New York Journal of Mathematics

New York J. Math. 9 (2003) 99–115.

On Hopf Galois structures and complete groups

Lindsay N. Childs

Abstract. Let L be a Galois extension of K, fields, with Galois group Γ. We obtain two results. First, if Γ = Hol(Z_{p^{e}}), we determine the number of Hopf Galois structures on L/K where the associated group of the Hopf algebra H is Γ (i.e. L ⊗_{K} H ≅ L[Γ]). Now let p be a safeprime, that is, p is a prime such that q = (p−1)/2 > 2 is also prime. If L/K is Galois with group Γ = Hol(Z_{p}), p a safeprime, then for every group G of cardinality p(p−1) there is an H-Hopf Galois structure on L/K where the associated group of H is G, and we count the structures.

Contents

1. Regular embeddings 100 2. Hol(Z_{p^{e}}) 102 3. Groups of order p(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 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 L ⊗_{K} H ≡ LG, 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 : Γ → Hol(G) ⊂ Perm(G). Here Perm(G) is the group of permutations of the set G, and Hol(G) is the normalizer of the left regular representation of G in Perm(G). One sees easily that Hol(G) contains the image of the right regular representation ρ : G → Perm(G) and also Aut(G); then Hol(G) = ρ(G) · Aut(G) and is isomorphic to the semidirect product G ⋊ Aut(G). The equivalence relation on

99^{ISSN 1076-9803/03}

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.

regular embeddings is by conjugation by elements of Aut(G) inside Hol(G): β ∼ β′ if there exists γ in Aut(G) so that for all g in G, γβ(g)γ^{−1} = β′(g).

Thus the number e(Γ, G) of H-Hopf Galois structures on L/K 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 ≅ S_{n}, then in [CC99] we counted the number of Hopf Galois structures on L/K 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 e(G, G) when G is the complete group Hol(Z_{p^{e}}), 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 L/K is Galois with Galois group Γ and is H-Hopf Galois where H has associated group G, then G ≅ Γ. Positive results in this direction include Byott’s original uniqueness theorem [By96]; Kohl’s Theorem [Ko98] that if Γ = Z_{p^{e}} then G ≅ Γ; 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 log_{p} of the exponent of G, then G ≅ 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 cardinality p(p − 1). We show that if L/K is a Galois extension with Galois group Γ = Hol(Z_{p}), then for each of the six groups G of cardinality p(p − 1) up to isomorphism, there is an H-Hopf Galois structure on L/K with associated group G, 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 extension L/K with Galois group Γ, and a group G of cardinality that of Γ, the number e(Γ, G) 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 Γ into Hol(G), the semidirect product of G and Aut(G). Here an embedding β : 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 of G under β(Γ) is all of G.

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

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

(b) finding monomorphisms β from Γ to Hol(G) by defining β 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 of β; (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 embeddings β;

(e) counting the representatives.

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

(g, α) · (g′, α′) = (gα(g′), αα′)

for g, g′ ∈ G, α, α′ ∈ Aut(G). Then G ≅ {(g, 1)} is a normal subgroup of Hol(G), and the projection π_{2} onto Aut(G) by π_{2}(g, α) = α is a homomorphism; however the projection π_{1} onto G, π_{1}(g, α) = g, is not. But since an element (g, α) viewed in Perm(G) acts on the identity element 1 of the set G 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) π_{1}β : Γ → G is 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 map C : G → Aut(G) is conjugation: C(g)(x) = gxg^{−1} for g, x ∈ G, and Z(G) is the center of G.

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 G ⋊ Inn(G), and, following [CC99], we may decompose β as follows: we have a homomorphism

j : G ⋊ Inn(G) → G × G

by j(g, C(h)) = (gh, h) for g, h ∈ G, with inverse sending (g, h) to (gh^{−1}, C(h)). Letting p_{i} : G × G → G be projection onto the ith factor, i = 1, 2, the homomorphism β yields homomorphisms β_{1} = p_{1}jβ and β_{2} = p_{2}jβ : Γ → G such that

β(γ) = (β_{1}(γ)β_{2}(γ)^{−1}, C(β_{2})).

Then β is regular iff

{π_{1}β(γ)|γ ∈ Γ} = {β_{1}(γ)β_{2}(γ)^{−1}|γ ∈ Γ} = G.

Thus whenever Z(G) = (1) and β(Γ) ⊂ G ⋊ Inn(G), we may describe β, and in particular the function π_{1}β : Γ → G, in terms of the homomorphisms β_{1}, β_{2} : Γ → G, namely,

π_{1}β(γ) = β_{1}(γ)β_{2}(γ)^{−1}.

A class of groups G where Z(G) = (1) and β(Γ) ⊂ G ⋊ Inn(G) is the class of complete groups, that is, finite groups G with Z(G) = (1) and O(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 any β : G → Hol(G) has image in G ⋊ Inn(G).

In [CC99] we let Γ = G and determined e(G, G), the number of regular embeddings 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}) where q = p^{e} and p is an odd prime.

2. Hol(Z_{p^{e}})

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 Γ = G. In this section we prove:

Theorem 2.1. If G = Hol(Z_{p^{e}}), p odd, then up to equivalence there are

e(G, G) = 2p^{e−1}ϕ(p^{e−1}) + 2p^{e}ϕ(p^{e−1})(ϕ(p − 1) − 1)

regular embeddings of G into Hol(G). Thus e(G, G) 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 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) = β_{1}(g)β_{2}(g)^{−1}C(β_{2}(g))

for homomorphisms β_{1}, β_{2} : G → G. So we begin by describing the homomorphisms from G to G.

Let α : G → G be a homomorphism. Then α is determined by

α(1, 1) = (m, c) of order dividing p^{e}, and
α(0, b) = (n, d) of order dividing p^{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 + c^{2} + … c^{s−1}), c^{s})

so c must have order dividing p^{e}, which implies that c ≡ 1 (mod p). Also, for α(0, b) to have order dividing p^{e−1}(p − 1) we require that d ≢ 1 (mod p) or p divides n. For if d ≡ 1 (mod p) one sees by induction that (n, d)^{p^{e−1}} = (p^{e−1}n′, 1) where p divides n′ iff p divides n. Thus if p does not divide n, then (n, d) has order p^{e}. On the other hand, if d ≢ 1 (mod p), then

(n, d)^{p−1} = (n (\frac{d^{p−1} − 1}{d − 1}) , d^{p−1})

and p divides d^{p−1} − 1 but not d − 1, so the order of (n, d) divides p^{e−1}(p − 1). Now we check the relation

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

Applying α yields:

(m(1 + c + … c^{b−1}), c^{b})(n, d) = (n, d)(m, c),

hence

(m(1 + c + … c^{b−1}) + c^{b}n, c^{b}d) = (n + dm, dc).

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 = (1 + pf)^{b−1}.

Since b − 1 is relatively prime to p and (1 + pf)^{p^{e−1}} = 1, it follows that 1 + pf = 1, hence f = 0 and c = 1.

Thus for any homomorphism α : G → G,

α(1, 1) = (m, 1)

for some m.

Since c = 1, the requirement

(m(1 + c + … c^{b−1}) + c^{b}n, c^{b}d) = (n + dm, dc)

becomes

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

which implies that bm = dm. Thus if m ≠ 0, then b ≡ d (mod p), and if m is a unit (i.e. relatively prime to p), then b = d. In the latter case, α(1, 1) = (m, 1) has order p^{e} and α(0, b) = (n, b) has order p^{e−1}(p − 1).

Clearly if α is an automorphism then m is relatively prime to p and so b = d. Conversely, if α(1, 1) = (m, 1) with m relatively prime to p, then α(0, b) = (n, b) for some n. Conjugating α by (h, c) in G yields

(h, c)α(1, 1)(h, c)^{−1} = (h, c)(m, 1)(−c^{−1}h, c^{−1})
= (h + cm − h, 1)
= (cm, 1).

We choose c so that cm = 1. Then we choose h so that

(0, b) = (h, c)α(0, b)(h, c)^{−1} = (h, c)(n, b)(−c^{−1}h, c^{−1})
= (h + cn, bc)(−c^{−1}h, c^{−1})
= (h + cn − bh, b)
= ((1 − b)h + cn, b) :

we set h = −(1 − b)^{−1}cn, possible because b has order p^{e−1}(p − 1) (mod p^{e}) and hence b ≢ 1 (mod p). With these choices of h and c, the homomorphism C(h, c)α : G → G is the identity on the generators (1, 1) and (0, b) of G, and so α = C(h, c)^{−1} is an automorphism of G.

Now we ask about regularity: for which pairs of endomorphisms (β_{1}, β_{2}) is

{β_{1}(g)β_{2}(g)^{−1}|g ∈ G} = G,

or equivalently, π_{1}β = β_{1} · β_{2}^{−1} 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}) for i = 1, 2.

If neither β_{1} nor β_{2} is an automorphism, then p divides m_{1} and m_{2}, so

\begin{aligned}β_{1}(p^{e−1}l, 1)β_{2}(p^{e−1}l, 1)^{−1} & = (p^{e−1}lm_{1}, 1)(−p^{e−1}lm_{2}, 1) \\ & = (0, 1)(0, 1) = (0, 1) \end{aligned}

for all l, and so β_{1} · β_{2}^{−1} is not 1-1.

Suppose both β_{1} and β_{2} are automorphisms. If m_{1} ≡ m_{2} (mod p) then

β_{1} · β_{2}^{−1}(p^{e−1}, 1) = (p^{e−1}m_{1} − p^{e−1}m_{2}, 1)
= (0, 1)
= β_{1} · β_{2}^{−1}(0, 1)

so β_{1} · β_{2}^{−1} is not 1-1. If m_{1} ≢ m_{2} (mod p), then let s(m_{1} − m_{2}) = n_{1} − n_{2}. Then

β_{1} · β_{2}^{−1}(0, b) = (n_{1} − n_{2}, 1)
= ((m_{1} − m_{2})s, 1)
= β_{1} · β_{2}^{−1}(s, 1),

so again, β_{1} · β_{2}^{−1} is not 1-1.

Thus for β to be regular, exactly one of β_{1} and β_{2} is an automorphism.

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

For g, h, k ∈ G we have

\begin{aligned}(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}). \end{aligned}

Thus if β(x) = (β_{1}(x)β_{2}(x)^{−1}, C(β_{2}(x))) for x ∈ G, then β ∼ β′ with

\begin{aligned}β′(x) & = (gβ_{1}(x)β_{2}(x)^{−1}g^{−1}, C(gβ_{2}(x)g^{−1})) \\ & = (gβ_{1}(x)g^{−1} · (gβ_{2}(x)g^{−1})^{−1}, C(gβ_{2}(x)g^{−1})) : \end{aligned}

we get from β to β′ by simultaneously conjugating β_{1} and β_{2} by g ∈ G.

Assume β_{2} is an automorphism, then, up to equivalence, we may assume that β_{2} is the identity automorphism on G and β_{1} is not an automorphism. Then e(G, G) will be twice the number of possible β_{1}, since the case where β_{1} is an automorphism and β_{2} is not is the same.

Returning to the regularity question, we ask, for which β_{1} is {β_{1}(g)g^{−1}} = G? Assume

\begin{aligned}β_{1}(1, 1) & = (m, 1) \\ β_{1}(0, b) & = (n, d) \end{aligned}

for some m divisible by p, some d ≠ 1 and some n. Then

β_{1}(l, b^{k}) = β_{1}(l, 1)β_{1}(0, b^{k})

= (lm, 1) (n (\frac{d^{k} − 1}{d − 1}) , d^{k})

= (lm + n (\frac{d^{k} − 1}{d − 1}) , d^{k}) .

For any h, r we want to find l, k so that

(∗)  β_{1}(l, b^{k})(l, b^{k})^{−1} = (lm + n (\frac{d^{k} − 1}{d − 1}) , d^{k}) (−b^{−k}l, b^{−k})

= (lm + n (\frac{d^{k} − 1}{d − 1}) − d^{k}b^{−k}l, d^{k}b^{−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} generates Z_{p^{e}}^{∗}. Now for β_{1} to be a homomorphism, we need bm = dm, and this is possible only if m = 0 or b ≡ d (mod p). But in the latter case, db^{−1} ≡ 1 (mod p), so cannot generate Z_{p^{e}}^{∗}. Thus if {β_{1}(g)g^{−1}} = G, then m = 0, and d = b^{f+1} such that b^{f} generates Z_{p^{e}}^{∗}.

Thus β_{1} is defined by

\begin{aligned}β_{1}(1, 1) & = (0, 1) \\ β_{1}(0, b) & = (n, d), \end{aligned}

and (∗) becomes

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

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} ≡ 1 (mod p), then p − 1 divides f + 1 and p divides n (or else β is not a homomorphism).

If d = b^{f+1} ≢ 1 (mod p), then n is arbitrary.

The first case gives p^{e−1} choices for n and ϕ(p^{e−1}) choices for f.

The second case gives p^{e} choices for n and

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

choices for f. The theorem follows. □

Corollary 2.2. If G = Hol(Z_{p}), then e(G, G) = 2(1 + p(ϕ(p − 1) − 1)).

Example 2.3. For p = 5 there are, up to equivalence, 2(1 + 5(2 − 1)) = 12 regular embeddings of G = Z_{5} ⋊ Z_{5}^{∗} into Hol(G). 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 β_{2} be the identity automorphism, then β_{1}(1, 1) = (0, 1) and β_{1}(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 groups G of order 2qp other than Γ, namely, Z_{2qp}, D_{qp}, D_{p} × Z_{q}, D_{q} × Z_{p}, and (Z_{p} ⋊ Z_{q}) × Z_{2} (where Z_{q} is identified as the subgroup of Aut(Z_{p}) 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 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 p be prime, p = 2q + 1. Then

Aut(D_{p}) = Aut(Z_{p} ⋊ Z_{q}) = Aut(Hol(Z_{p})) = Inn(Hol(Z_{p})).

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}^{∗} = Aut(Z_{p}) and G ⊂ Hol(Z_{p}). Let b generate Z_{a}. If α is an automorphism of G, then:

1. α(1, 1) = (m, 1) for p not dividing m, and

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

(b, b) = (0, b)(1, 1).) So there are p(p − 1) = |Hol(Z_{p})| automorphisms of G. Now if (l, d) is any element of Hol(Z_{p}), then conjugation by (l, d) is an automorphism of G, since

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

Hence the conjugation map C : Hol(Z_{p}) → Aut(G),

C(l, d)(m, b^{r}) = (l, d)(m, b^{r})(l, d)^{−1} = (l + dm − b^{r}l, b^{r}),

is defined. Then C is 1-1. For if C(l_{1}, d_{1}) = C(l_{2}, d_{2}) 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(p − 1).

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

Proof. If G has order p(p − 1) 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 ⋊_{α} Z_{2}.

If J ≅ Z_{pq} then G ≅ Z_{pq} ⋊_{α} Z_{2} where

α : Z_{2} → Aut(Z_{pq}) ≅ Z_{p−1} × Z_{q−1}

has four possibilities, α(−1) = (±1, ±1). These yield G ≅ D_{pq}, D_{p} × Z_{q}, D_{q} × Z_{p} and Z_{2pq}.

If J ≅ Z_{p} ⋊ Z_{q} ⊂ Hol(Z_{p}), then G ≅ J ⋊_{α} Z_{2} where α : Z_{2} → Aut(Z_{p} ⋊ Z_{q}) ≅ Inn(Hol(Z_{p})). If α is trivial, we obtain (Z_{p} ⋊ Z_{q}) × Z_{2}. Otherwise, the elements of order 2 in Inn(Hol(Z_{p})_{)} are of the form C(l, −1) for any l ∈ Z_{p}. Define

α_{l} : Z_{2} → Inn(Hol(Z_{p}))

by α_{l}(−1) = C(l, −1). One checks easily that

(Z_{p} ⋊ Z_{q}) ⋊_{α_{0}} Z_{2} ≅ Z_{p} ⋊ Z_{2q} = Hol(Z_{p})

by the map sending ((a, b), ϵ) to (a, ϵb) for a ∈ Z_{p}, b ∈ Z_{q} ⊂ Z_{p−1}, ϵ ∈ Z_{2} ⊂ Z_{p−1}. Now ⟨α_{0}⟩ = ⟨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_{p} ⋊ Z_{q}) ⋊_{α_{l}} Z_{2} ≅ (Z_{p} ⋊ Z_{q}) ⋊_{α_{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 Γ 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 Γ into Hol(G), 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 = 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 groups G of order p(p − 1) = 2pq. We did the case G = Γ in Theorem 2.1 and found that e(G, G) = 2 + 2p(q − 2). Here is the result for G ≅ Γ̸:

Theorem 4.1. Let Γ = Hol(Z_{p}), with p and q = (p − 1)/2 odd primes. Then:

(1) e(Γ, Z_{p} ⋊ Z_{q} × Z_{2}) = 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 that e(Γ, Z_{p} ⋊ Z_{q} × Z_{2}) = 2p(q − 1). Using Lemma 3.1 we have

Hol(Z_{p} ⋊ Z_{q} × Z_{2}) = Hol(Z_{p} ⋊ Z_{q}) × Hol(Z_{2})
= ((Z_{p} ⋊ Z_{q}) ⋊ Inn(Hol(Z_{p})) × Z_{2}
⊂ (Hol(Z_{p}) ⋊ Inn(Hol(Z_{p})) × Z_{2}
≅ Hol(Z_{p}) × Hol(Z_{p}) × Z_{2};

thus we may unwind any homomorphism β : Γ → ((Z_{p} ⋊ Z_{q}) ⋊ Inn(Hol(Z_{p})) × Z_{2} as in Section 1. If

β(γ) = ((m, b^{2r}), C(l, b^{s})) × ϵ

with ϵ = ±1, Z_{p}^{∗} = ⟨b⟩ and m, l are modulo p, then β(γ) maps to (m, b^{2r})(l, b^{s}) × (l, b^{s}) × ϵ under the map to Hol(Z_{p}) × Hol(Z_{p}) × Z_{2}. So β induces homomorphisms β_{1} : Γ → Hol(Z_{p}), defined by β_{1}(γ) = (m, b^{2r})(l, b^{s}), and β_{2} : Γ → Hol(Z_{p}), defined by β_{2}(γ) = (l, b^{s}). If we define β_{0} : Γ → Z_{2} by β_{0}(γ) = ϵ, then

β(γ) = (β_{1}(γ)β_{2}(γ)^{−1}, C(β_{2}(γ)) × β_{0}(γ).

Let γ ∈ Hol(Z_{p}), γ = (m, b^{t}). Then the only non-trivial homomorphism β_{0} is given by β_{0}(γ) = −1 if t is odd, and = 1 if t is even. (If β_{0}(γ) = 1 for all γ, then β will not be regular.) As for the possibilities for the homomorphisms β_{1} and β_{2}, we determined these in the proof of Theorem 2.1, namely:

1. If β_{i} is an automorphism, then β_{i}(1, 1) = (m, 1), m ≠ 0 and β_{i}(0, b) = (n, b) for any n.

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 any n, or β_{i}(0, b) = (0, 1).

As in the proof of Theorem 2.1, if both β_{1} and β_{2} are not automorphisms, or if both β_{1} and β_{2} are automorphisms, then β_{1} · β_{2}^{−1} is not 1-1, so β is not regular. Thus we can assume one is an automorphism and the other not. Assuming β_{2} is an automorphism, we can conjugate it by an automorphism of Hol(Z_{p}) to transform it to the identity map. Then

\begin{aligned}β_{1}(0, b)β_{2}(0, b)^{−1} & = (n, d)(0, b)^{−1} \\ & = (n, db^{−1}) : \end{aligned}

this must lie in Z_{p} ⋊ Z_{q}, which means that d = b^{r} must be an odd power of b. Then

\begin{aligned}β_{1} · β_{2}^{−1}(l, b^{s}) & = (n, d)^{s}(l, b^{s})^{−1} \\ & = (n (\frac{d^{s} − 1}{d − 1}) − l(db^{−1})^{s}, (db^{−1})^{s}) . \end{aligned}

For β to be regular, β_{1} · β_{2}^{−1} must map onto Z_{p} ⋊ Z_{q}, so we need that db^{−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, (db^{−1})^{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 (\frac{d^{s}−1}{d−1}) − l(db^{−1})^{s} = h. Hence for any suitable r and any n, β_{1} · β_{2}^{−1} maps Hol(Z_{p}) onto Z_{p} ⋊ Z_{q}.

Thus we have determined all regular β: we have p choices for n, and q −1 choices for d. Interchanging the roles of β_{1} and β_{2} yields the same result. Thus there are 2p(q − 1) regular embeddings of Hol(Z_{p}) into Hol(Z_{p} ⋊ Z_{q} × Z_{2}), as claimed.

(2): Proof that e(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_{q} ⋊ Inn(Hol(Z_{q})) → Hol(Z_{q}) × Hol(Z_{q})

is 1-1 and maps (m, ϵ)C(n, d) to (m, ϵ)(n, d) × (n, d). Suppose

β(1, 1) = (q, ϵ)C(n, d) × (m, c).

Now β(1, 1) must have order p (or else β is not 1-1), so (n, d) = (q, ϵ) = (0, 1), c = 1 and m ≠ 0, hence

β(1, 1) = (0, 1)C(0, 1) × (m, 1)

with m ≠ 0.

Suppose β(0, b) = (l, ϵ)C(k, d)×(s, c) in D_{q} ⋊Inn(Hol(Z_{q}))×Hol(Z_{p}) for l, ϵ, k, d mod q and s, c mod p. If ϵ = 1, then no element of D_{q} of the form (∗, −1) is hit by π_{1}(β), and so β is not regular. Thus ϵ = −1.

Applying β to the condition (b, 1)(0, b) = (0, b)(1, 1) yields c ≡ 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 or q in Hol(Z_{q}). But then d = 1 or −1.

Thus any regular embedding β satisfies:

1. β(1, 1) = (0, 1)C(0, 1) × (m, 1) with m ≠ 0 in Z_{p}, and

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

Now modify β by conjugating by

(0, 1)C(h, c) × (0, b^{r}) ∈ Aut(D_{q} × Z_{p}) ⊂ D_{q} ⋊ Inn(Hol(Z_{q})) × Hol(Z_{p}).

First, looking at β(1, 1):

\begin{aligned}((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^{r}m, 1). \end{aligned}

Now, looking at β(0, b):

\begin{aligned}((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^{r}k, b). \end{aligned}

Let β henceforth denote the conjugated embedding. Choose r so that b^{r}m = 1. Then

β(1, 1) = (0, 1)C(0, 1) × (1, 1).

We choose h, c in

β(0, b) = (2h + cn, −1)C(cl + h(1 − d), d) + (b^{r}k, b).

We have four possibilities.

If d = 1 and l = 0 we can choose c ≠ 0 and h so that

β(0, b) = (0, −1)C(0, 1) + (b^{r}k, b).

But then β is not regular, for π_{1}β does not map onto D_{q}.

If d = −1 and l = n, then we can choose c = 2, h = −n, but then

β(0, b) = (0, −1)C(0, −1) + (b^{r}k, b),

so β is not regular.

If d = 1 and l ≠ 0, choose c ≡ l^{−1} (mod q) and h so that 2h + cn ≡ 0, then

β(0, b) = (0, −1)C(1, 1) + (b^{r}k, b).

If d = −1 and l ≠ n, then we can choose c, h ≠ 0 with

2h + cn = 0
cl + 2h = 1,

giving

β(0, b) = (0, −1)C(1, −1) + (b^{r}k, b).

One verifies that

β(1, 1) = ((0, 1)C(0, 1) × (1, 1)
β(0, b) = ((0, −1)C(1, ±1) × (b^{r}k, b)

is regular, for any k. There are p choices for k, and hence, up to equivalence, there are 2p regular 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_{p} ⋊ Inn(Hol(Z_{p}) → Hol(Z_{p}) × Hol(Z_{p})

is 1-1 and maps (m, ϵ)C(n, d) to (m, ϵ)(n, d) × (n, d).

Suppose β(1, 1) = (m, ϵ)C(n, d)×(l, c). Then β(1, 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 Hol(Z_{q}), so (l, c) = (0, 1). Thus

β(1, 1) = (m, 1)C(n, 1) × (0, 1)

with m or n ≢ 0 (mod p).

Suppose β(0, b) = (l, ϵ)C(k, d) × (s, c), of order 2q. Then ϵ = −1 or else β is not regular. Also, (s, c) must have order dividing 2q in Hol(Z_{q}). But since β is regular, we must have (t, ∗) in the image of β for all t in Z_{q}, so c = 1 and s ≠ 0.

Applying β to the relation (b, 1)(0, b) = (0, b)(1, 1) yields

\begin{aligned}((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)). \end{aligned}

This yields no condition on s, but on the left components we obtain

(bm + 2bn + l, −1)C(bn + k, d) = (l − dm, −1)C(dn + k, d).

Thus

\begin{aligned}2bn + bm & = −dm \\ bn & = dn. \end{aligned}

If n ≠ 0, then b = d and n = −m. If n = 0, then d = −b. Thus

\begin{aligned}β(1, 1) & = (m, 1)C(−m, 1) × (0, 1) \\ β(0, b) & = (l, −1)C(k, b) × (s, 1), \end{aligned}

or

\begin{aligned}β(1, 1) & = (m, 1)C(0, 1) × (0, 1) \\ β(0, b) & = (l, −1)C(k, −b) × (s, 1). \end{aligned}

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

Now we modify β by an automorphism of D_{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).

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:

\begin{aligned}β(1, 1) & = (1, 1)C(−1, 1) × (0, 1) \\ β(0, b) & = (0, −1)C(k, b) × (1, 1), \end{aligned}

and

\begin{aligned}β(1, 1) & = (1, 1)C(0, 1) × (0, 1) \\ β(0, b) & = (0, −1)C(k, −b) × (1, 1). \end{aligned}

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 Hol(Z_{p}) into Hol(D_{p} × Z_{q}).

(4): Proof that e(Γ, Z_{2qp}) = p. Let

β : Hol(Z_{p}) → Hol(Z_{2pq}) ≅ Hol(Z_{p}) × Hol(Z_{q}) × Z_{2}

be a regular embedding. Then β(1, 1) has order p, so

β(1, 1) = (m, 1) × (0, 1) × 0

for some m ≢ 0 (mod p). Also, β(0, b) has order p − 1 = 2q, so

β(0, b) = (n, d) × (l, 1) × e

with d ≢ 1 (mod p). If e ≡ 0 (mod 2) or l ≡ 0 (mod q) then β is not regular, hence we have e ≡ 1 (mod 2) and l ≢ 0 (mod q).

The condition (b, 1)(0, b) = (0, b)(1, 1) implies that d ≡ b (mod p). Thus in Hol(Z_{2pq}),

\begin{aligned}β(1, 1) & = (2qm, 1) \\ β(0, b) & = (n, c) \end{aligned}

with p not dividing m, 2q not dividing n, and c ≡ b (mod p), c ≡ 1 (mod 2q).

Now we consider β under equivalence by automorphisms of Z_{2pq}. We conjugate β by (0, d) with d coprime to 2pq:

\begin{aligned}(0, d)(2qm, 1)(0, d^{−1}) & = (2dqm, 1) \\ (0, d)(n, c)(0, d^{−1}) & = (dn, c). \end{aligned}

Choose d so that

\begin{aligned}dm & ≡ 1 (mod p) \\ dn & ≡ 1 (mod 2q). \end{aligned}

Then after conjugating, β becomes:

\begin{aligned}β(1, 1) & = (2q, 1) \\ β(0, b) & = (1 + 2ql, c). \end{aligned}

We check that β is regular for any l modulo p. 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}).

Let a be any element of Z_{2pq}. To solve

a ≡ 2qn + (1 + 2ql)(1 + c + ⋯ + c^{k−1}) (mod 2pq)

for n and k it suffices to solve

\begin{aligned}a & ≡ 2qn + (1 + 2ql)(1 + c + ⋯ + c^{k−1}) (mod p) \\ a & ≡ 1 + c + ⋯ + c^{k−1} ≡ k (mod 2q). \end{aligned}

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 e(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_{pq} ⋊ Inn(Hol(Z_{pq})) → Hol(Z_{pq}) × Hol(Z_{pq})

by (m, ϵ)C(n, d) ↦ (m, ϵ)(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 into D_{p} × D_{q}. Let β(1, 1) = (m, ϵ)C(n, d), of order p. Then β(1, 1) maps to

(m + ϵd, ϵd) × (n, d) mod p, (m + ϵd, ϵd) × (n, d) mod q,

which must have order p. Hence

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

and

ϵd ≡ d ≡ 1 (mod p)
n or m + ϵn ≢ 0 (mod p).

Let β(0, b) = (l, ϵ)C(k, c). By regularity we must have ϵ = −1, or else π_{1}(β) maps into {(∗, 1)} ⊂ D_{pq}. Then β(0, b) maps to

(l − k, −c) × (k, c) mod p, (l − k, −c) × (k, c) mod q.

This has order 2q = p − 1, so we must have

c ≡ ±1 (mod q).

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

\begin{aligned}bqn & = cqn \\ cqm & = bqm + 2bqn. \end{aligned}

This yields no conditions modulo q, but modulo p, we have:

1. If n ≢ 0, then c = b and m = −n.

2. If n ≡ 0, then c = −b and m ≢ 0.

Now we look for a nice representative for β 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)^{−1}C((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)^{−1}C((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.

Modulo p:

1. If n ≢ 0, choose d so that dqn ≡ −1, then dqm ≡ −dqn ≡ 1 and c ≡ b.

2. If n ≡ 0, choose d so that dqm ≡ 1 and c ≡ −b.

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

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).

Modulo q:

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

where c ≡ ±1.

If c ≡ 1 and k ≢ 0, set dk ≡ 1 and

2g + dl ≡ 0.

Then

β(1, 1) = (0, 1)C(0, 1)

and

β(0, b) = (0, −1)C(1, 1)

or (if k ≡ 0)

β(0, b) = (0, −1)C(0, 1).

However, in this last case, β is not regular: π_{1}β maps to {(0, ±1)} ⊂ Hol(Z_{q}). If c ≡ −1, we have

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

If l ≡ k we can choose g so that 2g + dk = 2g + dl ≡ 0, but then β is not regular. Thus we must have l ≢ k, and then we may choose d ≠ 0 and g so 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)

modulo p and 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)

modulo q.

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,

β_{1}(l, b^{r})β_{2}(l, b^{r})^{−1} = (∗, (−1)^{r})

both modulo p and modulo q, and so β_{1} · β_{2}^{−1} maps to D_{pq} ⊂ Hol(Z_{p}). To show regularity, we need only show that modulo p every element of D_{p} is in the image of β_{1} · β_{2}^{−1}, and similarly modulo q.

Mod p: If

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

modulo p, then

β_{1}(l, b^{r})β_{2}(l, b^{r})^{−1} = (−k (\frac{1 − (−b)^{r}}{1 + b} ) + (−1)^{r} (−l + k (\frac{1 − b^{r}}{1 − b} )) , (−1)^{r})

and for r and l 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)

modulo p then

β_{1}(l, b^{r})β_{2}(l, b^{r})^{−1} = ((l + (−k)\frac{1 − b^{r}}{1 − b} ) + (−1)^{r}k (\frac{1 − (−b)^{r}}{1 + b} ) , (−1)^{r})

and again for r and l arbitrary we can obtain all elements (m, ±1) of D_{p}.

Mod q: If

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

modulo q, then

β_{1}(l, b^{r})β_{2}(l, b^{r})^{−1} = \begin{cases}(−r, 1) & if r is even, \\ (−1 + r, −1) & if r is odd. \end{cases}

Since 0 ≤ r < 2q and we’re working modulo q, we can obtain all elements (m, ±1) of D_{q}.

If

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

modulo q, then

β_{1}(l, b^{r})β_{2}(l, b^{r})^{−1} = \begin{cases}(−r, 1) & if r is even, \\ (1 − r, −1) & if r is odd. \end{cases}

Again, since 0 ≤ r < 2q and we’re working modulo q, we can obtain all elements (m, ±1) of D_{q}.

Thus

\{β_{1}(l, b^{r})β_{2}(l, b^{r})^{−1}\} = \begin{cases}D_{p} & modulo p \\ D_{q} & modulo q, \end{cases}

and so all four combinations of β modulo p and modulo q are regular. This completes the proof that e(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 > 5 a safeprime, then for every group G of cardinality that of Γ, 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.

On Hopf Galois structures and complete groups 115

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, p and q primes with q dividing p − 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 lc802@math.albany.edu http://math.albany.edu:8000/˜ lc802/

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