subgroup of order 2 in $G$, so $G$ is isomorphic to a semidirect product $J{⋊}_{\alpha }{Z}_2$.

If $J\cong Z_{pq}$ then $G\cong Z_{pq}{⋊}_{\alpha }{Z}_2$ where $$\alpha :Z_2\to \mathrm{Aut}\left(Z_{pq}\right)\cong Z_{p-1}\times Z_{q-1}$$ has four possibilities, $\alpha \left(-1\right)=\left(\pm 1,\pm 1\right)$. These yield $G\cong D_{pq},D_p\times Z_q,D_q\times Z_p$ and $Z_{2pq}$.

If $J\cong Z_p⋊Z_q\subset \mathrm{Hol}\left(Z_p\right)$, then $G\cong J{⋊}_{\alpha }{Z}_2$ where $\alpha :Z_2\to \mathrm{Aut}\left(Z_p⋊Z_q\right)\cong \mathrm{Inn}\left(\mathrm{Hol}\left(Z_p\right)\right)$. If $\alpha$ is trivial, we obtain $\left(Z_p⋊Z_q\right)\times Z_2$. Otherwise, the elements of order 2 in $\mathrm{Inn}\left(\mathrm{Hol}\left(Z_p\right)\right)$ are of the form $C\left(l,-1\right)$ for any $l\in Z_p$. Define $${\alpha }_l:Z_2\to \mathrm{Inn}\left(\mathrm{Hol}\left(Z_p\right)\right)$$ by ${\alpha }_l\left(-1\right)=C\left(l,-1\right)$. One checks easily that $$\left(Z_p⋊Z_q\right){⋊}_{{\alpha }_0}{Z}_2\cong Z_p⋊Z_{2q}=\mathrm{Hol}\left(Z_p\right)$$ by the map sending $\left(\left(a,b\right),\epsilon \right)$ to $\left(a,\epsilon b\right)$ for $a\in Z_p$, $b\in Z_q\subset Z_{p-1}$, $\epsilon \in Z_2\subset Z_{p-1}$. Now $\langle {\alpha }_0\rangle =\langle C\left(0,-1\right)\rangle$ and $\langle {\alpha }_l\rangle =\langle C\left(l,-1\right)\rangle$ are conjugate in $\mathrm{Inn}\left(\mathrm{Hol}\left(Z_p\right)\right)$ for $l\ne 0$ by $C\left(m,1\right)$ where $2m\equiv l\left(modp\right)$: $C\left(m,1\right)C\left(0,-1\right)C\left(-m,1\right)=C\left(2m,-1\right)$. It follows from [DF99], p. 186, Exercise 6 that $$\left(Z_p⋊Z_q\right){⋊}_{{\alpha }_l}{Z}_2\cong \left(Z_p⋊Z_q\right){⋊}_{{\alpha }_0}{Z}_2$$ for all $l$. Thus $J=Z_p⋊Z_q$ yields only two possible groups, up to isomorphism.

4.  Nonuniqueness

If we begin with the Galois group $\Gamma$ of a Galois extension $L/K$ of fields, then to determine the $K$-Hopf Galois structures on $L$, we need to count equivalence classes of regular embeddings of $\Gamma$ into $\mathrm{Hol}\left(G\right)$, where $G$ varies over isomorphism classes of groups of the same cardinality as $\Gamma$. This can be a formidable task for certain cardinalities!

In this section, we let $\Gamma =\mathrm{Hol}\left(Z_p\right)$, $p=2q+1$ a safeprime $>5$, and determine $e\left(\Gamma ,G\right)$, the number of regular embeddings of $\Gamma$ into $\mathrm{Hol}\left(G\right)$, for all six groups $G$ of order $p\left(p-1\right)=2pq$. We did the case $G=\Gamma$ in Theorem 2.1 and found that $e\left(G,G\right)=2+2p\left(q-2\right)$. Here is the result for $G\ncong \Gamma$:

Theorem 4.1.   Let $\Gamma =\mathrm{Hol}\left(Z_p\right)$, with $p$ and $q=\left(p-1\right)/2$ odd primes. Then:

1.   $e\left(\Gamma ,Z_p⋊Z_q\times Z_2\right)=2p\left(q-1\right)$;

2.   $e\left(\Gamma ,D_q\times Z_p\right)=2p$;

3.   $e\left(\Gamma ,D_p\times Z_q\right)=2p$;

4.   $e\left(\Gamma ,Z_{2qp}\right)=p$;

5.   $e\left(\Gamma ,D_{pq}\right)=4p$.

Proof. We do each case in turn, following the steps outlined in Section 1.

(1): Proof that $e\left(\Gamma ,Z_p⋊Z_q\times Z_2\right)=2p\left(q-1\right)$. Using Lemma 3.1 we have $$\begin{array}{cccc}& \hfill \mathrm{Hol}\left(Z_p⋊Z_q\times Z_2\right)& \hfill =\hfill & \mathrm{Hol}\left(Z_p⋊Z_q\right)\times \mathrm{Hol}\left(Z_2\right)\hfill \\ & \hfill & \hfill =\hfill & \left(\left(Z_p⋊Z_q\right)⋊\mathrm{Inn}\left(\mathrm{Hol}\left(Z_p\right)\right)\right)\times Z_2\hfill \\ & \hfill & \hfill \subset \hfill & \left(\mathrm{Hol}\left(Z_p\right)⋊\mathrm{Inn}\left(\mathrm{Hol}\left(Z_p\right)\right)\right)\times Z_2\hfill \\ & \hfill & \hfill \cong \hfill & \mathrm{Hol}\left(Z_p\right)\times \mathrm{Hol}\left(Z_p\right)\times Z_2;\hfill \end{array}$$ thus we may unwind any homomorphism $\beta :\Gamma \to \left(\left(Z_p⋊Z_q\right)⋊\mathrm{Inn}\left(\mathrm{Hol}\left(Z_p\right)\right)\right)\times Z_2$ as in Section 1. If $$\beta \left(\gamma \right)=\left(\left(m,b^{2r}\right),C\left(l,b^s\right)\right)\times \epsilon$$ with $\epsilon =\pm 1,{Z_p}^{*}=\langle b\rangle$ and $m,l$ are modulo $p$, then $\beta \left(\gamma \right)$ maps to $\left(m,b^{2r}\right)\left(l,b^s\right)\times \left(l,b^s\right)\times \epsilon$ under the map to $\mathrm{Hol}\left(Z_p\right)\times \mathrm{Hol}\left(Z_p\right)\times Z_2$. So $\beta$ induces homomorphisms ${\beta }_1:\Gamma \to \mathrm{Hol}\left(Z_p\right)$, defined by ${\beta }_1\left(\gamma \right)=\left(m,b^{2r}\right)\left(l,b^s\right)$, and ${\beta }_2:\Gamma \to \mathrm{Hol}\left(Z_p\right)$, defined by ${\beta }_2\left(\gamma \right)=\left(l,b^s\right)$. If we define ${\beta }_0:\Gamma \to Z_2$ by ${\beta }_0\left(\gamma \right)=\epsilon$, then $$\beta \left(\gamma \right)=\left({\beta }_1\left(\gamma \right){\beta }_2{\left(\gamma \right)}^{-1},C\left({\beta }_2\left(\gamma \right)\right)\times {\beta }_0\left(\gamma \right)\right).$$

Let $\gamma \in \mathrm{Hol}\left(Z_p\right),\gamma =\left(m,b^t\right)$. Then the only non-trivial homomorphism ${\beta }_0$ is given by ${\beta }_0\left(\gamma \right)=-1$ if $t$ is odd, and $=1$ if $t$ is even. (If ${\beta }_0\left(\gamma \right)=1$ for all $\gamma$, then $\beta$ will not be regular.) As for the possibilities for the homomorphisms ${\beta }_1$ and ${\beta }_2$, we determined these in the proof of Theorem 2.1, namely:

1. If ${\beta }_i$ is an automorphism, then ${\beta }_i\left(1,1\right)=\left(m,1\right)$, $m\ne 0$ and ${\beta }_i\left(0,b\right)=\left(n,b\right)$ for any $n$.

2. If ${\beta }_i$ is not an automorphism, then ${\beta }_i\left(1,1\right)=\left(0,1\right)$ and ${\beta }_i\left(0,b\right)=\left(n,d\right)$ for $d=b^r$, $r\ne 0$, and any $n$, or ${\beta }_i\left(0,b\right)=\left(0,1\right)$.

As in the proof of Theorem 2.1, if both ${\beta }_1$ and ${\beta }_2$ are not automorphisms, or if both ${\beta }_1$ and ${\beta }_2$ are automorphisms, then ${\beta }_1·{{\beta }_2}^{-1}$ is not 1-1, so $\beta$ is not regular. Thus we can assume one is an automorphism and the other not. Assuming ${\beta }_2$ is an automorphism, we can conjugate it by an automorphism of $\mathrm{Hol}\left(Z_p\right)$ to transform it to the identity map. Then $$\begin{array}{cccc}& \hfill {\beta }_1\left(0,b\right){\beta }_2{\left(0,b\right)}^{-1}& \hfill =\hfill & \left(n,d\right){\left(0,b\right)}^{-1}\hfill \\ & \hfill & \hfill =\hfill & \left(n,d{b}^{-1}\right):\hfill \end{array}$$ this must lie in $Z_p⋊Z_q$, which means that $d=b^r$ must be an odd power of $b$. Then $$\begin{array}{cccc}& \hfill {\beta }_1·{{\beta }_2}^{-1}\left(l,b^s\right)& \hfill =\hfill & {\left(n,d\right)}^s{\left(l,b^s\right)}^{-1}\hfill \\ & \hfill & \hfill =\hfill & \left(n\left(\frac{d^s-1}{d-1}\right)-l{\left(d{b}^{-1}\right)}^s,{\left(d{b}^{-1}\right)}^s\right)\text{.}\hfill \end{array}$$ For $\beta$ to be regular, ${\beta }_1·{{\beta }_2}^{-1}$ must map onto $Z_p⋊Z_q$, so we need that $d{b}^{-1}=b^{r-1}$ generates $Z_q$. Thus $r-1$ must be coprime to $q$. There are $q-1$ odd numbers $r<p$ with $r-1$ coprime to $q$.

For any such $r$, ${\left(d{b}^{-1}\right)}^s$ is a unit of $Z_p$, so given any $n,h$ in $Z_p$ there is some $l$ in $Z_p$ so that $n\left(\frac{d^s-1}{d-1}\right)-l{\left(d{b}^{-1}\right)}^s=h$. Hence for any suitable $r$ and any $n$, ${\beta }_1·{{\beta }_2}^{-1}$ maps $\mathrm{Hol}\left(Z_p\right)$ onto $Z_p⋊Z_q$.

Thus we have determined all regular $\beta$: we have $p$ choices for $n$, and $q-1$ choices for $d$. Interchanging the roles of ${\beta }_1$ and ${\beta }_2$ yields the same result. Thus there are $2p\left(q-1\right)$ regular embeddings of $\mathrm{Hol}\left(Z_p\right)$ into $\mathrm{Hol}\left(Z_p⋊Z_q\times Z_2\right)$, as claimed.

(2): Proof that $e\left(\Gamma ,D_q\times Z_p\right)=2p$. We seek regular embeddings $$\beta :\Gamma \to \mathrm{Hol}\left(D_q\times Z_p\right).$$ We have $$\mathrm{Hol}\left(D_q\times Z_p\right)\cong \mathrm{Hol}\left(D_q\right)\times \mathrm{Hol}\left(Z_p\right)$$ and by Lemma 3.1, the map $$\mathrm{Hol}\left(D_q\right)\cong D_q⋊\mathrm{Inn}\left(\mathrm{Hol}\left(Z_q\right)\right)\to \mathrm{Hol}\left(Z_q\right)\times \mathrm{Hol}\left(Z_q\right)$$ is 1-1 and maps $\left(m,\epsilon \right)C\left(n,d\right)$ to $\left(m,\epsilon \right)\left(n,d\right)\times \left(n,d\right)$. Suppose $$\beta \left(1,1\right)=\left(q,\epsilon \right)C\left(n,d\right)\times \left(m,c\right).$$ Now $\beta \left(1,1\right)$ must have order $p$ (or else $\beta$ is not 1-1), so $\left(n,d\right)=\left(q,\epsilon \right)=\left(0,1\right)$, $c=1$ and $m\ne 0$, hence $$\beta \left(1,1\right)=\left(0,1\right)C\left(0,1\right)\times \left(m,1\right)$$ with $m\ne 0$.

Suppose $\beta \left(0,b\right)=\left(l,\epsilon \right)C\left(k,d\right)\times \left(s,c\right)$ in $D_q⋊\mathrm{Inn}\left(\mathrm{Hol}\left(Z_q\right)\right)\times \mathrm{Hol}\left(Z_p\right)$ for $l,\epsilon ,k,d$ mod $q$ and $s,c$ mod $p$. If $\epsilon =1$, then no element of $D_q$ of the form $\left(*,-1\right)$ is hit by ${\pi }_1\left(\beta \right)$, and so $\beta$ is not regular. Thus $\epsilon =-1$.

Applying $\beta$ to the condition $\left(b,1\right)\left(0,b\right)=\left(0,b\right)\left(1,1\right)$ yields $c\equiv b\left(modp\right)$. Hence $\left(k,b\right)$ has order $p-1$. We require that $\left(n,-1\right)C\left(l,d\right)$ have order dividing $2q$ in $D_q⋊\mathrm{Inn}\left(\mathrm{Hol}\left(Z_q\right)\right)$, which maps 1-1 to $\mathrm{Hol}\left(Z_q\right)\times \mathrm{Hol}\left(Z_q\right)$ as noted above. Thus $\left(n-l,-d\right)$ and $\left(l,d\right)$ must have order 1, 2 or $q$ in $\mathrm{Hol}\left(Z_q\right)$. But then $d=1$ or $-1$.

Thus any regular embedding $\beta$ satisfies:

1. $\beta \left(1,1\right)=\left(0,1\right)C\left(0,1\right)\times \left(m,1\right)$ with $m\ne 0$ in $Z_p$, and

2. $\beta \left(0,b\right)=\left(n,-1\right)C\left(l,d\right)\times \left(k,b\right)$ with $d=\pm 1$.

Now modify $\beta$ by conjugating by $$\left(0,1\right)C\left(h,c\right)\times \left(0,b^r\right)\in \mathrm{Aut}\left(D_q\times Z_p\right)\subset D_q⋊\mathrm{Inn}\left(\mathrm{Hol}\left(Z_q\right)\right)\times \mathrm{Hol}\left(Z_p\right).$$

First, looking at $\beta \left(1,1\right)$: $$\begin{array}{cccc}& \hfill \left(\left(0,1\right)C\left(h,c\right)\times \left(0,b^r\right)\right)& \hfill \left(\left(0,1\right)C\left(0,1\right)\times \left(m,1\right)\right)\hfill & {\left(\left(0,1\right)C\left(h,c\right)\times \left(0,b^r\right)\right)}^{-1}\hfill \\ & \hfill & \hfill =\left(0,1\right)C\left(0,1\right)\times \left(b^{r}m,1\right)\text{.}\hfill & \hfill \end{array}$$

Now, looking at $\beta \left(0,b\right)$: $$\begin{array}{cccc}& \hfill \left(\left(0,1\right)C\left(h,c\right)\times \left(0,b^r\right)\right)& \hfill \left(\left(n,-1\right)C\left(l,d\right)\times \left(k,b\right)\right)\hfill & {\left(\left(0,1\right)C\left(h,c\right)\times \left(0,b^r\right)\right)}^{-1}\hfill \\ & \hfill & \hfill =\left(2h+cn,-1\right)C\left(cl+h-dh,d\right)\times \left(b^{r}k,b\right)\text{.}\hfill & \hfill \end{array}$$

Let $\beta$ henceforth denote the conjugated embedding. Choose $r$ so that $b^{r}m=1$. Then $$\beta \left(1,1\right)=\left(0,1\right)C\left(0,1\right)\times \left(1,1\right).$$ We choose $h,c$ in $$\beta \left(0,b\right)=\left(2h+cn,-1\right)C\left(cl+h\left(1-d\right),d\right)+\left(b^{r}k,b\right)\text{.}$$ We have four possibilities.

If $d=1$ and $l=0$ we can choose $c\ne 0$ and $h$ so that $$\beta \left(0,b\right)=\left(0,-1\right)C\left(0,1\right)+\left(b^{r}k,b\right).$$ But then $\beta$ is not regular, for ${\pi }_1\beta$ does not map onto $D_q$.

If $d=-1$ and $l=n$, then we can choose $c=2,h=-n$, but then $$\beta \left(0,b\right)=\left(0,-1\right)C\left(0,-1\right)+\left(b^{r}k,b\right),$$ so $\beta$ is not regular.

If $d=1$ and $l\ne 0$, choose $c\equiv l^{-1}\left(modq\right)$ and $h$ so that $2h+cn\equiv 0$, then $$\beta \left(0,b\right)=\left(0,-1\right)C\left(1,1\right)+\left(b^{r}k,b\right).$$

If $d=-1$ and $l\ne n$, then we can choose $c,h\ne 0$ with $$\begin{array}{cccc}& \hfill 2h+cn& \hfill =\hfill & 0\hfill \\ & \hfill cl+2h& \hfill =\hfill & 1,\hfill \end{array}$$ giving $$\beta \left(0,b\right)=\left(0,-1\right)C\left(1,-1\right)+\left(b^{r}k,b\right).$$ One verifies that $$\begin{array}{cccc}& \hfill \beta \left(1,1\right)& \hfill =\hfill & \left(0,1\right)C\left(0,1\right)\times \left(1,1\right)\hfill \\ & \hfill \beta \left(0,b\right)& \hfill =\hfill & \left(0,-1\right)C\left(1,\pm 1\right)\times \left(b^{r}k,b\right)\hfill \end{array}$$ is regular, for any $k$. There are $p$ choices for $k$, and hence, up to equivalence, there are $2p$ regular embeddings of $\mathrm{Hol}\left(Z_p\right)$ into $\mathrm{Hol}\left(D_q\times Z_p\right)$: $e\left(\mathrm{Hol}\left(Z_p\right),D_q\times Z_p\right)=2p$.

(3): Proof that $e\left(\Gamma ,D_p\times Z_q\right)=2p$. This argument is similar to the last case. We seek regular embeddings $$\beta :\mathrm{Hol}\left(Z_p\right)\to \mathrm{Hol}\left(D_p\times Z_q\right)\cong \mathrm{Hol}\left(D_p\right)\times \mathrm{Hol}\left(Z_q\right).$$ We have $$\mathrm{Hol}\left(D_p\times Z_q\right)\cong \mathrm{Hol}\left(D_p\right)\times \mathrm{Hol}\left(Z_q\right)$$ and by Lemma 3.1, the map $$\mathrm{Hol}\left(D_p\right)\cong D_p⋊\mathrm{Inn}\left(\mathrm{Hol}\left(Z_p\right)\right)\to \mathrm{Hol}\left(Z_p\right)\times \mathrm{Hol}\left(Z_p\right)$$ is 1-1 and maps $\left(m,\epsilon \right)C\left(n,d\right)$ to $\left(m,\epsilon \right)\left(n,d\right)\times \left(n,d\right)$.

Suppose $\beta \left(1,1\right)=\left(m,\epsilon \right)C\left(n,d\right)\times \left(l,c\right)$. Then $\beta \left(1,1\right)$ must have order $p$, so $\left(n,d\right)$ and $\left(m+\epsilon n,\epsilon d\right)$ have order dividing $p$. This implies $d=\epsilon =1$. Also $\left(l,c\right)$ has order dividing $p$ in $\mathrm{Hol}\left(Z_q\right)$, so $\left(l,c\right)=\left(0,1\right)$. Thus $$\beta \left(1,1\right)=\left(m,1\right)C\left(n,1\right)\times \left(0,1\right)$$ with $m$ or $n\not\equiv 0\left(modp\right)$.

Suppose $\beta \left(0,b\right)=\left(l,\epsilon \right)C\left(k,d\right)\times \left(s,c\right)$, of order $2q$. Then $\epsilon =-1$ or else $\beta$ is not regular. Also, $\left(s,c\right)$ must have order dividing $2q$ in $\mathrm{Hol}\left(Z_q\right)$. But since $\beta$ is regular, we must have $\left(t,*\right)$ in the image of $\beta$ for all $t$ in $Z_q$, so $c=1$ and $s\ne 0$.

Applying $\beta$ to the relation $\left(b,1\right)\left(0,b\right)=\left(0,b\right)\left(1,1\right)$ yields $$\begin{array}{cccc}& \hfill \left(\left(bm,1\right)C\left(bn,1\right)\times \left(0,1\right)\right)& \hfill \left(\left(l,-1\right)C\left(k,d\right)\times \left(s,1\right)\right)\hfill & \hfill \\ & \hfill & \hfill =\left(\left(l,-1\right)C\left(k,d\right)\times \left(s,1\right)\right)\left(\left(m,1\right)C\left(n,1\right)\times \left(0,1\right)\right)\text{.}\hfill & \hfill \end{array}$$ This yields no condition on $s$, but on the left components we obtain $$\left(bm+2bn+l,-1\right)C\left(bn+k,d\right)=\left(l-dm,-1\right)C\left(dn+k,d\right).$$ Thus $$\begin{array}{cccc}& \hfill 2bn+bm& \hfill =\hfill & -dm\hfill \\ & \hfill bn& \hfill =\hfill & dn\text{.}\hfill \end{array}$$ If $n\ne 0$, then $b=d$ and $n=-m$. If $n=0$, then $d=-b$. Thus $$\begin{array}{cccc}& \hfill \beta \left(1,1\right)& \hfill =\hfill & \left(m,1\right)C\left(-m,1\right)\times \left(0,1\right)\hfill \\ & \hfill \beta \left(0,b\right)& \hfill =\hfill & \left(l,-1\right)C\left(k,b\right)\times \left(s,1\right),\hfill \end{array}$$ or $$\begin{array}{cccc}& \hfill \beta \left(1,1\right)& \hfill =\hfill & \left(m,1\right)C\left(0,1\right)\times \left(0,1\right)\hfill \\ & \hfill \beta \left(0,b\right)& \hfill =\hfill & \left(l,-1\right)C\left(k,-b\right)\times \left(s,1\right)\text{.}\hfill \end{array}$$ One then sees easily that $\beta \left(0,b\right)$ has order dividing $2q$.

Now we modify $\beta$ by an automorphism of $D_p\times Z_q$.

If we conjugate the right factors by $\left(0,s^{-1}\right)$ in $\mathrm{Aut}\left(Z_q\right)$, then $\left(s,1\right)$ is transformed into $\left(1,1\right)$.

If we conjugate the left factors by $\left(0,1\right)C\left(g,c\right)$ in $\mathrm{Aut}\left(D_p\right)$, then $$\left(m,1\right)C\left(n,1\right)\text{ becomes }\left(cm,1\right)C\left(cn,1\right)$$ and $$\left(l,-1\right)C\left(k,\pm b\right)\text{ becomes }\left(2g+cl,-1\right)C\left(g+ck\mp bg,\pm b\right).$$ Choose $c$ so that $cm=1$. Then $cn=0$ or $-1$.

Choose $g$ so that $2g+cl=0$. Then we have the following representatives of equivalence classes of $\beta$: $$\begin{array}{cccc}& \hfill \beta \left(1,1\right)& \hfill =\hfill & \left(1,1\right)C\left(-1,1\right)\times \left(0,1\right)\hfill \\ & \hfill \beta \left(0,b\right)& \hfill =\hfill & \left(0,-1\right)C\left(k,b\right)\times \left(1,1\right),\hfill \end{array}$$ and $$\begin{array}{cccc}& \hfill \beta \left(1,1\right)& \hfill =\hfill & \left(1,1\right)C\left(0,1\right)\times \left(0,1\right)\hfill \\ & \hfill \beta \left(0,b\right)& \hfill =\hfill & \left(0,-1\right)C\left(k,-b\right)\times \left(1,1\right)\text{.}\hfill \end{array}$$ It is a routine check that for any $k$ modulo $p$, both $\beta$ are regular Thus we have $2p$ equivalence classes of regular embeddings of $\mathrm{Hol}\left(Z_p\right)$ into $\mathrm{Hol}\left(D_p\times Z_q\right)$.

(4): Proof that $e\left(\Gamma ,Z_{2qp}\right)=p$. Let $$\beta :\mathrm{Hol}\left(Z_p\right)\to \mathrm{Hol}\left(Z_{2pq}\right)\cong \mathrm{Hol}\left(Z_p\right)\times \mathrm{Hol}\left(Z_q\right)\times Z_2$$ be a regular embedding. Then $\beta \left(1,1\right)$ has order $p$, so $$\beta \left(1,1\right)=\left(m,1\right)\times \left(0,1\right)\times 0$$ for some $m\not\equiv 0\left(modp\right)$. Also, $\beta \left(0,b\right)$ has order $p-1=2q$, so $$\beta \left(0,b\right)=\left(n,d\right)\times \left(l,1\right)\times e$$ with $d\not\equiv 1\left(modp\right)$. If $e\equiv 0\left(mod2\right)$ or $l\equiv 0\left(modq\right)$ then $\beta$ is not regular, hence we have $e\equiv 1\left(mod2\right)$ and $l\not\equiv 0\left(modq\right)$.

The condition $\left(b,1\right)\left(0,b\right)=\left(0,b\right)\left(1,1\right)$ implies that $d\equiv b\left(modp\right)$. Thus in $\mathrm{Hol}\left(Z_{2pq}\right)$, $$\begin{array}{cccc}& \hfill \beta \left(1,1\right)& \hfill =\hfill & \left(2qm,1\right)\hfill \\ & \hfill \beta \left(0,b\right)& \hfill =\hfill & \left(n,c\right)\hfill \end{array}$$ with $p$ not dividing $m$, $2q$ not dividing $n$, and $c\equiv b\left(modp\right),c\equiv 1\left(mod2q\right)$.

Now we consider $\beta$ under equivalence by automorphisms of $Z_{2pq}$. We conjugate $\beta$ by $\left(0,d\right)$ with $d$ coprime to $2pq$: $$\begin{array}{cccc}& \hfill \left(0,d\right)\left(2qm,1\right)\left(0,d^{-1}\right)& \hfill =\hfill & \left(2dqm,1\right)\hfill \\ & \hfill \left(0,d\right)\left(n,c\right)\left(0,d^{-1}\right)& \hfill =\hfill & \left(dn,c\right)\text{.}\hfill \end{array}$$ Choose $d$ so that $$\begin{array}{cccc}& \hfill dm& \hfill \equiv \hfill & 1\left(modp\right)\hfill \\ & \hfill dn& \hfill \equiv \hfill & 1\left(mod2q\right)\text{.}\hfill \end{array}$$ Then after conjugating, $\beta$ becomes: $$\begin{array}{cccc}& \hfill \beta \left(1,1\right)& \hfill =\hfill & \left(2q,1\right)\hfill \\ & \hfill \beta \left(0,b\right)& \hfill =\hfill & \left(1+2ql,c\right)\text{.}\hfill \end{array}$$ We check that $\beta$ is regular for any $l$ modulo $p$. We have $$\begin{array}{cccc}& \hfill \beta \left(n,b^k\right)& \hfill =\hfill & \beta \left(n,1\right)\beta {\left(0,b\right)}^k\hfill \\ & \hfill & \hfill =\hfill & \left(2qn,1\right)\left(\left(1+2ql\right)\left(1+c+\dots +c^{k-1}\right),c^k\right)\hfill \\ & \hfill & \hfill =\hfill & \left(2qn+\left(1+2ql\right)\left(1+c+\dots +c^{k-1}\right),c^k\right)\text{.}\hfill \end{array}$$ Let $a$ be any element of $Z_{2pq}$. To solve $$a\equiv 2qn+\left(1+2ql\right)\left(1+c+\dots +c^{k-1}\right)\left(mod2pq\right)$$ for $n$ and $k$ it suffices to solve $$\begin{array}{cccc}& \hfill a& \hfill \equiv \hfill & 2qn+\left(1+2ql\right)\left(1+c+\dots +c^{k-1}\right)\left(modp\right)\hfill \\ & \hfill a& \hfill \equiv \hfill & 1+c+\dots +c^{k-1}\equiv k\left(mod2q\right)\text{.}\hfill \end{array}$$ Clearly for each $a$ modulo $2pq$ there are unique values for $k,n$ that solve these congruences. Thus for any $l$, $\beta$ is 1-1 and regular, and so $e\left(\mathrm{Hol}\left(Z_p\right),Z_{2pq}\right)=p$.

(5): Proof that $e\left(\Gamma ,D_{pq}\right)=4p$. Let $\beta :\mathrm{Hol}\left(Z_p\right)\to \mathrm{Hol}\left(D_{pq}\right)$ be a regular embedding. We have $$\mathrm{Hol}\left(D_{pq}\right)=D_{pq}⋊\mathrm{Inn}\left(\mathrm{Hol}\left(Z_{pq}\right)\right)\to \mathrm{Hol}\left(Z_{pq}\right)\times \mathrm{Hol}\left(Z_{pq}\right)$$ by $\left(m,\epsilon \right)C\left(n,d\right)\mapsto \left(m,\epsilon \right)\left(n,d\right)\times \left(n,d\right)$. Also, $$\mathrm{Hol}\left(Z_{pq}\right)\cong \mathrm{Hol}\left(Z_p\right)\times \mathrm{Hol}\left(Z_q\right)$$ by $\left(l,c\right)\mapsto \left(\left(l,c\right)mod\left(p\right),\left(l,c\right)mod\left(q\right)\right)$. Under this map, $D_{pq}$ maps into $D_p\times D_q$.

Let $\beta \left(1,1\right)=\left(m,\epsilon \right)C\left(n,d\right)$, of order $p$. Then $\beta \left(1,1\right)$ maps to $$\left(m+\epsilon d,\epsilon d\right)\times \left(n,d\right)modp,\left(m+\epsilon d,\epsilon d\right)\times \left(n,d\right)modq,$$ which must have order $p$. Hence $$\begin{array}{cccc}& \hfill \left(m+\epsilon d,\epsilon d\right)& \hfill \equiv \hfill & \left(0,1\right)\left(modq\right)\hfill \\ & \hfill \left(n,d\right)& \hfill \equiv \hfill & \left(0,1\right)\left(modq\right)\hfill \end{array}$$ and $$\begin{array}{cccc}& \hfill \epsilon d\equiv d& \hfill \equiv \hfill & 1\left(modp\right)\hfill \\ & \hfill n\text{ or }m+\epsilon n& \hfill \not\equiv \hfill & 0\left(modp\right)\text{.}\hfill \end{array}$$

Let $\beta \left(0,b\right)=\left(l,\epsilon \right)C\left(k,c\right)$. By regularity we must have $\epsilon =-1$, or else ${\pi }_1\left(\beta \right)$ maps into $\left\{\left(*,1\right)\right\}\subset D_{pq}$. Then $\beta \left(0,b\right)$ maps to $$\left(l-k,-c\right)\times \left(k,c\right)modp,\left(l-k,-c\right)\times \left(k,c\right)modq\text{.}$$ This has order $2q=p-1$, so we must have $$c\equiv \pm 1\left(modq\right).$$

From $\left(b,1\right)\left(0,b\right)=\left(0,b\right)\left(1,1\right)$ we obtain $$\left(bqm,1\right)C\left(bqn,1\right)\left(l,-1\right)C\left(k,c\right)=\left(l,-1\right)C\left(k,c\right)\left(qm,1\right)C\left(qn,1\right),$$ hence $$\left(bqm+2bqn+l,C\left(k+bqn,c\right)\right)=\left(l-cqm,-1\right)C\left(k+cqn,c\right)$$ and so $$\begin{array}{cccc}& \hfill bqn& \hfill =\hfill & cqn\hfill \\ & \hfill cqm& \hfill =\hfill & bqm+2bqn\text{.}\hfill \end{array}$$ This yields no conditions modulo $q$, but modulo $p$, we have:

1. If $n\not\equiv 0$, then $c=b$ and $m=-n$.

2. If $n\equiv 0$, then $c=-b$ and $m\not\equiv 0$.

Now we look for a nice representative for $\beta$ modulo conjugation by elements of $\mathrm{Inn}\left(\mathrm{Hol}\left(Z_{pq}\right)\right)$. For $\left(g,d\right)\in \mathrm{Hol}\left(Z_{pq}\right)$, $$\begin{array}{cccc}& \hfill C\left(g,d\right)\beta \left(1,1\right)C{\left(g,d\right)}^{-1}& \hfill =\hfill & \left(g,d\right)\left(qm,1\right){\left(g,d\right)}^{-1}C\left(\left(g,d\right)\left(qn,1\right){\left(g,d\right)}^{-1}\right)\hfill \\ & \hfill & \hfill =\hfill & \left(dqm,1\right)C\left(dqn,1\right),\hfill \\ & \hfill C\left(g,d\right)\beta \left(0,b\right)C{\left(g,d\right)}^{-1}& \hfill =\hfill & \left(g,d\right)\left(l,-1\right){\left(g,d\right)}^{-1}C\left(\left(g,d\right)\left(k,c\right){\left(g,d\right)}^{-1}\right)\hfill \\ & \hfill & \hfill =\hfill & \left(2g+dl,-1\right)C\left(g+dk-cg,c\right)\text{.}\hfill \end{array}$$ We may choose $d$ and $g$ modulo $pq$ by choosing them modulo $p$ and modulo $q$ separately.

Modulo $p$:

1. If $n\not\equiv 0$, choose $d$ so that $dqn\equiv -1$, then $dqm\equiv -dqn\equiv 1$ and $c\equiv b$.

2. If $n\equiv 0$, choose $d$ so that $dqm\equiv 1$ and $c\equiv -b$.

Choose $g$ so that $2g+dl\equiv 0$. Then, since $d\not\equiv 0$, $g-cg+dk$ is arbitrary, so (letting $\beta$ now denote the conjugated embedding) we have, modulo $p$: $$\beta \left(1,1\right)=\left(1,1\right)C\left(-1,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(k,b\right)$$ or $$\beta \left(1,1\right)=\left(1,1\right)C\left(0,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(k,-b\right).$$

Modulo $q$:

$$\begin{array}{cccc}& \hfill \beta \left(1,1\right)& \hfill =\hfill & \left(0,1\right)C\left(0,1\right)\hfill \\ & \hfill \beta \left(0,b\right)& \hfill =\hfill & \left(2g+dl,-1\right)C\left(\left(1-c\right)g+dk,c\right)\hfill \end{array}$$ where $c\equiv \pm 1$.

If $c\equiv 1$ and $k\not\equiv 0$, set $dk\equiv 1$ and $$2g+dl\equiv 0.$$ Then $$\beta \left(1,1\right)=\left(0,1\right)C\left(0,1\right)$$ and $$\beta \left(0,b\right)=\left(0,-1\right)C\left(1,1\right)$$ or (if $k\equiv 0$) $$\beta \left(0,b\right)=\left(0,-1\right)C\left(0,1\right).$$ However, in this last case, $\beta$ is not regular: ${\pi }_1\beta$ maps to $\left\{\left(0,\pm 1\right)\right\}\subset \mathrm{Hol}\left(Z_q\right)$.

If $c\equiv -1$, we have $$\beta \left(0,b\right)=\left(2g+dl,-1\right)C\left(2g+dk,-1\right).$$ If $l\equiv k$ we can choose $g$ so that $2g+dk=2g+dl\equiv 0$, but then $\beta$ is not regular. Thus we must have $l\not\equiv k$, and then we may choose $d\ne 0$ and $g$ so that $$\begin{array}{cccc}& \hfill 2g+dl& \hfill \equiv \hfill & 0\hfill \\ & \hfill 2g+dk& \hfill \equiv \hfill & 1\hfill \end{array}$$ and so $$\begin{array}{cccc}& \hfill \beta \left(1,1\right)& \hfill =\hfill & \left(0,1\right)C\left(0,1\right)\hfill \\ & \hfill \text{and}\beta \left(0,b\right)& \hfill =\hfill & \left(0,-1\right)C\left(-1,1\right)\text{.}\hfill \end{array}$$

To summarize, any regular embedding is equivalent to $$\beta \left(1,1\right)=\left(1,1\right)C\left(-1,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(k,b\right)$$ or $$\beta \left(1,1\right)=\left(1,1\right)C\left(0,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(k,-b\right)$$ modulo $p$ and to $$\beta \left(1,1\right)=\left(0,1\right)C\left(0,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(1,1\right)$$ or $$\beta \left(1,1\right)=\left(0,1\right)C\left(0,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(-1,1\right)$$ modulo $q$.

We show that all four combinations give regular embeddings of $\mathrm{Hol}\left(Z_p\right)$ to $\mathrm{Hol}\left(D_{pq}\right)$, and so, since $k$ is arbitrary in $Z_p$, we have $4p$ equivalence classes of regular embeddings.

Note that in every combination, $${\beta }_1\left(l,b^r\right){\beta }_2{\left(l,b^r\right)}^{-1}=\left(*,{\left(-1\right)}^r\right)$$ both modulo $p$ and modulo $q$, and so ${\beta }_1·{{\beta }_2}^{-1}$ maps to $D_{pq}\subset \mathrm{Hol}\left(Z_p\right)$. To show regularity, we need only show that modulo $p$ every element of $D_p$ is in the image of ${\beta }_1·{{\beta }_2}^{-1}$, and similarly modulo $q$.

Mod $p$: If $$\beta \left(1,1\right)=\left(1,1\right)C\left(-1,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(k,b\right)$$ modulo $p$, then $${\beta }_1\left(l,b^r\right){\beta }_2{\left(l,b^r\right)}^{-1}=\left(-k\left(\frac{1-{\left(-b\right)}^r}{1+b}\right)+{\left(-1\right)}^r\left(-l+k\left(\frac{1-b^r}{1-b}\right)\right),{\left(-1\right)}^r\right)$$ and for $r$ and $l$ arbitrary we can obtain all elements $\left(m,\pm 1\right)$ of $D_p$.

If $$\beta \left(1,1\right)=\left(1,1\right)C\left(0,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(k,-b\right)$$ modulo $p$ then $${\beta }_1\left(l,b^r\right){\beta }_2{\left(l,b^r\right)}^{-1}=\left(\left(l+\left(-k\right)\frac{1-b^r}{1-b}\right)+{\left(-1\right)}^{r}k\left(\frac{1-{\left(-b\right)}^r}{1+b}\right),{\left(-1\right)}^r\right)$$ and again for $r$ and $l$ arbitrary we can obtain all elements $\left(m,\pm 1\right)$ of $D_p$.

Mod $q$: If $$\beta \left(1,1\right)=\left(0,1\right)C\left(0,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(1,1\right)$$ modulo $q$, then $${\beta }_1\left(l,b^r\right){\beta }_2{\left(l,b^r\right)}^{-1}=\left\{\begin{array}{cc}\left(-r,1\right)\hfill & \text{ if }r\text{ is even, }\hfill \\ \left(-1+r,-1\right)\hfill & \text{ if }r\text{ is odd}\text{.}\hfill \end{array}\right.$$ Since $0\le r<2q$ and we're working modulo $q$, we can obtain all elements $\left(m,\pm 1\right)$ of $D_q$.

If $$\beta \left(1,1\right)=\left(0,1\right)C\left(0,1\right),\beta \left(0,b\right)=\left(0,-1\right)C\left(-1,1\right)$$ modulo $q$, then $${\beta }_1\left(l,b^r\right){\beta }_2{\left(l,b^r\right)}^{-1}=\left\{\begin{array}{cc}\left(-r,1\right)\hfill & \text{ if }r\text{ is even, }\hfill \\ \left(1-r,-1\right)\hfill & \text{ if }r\text{ is odd}\text{.}\hfill \end{array}\right.$$ Again, since $0\le r<2q$ and we're working modulo $q$, we can obtain all elements $\left(m,\pm 1\right)$ of $D_q$.

Thus $$\left\{{\beta }_1\left(l,b^r\right){\beta }_2{\left(l,b^r\right)}^{-1}\right\}=\left\{\begin{array}{cc}{D}_p\hfill & \text{modulo }p\hfill \\ D_q\hfill & \text{modulo }q,\hfill \end{array}\right.$$ and so all four combinations of $\beta$ modulo $p$ and modulo $q$ are regular. This completes the proof that $e\left(\mathrm{Hol}\left(Z_p\right),D_{pq}\right)=4p$.

Combining Theorems 2.1 and 4.1 yields:

Corollary 4.2.  

a)  If $L$ is a Galois extension of $K$, fields, with Galois group $\Gamma =\mathrm{Hol}\left(Z_p\right)$, $p>5$ a safeprime, then for every group $G$ of cardinality that of $\Gamma$, there is a $H$-Hopf Galois structure on $L/K$ where the associated group of $H$ is $G$.

b)  The number of Hopf Galois structures on $L/K$ is $2+3p+4pq$.

Remark 4.3.   N. Byott [By03b] has obtained the analogous result to Corollary 4.2(a) for both the cyclic group and the non-abelian group of order $pq$, $p$ and $q$ primes with $q$ dividing $p-1$. His approach is not to determine regular embeddings of $\Gamma$ into $\mathrm{Hol}\left(G\right)$ up to equivalence, but rather to determine the set of regular subgroups of $\mathrm{Hol}\left(G\right)$ whose intersection with $\mathrm{Aut}\left(G\right)$ is a given cardinality.