New York Journal of Mathematics, vol. 9 (2003)

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 = K Γ

, where

K Γ

acts on

L

via the natural action of the Galois group

Γ

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

K Γ

. The Hopf algebras

H

that arise have the property that

L \otimes_{K} H \equiv L G

, 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 \to Hol (G) \subset 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

Perm (G)

. One sees easily that

Hol (G)

contains the image of the right regular representation

ρ : G \to Perm (G)

and also

Aut (G)

; then

Hol (G) = ρ (G) \cdot Aut (G)

and is isomorphic to the semidirect product

G ⋊ Aut (G)

. The equivalence relation on regular embeddings is by conjugation by elements of

Aut (G)

inside

Hol (G)

β \sim β^{'}

if there exists

γ

Aut (G)

so that for all

g

G

γ β (g) γ^{- 1} = β^{'} (g)

Thus the number

e (Γ, G)

H

-Hopf Galois structures on

L / K

where the associated group of

H

G

depends only on

G

and the Galois group

Γ

, and reduces to a purely group-theoretic problem.

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 = 2 q + 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 $β : Γ \to Hol (G)$ 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

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

;

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 $G \times Aut (G)$ as sets, the operation is $(g, α) \cdot (g^{'}, α^{'}) = (g α (g^{'}), α α^{'})$ for $g, g^{'} \in G, α, α^{'} \in Aut (G)$ . 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 $β : Γ \to Hol (G)$ is the same as determining whether the function (non-homomorphism) $π_{1} β : Γ \to G$ 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 $1 \to Z (G) \to G \to Aut (G) \to O (G) \to 1$ where the map $C : G \to Aut (G)$ is conjugation: $C (g) (x) = g x g^{- 1}$ for $g, x \in G$ , and $Z (G)$ is the center of $G$ .

Suppose $Z (G) = (1)$ and the composition of the 1-1 homomorphism $β : Γ \to Hol (G)$ with the homomorphism $Hol (G) \to O (G)$ 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 $j : G ⋊ Inn (G) \to G \times G$ by $j (g, C (h)) = (g h, h)$ for $g, h \in G$ , with inverse sending $(g, h)$ to $(g h^{- 1}, C (h))$ . Letting $p_{i} : G \times G \to G$ be projection onto the $i$ th factor, $i = 1, 2$ , the homomorphism $β$ yields homomorphisms $β_{1} = p_{1} j β$ and $β_{2} = p_{2} j β : Γ \to G$ such that $β (γ) = (β_{1} (γ) β_{2} {(γ)}^{- 1}, C (β_{2})) .$ Then $β$ is regular iff ${π_{1} β (γ) | γ \in Γ} = {β_{1} (γ) β_{2} {(γ)}^{- 1} | γ \in Γ} = G .$ Thus whenever $Z (G) = (1)$ and $β (Γ) \subset G ⋊ Inn (G)$ , we may describe $β$ , and in particular the function $π_{1} β : Γ \to G$ , in terms of the homomorphisms $β_{1}, β_{2} : Γ \to G$ , namely, $π_{1} β (γ) = β_{1} (γ) β_{2} {(γ)}^{- 1} .$

A class of groups $G$ where $Z (G) = (1)$ and $β (Γ) \subset G ⋊ Inn (G)$ 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 $n \geq 3, n \neq 6$ , 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 $β : G \to Hol (G)$ 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 $S_{n}, n \geq 4$ . 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) = 2 p^{e - 1} ϕ (p^{e - 1}) + 2 p^{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

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 $β : G \to Hol (G)$ may be decomposed as $β (g) = β_{1} (g) β_{2} {(g)}^{- 1} C (β_{2} (g))$ for homomorphisms $β_{1}, β_{2} : G \to G$ . So we begin by describing the homomorphisms from $G$ to $G$ .

Let $α : G \to G$ be a homomorphism. Then $α$ is determined by $\begin{matrix} α (1, 1) & = & (m, c) of order dividing p^{e}, and \\ α (0, b) & = & (n, d) of order dividing p^{e - 1} (p - 1) . \end{matrix}$ If $α (1, 1)$ has order dividing $p^{e}$ , then $c = 1$ . To see this, first note that for any $s > 0$ , ${(m, c)}^{s} = (m (1 + c + c^{2} + \dots c^{s - 1}), c^{s})$ so $c$ must have order dividing $p^{e}$ , which implies that $c \equiv 1 (mod p)$ . 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 $d \equiv 1 (mod p)$ 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: $(m (1 + c + \dots c^{b - 1}), c^{b}) (n, d) = (n, d) (m, c),$ hence $(m (1 + c + \dots c^{b - 1}) + c^{b} n, c^{b} d) = (n + d m, d c) .$ Thus $c^{b} d = c d$ . Since $d$ is a unit modulo $p^{e}$ it follows that $c^{b} = c$ , hence $c^{b - 1} = 1$ . But since $c = 1 + p f$ for some $f$ , $1 = {(1 + p f)}^{b - 1} .$ Since $b - 1$ is relatively prime to $p$ and ${(1 + p f)}^{p^{e - 1}} = 1$ , it follows that $1 + p f = 1$ , hence $f = 0$ and $c = 1$ .

Thus for any homomorphism $α : G \to G$ , $α (1, 1) = (m, 1)$ for some $m$ .

Since $c = 1$ , the requirement $(m (1 + c + \dots c^{b - 1}) + c^{b} n, c^{b} d) = (n + d m, d c)$ becomes $(m b + n, d) = (n + d m, d)$ which implies that $b m = d m$ . Thus if $m \neq 0$ , then $b \equiv d (mod p)$ , 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 $\begin{matrix} (h, c) α (1, 1) {(h, c)}^{- 1} & = & (h, c) (m, 1) (- c^{- 1} h, c^{- 1}) \\ = & (h + c m - h, 1) \\ = & (c m, 1) . \end{matrix}$ We choose $c$ so that $c m = 1$ . Then we choose $h$ so that $\begin{matrix} (0, b) = (h, c) α (0, b) {(h, c)}^{- 1} & = & (h, c) (n, b) (- c^{- 1} h, c^{- 1}) \\ = & (h + c n, b c) (- c^{- 1} h, c^{- 1}) \\ = & (h + c n - b h, b) \\ = & ((1 - b) h + c n, b) : \end{matrix}$ we set $h = - {(1 - b)}^{- 1} c n$ , 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 $C (h, c) α : G \to G$ 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 ${β_{1} (g) β_{2} {(g)}^{- 1} | g \in G} = G,$ or equivalently, $π_{1} β = β_{1} \cdot {β_{2}}^{- 1}$ 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 $\begin{matrix} β_{1} (p^{e - 1} l, 1) β_{2} {(p^{e - 1} l, 1)}^{- 1} & = & (p^{e - 1} l m_{1}, 1) (- p^{e - 1} l m_{2}, 1) \\ = & (0, 1) (0, 1) = (0, 1) \end{matrix}$ for all $l$ , and so $β_{1} \cdot {β_{2}}^{- 1}$ is not 1-1.

Suppose both $β_{1}$ and $β_{2}$ are automorphisms. If $m_{1} \equiv m_{2} (mod p)$ then $\begin{matrix} β_{1} \cdot {β_{2}}^{- 1} (p^{e - 1}, 1) & = & (p^{e - 1} m_{1} - p^{e - 1} m_{2}, 1) \\ = & (0, 1) \\ = & β_{1} \cdot {β_{2}}^{- 1} (0, 1) \end{matrix}$ so $β_{1} \cdot {β_{2}}^{- 1}$ is not 1-1. If $m_{1} ≢ m_{2} (mod p)$ , then let $s (m_{1} - m_{2}) = n_{1} - n_{2}$ . Then $\begin{matrix} β_{1} \cdot {β_{2}}^{- 1} (0, b) & = & (n_{1} - n_{2}, 1) \\ = & ((m_{1} - m_{2}) s, 1) \\ = & β_{1} \cdot {β_{2}}^{- 1} (s, 1), \end{matrix}$ so again, $β_{1} \cdot {β_{2}}^{- 1}$ 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 $g, h, k \in G$ we have $\begin{matrix} (1, C (g)) (h, C (k)) (1, C {(g)}^{- 1}) & = & (C (g) (h), C (g) C (k) C {(g)}^{- 1}) \\ = & (g h g^{- 1}, C (g k g^{- 1})) . \end{matrix}$ Thus if $β (x) = (β_{1} (x) β_{2} {(x)}^{- 1}, C (β_{2} (x)))$ for $x \in G$ , then $β \sim β^{'}$ with $\begin{matrix} β^{'} (x) & = & (g β_{1} (x) β_{2} {(x)}^{- 1} g^{- 1}, C (g β_{2} (x) g^{- 1})) \\ = & (g β_{1} (x) g^{- 1} \cdot {(g β_{2} (x) g^{- 1})}^{- 1}, C (g β_{2} (x) g^{- 1})) : \end{matrix}$ we get from $β$ to $β^{'}$ by simultaneously conjugating $β_{1}$ and $β_{2}$ by $g \in G$ .

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 $\begin{matrix} β_{1} (1, 1) & = & (m, 1) \\ β_{1} (0, b) & = & (n, d) \end{matrix}$ for some $m$ divisible by $p$ , some $d \neq 1$ and some $n$ . Then $\begin{matrix} β_{1} (l, b^{k}) & = & β_{1} (l, 1) β_{1} (0, b^{k}) \\ = & (l m, 1) (n (\frac{d^{k} - 1}{d - 1}), d^{k}) \\ = & (l m + n (\frac{d^{k} - 1}{d - 1}), d^{k}) . \end{matrix}$ For any $h, r$ we want to find $l, k$ so that $\begin{matrix} (**) & β_{1} (l, b^{k}) {(l, b^{k})}^{- 1} & = & (l m + n (\frac{d^{k} - 1}{d - 1}), d^{k}) (- b^{- k} l, b^{- k}) \\ = & (l m + n (\frac{d^{k} - 1}{d - 1}) - d^{k} b^{- k} l, d^{k} b^{- k}) \\ = & (h, b^{r}) . \end{matrix}$ In order that for any $r$ , there is a $k$ so that $b^{r} = {(d b^{- 1})}^{k}$ , we require that $d b^{- 1}$ generates ${Z_{p^{e}}}^{*}$ . Now for $β_{1}$ to be a homomorphism, we need $b m = d m$ , and this is possible only if $m = 0$ or $b \equiv d (mod p)$ . But in the latter case, $d b^{- 1} \equiv 1 (mod p)$ , 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 $\begin{matrix} β_{1} (1, 1) & = & (0, 1) \\ β_{1} (0, b) & = & (n, d), \end{matrix}$ and (**) becomes $(n (1 + d + \dots + d^{k - 1}) - b^{f k} l, b^{f k}) = (h, b^{r}),$ which is solvable for all $n$ and for all $f$ such that $b^{f}$ generates ${Z_{p^{e}}}^{*}$ . We have two cases:

If $d = b^{f + 1} \equiv 1 (mod p)$ , 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

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 $p - 1 = 2 q$ . Then there are at least five non-isomorphic groups $G$ of order $2 q p$ other than $Γ$ , namely, $Z_{2 q p}, D_{q p}, D_{p} \times Z_{q}, D_{q} \times Z_{p}$ , and $(Z_{p} ⋊ Z_{q}) \times Z_{2}$ (where $Z_{q}$ is identified as the subgroup of $Aut (Z_{p})$ of index 2, and $D_{n}$ is the dihedral group of order $2 n$ ). The five groups are non-isomorphic because their centers have orders $2 p q, 1, q, p$ 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 $2 p q$ , 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 = 2 q + 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 $2 q = p - 1$ . Write $Z_{p}$ additively and view $Z_{a}$ as the cyclic subgroup of order $a$ inside ${Z_{p}}^{*} = Aut (Z_{p})$ and $G \subset Hol (Z_{p})$ . Let $b$ generate $Z_{a}$ . If $α$ is an automorphism of $G$ , then:

$α (1, 1) = (m, 1)$ for $p$ not dividing $m$ , and
$α (0, b) = (n, b)$ 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 $(l, d) (m, b^{r}) {(l, d)}^{- 1} = (l + d m - b^{r} l, b^{r}) \in G .$ Hence the conjugation map $C : Hol (Z_{p}) \to Aut (G)$ , $C (l, d) (m, b^{r}) = (l, d) (m, b^{r}) {(l, d)}^{- 1} = (l + d m - b^{r} l, b^{r}),$ 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 = 2 q + 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 $2 q$ , and hence a subgroup $K$ of order $q$ . Then $G_{p} K = J$ is a subgroup of $G$ of order $p q$ , 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 $J ≅ Z_{p q}$ then $G ≅ Z_{p q} ⋊_{α} Z_{2}$ where $α : Z_{2} \to Aut (Z_{p q}) ≅ Z_{p - 1} \times Z_{q - 1}$ has four possibilities, $α (- 1) = (\pm 1, \pm 1)$ . These yield $G ≅ D_{p q}, D_{p} \times Z_{q}, D_{q} \times Z_{p}$ and $Z_{2 p q}$ .

If $J ≅ Z_{p} ⋊ Z_{q} \subset Hol (Z_{p})$ , then $G ≅ J ⋊_{α} Z_{2}$ where $α : Z_{2} \to Aut (Z_{p} ⋊ Z_{q}) ≅ Inn (Hol (Z_{p}))$ . If $α$ is trivial, we obtain $(Z_{p} ⋊ Z_{q}) \times Z_{2}$ . Otherwise, the elements of order 2 in $Inn (Hol (Z_{p}))$ are of the form $C (l, - 1)$ for any $l \in Z_{p}$ . Define $α_{l} : Z_{2} \to Inn (Hol (Z_{p}))$ by $α_{l} (- 1) = C (l, - 1)$ . One checks easily that $(Z_{p} ⋊ Z_{q}) ⋊_{α_{0}} Z_{2} ≅ Z_{p} ⋊ Z_{2 q} = Hol (Z_{p})$ by the map sending $((a, b), ε)$ to $(a, ε b)$ for $a \in Z_{p}$ , $b \in Z_{q} \subset Z_{p - 1}$ , $ε \in Z_{2} \subset Z_{p - 1}$ . Now $〈 α_{0} 〉 = 〈 C (0, - 1) 〉$ and $〈 α_{l} 〉 = 〈 C (l, - 1) 〉$ are conjugate in $Inn (Hol (Z_{p}))$ for $l \neq 0$ by $C (m, 1)$ where $2 m \equiv l (mod p)$ : $C (m, 1) C (0, - 1) C (- m, 1) = C (2 m, - 1)$ . 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.

On Hopf Galois structures and complete groups

Lindsay N. Childs¹

Department of Mathematics and Statistics
University at Albany
Albany, NY 12222

ABSTRACT

1. Regular embeddings

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

3. Groups of order $p (p - 1)$

4. Nonuniqueness

On Hopf Galois structures and complete groups

Lindsay N. Childs1

Department of Mathematics and Statistics University at Albany Albany, NY 12222

ABSTRACT

1. Regular embeddings

2. HolZpe

3. Groups of order pp−1

4. Nonuniqueness

Lindsay N. Childs¹

Department of Mathematics and Statistics
University at Albany
Albany, NY 12222

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

3. Groups of order $p (p - 1)$