Regular Permutation Groups of Order mp

Regular Permutation Groups of Order mp Timothy Kohl Department of Mathematics and Statistics Boston University Boston, MA 02215 tkohl@math.bu.edu Abstract For Γ a group of order mp for p > m prime, we study the enumeration of the regular subgroups of P erm(γ) normalized by the left regular representation of Γ. Part of the interest in these is that such regular subgroups are in one-to-one correspondence with the Hopf-Galois structures on separable field extensions L/K with Γ = Gal(L/K). This is a sequel to the author s work on the case where m = 4. Key words: Hopf-Galois extension, Greither-Pareigis theory, regular subgroup, holomorph, block structure, wreath product MSC: 20B35, 16T05 Introduction Let L/K be a finite separable (but not necessarily normal) extension with normal closure L where Γ = Gal( L/K) and = Gal( L/L). In [5] it is shown that the Hopf Galois structures on L/K are in direct correspondence with the regular subgroups of N Perm(Γ/ ) normalized by λ(γ) Perm(Γ/ ). For the case where L/K is already Galois, Γ = Gal(L/K) and the Hopf-Galois structures on L/K correspond to regular subgroups N Perm(Γ) normalized by the left regular representation of Γ. The study of these has included the enumeration being carried out for various classes of groups, some based on structural properties of Γ such as completeness, 1

e.g. [2], or based on order, e.g. [1]. We shall consider a set of cases where order and structural properties both bear on the enumeration, namely the case where Γ (and therefore any N) has order mp for p > m where p is prime. This generalizes the author s work in [6] which dealt with the case m = 4 where for each p, the number of isomorphism classes was either four or five. For groups of order mp the number of classes can vary quite considerably. Nonetheless, all have structural similarities which can utilized to aid in the enumeration of the regular subgroups in question. 1 Preliminaries 1.1 Groups of order mp Although we will be working with permutation groups, let s make a few important observations about abstract groups of order mp which we will use to in the setting of permutation groups of order mp. First, if G has order mp with p > m then it has a unique p-sylow subgroup P and there exists a subgroup Q G with Q = m and by the Schur-Zassenhaus lemma, G = P τ Q with τ : Q Aut(P) induced by conjugation within G. If G is abelian then G = P Q and, if τ is the trivial map, then, also, G = P Q, for example if gcd(m, p 1) = 1. Part (a) of the following is trivial, and (b) is certainly standard, but no convenient reference could be found. Lemma 1.1: If G = mp with p prime and p > m with G = P τ Q with P, Q as above, then (a) If τ is trivial, that is G = P Q then p Aut(G). (b) If τ non-trivial,then p Aut(G) but p2 Aut(G). The point is that for G of order mp, Aut(G) has no p-sylow subgroup if G = P Q. For G = P τ Q for τ non-trivial Aut(G) has a unique p-sylow subgroup of order p. 2

1.2 Regular Subgroups For Γ a group of order mp with p > m we let B = Perm(Γ) and have λ(γ) B the left regular representation of Γ acting on itself, via λ(g)(g ) = gg. As observed above, Γ, hence λ(γ) too has a characteristic subgroup of order p. Definition 1.2: Let P λ(γ) be the unique p-sylow subgroup of λ(γ). As regularity is the central theme, we define it here. Definition 1.3: A subgroup N B is regular if any two of the following three conditions are satisfied: 1. Stab N (γ) = {n N n(γ) = γ } is the trivial group, for any γ Γ 2. N acts transitively on Γ 3. N = Γ Condition 1 above is sometimes, (e.g. [8]), used as the definition of a semiregular permutation group. This is a useful definition to keep in mind, since then a semi-regular subgroup with Γ elements must be regular. One useful fact to observe about regular permutation groups is that the cycle structure of elements must be balanced in the following way, which too is standard, but must be mentioned as it is used considerably in the computations in section 4. Lemma 1.4: If n N where N is a regular subgroup of B, then n k must act without fixed points if n k e N. That is, n is a product of cycles of all the same length and Supp(n) = Γ if n e N where Supp(n) = {γ Γ n(γ) γ}. The above gives us important restrictions on the possible generators of any such N. More importantly, the cycle structure has global implications as far as how such an N acts (set-wise) on Γ, specifically block structure, as we shall see first in 1.6 below. 3

We are not interested in all the regular subgroups of B, rather we consider those of the following type. Definition 1.5: 1. R(Γ) = {N B N is regular and λ(γ) Norm B (N)} 2. For [M] an isomorphism class of group of order Γ, let R(Γ, [M]) = {N R(Γ) N = M}. Our goal is to enumerate R(Γ) for Γ = mp and to do this we observe that R(Γ) is the union of the R(Γ, [M]) for each isomorphism class of group M of order mp. As indicated in the introduction, the groups in R(Γ) correspond in a 1-1 fashion to the Hopf-Galois structures on an (already) Galois extension L/K where Γ = Gal(L/K). This paper generalizes the work of the author in [6] by considering the case where Γ = mp rather than simply 4p. We observe that if λ(γ) normalizes a regular subgroup N B then for P(N) the unique subgroup of N of order p, λ(γ) must normalize P(N) since P(N) is characteristic in N. The regularity of λ(γ) and N implies a relationship between P and P(N). Proposition 1.6: If P λ(γ) is the unique subgroup of order p then P = π 1 π 2...π m where the π i are disjoint p-cycles. If P(N) is the unique subgroup of order p of a regular subgroup N B that is normalized by λ(γ) then P(N) = π a 1 1 π a 2 2... πm am where a i U p = F p for each i. Proof. This is a generalization of the argument in [6, Proposition 2.7]. The semi-regularity of P (and P(N) for any regular N) and the fact that if λ(γ) normalizes N then P must centralize P(N) shows that the underlying cycle structure involves the same π i that determine the generator of P and that the a i are non-zero. If N is any regular subgroup of B then N may be written as P(N)Q(N) where Q(N) is the complementary subgroup of P(N) in N. Moreover, as 4

Q(N) normalizes P(N) and [N : P(N)] = Q(N) = m then if {q 1, q 2,...,q m } is a transversal of P(N) in N (i.e. Q(N) itself, with q 1 = e N ), and if we define Π i = P(N)q i (e Γ ) the orbit of e Γ under the action of the elements in the coset P(N)q i, then we may partition Γ as Π 1 Π 2 Π m. As P(N) is normal in N, then the Π i are a system of blocks for N and we have the following: Proposition 1.7: Q(N) permutes the Π i in a fixed point free manner. Proof. Given q Q(N) then q Π i = qp(n)q i (e Γ ) and if qp(n)q i (e Γ ) = P(N)q i (e Γ ) then we have qp 1 q i (e Γ ) = p 2 q i (e Γ ) for some p 1, p 2 P(N). But, as N is a regular subgroup of B then qp 1 q i = p 2 q i which means q = p 2 p 1 1 which, since P(N) Q(N) = {1}, implies that q = 1. Corollary 1.8: Q(N) is isomorphic to a regular subgroup of S m embedded via the action on the Π i above. Now, if Γ = P(Γ)Q(Γ) where P(Γ) = φ is the order p subgroup and Q(Γ) = {µ 1, µ 2,...,µ m } is the complementary subgroup, where we may assume µ 1 = e Γ, then P = λ(p(γ)) = λ(φ) = π 1 π 2 π m (as in 1.6) and π 1 π 2...π m = (µ 1, φµ 1,...,φ p 1 µ 1 )(µ 2, φµ 2,..., φ p 1 µ 2 ) (µ m, φµ m,...,φ p 1 µ m ) and, in the notation above, with N = λ(γ), we have Π i = Pλ(µ i )(e Γ ) = Pλ(µ i )(µ 1 ) = P(µ i ) = Orbit (µi,φµ i,...,φ p 1 µ i ) (µ i ) = Supp(π i ) The point being that, in light of 1.6, if N is a regular subgroup of B = Perm(Γ) normalized by λ(γ) then, with P = π 1...π m and P(N) = 5

π a 1 1...π am m, then Π i = Supp(π i ) and Q(N) acts as a regular subgroup of Perm({Π 1,...,Π m }). Moreover, we may write λ(γ) = PQ where the nontrivial action of λ(γ) on P comes from the action of Q = λ(q(γ)) which permutes the Π i in a fixed-point free manner. 2 Characters and the enumeration of P(N) To determine those regular N normalized by λ(γ) we first need to determine the possible P(N) that may arise. Amongst other things this will allow us to show that [6, Theorem 4.4] (for m = 4) is true for all m in fact, namely that N R(Γ) implies that N Norm B (P). We make the following definition: Definition 2.1: Given π 1...π m as above, and (i 1,...,i m ) F m p [i 1,..., i m ] = π i 1 1...πm im. we let The rationale for this is that if V = π 1, π 2,...,π m B then V = F m p and as such, π 1 π 2 π m = [1, 1,...1] (1, 1,..., 1) F m p and similarly if P(N) = [a 1, a 2,...,a m ] then [a 1, a 2,..., a m ] corresponds to a vector (a 1, a 2,...,a m ) F m p with all non-zero coordinates. Even though V is a multiplicative group, it is convenient to use this additive notation for elements of V in particular the generators of potential P(N) V since the action(s) by the complementary group(s) of order m are linear. If N R(Γ) with order p subgroup P(N), then since P(N) is characteristic in N and Γ Norm B (N) then λ(γ) normalizes P(N). Based upon this we can determine what restrictions are placed on the a i. One should note that what we re going to determine is the set of all semi-regular order p subgroups of B that are normalized by λ(γ). (The determination of whether a given N R(Γ) has one of these as its order p subgroup is something we will deal with later.) As noted above, since λ(γ) = PQ where Q has order m then we consider γ[a 1,...,a m ]γ 1 where γ Q. As [a 1,...,a m ] must be 6

normalized then γ[a 1,...,a m ]γ 1 = d γ [a 1,...,a m ] = [d γ a 1,...,d γ a m ] for some d γ U p. Also note that, since P is normal in λ(γ), then for any γ λ(γ) we have γπ i γ 1 = π cγ γ(i) where c γ U p, c γ γ = c γ c γ, and γ permutes the indices {1,..., m}. If we restrict to Q then by 1.7 we have an action by a regular permutation group on the set of indices {1,...,m}. If Γ is direct product (abelian or not) then of course c γ = 1 for all γ and if Γ is not a direct product then c γ = τ(γ) U p. As U p = F p then one can view τ as a linear character of Q. Likewise the association γ d γ is also a character. So now, given [a 1,..., a m ] we have γ[a 1,..., a m ]γ 1 = c γ [a γ 1 (1),...,a γ 1 (m)] which implies that c γ a γ 1 (i) = d γ a i for each i. Rewriting this with i = γ(j) we have c γ a j = d γ a γ(j), that is a γ(j) = ( cγ d γ ) a j Now, the γ Q act (on the indices) as a regular subgroup of S m. Moreover, since all the a i are non-zero then, as [a 1,...,a m ] generates a subgroup of order p, we may assume that a 1 = 1 as this subgroup will certainly contain a unique vector with the first entry being 1. Therefore, for each γ Q, ( ) cγ a γ(1) = a 1 d γ ( ) cγ = d γ and by regularity of Q the orbit of 1 is all of {1,...,m} so that the a i are determined by the action of Q on {1,...,m} and the mapping γ c γ /d γ 7

which is clearly a homomorphism (i.e. another linear character) as well. If we define ˆv i = [0,...,1,..., 0] = π i then we assert the following. Theorem 2.2: Any semi-regular subgroup of V of order p that is normalized by Q is generated by ˆp χ = γ Qχ(γ)ˆv γ(1) where χ : Q U p = F p is a linear character of Q. Proof. By the preceding discussion, we know that if P(N) is normalized by Q then it is of the form asserted in the theorem. Now we show that any such ˆp χ generates a semi-regular order p subgroup of V normalized by Q. If µ λ(γ) then µ acts on π by raising π to the power τ(µ) where τ is the structure map which defines λ(γ), which includes the possibility that τ is trivial. Observe first that for µ λ(γ) ˆp χ = γ Q χ(γ)ˆv γ(1) = γ Q χ(µγ)ˆv µγ(1) = χ(µ) γ Qχ(γ)ˆv µγ(1) and so µˆp χ µ 1 = γ Q χ(γ)(µˆv γ(1) µ 1 ) = γ Q χ(γ)τ(µ)ˆv µγ(1) = τ(µ) γ Q χ(γ)ˆv (µγ(1)) = τ(µ)χ(µ) 1ˆp χ and so ˆp χ is normalized by µ. 8

Note, since P is generated by π = π 1 π 2 π m which corresponds to [1, 1,, 1] then, in the above notation, π corresponds to the trivial character ι : Q U p, i.e. P = ˆp ι. So to determine the possible choices for P(N) we need to determine the linear characters of groups of order m in the field F p. This can make for some interesting possibilities since congruence conditions between m and p as well as structural facts about the groups of order m dictate what characters arise. For example, with m = 4, as in [6], the possibilities for P(N) were denoted P 1,...,P 6 and although they weren t determined via the above result, we can recast them in terms of the linear characters of groups of order 4 into F p. For Q = C 2 C 2 = x, y we have 1 x y xy χ 1 1 1 1 1 χ 2 1 1-1 -1 χ 3 1-1 1-1 χ 4 1-1 -1 1 and for Q = C 4 = x we have, for all p, 1 x x 2 x 3 ψ 1 1 1 1 1 ψ 2 1-1 1-1 and if p 1(mod 4) then there exists an element ζ U p of order 4 (with inverse ζ) and subsequently two additional characters 1 x x 2 x 3 ψ 3 1 ζ -1 ζ ψ 4 1 ζ -1 ζ That is ˆp χ1 = [1, 1, 1, 1], ˆp χ2 = [1, 1, 1, 1], ˆp χ3 = [1, 1, 1, 1], ˆp χ4 = [1, 1, 1, 1] which parallels [6, Definition 2.9,Proposition 2.10], where (up to ordering of the π i ) the P 1, P 2, P 3 and P 4 are generated by the ˆp χi. A similar observation holds for the case of C 4 where P 1 = ˆp ψ1, P 2 = ˆp ψ2, P 5 = ˆp ψ3, and P 6 = ˆp ψ4 where the authors reuse of P 1 and P 2 therein was a slight abuse of notation. We shall return to this when we consider calculations for other m and p in section 4. In the next section we shall need the following fact about these ˆp χ which can be thought of as a modified version of orthogonality of characters. 9

Lemma 2.3: For ι the trivial character of Q and χ any other linear character of Q (into U p = F p ) one has ˆp ι ˆp χ = 0. Proof. The basic fact is that for χ a non-trivial linear character, ˆp ι ˆp χ = χ(γ) = [U p : T] x γ Q x T where T = im(χ) U p and since U p is cyclic, so is T, and therefore the sum over the elements of T is zero mod p. 3 Normalizers and Holomorphs Definition 3.1: For Γ an abstract group, Hol(Γ) the holomorph of Γ is the semi-direct product Γ Aut(Γ). Equivalently, if λ(γ) is the left-regular representation of Γ in B then we define Hol(Γ) = Norm B (λ(γ)) and have Hol(Γ) = ρ(γ)aut(γ) where ρ : Γ B is the right regular representation given by ρ(g)(g ) = g g 1 and Aut(Γ) is embedded in the obvious way as a subgroup of B. As Aut(Γ) is naturally embedded in Norm B (Γ) as those elements which fix the identity we observe the following mild generalization for any other regular subgroup N of B. Lemma 3.2: [6, Proposition 3.6] For N a regular subgroup of B, Norm B (N) = Hol(N) where Hol(N) is viewed as NA z with A z = {δ Norm B (N) δ(z) = z} for any chosen z Γ. The uniqueness of the order p subgroup gives rise to the following generalization of [6, 4.1,4.3] regarding the p-sylow subgroup of the normalizer. Proposition 3.3: If N is a regular subgroup of B then the p-sylow subgroup of Norm B (N) is isomorphic to C p if N is a cross product of two groups, otherwise it is isomorphic to C p C p. 10

Proof. This is a direct consequence of 1.1 and 3.2 above. Specifically, if G has order mp with p > m where the p-sylow subgroup of G is x then the p-sylow subgroup of Hol(G) is either (x, id G ) or (x, id G ), (e G, c x ) where c x is conjugation by x. Moreover, we have the following: Proposition 3.4:[6, 3.8] If N is a regular subgroup of B and M Norm B (N) where M is also regular, then Norm B (N) Norm B (M) and, if Aut(N) = Aut(M) as abstract groups, then Norm B (N) = Norm B (M). The virtue of this is that it gives a criterion for two regular subgroups to have the same normalizer. This is important in that it also gives us information on the p-sylow subgroup of Norm B (N) in terms of that of N. To this end we need to discuss the notion of the opposite group to a given subgroup of B. Definition 3.5: If N B is a regular subgroup, then let N opp = Cent B (N), the centralizer of N in B. For example, λ(γ) opp = ρ(γ) where ρ : Γ B is the right regular representation mentioned above. Several important facts about the opposite group are: Proposition 3.6:[6, 3.2,3.3,3.4,3.5,3.9] If N is a regular subgroup of B then: (a) N opp is also a regular subgroup. (b) N N opp = Z(N) (c) If N is abelian then N = N opp. (d) (N opp ) opp = N (e) Norm B (N) = Norm B (N opp ) (f) N R(Γ) if and only if N opp R(Γ) The upshot of this is the following refinement of 3.3, which for the special 11

case m = 4 is given in [6, Proposition 4.3]. Recall that P = P(λ(Γ)). Proposition 3.7: For regular N of order mp, both N and N opp have unique p-sylow subgroups P(N) and P(N opp ) and (a) If N is of the form P(N) Q(N) then P(N) = P(N opp ) and if N R(Γ), then P(N) = P = P(λ(Γ)). (b) If N = P(N) τ Q(N), where τ is non-trivial, then P(N) P(N opp ) and so the p-sylow subgroup of Norm B (N) is P(N) P(N opp ). Moreover, either P(N) = P or P(N opp ) = P. Proof. For part (a) we already observed that Norm B (N) has a p-sylow subgroup of order p and since P(N) and P(N opp ) both normalize N then they must be identical. If λ(γ) Norm B (N) then obviously P P(N) whence P = P(N). For (b), we know that the p-sylow subgroup of Norm B (N) is isomorphic to C p C p and, since N is non-abelian, Z(N) has no p-torsion and therefore N N opp is trivial and since P(N) centralizes P(N opp ), P(N) P(N opp ) is elementary abelian of order p 2. Now if λ(γ) Norm B (N) then P P(N) P(N opp ) but the question is how P is embedded? For this we recall 2.2 which implies that P(N) = ˆp χ1 and P(N opp ) = ˆp χ2 for linear characters χ 1, χ 2 : Q U p where Q = Q(λ(Γ)). Since P = ˆp ι then we must have ˆp ι = rˆp χ1 + sˆp χ2 and by using 2.3, we find that either r = 1, s = 0 or viceversa, that is ˆp χ1 = ˆp ι or ˆp χ2 = ˆp ι, that is P(N) = P or P(N opp ) = P. As a consequence we have the following which is a generalization (and synthesis) of [6, Theorem 4.4 and Proposition 5.7]. Theorem 3.8: Given Γ with associated P, and N R(Γ) then (a) If N is of the form P(N) Q(N) then N Cent B (P) (b) If N is of the form P(N) τ Q(N) then either N Cent B (P) or N opp Cent B (P) (c) N Norm B (P) 12

As such, if P = P 1, P 2,...,P k are the possible P(N) and if N is a direct product (with P(N) as a factor) then N R(Γ, [M]) implies that P(N) = P = P(N opp ) while if N is a semi-direct product then P(N) P(N opp ) and {N R(Γ, [M]) P(N) = P1 } k = {N R(Γ, [M]) P(N) = Pi } which is useful from a purely enumerative standpoint since, if one wants only R(Γ, [M]), one need not actually determine all the different P i nor even how many there are in the first place since they are mirrored by those containing P 1 = P. From this we can an indication of how the N are distributed within R(Γ). Corollary 3.9: If N R(Γ) is of the from P(N) Q(N) and non-abelian then P(N) = P = P(N opp ) but N N opp. As such R(Γ, [N]) is even. As to the structure of Cent B (P) and Norm B (P) both act to preserve the block structure induced on Γ, specifically Cent B (P) preserves not only the blocks Π i = Supp(π i ) (in fact they are the maximal blocks) but also the ordering of the elements within each block in parallel, as elucidated in [6, Propositions 5.3,5.4,5.5]. The normalizer is a twisted wreath product, in this case, generalizing [6, Theorem 5.6], where Aut(C p ) = U p = (Z/pZ) = u : Theorem 3.10: i=2 where Norm B (P) = (C m p Aut(C p)) S m (ˆb, u s, β)(â, u r, α) = (ˆb + u s β(â), u s+r, βα) where the elements of S m act by coordinate shift on the vectors in the first coordinate and the elements of Aut(C p ) = U p = u act by scalar multiplication. An alternate view of Norm B (P), one which is in sync with the additive notation in and preceding 2.2, is this 13

Proposition 3.11: If U, S GL m (F p ) are respectively, the groups of m m non-zero scalar matrices, m m permutation matrices, then Norm B (P) = F m p US AGL m(f p ) Proof. The point is that the Cp m factor in 3.10 is V, the group generated by the π i that appear in the factorization of P into disjoint p-cycles. The action given by Aut(C p ) on a vector in V is by a scalar matrix u r I for u r U p and the action by S m is equivalent to the action of an m m permutation matrix. A monomial or generalized permutation matrix is one which consists of exactly one non-zero entry in each row and column. The group US consists of those monomial matrices where the non-zero entries are all the same. As far as computational necessity is concerned, since any N in R(Γ) will be a subgroup of Norm B (P), we shall look for N generated by elements of this semi-direct product. For convenience, we note the following operational facts: k 1 (ˆb, u s, β) k = ( u is β i (ˆb), u sk, β k ) (1) i=0 (ˆb, u s, β)(â, u r, α)(ˆb, u s, β) 1 = (ˆb + u s β(â) u r (βαβ 1 )(ˆb), u r, βαβ 1 ) (2) (ˆb, u s, β)(ˆp ι, 1, I)(ˆb, u s, β) 1 = (u sˆp ι, 1, I) (3) As the N must be regular subgroups, they must act without fixed points. We recall [6, Proposition 5.9] by noting that ([a 1,...,a m ], u r, α) acts fixed point freely (viewed as an element of the normalizer of P contained in B = Perm(Γ)) provided either or r = 0 and if a i = 0 then α(i) i (4) r 0 and α fixes no element of {1...m} (5) and based on this we can make another simple but useful observation when determining elements of a given power that act without fixed points. 14

Lemma 3.12: If (â, u r, α) belongs to a regular subgroup of Norm B (P) then α e = I implies (u r ) e = 1. Proof. If (â, u r, α) e = (â, u re, α e ) = (â, u re, I) and u re 1 then by (5) (â, u re, I) has a fixed point. As a result, if (â, u r, α) belongs to a regular subgroup of Norm B (P) then the order of α cannot be a proper divisor of the order of u r. With the above description of Norm B (P), one can do things like calculate powers of elements and make determinations as to their order. However, one must first go back to the definition of R(Γ) namely that it consists of the those N Perm(Γ) normalized by λ(γ), the left regular representation of Γ in P erm(γ) which means that one must explicitly calculate the left regular representations of the different Γ that may arise and then, since P λ(γ) Norm B (P), compute the images of the generators of λ(γ) inside this twisted wreath product and then determine the possible subgroups N of Norm B (P) that are normalized by λ(γ). In [6] this was done for the groups of order 4p to enumerate R(Γ, [M]) for the four (if p 3(mod 4)) or five (if p 1(mod 4)) different classes of groups. From the point of view of a complete classification, we not only obtain the enumeration of R(Γ, [M]) but a descent ready means of computing the Hopf algebra H = L[N] Γ once given the action of Γ on a particular L/K with Galois group Γ. However for the purpose of more easily enumerating R(Γ), we will take a different approach, motivated by the following simple yet, in the author s opinion, extremely important property about regular permutation groups, which is mentioned in passing in [3]. Proposition 3.13: If N, N are regular subgroups of S n which are isomorphic (as abstract groups) then they are, in fact, conjugate as subgroups of S n. This being the case we need not view B as Perm(Γ) for different Γ but rather as S mp = Perm({1,.., mp}) and if there are r distinct isomorphism classes of groups of order mp we shall choose representative regular Γ 1,...,Γ r B and enumerate R(Γ i, [M]), all as subgroups of this fixed B. Since N Norm B (Γ) if and only if σγσ 1 Norm B (σnσ 1 ) then R(Γ, [M]) = R(σΓσ 1, [M]). We will unify our approach still further 15

by defining P = π 1 π 2 π m where π i = (1 + (i 1)p...ip) and choose Q 1,..., Q s to be distinct isomorphism classes of subgroups of B of order m (embedded in Norm B (P)) which act regularly on the Π k as elucidated in 1.7 and 1.8. With this we can determine the linear characters {χ i } of all these Q j and from these, using 2.2, let {Γ t } be the distinct isomorphism classes of subgroups of B amongst ( ˆp χi Q j ) opp as these will all contain P as their order p subgroups and all be contained in Norm B (P). Moreover, these semi-regular ˆp χ will all be generated by products of powers of π 1,...,π m the point being that characters of different Q j can give rise to the same possible ˆp χ. For some classes of groups (such as where m is prime) the choice of the Γ t is relatively simple, but for others, it will be a matter of determining the unique Γ t that arise due to the aforementioned algorithm. 4 Groups of Order (2q+1)2q Here we consider the specifics of applying this plan for computing R(Γ). Here we consider groups of order mp where p = 2q + 1 with q a prime (making p a so-called safe prime and m = 2q = φ(p). Such groups were explored in some detail in [2]. In [2] (and [7, Example 8.7, p.133+] for q = 3),it is observed that there are six isomorphism classes of groups of order p(p 1) where p = 2q + 1 for q prime, namely C mp = C p C 2q C p D q = C p (C q C 2 ) F C 2 = (C p C q ) C 2 (F is Frobenius) D p C q = (C p C 2 ) C q D pq = C pq C 2 = C p D q = C p (C q C 2 ) Hol(C p ) = C p C 2q which can be presented as follows: 16

x, y x p = y 2q = 1 x, a, b x p = a q = b 2 = 1; bx = xb; ax = xa; bab 1 = a 1 x, y x p = y 2q = 1; yxy 1 = x 2 x, y x p = y 2q = 1; yxy 1 = x 1 x, a, b x p = a q = b 2 = 1; bxb 1 = x 1 ; ax = xa; bab 1 = a 1 x, y x p = y 2q = 1; yxy 1 = x u where u = F p = U p = Aut(C p ). where C mp and C p D q are direct products of the order p subgroup by the order m subgroup, either C m or D q. For F C 2, D p C q and Hol(C p ), C p is extended by C m in three different ways, depending on the homomorphism from C m to Aut(C p ). Depending on the homomorphism, one or the other generators of order q and 2 centralizes the order p generator. The group D pq is the only non-trivial semi-direct product of C p and D q, where the order q generator in D q centralizes. In this setting, we will work within B = S mp and set P = π 1 π m where π i = (1+(i 1)p... ip) and so Norm B (P) is isomorphic to the group of 3-tuples (â, u s, α) where â = [a 1,...,a m ] with a i F p, u = U p and α S m. As mentioned above, we consider groups Q j of order m, viewed as subgroups of Norm B (P) acting regularly on {Π 1,..., Π m }. These in turn can be viewed as subgroups of this group of 3-tuples. That is, embedded as regular subgroups of {(ˆ0, 1, α) α S m } = S m. That is, for a regular subgroup of S m (which by a slight abuse of notation we shall also call Q) Q α (ˆ0, 1, α) so we shall identify these Q initially as these regular subgroups of S m. We choose them as follows, let Q 1 = σ = C m and Q 2 = σ 2, δ = D q where σ = (1, 4, 5,...2q 1, 2, 3, 6,..., 2q) σ 2 = (1, 3, 5,..., 2q 1)(2, 4, 6,..., 2q) which we denote σ L σ R δ = (1, 2)(3, 2q)(4, 2q 1) (q, q + 3)(q + 1, q + 2) 17

which are regular subgroups of S m. We take a moment to observe that Q opp 1 = Q 1 of course and that Q opp 2 = σ L σ 1 R, σq where σ q = (1, 2)(3, 4) (2q 1, 2q). For the possible order p subgroups of N R(Γ) we consider linear characters ψ i : C m F p and χ i : D q F p. where ψ i(σ) = u i for i = 0..(m 1) and where χ 0 = ι the trivial character and χ 1 (σ 2 ) = u q and χ 1 (δ) = u 0 = 1. With 2.2 in mind, we have therefore the semiregular subgroups of Norm B (P) normalized by these Q. With ψ i and χ i as above for Q 1,Q 2, we define P i = ˆp ψi for i = 0...m 1 and observe that ˆp χ0 = ˆp ψ0 = P 0 = [1, 1,..., 1] = P and that ˆp χ1 = ˆp ψq = P q = [1, 1, 1, 1,..., 1, 1]. One should also observe (and this is tacit in the discussion preceding 2.2) that given a character χ : Q F p the resulting ˆp χ is an eigenvector under the action of every element of Q. This lends itself very naturally to the construction of P i Q j in Norm B (P) since the first component of each element is identifiable with a vector in F m p. For the ˆp ψ i and ˆp χi defined above we have the following: Lemma 4.1: For σ the generator of Q 1 and σ 2, δ the generators of Q 2 given above: σ(ˆp ψi ) = u 2q iˆp ψi for each i σ 2 (ˆp χ1 ) = ˆp χ1 δ(ˆp χ1 ) = u qˆp χ1 If we consider P i Q j for each P i and Q j above then each such group centralizes P. As such (P i Q j ) opp will contain P. Now it should be observed that not all the (P i Q j ) will give distinct isomorphism classes, but given the presentations for the different isomorphism classes of groups of order mp together with 4.1 above we have the following: Proposition 4.2: With Q i, P j as defined we have P 0 Q 1 = Cp C m P e Q 1 = F C2 for e even, e 0 18

P e Q 1 = Hol(Cp ) for e odd, e q P q Q 1 = Dp C q P 0 Q 2 = Dq C p P q Q 2 = Dpq The following needs to be observed in order to proceed. Proposition 4.3: Each group P i Q j above is a regular subgroup of Norm B (P). Proof. That each P i Q j is a regular subgroup of order mp is due to the fact that each P i is generated by the vectors induced by the characters of the given Q j as in 4.1. To show regularity we show that each such group acts fixed point freely. This is relatively straightforward since a given element is of the form (ˆv, 1, α) where ˆv P i and α Q j viewed as a fixed point free permutation of the blocks Π 1,...,Π 2q. As such, for t {1,...,mp} one has that (ˆv, 1, α)(t) = ˆv(α(t)) and if t Π k for some k then if α = I then this equals ˆv(t) so if ˆv ˆ0 then ˆv(t) t. If α I then α(t) = t in some other block Π k and so, regardless of whether ˆv = ˆ0 we again have (ˆv, 1, α)(t) t. The point is that the only element of P i Q j with fixed points is the identity element (ˆ0, 1, I). With this in mind we choose the six Γ as opposites of the distinct isomorphism classes of the above P i Q j. As to the nature of these opposite groups, one can determine them based on the fact that each will contain P = P 0 as well as basic facts about the presentations of these groups. The choice of Γ is somewhat flexible except for the convention that P(Γ) = P 0 = P. For those groups for which the order p subgroup is cen- 19

tralized, C p C m and D q C p, we have Γ = C p C 2q = (P 0 Q 1 ) opp = P 0 Q 1 = P 0 (ˆ0, 1, σ) Γ = C p D q = (P 0 Q 2 ) opp = P 0 (ˆ0, 1, σ q ), (ˆ0, 1, σ L σ 1 R ) For D p C q and D pq we shall set Γ = D p C q = (P q Q 1 ) opp = P 0 (ˆ0, u q, σ) Γ = D pq = (P q Q 2 ) opp = P 0 (ˆ0, u q, σ q ), (ˆ0, 1, σ L σ 1 R ) For F C 2 and Hol(C p ) there is a slight arbitrariness as to the possible choices of Γ since they can be the opposites of any of the q 1 choices of P e Q 1. The only effect will be a slight asymmetry in the distribution of the P(N). So we may as well choose Γ = F C 2 = (P 2 Q 1 ) opp = P 0 (ˆ0, u 2, σ) Γ = Hol(C p ) = (P 1 Q 1 ) opp = P 0 (ˆ0, u, σ) One should note something interesting about the permutation components of these opposite groups, namely that they are precisely the opposites of the Q j that we identified above. We will need the following technical fact, which can be viewed as a small application of 3.10. Lemma 4.4: If x = (a 1, a 2,...,a q ), y = (b 1, b 2,...,b q ) are elements with disjoint support in S 2q = Perm({1,..., 2q}) then Norm S2q ( xy ) contains 2qφ(q) elements z of order 2q, half of which centralize xy and are such that z 2 = (xy) 2 and the other half invert xy and satisfy z 2 = (xy 1 ) 2. Also, Norm S2q ( xy ) contains two subgroups isomorphic to D q, which are opposites of each other, one of which is contained in Cent S2q (xy). 20

Proof. First we observe by 3.10 that Norm S2q ( xy ) is a twisted wreath product, specifically it is isomorphic to ((C q C q ) ) S 2 which of course can be viewed as a subgroup of AGL 2 (F q ) where represents multiplication of vectors by scalars in F q. To that end, one may readily count how many elements have order 2q. In particular, since a typical element is a 3-tuple (â, u r, α) with â = (a 1, a 2 ) F 2 q, u = F q and α S 2, then one may show that (â, u r, α) = 2q provided that α = (1, 2), and either a 1 a 2 if u r = 1 or a 1 a 2 if u r = 1. This yields precisely 2(q 2 q) = 2q(q 1) = 2qφ(q) elements as claimed. We can exhibit the particular elements of order 2q (as elements in S 2q ) as follows. Let t 0 = (a 1, b 1 )(a 2, b 2 ) (a q, b q ) t 1 = (a 1, b 2 )(a 2, b 3 ) (a q, b 1 ). t q 1 = (a 1, b q )(a 2, b 1 ) (a q, b q 1 ) τ 0 = (a 1, b 1 )(a 2, b q ) (a q, b 2 ) τ 1 = (a 1, b 2 )(a 2, b 1 ) (a q, b 3 ). τ q 1 = (a 1, b q )(a 2, b q 1 ) (a q, b 1 ) and consider the elements xyt i and xy 1 τ i. One may verify that each t i centralizes x and y so that xyt i centralizes xy and that τ i xτ 1 i = y 1 and τ i yτ 1 i = x 1 so that therefore xy 1 τ i inverts xy. Each of these elements has order 2q (and is therefore a 2q-cycle) and each generates a distinct subgroup of order 2q. Moreover (xyt i ) 2 = (xy) 2 xy while (xy 1 τ i ) 2 = (xy 1 ) 2 xy 1. The conclusion we get is that if a 2q-cycle z inverts or centralizes xy then z 2 xy 1 or xy. The groups xy 1, t i for each i are all equal and isomorphic to D q and xy, τ i are all equal and isomorphic to D q, the former lies within Cent S2q (α) while the latter does not. Moreover xy 1, t i opp = xy, τ i since each clearly centralizes the other. With this framework in place, we can now list the sizes of R(Γ, [M]) for the 36 possible pairings of Γ and [M] and then show specific generators for some of the different Γ and M. 21

Theorem 4.5: The cardinalities of R(Γ, [M]) are as follows, where we define R(Γ, [M]; P i ) to be those N with P(N) = P i and we only include those R(Γ, [M]; P i ) which are non-empty. R(C mp ) R(C mp, [C mp ]; P 0 ) = 1 R(C mp, [C p D q ]; P 0 ) = 2 R(C mp, [F C 2 ]; P 0 ) = q 1 R(C mp, [F C 2 ]; P i ) = 1 for i = 2, 4,...2q R(C mp, [D p C q ]; P 0 ) = 1 R(C mp, [D p C q ]; P q ) = 1 R(C mp, [D pq ]; P 0 ) = 2 and R(C mp, [D pq ]; P q ) = 2 R(C mp, [Hol(C p )]; P 0 ) = q 1 R(C mp, [Hol(C p )]; P i ) = 1 for i {1, 3,..., 2q 1} {q} R(C p D q ) R(C p D q, [C mp ]; P 0 ) = q R(C p D q, [C p D q ]; P 0 ) = 2 R(C p D q, [F C 2 ]) = 0 R(C p D q, [D p C q ]; P 0 ) = q R(C p D q, [D p C q ]; P q ) = q R(C p D q, [D pq ]; P 0 ) = 2 and R(C p D q, [D pq ]; P q ) = 2 R(C p D q, [Hol(C p )]) = 0 R(F C 2 ) R(F C 2, [C mp ]; P 0 ) = p R(F C 2, [C p D q ]; P 0 ) = 2p R(F C 2, [F C 2 ]; P 0 ) = p(q 2) + 1 R(F C 2, [F C 2 ]; P 2 ) = 1 R(F C 2, [F C 2 ]; P i ) = p for i = 4,...,2q 22

R(F C 2, [D p C q ]; P 0 ) = p R(F C 2, [D p C q ]; P q ) = p R(F C 2, [D pq ]; P 0 ) = 2p and R(F C 2, [D pq ]; P q ) = 2p R(F C 2, [Hol(C p )]; P 0 ) = p(q 1) R(F C 2, [Hol(C p )]; P i ) = p for i {1, 3,..., 2q 1} {q} R(D p C q ) R(D p C q, [C mp ]; P 0 ) = p R(D p C q, [C p D q ]; P 0 ) = 2p R(D p C q, [F C 2 ]; P 0 ) = p(q 1) R(D p C q, [F C 2 ]; P i ) = p for i = 2, 4,...2q R(D p C q, [D p C q ]; P 0 ) = 1 R(D p C q, [D p C q ]; P q ) = 1 R(D p C q, [D pq ]; P 0 ) = 2 and R(D p C q, [D pq ]; P q ) = 2 R(D p C q, [Hol(C p )]; P 0 ) = p(q 1) R(D p C q, [Hol(C p )]; P i ) = p for i {1, 3,..., 2q 1} {q} R(D pq ) R(D pq, [C mp ]; P 0 ) = qp R(D pq, [C p D q ]; P 0 ) = 2p R(D pq, [F C 2 ]) = 0 R(D pq, [D p C q ]; P 0 ) = q R(D pq, [D p C q ]; P q ) = q R(D pq, [D pq ]; P 0 ) = 2 and R(D pq, [D pq ]; P q ) = 2 R(D pq, [Hol(C p )]) = 0 R(Hol(C p )) R(Hol(C p ), [C mp ]; P 0 ) = p R(Hol(C p ), [C p D q ]; P 0 ) = 2p R(Hol(C p ), [F C 2 ]; P 0 ) = p(q 1) 23

R(Hol(C p ), [F C 2 ]; P i ) = p for i = 2, 4,...2q R(Hol(C p ), [D p C q ]; P 0 ) = p R(Hol(C p ), [D p C q ]; P q ) = p R(Hol(C p ), [D pq ]; P 0 ) = 2p and R(Hol(C p ), [D pq ]; P q ) = 2p R(Hol(C p ), [Hol(C p )]; P 0 ) = p(q 2) + 1 R(Hol(C p ), [Hol(C p )]; P 1 ) = 1 R(Hol(C p ), [Hol(C p )]; P i ) = p for i {3,..., 2q 1} {q} Proof. As in [6] the determination of R(Γ, [M]) is based on determining the generators of regular subgroups of N Norm B (P) (viewed as 3-tuples (â, u r, α)) with various P i as P(N) and then using the requirement that they be normalized by Γ to narrow down the possibilities. Given that many of the cases involve similar sorts of calculations we shall not demonstrate the enumeration of the 36 cases above, but will give a sample to indicate the basic techniques. R(C mp, [C p D q ]) Here Γ is generated by ˆp ψ0 = ˆp ι = [1, 1,..., 1], where of course P(Γ) = ˆp ψ0 = P 0 = P, and (ˆ0, 1, σ) where σ = (1, 4, 5,...2q 1, 2, 3, 6,..., 2q). If N R(C mp, [C p D q ]) where P(N) = P 0 then it has two other generators (â, u r, α) of order q and (ˆb, u s, β) of order 2 (which give the copy of D q inside N), where the order 2 generator inverts the order q generator and both centralize P(N) = P 0. By (3) we must therefore have r = s = 0. As such we must have α = q and β = 2 otherwise these elements would not have order q and 2. Since α = q then α is either a q-cycle or a product of two q-cycles. Since (â, 1, α ) = q then â + α(â) + + α q 1 (â) = ˆ0 and so if α were a q-cycle then if k is not in the support of α then this would imply that qa k = 0 where a k is the k-th component of â which would (since a k F p ) imply that a k = 0, and similarly for any other component of â not moved by α. But then by (4) we would have that (â, 1, α) has fixed points. 24

As such, α must be a product of two q-cycles and similarly β must be a product of q 2-cycles. Now we consider the role of Γ. Since Γ normalizes N then in particular (ˆ0, 1, σ) must normalize N and since the copy of D q is characteristic in N then it must be normalized by this element as well. Since (â, 1, α) is normal in the D q component, then we must have (ˆ0, 1, σ)(â, 1, α)(ˆ0, 1, σ 1 ) = (σ(â), 1, σασ 1 ) = (â + + α k 1 (â), 1, α k ) for some k 0 which implies that σ Norm S2q ( α ). But since α has order q and σ has order m = 2q then σ must either centralize or invert α. As such, (σ(â), 1, σασ 1 ) is either (â, 1, α) or ( α 1 (â), 1, α 1 ) If σ(â) = â then â = a[1, 1,..., 1] = aˆp ψ0 and so certainly α(â) = â which, since â + α(â) + + α q 1 (â) = qâ = ˆ0, means that â = ˆ0. And, if σ(â) = α 1 (â) we have ασ(â) = â but by 4.4 ασ is a 2q cycle which means that half of the components of â are the negatives of the other half (i.e. the components of â consist of q occurrences of a and q occurrences of a for some a). However, since (â, 1, α) has order q then â+α(â)+ +α q 1 (â) = ˆ0 which means that for each coordinate a i of â that a i + a α 1 (i) + + a α q 1 (i) = 0. However this implies that â = ˆ0 since there is no way to choose an odd number of elements from {a, a} and have them add up to 0 unless a = 0. One should note that, by 4.4, since σ Norm( α ) where α = xy (a product of disjoint q-cycles) then either σ = (xyt i ) k (centralizing α) or (xy 1 τ i ) k for (k, 2q) = 1 (inverting α). As such, either σ 2 = σ L σ R = x 2k y 2k or x 2k y 2k. And since (k, 2q) = 1 then (2k, q) = 1 and since (ˆ0, 1, α) n = (ˆ0, 1, α n ), then we may assume that, in fact, either α = σ L σ R or α = σ L σ 1 R. As (ˆb, 1, β) inverts (â, 1, α) = (ˆ0, 1, α) then (ˆ0, 1, α 1 ) = (ˆb βαβ 1 (ˆb), 1, βαβ 1 ) and so βαβ 1 = α 1 and therefore α, β gives a copy of D q embedded in S 2q. Again, by 4.4, α, β must be one of the two copies of D q in Norm S2q ( α ). Since α is either σ L σ R or σ L σ 1 R then α, β is either Q 2 or Q opp 2. In either case, looking at the first coordinate, we get ˆb α 1 (ˆb) = ˆ0 so that α(ˆb) = ˆb. Since (ˆb, 1, β) has order 2 then β(ˆb) = ˆb. If we conjugate (ˆb, 1, β) by (ˆ0, 1, σ) we get (σ(ˆb), 1, σβσ 1 ) which must be one of the order 2 elements 25

of N, namely (ˆ0, 1, α) k (ˆb, 1, β) = (α k (ˆb), 1, α k ) = (ˆb, 1, α k ). This implies that σ(ˆb) = ˆb which means that ˆb = bˆp ψ0 and since β(ˆb) = ˆb then ˆb must be ˆ0. As such, the distinct choices of N correspond to the distinct possibilities for α, β, namely Q 2 or (Q 2 ) opp. By 3.8 there are no others with other P i as P(N) since the order p subgroup is a direct factor. R(Hol(C p ), [Hol(C p )]) Here Γ = P 0 (ˆ0, u, σ) and we start by considering those N normalized by Γ with P(N) = P 0. In N there is an order m = 2q element which conjugates ˆp ψ0 to uˆp ψ0 (i.e. the order m = φ(p) automorphism of C p ) which we shall denote by ( ˆd, u r, δ). Since this element conjugates ˆp ψ0 to uˆp ψ0 then r = 1 of course, and since (ˆd, u, δ) = m then we must have by 3.12 that δ = m and in fact that δ is an m-cycle or else ( ˆd, u, δ) would have fixed points. Now, since P(N) = P 0 and P(Γ) = P 0 too, then Γ normalizes N if and only if (ˆ0, u, σ) conjugates ( ˆd, u, δ) to another element of order m in N. In Hol(C p ) there are (q 1)p elements of order m = 2q, which can be seen by observing that if Hol(C p ) = g, h where g = p and h = m then the elements of order m are of the form g i h j where gcd(j, m) = 1. So in N, if we conjugate ( ˆd, u, δ) by (ˆ0, u, σ) Γ we get (ˆ0, u, σ)(ˆd, u, δ)(ˆ0, u 1, σ 1 ) = (uσ(ˆd), u, σδσ 1 ) which must therefore be (kˆp ψ0 + ˆd, u, δ) for some k. As such, δ must equal σ r where gcd(r, m) = 1 since the only elements of order m in S m that normalize x for x = (a 1, a 2,...,a m ) are the generators of x. Now, the question is, what are the possible choices for ˆd F m p? Since ( ˆd, u, σ r ) = m then we have ˆd + uσ r ( ˆd) + u 2 σ 2r ( ˆd) + + u m 1 σ (m 1)r ( ˆd) = ˆ0 (6) which, since ˆd = [d 1, d 2,..., d m ] is a vector in F m p the above yields (ostensibly) m equations in the d i. However, one can verify that these equations are all scalar multiples of each other, as such we obtain d 1 + ud σ r (1) + u 2 d σ 2r (1) + + u (m 1) d σ (m 1)r (1) = 0 (7) One should note that if we replace any ˆd by ˆd + kˆp ψ0 = ˆd + [k, k,...,k] that (7) still holds since the sum over all the elements of U p = F p is 0 mod 26

p. That N is normalized by Γ, in particular (ˆ0, u, σ), implies, as indicated above, that [ud σ 1 (1), ud σ 2 (1),...,ud σ (m 1) (1)] = [d 1, d 2,...,d m ] + k[1, 1,..., 1] which, component by component yields the recurrence relation ud σ 1 (i) = d i + k whose values at the indices {1,...,m} are only a function of d 1 and k. If we iterate this over subsequent powers of σ 1, using the fact that σ acts transitively on {1,...,m}, we find that d σ t (1) = 1 ( ) d u t 1 + ut 1 u 1 k (8) So (8) gives a parametrization, keyed to (d 1, k) F p F p of the possible ˆd, but we must factor in (7) as well, in particular we need to determine when ( ( )) ( ( )) 1 d 1 +u d u r 1 + ur 1 1 + +u m 1 d u 1 u (m 1)r 1 + u(m 1)r 1 (9) u 1 equals 0, which can be better understood if we set v = u r for then the above becomes ( d 1 1 + u ) v + + um 1 + k ( ) u um 1 (v 1) + + v m 1 u 1 v v m 1 (vm 1 1) ( ) (10) where, if u 1 then 1 + u + + um 1 = 0 and where v v v m 1 ( k u u 1 v = k u 1 ) v m 1 (vm 1 1) um 1 (v 1) + + ( (u + u 2 + + u m 1) ( u v + u2 v + + um 1 2 v m 1 )) which, if also u/v 1, equals (k/u 1)( 1 ( 1)) = 0, for any (d 1, k). If u/v = 1 then (10) becomes (d 1 m) (mk/(u 1)) which is zero only when d 1 = k/(u 1). Since u/v = u 1 r then we have the following possibilities. 27

If r 1 then there are p 2 choices for ˆd such that ( ˆd, u, σ r ) and since all other order m elements in N (with u as their middle coordinate and σ r as their permutation coordinate) are of the form (lˆp ψ0 + ˆd, u, σ r ) (i.e. p of them) then for r 1 these p 2 possible ˆd must be distributed in p different N. For r = 1 we get only p choices for ˆd which means all p different ( ˆd, u, σ) must lie within the same N. As such there are (q 2)p + 1 choices of N with P(N) = P 0. For those N such that P(N) P 0 we have that two generators, ˆp ψi and ( ˆd, u s, δ) where the former has order p of course and the latter order m = 2q = φ(p). As before, regularity forces δ to be an m-cycle. Moreover, since N must be normalized by Γ = ˆp ψ0, (ˆ0, u, σ) then we again conclude that δ = σ r for some r relatively prime to m. As such we have ( ˆd, u s, σ r )(ˆp ψi, 1, I)(ˆd, u s, σ r ) 1 = (u s σ r (ˆp ψ0 ), 1, I) = (u s u (m i)r (ˆp ψ0 ), 1, I) which by 4.1 must equal (uˆp ψ0, 1, I). As such we must have s ir 1 (mod m). Also, and this wasn t necessary to check for when P(N) = P 0, we must determine when ˆp ψ0 normalizes N. Specifically, (ˆp ψ0, 1, I)(ˆd, u s, σ r )( ˆp ψ0, 1, I) = ((1 u s )ˆp ψ0 + ˆd, u s, σ r ) which must equal (kˆp ψi + ˆd, u s, σ r ) which can only happen if k = 0 and u s = 1, that is s = 0. As such, we must have ir 1(mod m), so that N = ˆp ψi, ( ˆd, 1, σ r ). As above, we must determine when ( ˆd, 1, σ r ) has order m. As above we get a single equation d 1 + d σ r (1) + d σ 2r (1) + + d σ (m 1)r (1) = 0 (11) Now, if ˆp ψi = [v 1, v 2,...,v m ] then, by 2.2 and the nature of ψ i as given at the beginning, we have that v σ t (i) = u ti. As such, the condition that uσ(ˆd) = kˆp ψi + ˆd gives a recurrence relation which can be solved to give 28

an explicit formula for each component of ˆd, again because the orbit of 1 under the powers of σ is all of {1,...,m}: ( t ) d σ t (1) = u t d 1 k u li+(t l) The order condition in (11) yields at first a somewhat complicated collection of geometric sums which (viewing r = m r(mod m)) requires ( ( m 1 ) d 1 1 + u m r + + u m (m 1)r) n k u li+(m nr l) = 0 The first parenthesized expression is 0 for any (d 1, k) F p F p and the second parenthesized expression is 0 for any (d 1, k) provided i 1, and if i = 1 then r = 1 and the above is therefore km/(u 1) which is therefore 0 only if k = 0. The enumeration of R(Hol(C p )) is in agreement with [2, Theorems 2.1 and 4.1]. R(C p D q, [F C 2 ]) This case contrasts markedly from the previous one in that this class is empty. Here Γ is P 0 (ˆ0, 1, σ 2 ), (ˆ0, 1, δ) and if N is isomorphic to F C 2 and contains P 0 then it also contains a generator of order 2q of the form (â, u r, α). Since (â, u r, α)(ˆp ψ0, 1, I)(â, u r, α) 1 = (2ˆp ψ0, 1, I) then u r = 2 More importantly, we must have that α has order 2q and, again by regularity concerns, must therefore be a 2q-cycle. If Γ normalizes N then conjugating (â, 2, α) by (ˆ0, 1, σ 2 ) and (ˆ0, 1, δ) must yield other order 2q elements of N, in particular (ˆp χ0 + â, 2, α) which implies that σ 2, δ Cent S2q (α). However, since α is a 2q-cycle then Cent S2q (α) = α but σ 2, δ = D q which means no such α exists. As such, there are no N R(C p D q, [F C 2 ]) with P(N) = P 0 and therefore none with P(N) P 0 either. Note, by nearly identical arguments one can show that R(C p D q, [Hol(C p )]), R(D pq, [F C 2 )]), R(D pq, [Hol(C p )]), are all zero too. In summary we have the following table, where R(Γ i, [Γ j ]) is the entry in the i th row and j th column: 29 l=1 n=1 l=1

C mp C p D q F C 2 C q D p D pq Hol(C p ) C mp 1 2 2(q-1) 2 4 2(q-1) C p D q q 2 0 2q 4 0 F C 2 p 2p 2(p(q-2)+1) 2p 4p 2p(q-1) D p C q p 2p 2p(q-1) 2 4 2p(q-1) D pq qp 2p 0 2q 4 0 Hol(C p ) p 2p 2p(q-1) 2p 4p 2(p(q-2)+1) 5 Computer Calculations The description of these N as subgroups of Norm B (P), which has this computationally convenient structure, lends itself very well to being implemented in a computer algebra system such as GAP [4]. This was done by the author in the development of this work, especially in gathering empirical information about some specific cases, for example with m = 6 and p = 7. References [1] N.P. Byott. Hopf-galois structures on galois field extensions of degree pq. J. Pure Appl. Algebra, 188(1-3):45 57, 2004. [2] L.N. Childs. On hopf galois structures and complete groups. New York J. Mathematics, 9:99 116, 2003. [3] J. Dixon. Maximal abelian subgroups of the symmetric groups. Canadian J. Math, 23:426 438, 1971. [4] The GAP Group. GAP Groups, Algorithms, and Programming, Version 4.3, 2002. (http://www.gap-system.org). [5] C. Greither and B. Pareigis. Hopf galois theory for separable field extensions. J. Algebra, 106:239 258, 1987. [6] T. Kohl. Groups of order 4p, twisted wreath products and hopf-galois theory. J. Algebra, 314:42 74, 2007. 30

[7] J. Moody. Groups for Undergraduates. World Scientific, Singapore, 1994. [8] H. Weilandt. Finite Permutation Groups. Academic Press, New York, 1964. 31