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.
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 .
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 . We did the case in Theorem 2.1 and found that . Here is the result for :
Proof. We do each case in turn, following the steps outlined in Section 1.
(1): Proof that . Using Lemma 3.1 we have thus we may unwind any homomorphism as in Section 1. If with and are modulo , then maps to under the map to . So induces homomorphisms , defined by , and , defined by . If we define by , then
Let . Then the only non-trivial homomorphism is given by if is odd, and if is even. (If for all , then will not be regular.) As for the possibilities for the homomorphisms and , we determined these in the proof of Theorem 2.1, namely:
If is an automorphism, then , and for any .
If is not an automorphism, then and for , , and any , or .
As in the proof of Theorem 2.1, if both and are not automorphisms, or if both and are automorphisms, then is not 1-1, so is not regular. Thus we can assume one is an automorphism and the other not. Assuming is an automorphism, we can conjugate it by an automorphism of to transform it to the identity map. Then this must lie in , which means that must be an odd power of . Then For to be regular, must map onto , so we need that generates . Thus must be coprime to . There are odd numbers with coprime to .
For any such , is a unit of , so given any in there is some in so that . Hence for any suitable and any , maps onto .
Thus we have determined all regular : we have choices for , and choices for . Interchanging the roles of and yields the same result. Thus there are regular embeddings of into , as claimed.
Suppose in for mod and mod . If , then no element of of the form is hit by , and so is not regular. Thus .
Applying to the condition yields . Hence has order . We require that have order dividing in , which maps 1-1 to as noted above. Thus and must have order 1, 2 or in . But then or .
Thus any regular embedding satisfies:
with in , and
Now modify by conjugating by
First, looking at :
Now, looking at :
Let henceforth denote the conjugated embedding. Choose so that . Then We choose in We have four possibilities.
If and we can choose and so that But then is not regular, for does not map onto .
If and , then we can choose , but then so is not regular.
If and , choose and so that , then
If and , then we can choose with giving One verifies that is regular, for any . There are choices for , and hence, up to equivalence, there are regular embeddings of into : .
Suppose . Then must have order , so and have order dividing . This implies . Also has order dividing in , so . Thus with or .
Suppose , of order . Then or else is not regular. Also, must have order dividing in . But since is regular, we must have in the image of for all in , so and .
Applying to the relation yields This yields no condition on , but on the left components we obtain Thus If , then and . If , then . Thus or One then sees easily that has order dividing .
Now we modify by an automorphism of .
If we conjugate the right factors by in , then is transformed into .
If we conjugate the left factors by in , then and Choose so that . Then or .
Choose so that . Then we have the following representatives of equivalence classes of : and It is a routine check that for any modulo , both are regular Thus we have equivalence classes of regular embeddings of into .
(4): Proof that . Let be a regular embedding. Then has order , so for some . Also, has order , so with . If or then is not regular, hence we have and .
The condition implies that . Thus in , with not dividing , not dividing , and .
Now we consider under equivalence by automorphisms of . We conjugate by with coprime to : Choose so that Then after conjugating, becomes: We check that is regular for any modulo . We have Let be any element of . To solve for and it suffices to solve Clearly for each modulo there are unique values for that solve these congruences. Thus for any , is 1-1 and regular, and so .
(5): Proof that . Let be a regular embedding. We have by . Also, by . Under this map, maps into .
Let , of order . Then maps to which must have order . Hence and
Let . By regularity we must have , or else maps into . Then maps to This has order , so we must have
From we obtain hence and so This yields no conditions modulo , but modulo , we have:
If , then and .
If , then and .
Now we look for a nice representative for modulo conjugation by elements of . For , We may choose and modulo by choosing them modulo and modulo separately.
If , choose so that , then and .
If , choose so that and .
Choose so that . Then, since , is arbitrary, so (letting now denote the conjugated embedding) we have, modulo : or
If and , set and Then and or (if ) However, in this last case, is not regular: maps to .
If , we have If we can choose so that , but then is not regular. Thus we must have , and then we may choose and so that and so
To summarize, any regular embedding is equivalent to or modulo and to or modulo .
We show that all four combinations give regular embeddings of to , and so, since is arbitrary in , we have equivalence classes of regular embeddings.
Note that in every combination, both modulo and modulo , and so maps to . To show regularity, we need only show that modulo every element of is in the image of , and similarly modulo .
Mod : If modulo , then and for and arbitrary we can obtain all elements of .
If modulo then and again for and arbitrary we can obtain all elements of .
Mod : If modulo , then Since and we're working modulo , we can obtain all elements of .
If modulo , then Again, since and we're working modulo , we can obtain all elements of .
Thus and so all four combinations of modulo and modulo are regular. This completes the proof that .
a) If is a Galois extension of , fields, with Galois group , a safeprime, then for every group of cardinality that of , there is a -Hopf Galois structure on where the associated group of is .
b) The number of Hopf Galois structures on is .