On Hopf Galois structures and complete groups

Lindsay N. Childs1

Department of Mathematics and Statistics
University at Albany
Albany, NY 12222


Let L be a Galois extension of K, fields, with Galois group Γ. We obtain two results. First, if Γ=HolZpe, we determine the number of Hopf Galois structures on LK where the associated group of the Hopf algebra H is Γ (i.e. LKHLΓ). Now let p be a safeprime, that is, p is a prime such that q=p12>2 is also prime. If LK is Galois with group Γ=HolZp, p a safeprime, then for every group G of cardinality pp1 there is an H-Hopf Galois structure on LK where the associated group of H is G, and we count the structures.

Let L be a Galois extension of K, fields, with finite Galois group Γ. Then L is an H-Hopf Galois extension of K for H=KΓ, where KΓ acts on L via the natural action of the Galois group Γ on L. Greither and Pareigis [GP87] showed that for many Galois groups Γ, L is also an H-Hopf Galois extension of K for H a cocommutative K-Hopf algebra other than KΓ. The Hopf algebras H that arise have the property that LKHLG, the group ring over L of a group G of the same cardinality as Γ. We call G the associated group of H.

From [By96] we know that for H a cocommutative K-Hopf algebra, H-Hopf Galois structures on L correspond bijectively to equivalence classes of regular embeddings β:ΓHolGPermG. Here PermG is the group of permutations of the set G, and HolG is the normalizer of the left regular representation of G in PermG. One sees easily that HolG contains the image of the right regular representation ρ:GPermG and also AutG; then HolG=ρG·AutG and is isomorphic to the semidirect product GAutG. The equivalence relation on regular embeddings is by conjugation by elements of AutG inside HolG: ββ if there exists γ in AutG so that for all g in G, γβgγ1=βg.

Thus the number eΓ,G of H-Hopf Galois structures on LK where the associated group of H is G depends only on G and the Galois group Γ, and reduces to a purely group-theoretic problem.

If L is a Galois extension of K with Galois group Γ non-abelian simple or Sn, then in [CC99] we counted the number of Hopf Galois structures on LK with associated group G=Γ by "unwinding" regular embeddings. The unwinding idea applies more generally when G is a complete group, i. e. has trivial center and trivial outer automorphism group. In this paper we apply this unwinding idea to determine eG,G when G is the complete group HolZpe, p an odd prime.

One theme of research on Hopf Galois structures on Galois extensions of fields is to determine to what extent it is true that if LK is Galois with Galois group Γ and is H-Hopf Galois where H has associated group G, then GΓ. Positive results in this direction include Byott's original uniqueness theorem [By96]; Kohl's Theorem [Ko98] that if Γ=Zpe then GΓ; Byott's recent result [By03a], complementing [CC99], that if Γ is non-abelian simple then G=Γ; and Featherstonhaugh's recent result [Fe03] that if G and Γ are abelian p-groups with p sufficiently large compared to the p-rank of G and logp of the exponent of G, then GΓ. (A survey of results before 2000 in this area may be found in Chapter 2 of [C00]; Chapter 0 of [C00] describes how Hopf Galois structures on Galois extensions of local fields relate to local Galois module theory of wildly ramified extensions.)

In the last two sections of this paper we consider this uniqueness question for Γ=HolZp when p is a safeprime, that is, p=2q+1 where p and q are odd primes. (The terminology “safeprime” arises in connection with factoring large numbers related to cryptography—see [C95, pp. 411-413].) Then there are exactly six isomorphism classes of groups of cardinality pp1. We show that if LK is a Galois extension with Galois group Γ=HolZp, then for each of the six groups G of cardinality pp1 up to isomorphism, there is an H-Hopf Galois structure on LK with associated group G, and we count their number. Thus Γ=HolZp yields an example of the opposite extreme to the uniqueness results listed above.

1.  Regular embeddings

Given a Galois extension LK with Galois group Γ, and a group G of cardinality that of Γ, the number eΓ,G of Hopf Galois structures on LK whose Hopf algebra H has assocated group G is equal to the number of equivalence classes of regular embeddings of Γ into HolG, the semidirect product of G and AutG. Here an embedding β:ΓHolG is regular if, when viewing HolG inside PermG via g,αx=αxg1, the orbit of the identity element 1 of G under βΓ is all of G.

Obtaining eΓ,G in any particular case involves a number of steps:

Of these tasks, checking regularity is the least natural. If we view HolG, the semidirect product of G and AutG, as G×AutG as sets, the operation is g,α·g,α=gαg,αα for g,gG,α,αAutG. Then G{g,1} is a normal subgroup of HolG, and the projection π2 onto AutG by π2g,α=α is a homomorphism; however the projection π1 onto G, π1g,α=g, is not. But since an element g,α viewed in PermG acts on the identity element 1 of the set G by g,α1=α1g1=g1, checking regularity of a 1-1 homomorphism β:ΓHolG is the same as determining whether the function (non-homomorphism) π1β:ΓG is bijective.

Let InnG be the group of inner automorphisms of G, then InnG is normal in AutG and AutGInnG=OG, the outer automorphism group, fits in the exact sequence 1ZGGAutGOG1 where the map C:GAutG is conjugation: Cgx=gxg1 for g,xG, and ZG is the center of G.

Suppose ZG=1 and the composition of the 1-1 homomorphism β:ΓHolG with the homomorphism HolGOG yields a trivial homomorphism from Γ to OG. Then β maps into GInnG, and, following [CC99], we may decompose β as follows: we have a homomorphism j:GInnGG×G by jg,Ch=gh,h for g,hG, with inverse sending g,h to gh1,Ch. Letting pi:G×GG be projection onto the ith factor, i=1,2, the homomorphism β yields homomorphisms β1=p1jβ and β2=p2jβ:ΓG such that βγ=β1γβ2γ1,Cβ2. Then β is regular iff {π1βγ|γΓ}={β1γβ2γ1|γΓ}=G. Thus whenever ZG=1 and βΓGInnG, we may describe β, and in particular the function π1β:ΓG, in terms of the homomorphisms β1,β2:ΓG, namely, π1βγ=β1γβ2γ1.

A class of groups G where ZG=1 and βΓGInnG is the class of complete groups, that is, finite groups G with ZG=1 and OG=1. The best-known examples of finite complete groups are AutA where A is simple and non-abelian, Sn for n3,n6, and HolZm where m is odd. (See [Sch65, III.4.u-w] .) Another class is the class of simple groups, for if G is simple, then OG is solvable, so any β:GHolG has image in GInnG.

In [CC99] we let Γ=G and determined eG,G, the number of regular embeddings of G to HolG, when G is simple non-abelian or Sn,n4. In the next section we examine the case where Γ=G=HolZq where q=pe and p is an odd prime.

2.  HolZpe

Let G=ZpeZpe*, the holomorph of Zpe. Let b be a number <pe that generates Zpe*. Let Γ=G. In this section we prove:

Theorem 2.1.   If G=HolZpe, p odd, then up to equivalence there are eG,G=2pe1ϕpe1+2peϕpe1ϕp11 regular embeddings of G into HolG. Thus eG,G is the number of H-Hopf Galois structures on a Galois extension LK 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 HolG. Since G is complete, we know from the last section that any homomorphism β:GHolG may be decomposed as βg=β1gβ2g1Cβ2g for homomorphisms β1,β2:GG. So we begin by describing the homomorphisms from G to G.

Let α:GG be a homomorphism. Then α is determined by  α1,1=m,cof order dividingpe,and α0,b=n,dof order dividingpe1p1. If α1,1 has order dividing pe, then c=1. To see this, first note that for any s>0, m,cs=m1+c+c2+cs1,cs so c must have order dividing pe, which implies that c1modp. Also, for α0,b to have order dividing pe1p1 we require that d1modp or p divides n. For if d1modp one sees by induction that n,dpe1=pe1n,1 where p divides n iff p divides n. Thus if p does not divide n, then n,d has order pe. On the other hand, if d1modp, then n,dp1=ndp11d1,dp1 and p divides dp11 but not d1, so the order of n,d divides pe1p1.

Now we check the relation b,10,b=b,b=0,b1,1. Applying α yields: m1+c+cb1,cbn,d=n,dm,c, hence m1+c+cb1+cbn,cbd=n+dm,dc. Thus cbd=cd. Since d is a unit modulo pe it follows that cb=c, hence cb1=1. But since c=1+pf for some f, 1=1+pfb1. Since b1 is relatively prime to p and 1+pfpe1=1, it follows that 1+pf=1, hence f=0 and c=1.

Thus for any homomorphism α:GG, α1,1=m,1 for some m.

Since c=1, the requirement m1+c+cb1+cbn,cbd=n+dm,dc becomes mb+n,d=n+dm,d which implies that bm=dm. Thus if m0, then bdmodp, and if m is a unit (i.e. relatively prime to p), then b=d. In the latter case, α1,1=m,1 has order pe and α0,b=n,b has order pe1p1.

Clearly if α is an automorphism then m is relatively prime to p and so b=d. Conversely, if α1,1=m,1 with m relatively prime to p, then α0,b=n,b for some n. Conjugating α by h,c in G yields  h,cα1,1h,c1=h,cm,1c1h,c1 =h+cmh,1 =cm,1. We choose c so that cm=1. Then we choose h so that  0,b=h,cα0,bh,c1=h,cn,bc1h,c1 =h+cn,bcc1h,c1 =h+cnbh,b =1bh+cn,b: we set h=1b1cn, possible because b has order pe1p1modpe and hence b1modp. With these choices of h and c, the homomorphism Ch,cα:GG is the identity on the generators 1,1 and 0,b of G, and so α=Ch,c1 is an automorphism of G.

Now we ask about regularity: for which pairs of endomorphisms β1,β2 is {β1gβ2g1|gG}=G, or equivalently, π1β=β1·β21 is a 1-1 function from G to G? Let βi1,1=mi,1,βi0,b=ni,di for i=1,2.

If neither β1 nor β2 is an automorphism, then p divides m1 and m2, so  β1pe1l,1β2pe1l,11=pe1lm1,1pe1lm2,1 =0,10,1=0,1 for all l, and so β1·β21 is not 1-1.

Suppose both β1 and β2 are automorphisms. If m1m2modp then  β1·β21pe1,1=pe1m1pe1m2,1 =0,1 =β1·β210,1 so β1·β21 is not 1-1. If m1m2modp, then let sm1m2=n1n2. Then  β1·β210,b=n1n2,1 =m1m2s,1 =β1·β21s,1, so again, β1·β21 is not 1-1.

Thus for β to be regular, exactly one of β1 and β2 is an automorphism.

We return to looking at regularity after we look at equivalence by AutG=InnG inside HolG.

For g,h,kG we have  1,Cgh,Ck1,Cg1=Cgh,CgCkCg1 =ghg1,Cgkg1. Thus if βx=β1xβ2x1,Cβ2x for xG, then ββ with  βx=gβ1xβ2x1g1,Cgβ2xg1 =gβ1xg1·gβ2xg11,Cgβ2xg1: we get from β to β by simultaneously conjugating β1 and β2 by gG.

Assume β2 is an automorphism, then, up to equivalence, we may assume that β2 is the identity automorphism on G and β1 is not an automorphism. Then eG,G will be twice the number of possible β1, since the case where β1 is an automorphism and β2 is not is the same.

Returning to the regularity question, we ask, for which β1 is {β1gg1}=G?

Assume  β11,1=m,1 β10,b=n,d for some m divisible by p, some d1 and some n. Then  β1l,bk=β1l,1β10,bk =lm,1ndk1d1,dk =lm+ndk1d1,dk. For any h,r we want to find l,k so that (**)β1l,bkl,bk1=lm+ndk1d1,dkbkl,bk =lm+ndk1d1dkbkl,dkbk =h,br. In order that for any r, there is a k so that br=db1k, we require that db1 generates Zpe*. Now for β1 to be a homomorphism, we need bm=dm, and this is possible only if m=0 or bdmodp. But in the latter case, db11modp, so cannot generate Zpe*. Thus if {β1gg1}=G, then m=0, and d=bf+1 such that bf generates Zpe*.

Thus β1 is defined by  β11,1=0,1 β10,b=n,d, and (**) becomes n1+d++dk1bfkl,bfk=h,br, which is solvable for all n and for all f such that bf generates Zpe* . We have two cases:

If d=bf+11modp, then p1 divides f+1 and p divides n (or else β is not a homomorphism).

If d=bf+11modp, then n is arbitrary.

The first case gives pe1 choices for n and ϕpe1 choices for f.

The second case gives pe choices for n and ϕpe1p1ϕpe1=ϕpe1ϕp11 choices for f. The theorem follows.

Corollary 2.2.   If G=HolZp, then eG,G=21+pϕp11.

Example 2.3.   For p=5 there are, up to equivalence, 21+521=12 regular embeddings of G=Z5Z5* into HolG. If we choose b=2 then b and b3 generate Z5*, so d=b0=1 or d=b2=4. Thus if we let β2 be the identity automorphism, then β11,1=0,1 and β10,2=n,4, n=0,1,2,3,4, or 0,1.

3.  Groups of order pp1

Let Γ=HolZp with p an odd prime, as above, and let p1=2q. Then there are at least five non-isomorphic groups G of order 2qp other than Γ, namely, Z2qp,Dqp,Dp×Zq,Dq×Zp, and ZpZq×Z2 (where Zq is identified as the subgroup of AutZp of index 2, and Dn 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 Γ=HolZp, 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 AutDp=AutZpZq=AutHolZp=InnHolZp.

Proof. Let G=ZpZa with a=2,q or 2q=p1. Write Zp additively and view Za as the cyclic subgroup of order a inside Zp*=AutZp and GHolZp. Let b generate Za. If α is an automorphism of G, then:

  1. α1,1=m,1 for p not dividing m, and

  2. α0,b=n,b for any n (as one sees by applying α to the relation b,10,b=b,b=0,b1,1.)

So there are pp1=|HolZp| automorphisms of G. Now if l,d is any element of HolZp, then conjugation by l,d is an automorphism of G, since l,dm,brl,d1=l+dmbrl,brG. Hence the conjugation map C:HolZpAutG, Cl,dm,br=l,dm,brl,d1=l+dmbrl,br, is defined. Then C is 1-1. For if Cl1,d1=Cl2,d2 on G, then l1+d1mbrl1=l2+d2mbrl2 for all m,r: in particular for m=1,r=0 we get d1=d2 and for m=0,r=1 we get l1=l2. 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 pp1.

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

Proof. If G has order pp1 then by the first Sylow Theorem, G has a unique normal subgroup Gp of order p. By Schur's Theorem, Gp has a complementary subgroup of order 2q, and hence a subgroup K of order q. Then GpK=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 JαZ2.

If JZpq then GZpqαZ2 where α:Z2AutZpqZp1×Zq1 has four possibilities, α1=±1,±1. These yield GDpq,Dp×Zq,Dq×Zp and Z2pq.

If JZpZqHolZp, then GJαZ2 where α:Z2AutZpZqInnHolZp. If α is trivial, we obtain ZpZq×Z2. Otherwise, the elements of order 2 in InnHolZp are of the form Cl,1 for any lZp. Define αl:Z2InnHolZp by αl1=Cl,1. One checks easily that ZpZqα0Z2ZpZ2q=HolZp by the map sending a,b,ε to a,εb for aZp, bZqZp1, εZ2Zp1. Now α0=C0,1 and αl=Cl,1 are conjugate in InnHolZp for l0 by Cm,1 where 2mlmodp: Cm,1C0,1Cm,1=C2m,1. It follows from [DF99], p. 186, Exercise 6 that ZpZqαlZ2ZpZqα0Z2 for all l. Thus J=ZpZq yields only two possible groups, up to isomorphism.

4.  Nonuniqueness

If we begin with the Galois group Γ of a Galois extension LK of fields, then to determine the K-Hopf Galois structures on L, we need to count equivalence classes of regular embeddings of Γ into HolG, where G varies over isomorphism classes of groups of the same cardinality as Γ. This can be a formidable task for certain cardinalities!

In this section, we let Γ=HolZp, p=2q+1 a safeprime >5, and determine eΓ,G, the number of regular embeddings of Γ into HolG, for all six groups G of order pp1=2pq. We did the case G=Γ in Theorem 2.1 and found that eG,G=2+2pq2. Here is the result for GΓ:

Theorem 4.1.   Let Γ=HolZp, with p and q=p12 odd primes. Then:

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

(1): Proof that eΓ,ZpZq×Z2=2pq1. Using Lemma 3.1 we have  HolZpZq×Z2=HolZpZq×HolZ2 =ZpZqInnHolZp×Z2 HolZpInnHolZp×Z2 HolZp×HolZp×Z2; thus we may unwind any homomorphism β:ΓZpZqInnHolZp×Z2 as in Section 1. If βγ=m,b2r,Cl,bs×ε with ε=±1,Zp*=b and m,l are modulo p, then βγ maps to m,b2rl,bs×l,bs×ε under the map to HolZp×HolZp×Z2. So β induces homomorphisms β1:ΓHolZp, defined by β1γ=m,b2rl,bs, and β2:ΓHolZp, defined by β2γ=l,bs. If we define β0:ΓZ2 by β0γ=ε, then βγ=β1γβ2γ1,Cβ2γ×β0γ.

Let γHolZp,γ=m,bt. Then the only non-trivial homomorphism β0 is given by β0γ=1 if t is odd, and =1 if t is even. (If β0γ=1 for all γ, then β will not be regular.) As for the possibilities for the homomorphisms β1 and β2, we determined these in the proof of Theorem 2.1, namely:

  1. If βi is an automorphism, then βi1,1=m,1, m0 and βi0,b=n,b for any n.

  2. If βi is not an automorphism, then βi1,1=0,1 and βi0,b=n,d for d=br, r0, and any n, or βi0,b=0,1.

As in the proof of Theorem 2.1, if both β1 and β2 are not automorphisms, or if both β1 and β2 are automorphisms, then β1·β21 is not 1-1, so β is not regular. Thus we can assume one is an automorphism and the other not. Assuming β2 is an automorphism, we can conjugate it by an automorphism of HolZp to transform it to the identity map. Then  β10,bβ20,b1=n,d0,b1 =n,db1: this must lie in ZpZq, which means that d=br must be an odd power of b. Then  β1·β21l,bs=n,dsl,bs1 =nds1d1ldb1s,db1s. For β to be regular, β1·β21 must map onto ZpZq, so we need that db1=br1 generates Zq. Thus r1 must be coprime to q. There are q1 odd numbers r<p with r1 coprime to q.

For any such r, db1s is a unit of Zp, so given any n,h in Zp there is some l in Zp so that nds1d1ldb1s=h. Hence for any suitable r and any n, β1·β21 maps HolZp onto ZpZq.

Thus we have determined all regular β: we have p choices for n, and q1 choices for d. Interchanging the roles of β1 and β2 yields the same result. Thus there are 2pq1 regular embeddings of HolZp into HolZpZq×Z2, as claimed.

(2): Proof that eΓ,Dq×Zp=2p. We seek regular embeddings β:ΓHolDq×Zp. We have HolDq×ZpHolDq×HolZp and by Lemma 3.1, the map HolDqDqInnHolZqHolZq×HolZq is 1-1 and maps m,εCn,d to m,εn,d×n,d. Suppose β1,1=q,εCn,d×m,c. Now β1,1 must have order p (or else β is not 1-1), so n,d=q,ε=0,1, c=1 and m0, hence β1,1=0,1C0,1×m,1 with m0.

Suppose β0,b=l,εCk,d×s,c in DqInnHolZq×HolZp for l,ε,k,d mod q and s,c mod p. If ε=1, then no element of Dq of the form *,1 is hit by π1β, and so β is not regular. Thus ε=1.

Applying β to the condition b,10,b=0,b1,1 yields cbmodp. Hence k,b has order p1. We require that n,1Cl,d have order dividing 2q in DqInnHolZq, which maps 1-1 to HolZq×HolZq as noted above. Thus nl,d and l,d must have order 1, 2 or q in HolZq. But then d=1 or 1.

Thus any regular embedding β satisfies:

  1. β1,1=0,1C0,1×m,1 with m0 in Zp, and

  2. β0,b=n,1Cl,d×k,b with d=±1.

Now modify β by conjugating by 0,1Ch,c×0,brAutDq×ZpDqInnHolZq×HolZp.

First, looking at β1,1:  0,1Ch,c×0,br0,1C0,1×m,10,1Ch,c×0,br1 =0,1C0,1×brm,1.

Now, looking at β0,b:  0,1Ch,c×0,brn,1Cl,d×k,b0,1Ch,c×0,br1 =2h+cn,1Ccl+hdh,d×brk,b.

Let β henceforth denote the conjugated embedding. Choose r so that brm=1. Then β1,1=0,1C0,1×1,1. We choose h,c in β0,b=2h+cn,1Ccl+h1d,d+brk,b. We have four possibilities.

If d=1 and l=0 we can choose c0 and h so that β0,b=0,1C0,1+brk,b. But then β is not regular, for π1β does not map onto Dq.

If d=1 and l=n, then we can choose c=2,h=n, but then β0,b=0,1C0,1+brk,b, so β is not regular.

If d=1 and l0, choose cl1modq and h so that 2h+cn0, then β0,b=0,1C1,1+brk,b.

If d=1 and ln, then we can choose c,h0 with  2h+cn=0 cl+2h=1, giving β0,b=0,1C1,1+brk,b. One verifies that  β1,1=0,1C0,1×1,1 β0,b=0,1C1,±1×brk,b is regular, for any k. There are p choices for k, and hence, up to equivalence, there are 2p regular embeddings of HolZp into HolDq×Zp: eHolZp,Dq×Zp=2p.

(3): Proof that eΓ,Dp×Zq=2p. This argument is similar to the last case. We seek regular embeddings β:HolZpHolDp×ZqHolDp×HolZq. We have HolDp×ZqHolDp×HolZq and by Lemma 3.1, the map HolDpDpInnHolZpHolZp×HolZp is 1-1 and maps m,εCn,d to m,εn,d×n,d.

Suppose β1,1=m,εCn,d×l,c. Then β1,1 must have order p, so n,d and m+εn,εd have order dividing p. This implies d=ε=1. Also l,c has order dividing p in HolZq, so l,c=0,1. Thus β1,1=m,1Cn,1×0,1 with m or n0modp.

Suppose β0,b=l,εCk,d×s,c, of order 2q. Then ε=1 or else β is not regular. Also, s,c must have order dividing 2q in HolZq. But since β is regular, we must have t,* in the image of β for all t in Zq, so c=1 and s0.

Applying β to the relation b,10,b=0,b1,1 yields  bm,1Cbn,1×0,1l,1Ck,d×s,1 =l,1Ck,d×s,1m,1Cn,1×0,1. This yields no condition on s, but on the left components we obtain bm+2bn+l,1Cbn+k,d=ldm,1Cdn+k,d. Thus  2bn+bm=dm bn=dn. If n0, then b=d and n=m. If n=0, then d=b. Thus  β1,1=m,1Cm,1×0,1 β0,b=l,1Ck,b×s,1, or  β1,1=m,1C0,1×0,1 β0,b=l,1Ck,b×s,1. One then sees easily that β0,b has order dividing 2q.

Now we modify β by an automorphism of Dp×Zq.

If we conjugate the right factors by 0,s1 in AutZq, then s,1 is transformed into 1,1.

If we conjugate the left factors by 0,1Cg,c in AutDp, then m,1Cn,1 becomes cm,1Ccn,1 and l,1Ck,±b becomes 2g+cl,1Cg+ckbg,±b. 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 β:  β1,1=1,1C1,1×0,1 β0,b=0,1Ck,b×1,1, and  β1,1=1,1C0,1×0,1 β0,b=0,1Ck,b×1,1. It is a routine check that for any k modulo p, both β are regular Thus we have 2p equivalence classes of regular embeddings of HolZp into HolDp×Zq.

(4): Proof that eΓ,Z2qp=p. Let β:HolZpHolZ2pqHolZp×HolZq×Z2 be a regular embedding. Then β1,1 has order p, so β1,1=m,1×0,1×0 for some m0modp. Also, β0,b has order p1=2q, so β0,b=n,d×l,1×e with d1modp. If e0mod2 or l0modq then β is not regular, hence we have e1mod2 and l0modq.

The condition b,10,b=0,b1,1 implies that dbmodp. Thus in HolZ2pq,  β1,1=2qm,1 β0,b=n,c with p not dividing m, 2q not dividing n, and cbmodp,c1mod2q.

Now we consider β under equivalence by automorphisms of Z2pq. We conjugate β by 0,d with d coprime to 2pq:  0,d2qm,10,d1=2dqm,1 0,dn,c0,d1=dn,c. Choose d so that  dm1modp dn1mod2q. Then after conjugating, β becomes:  β1,1=2q,1 β0,b=1+2ql,c. We check that β is regular for any l modulo p. We have  βn,bk=βn,1β0,bk =2qn,11+2ql1+c++ck1,ck =2qn+1+2ql1+c++ck1,ck. Let a be any element of Z2pq. To solve a2qn+1+2ql1+c++ck1mod2pq for n and k it suffices to solve  a2qn+1+2ql1+c++ck1modp a1+c++ck1kmod2q. Clearly for each a modulo 2pq there are unique values for k,n that solve these congruences. Thus for any l, β is 1-1 and regular, and so eHolZp,Z2pq=p.

(5): Proof that eΓ,Dpq=4p. Let β:HolZpHolDpq be a regular embedding. We have HolDpq=DpqInnHolZpqHolZpq×HolZpq by m,εCn,dm,εn,d×n,d. Also, HolZpqHolZp×HolZq by l,cl,cmodp,l,cmodq. Under this map, Dpq maps into Dp×Dq.

Let β1,1=m,εCn,d, of order p. Then β1,1 maps to m+εd,εd×n,dmodp,m+εd,εd×n,dmodq, which must have order p. Hence  m+εd,εd0,1modq n,d0,1modq and  εdd1modp n or m+εn0modp.

Let β0,b=l,εCk,c. By regularity we must have ε=1, or else π1β maps into {*,1}Dpq. Then β0,b maps to lk,c×k,cmodp,lk,c×k,cmodq. This has order 2q=p1, so we must have c±1modq.

From b,10,b=0,b1,1 we obtain bqm,1Cbqn,1l,1Ck,c=l,1Ck,cqm,1Cqn,1, hence bqm+2bqn+l,Ck+bqn,c=lcqm,1Ck+cqn,c and so  bqn=cqn cqm=bqm+2bqn. This yields no conditions modulo q, but modulo p, we have:

  1. If n0, then c=b and m=n.

  2. If n0, then c=b and m0.

Now we look for a nice representative for β modulo conjugation by elements of InnHolZpq. For g,dHolZpq,  Cg,dβ1,1Cg,d1=g,dqm,1g,d1Cg,dqn,1g,d1 =dqm,1Cdqn,1, Cg,dβ0,bCg,d1=g,dl,1g,d1Cg,dk,cg,d1 =2g+dl,1Cg+dkcg,c. We may choose d and g modulo pq by choosing them modulo p and modulo q separately.

Modulo p:

  1. If n0, choose d so that dqn1, then dqmdqn1 and cb.

  2. If n0, choose d so that dqm1 and cb.

Choose g so that 2g+dl0. Then, since d0, gcg+dk is arbitrary, so (letting β now denote the conjugated embedding) we have, modulo p: β1,1=1,1C1,1,β0,b=0,1Ck,b or β1,1=1,1C0,1,β0,b=0,1Ck,b.

Modulo q:

 β1,1=0,1C0,1 β0,b=2g+dl,1C1cg+dk,c where c±1.

If c1 and k0, set dk1 and 2g+dl0. Then β1,1=0,1C0,1 and β0,b=0,1C1,1 or (if k0) β0,b=0,1C0,1. However, in this last case, β is not regular: π1β maps to {0,±1}HolZq.

If c1, we have β0,b=2g+dl,1C2g+dk,1. If lk we can choose g so that 2g+dk=2g+dl0, but then β is not regular. Thus we must have lk, and then we may choose d0 and g so that  2g+dl0 2g+dk1 and so  β1,1=0,1C0,1 andβ0,b=0,1C1,1.

To summarize, any regular embedding is equivalent to β1,1=1,1C1,1,β0,b=0,1Ck,b or β1,1=1,1C0,1,β0,b=0,1Ck,b modulo p and to β1,1=0,1C0,1,β0,b=0,1C1,1 or β1,1=0,1C0,1,β0,b=0,1C1,1 modulo q.

We show that all four combinations give regular embeddings of HolZp to HolDpq, and so, since k is arbitrary in Zp, we have 4p equivalence classes of regular embeddings.

Note that in every combination, β1l,brβ2l,br1=*,1r both modulo p and modulo q, and so β1·β21 maps to DpqHolZp. To show regularity, we need only show that modulo p every element of Dp is in the image of β1·β21, and similarly modulo q.

Mod p: If β1,1=1,1C1,1,β0,b=0,1Ck,b modulo p, then β1l,brβ2l,br1=k1br1+b+1rl+k1br1b,1r and for r and l arbitrary we can obtain all elements m,±1 of Dp.

If β1,1=1,1C0,1,β0,b=0,1Ck,b modulo p then β1l,brβ2l,br1=l+k1br1b+1rk1br1+b,1r and again for r and l arbitrary we can obtain all elements m,±1 of Dp.

Mod q: If β1,1=0,1C0,1,β0,b=0,1C1,1 modulo q, then β1l,brβ2l,br1=r,1 if r is even, 1+r,1 if r is odd. Since 0r<2q and we're working modulo q, we can obtain all elements m,±1 of Dq.

If β1,1=0,1C0,1,β0,b=0,1C1,1 modulo q, then β1l,brβ2l,br1=r,1 if r is even, 1r,1 if r is odd. Again, since 0r<2q and we're working modulo q, we can obtain all elements m,±1 of Dq.

Thus {β1l,brβ2l,br1}=Dpmodulo pDqmodulo q, and so all four combinations of β modulo p and modulo q are regular. This completes the proof that eHolZp,Dpq=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 Γ=HolZp, p>5 a safeprime, then for every group G of cardinality that of Γ, there is a H-Hopf Galois structure on LK where the associated group of H is G.

b)  The number of Hopf Galois structures on LK 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 p1. His approach is not to determine regular embeddings of Γ into HolG up to equivalence, but rather to determine the set of regular subgroups of HolG whose intersection with AutG is a given cardinality.


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  1. * I wish to thank Union College and Auburn University Montgomery for their hospitality.