Let be a Galois extension of , fields, with Galois group . We obtain two results. First, if , we determine the number of Hopf Galois structures on where the associated group of the Hopf algebra is (i.e. ). Now let be a safeprime, that is, is a prime such that is also prime. If is Galois with group , a safeprime, then for every group of cardinality there is an -Hopf Galois structure on where the associated group of is , and we count the structures.
Let be a Galois extension of , fields, with finite Galois group . Then is an -Hopf Galois extension of for , where acts on via the natural action of the Galois group on . Greither and Pareigis [GP87] showed that for many Galois groups , is also an -Hopf Galois extension of for a cocommutative -Hopf algebra other than . The Hopf algebras that arise have the property that , the group ring over of a group of the same cardinality as . We call the associated group of .
From [By96] we know that for a cocommutative -Hopf algebra, -Hopf Galois structures on correspond bijectively to equivalence classes of regular embeddings . Here is the group of permutations of the set , and is the normalizer of the left regular representation of in . One sees easily that contains the image of the right regular representation and also ; then and is isomorphic to the semidirect product . The equivalence relation on regular embeddings is by conjugation by elements of inside : if there exists in so that for all in , .
Thus the number of -Hopf Galois structures on where the associated group of is depends only on and the Galois group , and reduces to a purely group-theoretic problem.
If is a Galois extension of with Galois group non-abelian simple or , then in [CC99] we counted the number of Hopf Galois structures on with associated group by "unwinding" regular embeddings. The unwinding idea applies more generally when is a complete group, i. e. has trivial center and trivial outer automorphism group. In this paper we apply this unwinding idea to determine when is the complete group , 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 is Galois with Galois group and is -Hopf Galois where has associated group , then . Positive results in this direction include Byott's original uniqueness theorem [By96]; Kohl's Theorem [Ko98] that if then ; Byott's recent result [By03a], complementing [CC99], that if is non-abelian simple then ; and Featherstonhaugh's recent result [Fe03] that if and are abelian -groups with sufficiently large compared to the -rank of and of the exponent of , then . (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 when is a safeprime, that is, where and 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 . We show that if is a Galois extension with Galois group , then for each of the six groups of cardinality up to isomorphism, there is an -Hopf Galois structure on with associated group , and we count their number. Thus yields an example of the opposite extreme to the uniqueness results listed above.
Given a Galois extension with Galois group , and a group of cardinality that of , the number of Hopf Galois structures on whose Hopf algebra has assocated group is equal to the number of equivalence classes of regular embeddings of into , the semidirect product of and . Here an embedding is regular if, when viewing inside via , the orbit of the identity element of under is all of .
Obtaining in any particular case involves a number of steps:
Of these tasks, checking regularity is the least natural. If we view , the semidirect product of and , as as sets, the operation is for . Then is a normal subgroup of , and the projection onto by is a homomorphism; however the projection onto , , is not. But since an element viewed in acts on the identity element of the set by , checking regularity of a 1-1 homomorphism is the same as determining whether the function (non-homomorphism) is bijective.
Let be the group of inner automorphisms of , then is normal in and , the outer automorphism group, fits in the exact sequence where the map is conjugation: for , and is the center of .
Suppose and the composition of the 1-1 homomorphism with the homomorphism yields a trivial homomorphism from to . Then maps into , and, following [CC99], we may decompose as follows: we have a homomorphism by for , with inverse sending to . Letting be projection onto the th factor, , the homomorphism yields homomorphisms and such that Then is regular iff Thus whenever and , we may describe , and in particular the function , in terms of the homomorphisms , namely,
A class of groups where and is the class of complete groups, that is, finite groups with and . The best-known examples of finite complete groups are where is simple and non-abelian, for , and where is odd. (See [Sch65, III.4.u-w] .) Another class is the class of simple groups, for if is simple, then is solvable, so any has image in .
In [CC99] we let and determined , the number of regular embeddings of to , when is simple non-abelian or . In the next section we examine the case where where and is an odd prime.
Let , the holomorph of . Let be a number that generates . Let . In this section we prove:
Proof. We wish to find equivalence classes of regular embeddings of in . Since is complete, we know from the last section that any homomorphism may be decomposed as for homomorphisms . So we begin by describing the homomorphisms from to .
Let be a homomorphism. Then is determined by If has order dividing , then . To see this, first note that for any , so must have order dividing , which implies that . Also, for to have order dividing we require that or divides . For if one sees by induction that where divides iff divides . Thus if does not divide , then has order . On the other hand, if , then and divides but not , so the order of divides .
Now we check the relation Applying yields: hence Thus . Since is a unit modulo it follows that , hence . But since for some , Since is relatively prime to and , it follows that , hence and .
Thus for any homomorphism , for some .
Since , the requirement becomes which implies that . Thus if , then , and if is a unit (i.e. relatively prime to ), then . In the latter case, has order and has order .
Clearly if is an automorphism then is relatively prime to and so . Conversely, if with relatively prime to , then for some . Conjugating by in yields We choose so that . Then we choose so that we set , possible because has order and hence . With these choices of and , the homomorphism is the identity on the generators and of , and so is an automorphism of .
Now we ask about regularity: for which pairs of endomorphisms is or equivalently, is a 1-1 function from to ? Let for .
If neither nor is an automorphism, then divides and , so for all , and so is not 1-1.
Suppose both and are automorphisms. If then so is not 1-1. If , then let . Then so again, is not 1-1.
Thus for to be regular, exactly one of and is an automorphism.
We return to looking at regularity after we look at equivalence by inside .
For we have Thus if for , then with we get from to by simultaneously conjugating and by .
Assume is an automorphism, then, up to equivalence, we may assume that is the identity automorphism on and is not an automorphism. Then will be twice the number of possible , since the case where is an automorphism and is not is the same.
Returning to the regularity question, we ask, for which is ?
Assume for some divisible by , some and some . Then For any we want to find so that In order that for any , there is a so that , we require that generates . Now for to be a homomorphism, we need , and this is possible only if or . But in the latter case, , so cannot generate . Thus if , then , and such that generates .
Thus is defined by and (**) becomes which is solvable for all and for all such that generates . We have two cases:
If , then divides and divides (or else is not a homomorphism).
If , then is arbitrary.
The first case gives choices for and choices for .
The second case gives choices for and choices for . The theorem follows.
Let with an odd prime, as above, and let . Then there are at least five non-isomorphic groups of order other than , namely, , and (where is identified as the subgroup of of index 2, and is the dihedral group of order ). The five groups are non-isomorphic because their centers have orders and respectively.
If is also an odd prime, then is called a Sophie Germain prime, resp. is called a safeprime, and in that case, these five groups, together with , are the only groups of order , up to isomorphism. To see this, we first obtain the following lemma, needed also in the next section:
Proof. Let with or . Write additively and view as the cyclic subgroup of order inside and . Let generate . If is an automorphism of , then:
for not dividing , and
for any (as one sees by applying to the relation .)
So there are automorphisms of . Now if is any element of , then conjugation by is an automorphism of , since Hence the conjugation map , is defined. Then is 1-1. For if on , then for all : in particular for we get and for we get . Thus is an isomorphism.
This is probably well-known, but we sketch a proof for the reader's convenience.
Proof. If has order then by the first Sylow Theorem, has a unique normal subgroup of order . By Schur's Theorem, has a complementary subgroup of order , and hence a subgroup of order . Then is a subgroup of of order , hence normal in . By Schur's Theorem again, has a complementary subgroup of order 2 in , so is isomorphic to a semidirect product .
If then where has four possibilities, . These yield and .
If , then where . If is trivial, we obtain . Otherwise, the elements of order 2 in are of the form for any . Define by . One checks easily that by the map sending to for , , . Now and are conjugate in for by where : . It follows from [DF99], p. 186, Exercise 6 that for all . Thus yields only two possible groups, up to isomorphism.
If we begin with the Galois group of a Galois extension of fields, then to determine the -Hopf Galois structures on , we need to count equivalence classes of regular embeddings of into , where varies over isomorphism classes of groups of the same cardinality as . This can be a formidable task for certain cardinalities!
In this section, we let , a safeprime , and determine , the number of regular embeddings of into , for all six groups of order