1. Regular embeddings

New York Journal of Mathematics

New York J. Math. 9(2003) 99–115.

On Hopf Galois structures and complete groups

Lindsay N. Childs

Abstract. LetLbe a Galois extension ofK, fields, with Galois group Γ. We obtain two results. First, if Γ = Hol(Zp^e), we determine the number of Hopf Galois structures onL/Kwhere the associated group of the Hopf algebraH is Γ (i.e. L⊗KH∼=L[Γ]). Now letpbe a safeprime, that is,pis a prime such thatq= (p−1)/2>2 is also prime. IfL/Kis Galois with group Γ = Hol(Zp), pa safeprime, then for every groupGof cardinalityp(p−1) there is anH-Hopf Galois structure onL/Kwhere the associated group ofH isG, and we count the structures.

Contents

1. Regular embeddings 100

2. Hol(Z_pe) 102

3. Groups of orderp(p−1) 105

4. Nonuniqueness 106

References 115

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 onL. Greither and Pareigis [GP87] showed that for many Galois groups Γ, L is also anH-Hopf Galois extension of K forH a cocommutativeK-Hopf algebra other thanKΓ. The Hopf algebrasH that arise have the property thatL⊗_KH ≡LG, the group ring overL of a groupGof the same cardinality as Γ. We callGthe associated group ofH.

From [By96] we know that for H a cocommutative K-Hopf algebra, H-Hopf Galois structures onL correspond bijectively to equivalence classes of regular em- beddingsβ: Γ→Hol(G)⊂Perm(G). Here Perm(G) is the group of permutations of the setG, and Hol(G) is the normalizer of the left regular representation ofGin Perm(G). One sees easily that Hol(G) contains the image of the right regular rep- resentationρ:G→Perm(G) and also Aut(G); then Hol(G) =ρ(G)·Aut(G) and is isomorphic to the semidirect productGAut(G). The equivalence relation on

Received July 22, 2003.

Mathematics Subject Classification. 12F, 16W.

Key words and phrases. Hopf Galois structure, complete group, holomorph.

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

ISSN 1076-9803/03

99

regular embeddings is by conjugation by elements of Aut(G) inside Hol(G): β ∼β if there existsγ in Aut(G) so that for allg inG,γβ(g)γ⁻¹=β(g).

Thus the numbere(Γ, G) ofH-Hopf Galois structures on L/K where the asso- ciated group ofH isGdepends only on Gand the Galois group Γ, and reduces to a purely group-theoretic problem.

IfLis a Galois extension ofK with Galois group Γ non-abelian simple or∼=S_n, then in [CC99] we counted the number of Hopf Galois structures on L/K with associated groupG= 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 determinee(G, G) whenGis the complete group Hol(Z_pe),pan 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 ifL/Kis Galois with Galois group Γ and isH-Hopf Galois whereH has associated groupG, thenG∼= Γ. Positive results in this direction include Byott’s original uniqueness theorem [By96]; Kohl’s Theorem [Ko98] that if Γ =Z_pe then G∼= Γ; Byott’s recent result [By03a], complementing [CC99], that if Γ is non-abelian simple thenG= Γ; and Featherstonhaugh’s recent result [Fe03] that ifGand Γ are abelianp-groups withpsufficiently large compared to thep-rank ofGand log_pof the exponent ofG, thenG∼= Γ. (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 = Hol(Z_p) 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 cardinalityp(p−1). We show that ifL/K is a Galois extension with Galois group Γ = Hol(Z_p), then for each of the six groups Gof cardinality p(p−1) up to isomorphism, there is anH-Hopf Galois structure onL/K with associated groupG, and we count their number. Thus Γ = Hol(Z_p) yields an example of the opposite extreme to the uniqueness results listed above.

1. Regular embeddings

Given a Galois extensionL/K with Galois group Γ, and a groupGof cardinality that of Γ, the numbere(Γ, G) of Hopf Galois structures onL/Kwhose Hopf algebra H has assocated group Gis equal to the number of equivalence classes of regular embeddings of Γ into Hol(G), the semidirect product of Gand Aut(G). Here an embeddingβ : Γ→Hol(G) is regular if, when viewing Hol(G) inside Perm(G) via (g, α)(x) =α(x)g⁻¹, the orbit of the identity element 1 ofGunderβ(Γ) is all ofG.

Obtaininge(Γ, G) in any particular case involves a number of steps:

(a) determining Aut(G), hence Hol(G);

(b) finding monomorphismsβfrom Γ to Hol(G) by definingβ on generators of Γ and checking β on the relations among those generators: in particular, it is helpful to know the orders of elements of Hol(G) in order to choose where to send the generators of Γ;

(c) checking for regularity ofβ;

(d) simplifying β under the equivalence relation of conjugation by elements of Aut(G) inside Hol(G)—that is, finding a complete set of representatives for the equivalence classes of regular embeddingsβ;

(e) counting the representatives.

Of these tasks, checking regularity is the least natural. If we view Hol(G), the semidirect product ofGand Aut(G), as G×Aut(G) as sets, the operation is

(g, α)·(g, α) = (gα(g), αα)

forg, g ∈G, α, α ∈Aut(G). ThenG∼={(g,1)} is a normal subgroup of Hol(G), and the projection π₂ onto Aut(G) byπ₂(g, α) =αis a homomorphism; however the projectionπ₁ ontoG, π₁(g, α) =g, is not. But since an element (g, α) viewed in Perm(G) acts on the identity element 1 of the setG by (g, α)(1) =α(1)g⁻¹ = g⁻¹, checking regularity of a 1-1 homomorphism β : Γ → Hol(G) is the same as determining whether the function (non-homomorphism)π₁β : Γ→Gis bijective.

Let Inn(G) be the group of inner automorphisms of G, then Inn(G) is normal in Aut(G) and Aut(G)/Inn(G) =O(G), the outer automorphism group, fits in the exact sequence

1→Z(G)→G→Aut(G)→O(G)→1

where the mapC:G→Aut(G) is conjugation: C(g)(x) =gxg⁻¹forg, x∈G, and Z(G) is the center ofG.

Suppose Z(G) = (1) and the composition of the 1-1 homomorphism β : Γ → Hol(G) with the homomorphism Hol(G) → O(G) yields a trivial homomorphism from Γ to O(G). Then β maps into GInn(G), and, following [CC99], we may decomposeβ as follows: we have a homomorphism

j:GInn(G)→G×G

by j(g, C(h)) = (gh, h) forg, h ∈ G, with inverse sending (g, h) to (gh⁻¹, C(h)).

Lettingp_i :G×G→Gbe projection onto the ith factor, i= 1,2, the homomor- phismβ yields homomorphisms β₁=p₁jβ andβ₂=p₂jβ: Γ→Gsuch that

β(γ) = (β₁(γ)β₂(γ)⁻¹, C(β₂)).

Thenβ is regular iff

{π₁β(γ)|γ∈Γ}={β₁(γ)β₂(γ)⁻¹|γ∈Γ}=G.

Thus whenever Z(G) = (1) andβ(Γ) ⊂ GInn(G), we may describe β, and in particular the functionπ₁β : Γ→G, in terms of the homomorphismsβ₁, β₂: Γ→ G, namely,

π₁β(γ) =β₁(γ)β₂(γ)⁻¹.

A class of groups G where Z(G) = (1) andβ(Γ) ⊂GInn(G) is the class of complete groups, that is, finite groupsG with Z(G) = (1) andO(G) = (1). The best-known examples of finite complete groups are Aut(A) where A is simple and non-abelian, S_n for n ≥ 3, n = 6, and Hol(Z_m) 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(G) is solvable, so anyβ:G→Hol(G) has image inGInn(G).

In [CC99] we let Γ =Gand determinede(G, G), the number of regular embed- dings of G to Hol(G), when G is simple non-abelian or S_n, n ≥ 4. In the next

section we examine the case where Γ =G= Hol(Z_q) whereq=p^e andpis an odd prime.

2. Hol( Z

)

LetG=Z_peZ_p^∗e, the holomorph ofZ_pe. Letbbe a number< p^ethat generates Z_p^∗e. Let Γ =G. In this section we prove:

Theorem 2.1. If G= Hol(Z_pe),podd, then up to equivalence there are e(G, G) = 2p^e−1φ(p^e−1) + 2p^eφ(p^e−1)(φ(p−1)−1)

regular embeddings ofGintoHol(G). Thuse(G, G)is the number ofH-Hopf Galois structures on a Galois extension L/K with Galois group G where the associated group of H isG.

Proof. We wish to find equivalence classes of regular embeddings ofGin Hol(G).

Since G is complete, we know from the last section that any homomorphism β : G→Hol(G) may be decomposed as

β(g) =β₁(g)β₂(g)⁻¹C(β₂(g))

for homomorphismsβ₁, β₂:G→G. So we begin by describing the homomorphisms fromGtoG.

Letα:G→Gbe a homomorphism. Thenαis determined by α(1,1) = (m, c) of order dividingp^e, and α(0, b) = (n, d) of order dividingp^e−1(p−1).

If α(1,1) has order dividing p^e, then c = 1. To see this, first note that for any s >0,

(m, c)^s= (m(1 +c+c²+. . . c^s−1), c^s)

socmust have order dividingp^e, which implies thatc≡1 (modp). Also, forα(0, b) to have order dividingp^e−1(p−1) we require thatd ≡1 (modp) or pdivides n.

For ifd ≡1 (modp) one sees by induction that (n, d)^pe−1 = (p^e−1n,1) where p dividesn iffpdividesn. Thus ifpdoes not dividen, then (n, d) has orderp^e. On the other hand, ifd ≡1 (modp), then

(n, d)^p−1=

n

d^p−1−1 d−1

, d^p−1

andpdividesd^p−1−1 but not d−1, so the order of (n, d) dividesp^e−1(p−1).

Now we check the relation

(b,1)(0, b) = (b, b) = (0, b)(1,1).

Applyingαyields:

(m(1 +c+. . . c^b−1), c^b)(n, d) = (n, d)(m, c), hence

(m(1 +c+. . . c^b−1) +c^bn, c^bd) = (n+dm, dc).

Thusc^bd=cd. Sincedis a unit modulo p^e it follows thatc^b =c, hencec^b−1= 1.

But sincec= 1 +pf for somef,

1 = (1 +pf)^b−1.

Sinceb−1 is relatively prime topand (1 +pf)^pe−1 = 1, it follows that 1 +pf = 1, hencef = 0 andc= 1.

Thus for any homomorphismα:G→G, α(1,1) = (m,1) for somem.

Sincec= 1, the requirement

(m(1 +c+. . . c^b−1) +c^bn, c^bd) = (n+dm, dc) becomes

(mb+n, d) = (n+dm, d)

which implies that bm=dm. Thus if m = 0, thenb ≡d (modp), and if m is a unit (i.e. relatively prime top), thenb=d. In the latter case,α(1,1) = (m,1) has orderp^eandα(0, b) = (n, b) has orderp^e−1(p−1).

Clearly if αis an automorphism then m is relatively prime to pand so b =d.

Conversely, if α(1,1) = (m,1) with m relatively prime top, then α(0, b) = (n, b) for somen. Conjugatingαby (h, c) inGyields

(h, c)α(1,1)(h, c)⁻¹= (h, c)(m,1)(−c⁻¹h, c⁻¹)

= (h+cm−h,1)

= (cm,1).

We choosec so thatcm= 1. Then we choose hso that

(0, b) = (h, c)α(0, b)(h, c)⁻¹= (h, c)(n, b)(−c⁻¹h, c⁻¹)

= (h+cn, bc)(−c⁻¹h, c⁻¹)

= (h+cn−bh, b)

= ((1−b)h+cn, b) :

we set h=−(1−b)⁻¹cn, possible because b has order p^e−1(p−1) (mod p^e) and henceb ≡1 (modp). With these choices ofhandc, the homomorphismC(h, c)α: G→Gis the identity on the generators (1,1) and (0, b) ofG, and soα=C(h, c)⁻¹ is an automorphism ofG.

Now we ask about regularity: for which pairs of endomorphisms (β₁, β₂) is {β₁(g)β₂(g)⁻¹|g∈G}=G,

or equivalently, π₁β = β₁·β⁻¹₂ is a 1-1 function from G to G? Let β_i(1,1) = (m_i,1), β_i(0, b) = (n_i, d_i) fori= 1,2.

If neitherβ₁ norβ₂is an automorphism, thenpdividesm₁ andm₂, so β₁(p^e−1l,1)β₂(p^e−1l,1)⁻¹= (p^e−1lm₁,1)(−p^e−1lm₂,1)

= (0,1)(0,1) = (0,1) for alll, and soβ₁·β₂⁻¹ is not 1-1.

Suppose bothβ₁ andβ₂ are automorphisms. Ifm₁≡m₂ (modp) then β₁·β⁻¹₂ (p^e−1,1) = (p^e−1m₁−p^e−1m₂,1)

= (0,1)

=β₁·β⁻¹₂ (0,1)

soβ₁·β₂⁻¹ is not 1-1. Ifm₁ ≡m₂ (modp), then lets(m₁−m₂) =n₁−n₂. Then β₁·β₂⁻¹(0, b) = (n₁−n₂,1)

= ((m₁−m₂)s,1)

=β₁·β₂⁻¹(s,1), so again,β₁·β₂⁻¹ is not 1-1.

Thus forβ to be regular, exactly one ofβ₁andβ₂ is an automorphism.

We return to looking at regularity after we look at equivalence by Aut(G) = Inn(G) inside Hol(G).

Forg, h, k∈Gwe have

(1, C(g))(h, C(k))(1, C(g)⁻¹) = (C(g)(h), C(g)C(k)C(g)⁻¹)

= (ghg⁻¹, C(gkg⁻¹).

Thus ifβ(x) = (β₁(x)β₂(x)⁻¹, C(β₂(x))) forx∈G, thenβ ∼β with β(x) = (gβ₁(x)β₂(x)⁻¹g⁻¹, C(gβ₂(x)g⁻¹))

= (gβ₁(x)g⁻¹·(gβ₂(x)g⁻¹)⁻¹, C(gβ₂(x)g⁻¹)) : we get fromβ to β by simultaneously conjugatingβ₁andβ₂ byg∈G.

Assumeβ₂is an automorphism, then, up to equivalence, we may assume thatβ₂ is the identity automorphism onGandβ₁is not an automorphism. Thene(G, G) will be twice the number of possibleβ₁, since the case whereβ₁is an automorphism andβ₂is not is the same.

Returning to the regularity question, we ask, for whichβ₁ is{β₁(g)g⁻¹}=G?

Assume

β₁(1,1) = (m,1) β₁(0, b) = (n, d)

for somemdivisible by p, somed= 1 and somen. Then β₁(l, b^k) =β₁(l,1)β₁(0, b^k)

= (lm,1)

n

d^k−1 d−1

, d^k

=

lm+n

d^k−1 d−1

, d^k

. For anyh, rwe want to findl, k so that

β₁(l, b^k)(l, b^k)⁻¹=

lm+n

d^k−1 d−1

, d^k

(−b^−kl, b^−k) (∗)

=

lm+n

d^k−1 d−1

−d^kb^−kl, d^kb^−k

= (h, b^r).

In order that for any r, there is a k so that b^r = (db⁻¹)^k, we require that db⁻¹ generatesZ_p^∗e. Now forβ₁ to be a homomorphism, we needbm=dm, and this is possible only ifm= 0 orb≡d (modp). But in the latter case,db⁻¹≡1 (modp), so cannot generateZ_p^∗e. Thus if{β₁(g)g⁻¹}=G, thenm= 0, and d=b^f+1 such thatb^f generatesZ_p^∗e.

Thusβ₁ is defined by

β₁(1,1) = (0,1) β₁(0, b) = (n, d), and (∗) becomes

(n(1 +d+· · ·+d^k−1)−b^{f k}l, b^{f k}) = (h, b^r),

which is solvable for alln and for allf such thatb^f generatesZ_p^∗e . We have two cases:

Ifd=b^f+1 ≡1 (modp), then p−1 divides f+ 1 and pdivides n(or elseβ is not a homomorphism).

Ifd=b^f+1 ≡1 (modp), thennis arbitrary.

The first case givesp^e−1 choices fornandφ(p^e−1) choices forf. The second case givesp^echoices fornand

φ(p^e−1(p−1))−φ(p^e−1) =φ(p^e−1)(φ(p−1)−1)

choices forf. The theorem follows.

Corollary 2.2. If G=Hol(Z_p), thene(G, G) = 2(1 +p(φ(p−1)−1)).

Example 2.3. Forp= 5 there are, up to equivalence, 2(1 + 5(2−1)) = 12 regular embeddings ofG=Z₅Z₅^∗into Hol(G). If we chooseb= 2 thenbandb³generate Z₅^∗, sod=b⁰= 1 or d=b²= 4. Thus if we letβ₂ be the identity automorphism, thenβ₁(1,1) = (0,1) and β₁(0,2) = (n,4), n= 0,1,2,3,4, or (0,1).

3. Groups of order p ( p − 1)

Let Γ = Hol(Z_p) with p an odd prime, as above, and let p−1 = 2q. Then there are at least five non-isomorphic groupsGof order 2qpother than Γ, namely, Z_2qp, D_qp, D_p×Z_q, D_q ×Z_p, and (Z_p Z_q)×Z₂ (where Z_q is identified as the subgroup of Aut(Z_p) of index 2, andD_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 Γ = Hol(Z_p), 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 pbe prime,p= 2q+ 1. Then

Aut(D_p) = Aut(Z_pZ_q) = Aut(Hol(Z_p)) = Inn(Hol(Z_p)).

Proof. LetG=Z_pZ_a witha= 2, qor 2q=p−1. WriteZ_padditively and view Z_a as the cyclic subgroup of orderainsideZ_p^∗= Aut(Z_p) andG⊂Hol(Z_p). Let b generateZ_a. Ifαis an automorphism ofG, then:

1. α(1,1) = (m,1) forpnot dividingm, and

2. α(0, b) = (n, b) for anyn(as one sees by applyingαto the relation (b,1)(0, b) = (b, b) = (0, b)(1,1).)

So there arep(p−1) =|Hol(Z_p)|automorphisms ofG. Now if (l, d) is any element of Hol(Z_p), then conjugation by (l, d) is an automorphism ofG, since

(l, d)(m, b^r)(l, d)⁻¹= (l+dm−b^rl, b^r)∈G.

Hence the conjugation mapC: Hol(Z_p)→Aut(G),

C(l, d)(m, b^r) = (l, d)(m, b^r)(l, d)⁻¹= (l+dm−b^rl, b^r), is defined. ThenC is 1-1. For ifC(l₁, d₁) =C(l₂, d₂) onG, then

l₁+d₁m−b^rl₁=l₂+d₂m−b^rl₂

for allm, r: in particular form= 1, r= 0 we getd₁=d₂ and form= 0, r= 1 we

getl₁=l₂. ThusC is an isomorphism.

Proposition 3.2. If q andp= 2q+ 1 are odd primes, then, up to isomorphism, there are exactly six groups of orderp(p−1).

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

Proof. IfG has orderp(p−1) then by the first Sylow Theorem, Ghas 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_pK =J is a subgroup of Gof orderpq, 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 productJαZ₂.

IfJ ∼=Z_pqthen G∼=Z_pqαZ₂ where

α:Z₂→Aut(Z_pq)∼=Z_p−1×Z_q−1

has four possibilities,α(−1) = (±1,±1). These yieldG∼=D_pq, D_p×Z_q, D_q ×Z_p andZ_2pq.

IfJ ∼=Z_pZ_q ⊂Hol(Z_p), then G∼=J αZ₂where α:Z₂ →Aut(Z_pZ_q)∼= Inn(Hol(Z_p)). Ifαis trivial, we obtain (Z_pZ_q)×Z₂. Otherwise, the elements of order 2 in Inn(Hol(Z_p)₎ are of the formC(l,−1) for any l∈Z_p. Define

α_l:Z₂→Inn(Hol(Z_p)) byα_l(−1) =C(l,−1). One checks easily that

(Z_pZ_q)_α0Z₂∼=Z_pZ_2q = Hol(Z_p)

by the map sending ((a, b), ) to (a, b) fora∈Z_p,b∈Z_q ⊂Z_p−1,∈Z₂⊂Z_p−1. Now α₀ = C(0,−1) and α_l = C(l,−1) are conjugate in Inn(Hol(Z_p)) for l= 0 by C(m,1) where 2m≡l (mod p): C(m,1)C(0,−1)C(−m,1) =C(2m,−1).

It follows from [DF99], p. 186, Exercise 6 that

(Z_pZ_q)αlZ₂∼= (Z_pZ_q)α0Z₂

for alll. ThusJ =Z_pZ_q yields only two possible groups, up to isomorphism.

4. Nonuniqueness

If we begin with the Galois group Γ of a Galois extensionL/K of fields, then to determine theK-Hopf Galois structures onL, we need to count equivalence classes of regular embeddings of Γ into Hol(G), where Gvaries 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 = Hol(Z_p),p= 2q+ 1 a safeprime>5, and determine e(Γ, G), the number of regular embeddings of Γ into Hol(G), for all six groupsG of order p(p−1) = 2pq. We did the caseG = Γ in Theorem 2.1 and found that e(G, G) = 2 + 2p(q−2). Here is the result forG ∼= Γ:

Theorem 4.1. Let Γ = Hol(Z_p), withpandq= (p−1)/2odd primes. Then:

(1) e(Γ, Z_pZ_q×Z₂) = 2p(q−1);

(2) e(Γ, D_q×Z_p) = 2p;

(3) e(Γ, D_p×Z_q) = 2p;

(4) e(Γ, Z_2qp) =p;

(5) e(Γ, D_pq) = 4p.

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

(1): Proof thate(Γ, Z_pZ_q×Z₂) = 2p(q−1). Using Lemma 3.1 we have Hol(Z_pZ_q×Z₂) = Hol(Z_pZ_q)×Hol(Z₂)

= ((Z_pZ_q)Inn(Hol(Z_p))×Z₂

⊂(Hol(Z_p)Inn(Hol(Z_p))×Z₂

∼= Hol(Z_p)×Hol(Z_p)×Z₂;

thus we may unwind any homomorphismβ : Γ→((Z_pZ_q)Inn(Hol(Z_p))×Z₂ as in Section 1. If

β(γ) = ((m, b^2r), C(l, b^s))×

with =±1, Z_p^∗ =b andm, l are modulo p, thenβ(γ) maps to (m, b^2r)(l, b^s)× (l, b^s)×under the map to Hol(Z_p)×Hol(Z_p)×Z₂. Soβ induces homomorphisms β₁: Γ→Hol(Z_p), defined byβ₁(γ) = (m, b^2r)(l, b^s), andβ₂: Γ→Hol(Z_p), defined byβ₂(γ) = (l, b^s). If we defineβ₀: Γ→Z₂ byβ₀(γ) =, then

β(γ) = (β₁(γ)β₂(γ)⁻¹, C(β₂(γ))×β₀(γ).

Let γ ∈ Hol(Z_p), γ = (m, b^t). Then the only non-trivial homomorphism β₀ is given byβ₀(γ) =−1 iftis odd, and = 1 iftis even. (If β₀(γ) = 1 for allγ, thenβ will not be regular.) As for the possibilities for the homomorphismsβ₁ andβ₂, we determined these in the proof of Theorem 2.1, namely:

1. Ifβ_i is an automorphism, thenβ_i(1,1) = (m,1),m= 0 and β_i(0, b) = (n, b) for anyn.

2. If β_i is not an automorphism, then β_i(1,1) = (0,1) and β_i(0, b) = (n, d) for d=b^r,r= 0, and anyn, orβ_i(0, b) = (0,1).

As in the proof of Theorem 2.1, if bothβ₁ andβ₂ are not automorphisms, or if both β₁ and β₂ are automorphisms, then β₁·β₂⁻¹ is not 1-1, so β is not regular.

Thus we can assume one is an automorphism and the other not. Assumingβ₂is an automorphism, we can conjugate it by an automorphism of Hol(Z_p) to transform it to the identity map. Then

β₁(0, b)β₂(0, b)⁻¹= (n, d)(0, b)⁻¹

= (n, db⁻¹) :

this must lie inZ_pZ_q, which means thatd=b^rmust be an odd power ofb. Then β₁·β₂⁻¹(l, b^s) = (n, d)^s(l, b^s)⁻¹

=

n

d^s−1 d−1

−l(db⁻¹)^s,(db⁻¹)^s

.

Forβ to be regular,β₁·β₂⁻¹ must map ontoZ_pZ_q, so we need thatdb⁻¹=b^r−1 generatesZ_q. Thusr−1 must be coprime toq. There areq−1 odd numbersr < p withr−1 coprime to q.

For any suchr, (db⁻¹)^sis a unit ofZ_p, so given anyn, hinZ_p there is somel in Z_p so thatn

d^s−1 d−1

−l(db⁻¹)^s=h. Hence for any suitable rand anyn,β₁·β₂⁻¹ maps Hol(Z_p) ontoZ_pZ_q.

Thus we have determined all regularβ: we havepchoices forn, andq−1 choices ford. Interchanging the roles ofβ₁ andβ₂ yields the same result. Thus there are 2p(q−1) regular embeddings of Hol(Z_p) into Hol(Z_pZ_q×Z₂), as claimed.

(2): Proof thate(Γ, D_q×Z_p) = 2p. We seek regular embeddings β : Γ→Hol(D_q×Z_p).

We have

Hol(D_q×Z_p)∼= Hol(D_q)×Hol(Z_p) and by Lemma 3.1, the map

Hol(D_q)∼=D_qInn(Hol(Z_q))→Hol(Z_q)×Hol(Z_q) is 1-1 and maps (m, )C(n, d) to (m, )(n, d)×(n, d). Suppose

β(1,1) = (q, )C(n, d)×(m, c).

Nowβ(1,1) must have orderp(or elseβis not 1-1), so (n, d) = (q, ) = (0,1),c= 1 andm= 0, hence

β(1,1) = (0,1)C(0,1)×(m,1) withm= 0.

Supposeβ(0, b) = (l, )C(k, d)×(s, c) inD_qInn(Hol(Z_q))×Hol(Z_p) forl, , k, d modqands, cmodp. If= 1, then no element ofD_q of the form (∗,−1) is hit by π₁(β), and so β is not regular. Thus=−1.

Applyingβto the condition (b,1)(0, b) = (0, b)(1,1) yieldsc≡b (mod p). Hence (k, b) has order p−1. We require that (n,−1)C(l, d) have order dividing 2q in D_q Inn(Hol(Z_q), which maps 1-1 to Hol(Z_q)×Hol(Z_q) as noted above. Thus (n−l,−d) and (l, d) must have order 1, 2 orq in Hol(Z_q). But then d= 1 or−1.

Thus any regular embeddingβ satisfies:

1. β(1,1) = (0,1)C(0,1)×(m,1) with m= 0 inZ_p, and 2. β(0, b) = (n,−1)C(l, d)×(k, b) withd=±1.

Now modifyβ by conjugating by

(0,1)C(h, c)×(0, b^r)∈Aut(D_q×Z_p)⊂D_qInn(Hol(Z_q))×Hol(Z_p).

First, looking atβ(1,1):

((0,1)C(h, c)×(0, b^r)))((0,1)C(0,1)×(m,1))((0,1)C(h, c)×(0, b^r))⁻¹)

= ((0,1)C(0,1)×(b^rm,1).

Now, looking atβ(0, b):

((0,1)C(h, c)×(0, b^r)))((n,−1)C(l, d)×(k, b))((0,1)C(h, c)×(0, b^r))⁻¹)

= (2h+cn,−1)C(cl+h−dh, d)×(b^rk, b).

Letβ henceforth denote the conjugated embedding. Chooserso thatb^rm= 1.

Then

β(1,1) = (0,1)C(0,1)×(1,1).

We chooseh, cin

β(0, b) = (2h+cn,−1)C(cl+h(1−d), d) + (b^rk, b).

We have four possibilities.

Ifd= 1 andl= 0 we can choosec= 0 andhso that β(0, b) = (0,−1)C(0,1) + (b^rk, b).

But thenβ is not regular, forπ₁β does not map ontoD_q.

Ifd=−1 andl=n, then we can choosec= 2, h=−n, but then β(0, b) = (0,−1)C(0,−1) + (b^rk, b),

soβ is not regular.

Ifd= 1 andl= 0, choosec≡l⁻¹ (modq) andhso that 2h+cn≡0, then β(0, b) = (0,−1)C(1,1) + (b^rk, b).

Ifd=−1 andl=n, then we can choosec, h= 0 with 2h+cn= 0

cl+ 2h= 1, giving

β(0, b) = (0,−1)C(1,−1) + (b^rk, b).

One verifies that

β(1,1) = ((0,1)C(0,1)×(1,1) β(0, b) = ((0,−1)C(1,±1)×(b^rk, b)

is regular, for anyk. There arepchoices fork, and hence, up to equivalence, there are 2pregular embeddings of Hol(Z_p) into Hol(D_q×Z_p): e(Hol(Z_p), D_q×Z_p) = 2p.

(3): Proof that e(Γ, D_p×Z_q) = 2p. This argument is similar to the last case.

We seek regular embeddings

β: Hol(Z_p)→Hol(D_p×Z_q)∼= Hol(D_p)×Hol(Z_q).

We have

Hol(D_p×Z_q)∼= Hol(D_p)×Hol(Z_q) and by Lemma 3.1, the map

Hol(D_p)∼=D_pInn(Hol(Z_p)→Hol(Z_p)×Hol(Z_p) is 1-1 and maps (m, )C(n, d) to (m, )(n, d)×(n, d).

Supposeβ(1,1) = (m, )C(n, d)×(l, c). Thenβ(1,1) must have orderp, so (n, d) and (m+n, d) have order dividingp. This impliesd== 1. Also (l, c) has order dividingpin Hol(Z_q), so (l, c) = (0,1). Thus

β(1,1) = (m,1)C(n,1)×(0,1) withmorn ≡0 (modp).

Supposeβ(0, b) = (l, )C(k, d)×(s, c), of order 2q. Then=−1 or elseβ is not regular. Also, (s, c) must have order dividing 2qin Hol(Z_q). But sinceβ is regular, we must have (t,∗) in the image ofβ for alltin Z_q, so c= 1 ands= 0.

Applyingβ to the relation (b,1)(0, b) = (0, b)(1,1) yields ((bm,1)C(bn,1)×(0,1))((l,−1)C(k, d)×(s,1))

= ((l,−1)C(k, d)×(s,1))((m,1)C(n,1)×(0,1)).

This yields no condition ons, but on the left components we obtain (bm+ 2bn+l,−1)C(bn+k, d) = (l−dm,−1)C(dn+k, d).

Thus

2bn+bm=−dm bn=dn.

Ifn= 0, thenb=dandn=−m. Ifn= 0, then d=−b. Thus β(1,1) = (m,1)C(−m,1)×(0,1) β(0, b) = (l,−1)C(k, b)×(s,1), or

β(1,1) = (m,1)C(0,1)×(0,1) β(0, b) = (l,−1)C(k,−b)×(s,1).

One then sees easily thatβ(0, b) has order dividing 2q.

Now we modifyβ by an automorphism ofD_p×Z_q.

If we conjugate the right factors by (0, s⁻¹) in Aut(Z_q), then (s,1) is transformed into (1,1).

If we conjugate the left factors by (0,1)C(g, c) in Aut(D_p), then (m,1)C(n,1) becomes (cm,1)C(cn,1) and

(l,−1)C(k,±b) becomes (2g+cl,−1)C(g+ck∓bg,±b).

Choosecso thatcm= 1. Thencn= 0 or−1.

Choose g so that 2g+cl = 0. Then we have the following representatives of equivalence classes ofβ:

β(1,1) = (1,1)C(−1,1)×(0,1) β(0, b) = (0,−1)C(k, b)×(1,1), and

β(1,1) = (1,1)C(0,1)×(0,1) β(0, b) = (0,−1)C(k,−b)×(1,1).

It is a routine check that for anyk modulop, bothβ are regular Thus we have 2p equivalence classes of regular embeddings of Hol(Z_p) into Hol(D_p×Z_q).

(4): Proof thate(Γ, Z_2qp) =p. Let

β: Hol(Z_p)→Hol(Z_2pq)∼= Hol(Z_p)×Hol(Z_q)×Z₂ be a regular embedding. Thenβ(1,1) has orderp, so

β(1,1) = (m,1)×(0,1)×0

for somem ≡0 (modp). Also, β(0, b) has orderp−1 = 2q, so β(0, b) = (n, d)×(l,1)×e

with d ≡ 1 (modp). If e ≡ 0 (mod 2) or l ≡ 0 (modq) then β is not regular, hence we havee≡1 (mod 2) andl ≡0 (modq).

The condition (b,1)(0, b) = (0, b)(1,1) implies that d ≡ b (mod p). Thus in Hol(Z_2pq),

β(1,1) = (2qm,1) β(0, b) = (n, c)

withpnot dividingm, 2qnot dividingn, andc≡b (modp), c≡1 (mod 2q).

Now we considerβ under equivalence by automorphisms ofZ_2pq. We conjugate β by (0, d) withdcoprime to 2pq:

(0, d)(2qm,1)(0, d⁻¹) = (2dqm,1) (0, d)(n, c)(0, d⁻¹) = (dn, c).

Choosedso that

dm≡1 (modp) dn≡1 (mod 2q).

Then after conjugating,β becomes:

β(1,1) = (2q,1) β(0, b) = (1 + 2ql, c).

We check thatβ is regular for anyl modulop. We have β(n, b^k) =β(n,1)β(0, b)^k

= (2qn,1)((1 + 2ql)(1 +c+· · ·+c^k−1), c^k)

= (2qn+ (1 + 2ql)(1 +c+· · ·+c^k−1), c^k).

Letabe any element ofZ_2pq. To solve

a≡2qn+ (1 + 2ql)(1 +c+· · ·+c^k−1) (mod 2pq) fornandkit suffices to solve

a≡2qn+ (1 + 2ql)(1 +c+· · ·+c^k−1) (modp) a≡1 +c+· · ·+c^k−1≡k (mod 2q).

Clearly for each a modulo 2pq there are unique values for k, n that solve these congruences. Thus for anyl,β is 1-1 and regular, and soe(Hol(Z_p), Z_2pq) =p.

(5): Proof that e(Γ, D_pq) = 4p. Let β : Hol(Z_p) → Hol(D_pq) be a regular embedding. We have

Hol(D_pq) =D_pqInn(Hol(Z_pq))→Hol(Z_pq)×Hol(Z_pq) by (m, )C(n, d)→(m, )(n, d)×(n, d). Also,

Hol(Z_pq)∼= Hol(Z_p)×Hol(Z_q)

by (l, c)→((l, c)mod(p),(l, c)mod(q)). Under this map, D_pq maps intoD_p×D_q. Letβ(1,1) = (m, )C(n, d), of order p. Thenβ(1,1) maps to

(m+d, d)×(n, d) modp,(m+d, d)×(n, d) modq, which must have orderp. Hence

(m+d, d)≡(0,1) (mod q) (n, d)≡(0,1) (mod q)

and

d≡d≡1 (modp) norm+n ≡0 (modp).

Let β(0, b) = (l, )C(k, c). By regularity we must have = −1, or else π₁(β) maps into{(∗,1)} ⊂D_pq. Thenβ(0, b) maps to

(l−k,−c)×(k, c) modp,(l−k,−c)×(k, c) modq.

This has order 2q=p−1, so we must have c≡ ±1 (modq).

From (b,1)(0, b) = (0, b)(1,1) we obtain

(bqm,1)C(bqn,1)(l,−1)C(k, c) = (l,−1)C(k, c)(qm,1)C(qn,1), hence

(bqm+ 2bqn+l, C(k+bqn, c) = (l−cqm,−1)C(k+cqn, c) and so

bqn=cqn

cqm=bqm+ 2bqn.

This yields no conditions moduloq, but modulop, we have:

1. Ifn ≡0, thenc=bandm=−n.

2. Ifn≡0, thenc=−bandm ≡0.

Now we look for a nice representative forβ modulo conjugation by elements of Inn(Hol(Z_pq)). For (g, d)∈Hol(Z_pq),

C(g, d)β(1,1)C(g, d)⁻¹= (g, d)(qm,1)(g, d)⁻¹C((g, d)(qn,1)(g, d)⁻¹)

= (dqm,1)C(dqn,1),

C(g, d)β(0, b)C(g, d)⁻¹= (g, d)(l,−1)(g, d)⁻¹C((g, d)(k, c)(g, d)⁻¹)

= (2g+dl,−1)C(g+dk−cg, c).

We may choose d and g modulo pq by choosing them modulo p and modulo q separately.

Modulop:

1. Ifn ≡0, choosedso thatdqn≡ −1, thendqm≡ −dqn≡1 andc≡b.

2. Ifn≡0, choosedso thatdqm≡1 andc≡ −b.

Choose g so that 2g+dl ≡0. Then, since d ≡ 0,g−cg+dk is arbitrary, so (lettingβ now denote the conjugated embedding) we have, modulop:

β(1,1) = (1,1)C(−1,1), β(0, b) = (0,−1)C(k, b) or

β(1,1) = (1,1)C(0,1), β(0, b) = (0,−1)C(k,−b).

Moduloq:

β(1,1) = (0,1)C(0,1)

β(0, b) = (2g+dl,−1)C((1−c)g+dk, c) wherec≡ ±1.

Ifc≡1 andk ≡0, setdk≡1 and

2g+dl≡0.

Then

β(1,1) = (0,1)C(0,1) and

β(0, b) = (0,−1)C(1,1) or (ifk≡0)

β(0, b) = (0,−1)C(0,1).

However, in this last case,β is not regular: π₁β maps to{(0,±1)} ⊂Hol(Z_q).

Ifc≡ −1, we have

β(0, b) = (2g+dl,−1)C((2g+dk,−1).

Ifl≡kwe can choose gso that 2g+dk= 2g+dl≡0, but thenβ is not regular.

Thus we must havel ≡k, and then we may choosed= 0 and gso that 2g+dl≡0

2g+dk≡1 and so

β(1,1) = (0,1)C(0,1) and

β(0, b) = (0,−1)C(−1,1).

To summarize, any regular embedding is equivalent to

β(1,1) = (1,1)C(−1,1), β(0, b) = (0,−1)C(k, b) or

β(1,1) = (1,1)C(0,1), β(0, b) = (0,−1)C(k,−b) modulopand to

β(1,1) = (0,1)C(0,1), β(0, b) = (0,−1)C(1,1) or

β(1,1) = (0,1)C(0,1), β(0, b) = (0,−1)C(−1,1) moduloq.

We show that all four combinations give regular embeddings of Hol(Z_p) to Hol(D_pq), and so, since k is arbitrary in Z_p, we have 4p equivalence classes of regular embeddings.

Note that in every combination,

β₁(l, b^r)β₂(l, b^r)⁻¹= (∗,(−1)^r)

both modulop and modulo q, and so β₁·β₂⁻¹ maps to D_pq ⊂ Hol(Z_p). To show regularity, we need only show that modulopevery element of D_p is in the image ofβ₁·β₂⁻¹, and similarly moduloq.

Modp: If

β(1,1) = (1,1)C(−1,1), β(0, b) = (0,−1)C(k, b) modulop, then

β₁(l, b^r)β₂(l, b^r)⁻¹=

−k

1−(−b)^r 1 +b

+ (−1)^r

−l+k

1−b^r 1−b

,(−1)^r

and forrandl arbitrary we can obtain all elements (m,±1) of D_p. If

β(1,1) = (1,1)C(0,1), β(0, b) = (0,−1)C(k,−b) modulopthen

β₁(l, b^r)β₂(l, b^r)⁻¹=

l+ (−k)1−b^r 1−b

+ (−1)^rk

1−(−b)^r 1 +b

,(−1)^r

and again forrandl arbitrary we can obtain all elements (m,±1) of D_p. Modq: If

β(1,1) = (0,1)C(0,1), β(0, b) = (0,−1)C(1,1) moduloq, then

β₁(l, b^r)β₂(l, b^r)⁻¹=

(−r,1) ifris even, (−1 +r,−1) ifris odd.

Since 0≤r <2qand we’re working modulo q, we can obtain all elements (m,±1) ofD_q.

If

β(1,1) = (0,1)C(0,1), β(0, b) = (0,−1)C(−1,1) moduloq, then

β₁(l, b^r)β₂(l, b^r)⁻¹=

(−r,1) ifris even, (1−r,−1) ifris odd.

Again, since 0 ≤r < 2q and we’re working moduloq, we can obtain all elements (m,±1) ofD_q.

Thus

{β₁(l, b^r)β₂(l, b^r)⁻¹}=

D_p modulop D_q moduloq,

and so all four combinations ofβ modulopand modulo qare regular. This com-

pletes the proof thate(Hol(Z_p), D_pq) = 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 = Hol(Z_p),p >5a safeprime, then for every group Gof cardinality that of Γ, there is a H-Hopf Galois structure onL/K where the associated group of H isG.

b) The number of Hopf Galois structures onL/K is2 + 3p+ 4pq.

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

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Department of Mathematics and Statistics, University at Albany, Albany, NY 12222 [email protected] http://math.albany.edu:8000/˜lc802/

This paper is available via http://nyjm.albany.edu:8000/j/2003/9-8.html.