Also available at http://amc.imfm.si
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 2 (2009) 59–76
On overgroups of regular abelian p-groups
Edward Dobson
Department of Mathematics and Statistics
Mississippi State University
PO Drawer MA Mississippi State, MS 39762, USA
Received 3 July 2008, accepted 28 January 2009, published online 26 March 2009
Abstract
Let G be a transitive group of odd prime-power degree whose Sylow p-subgroup P
is abelian of rank t. We show that if p > 2t−1, then G has a normal subgroup that is a
direct product of t permutation groups of smaller degree that are either cyclic or doubly-
transitive simple groups. As a consequence, we determine the full automorphism group of
a Cayley digraph of an abelian group with rank two such that the Sylow p-subgroup of the
full automorphism group is abelian.
Keywords: Cayley graph, abelian group, regular group, p-group
Math. Subj. Class.: 20B10, 05C25
1 Introduction
In this paper, we study transitive groups of odd prime-power degree whose Sylow p-
subgroup is abelian with rank t. In the case of t = 2, we explicitly determine them
(Corollary 6.5). Our main motivation is to determine the full automorphism group Aut(Γ)
of Cayley digraphs Γ of abelian groups G such that the full Sylow p-subgroup of the au-
tomorphism group is abelian (and hence GL and also as small as possible). This is an
important case in trying to determine the full automorphism group of arbitrary Cayley di-
graphs of G, as a “large” Sylow p-subgroup usually has quite a bit of structure (and so
in many cases it’s overgroups can be determined). This occurs because if Aut(Γ) has a
large Sylow p-subgroup, then usually Γ has many edges that occur in a structured way -
for example Γ can be written as a wreath product or perhaps a “semiwreath” product. As
a concrete example of this type of behavior, it has been shown that a Cayley digraph Γ
of an abelian group G of order p2 has automorphism group with Sylow p-subgroup either
isomorphic to G or to Zp o Zp [1, 10]. In the latter case, a classical result of Sabidussi [18]
E-mail address: dobson@math.msstate.edu (Edward Dobson)
Copyright c© 2009 DMFA
60 Ars Math. Contemp. 2 (2009) 59–76
gives Aut(Γ), so determining Aut(Γ) reduces to the case where the Sylow p-subgroup of
Aut(Γ) is G.
Historically, there has been some work on explicitly determining the transitive groups
of prime power degree given that its Sylow p-subgroup is abelian. Chronologically, the first
such result was obtained by Burnside [2] in 1901, who showed that a transitive group of
prime degree p is either doubly-transitive or contains a normal Sylow p-subgroup (which
is necessarily cyclic as the degree is prime). The author generalized this result [7] and
showed that a transitive group of odd prime-power degree pk such that every minimal
transitive subgroup is cyclic is either doubly-transitive or has a normal Sylow p-subgroup.
Building on work of Wielandt [20], Jones [14] showed in 1979 that a transitive group of
degree p2 with elementary abelian Sylow p-subgroup contains a normal subgroup that is
permutation isomorphic to a direct product of two groups, with the canonical action, each
of which is either cyclic or doubly-transitive and simple (and so this result of Jones can be
thought of as a generalization of Burnside’s result mentioned above). In this paper, we will
first generalize Jones’ result to an odd prime-power pk, k ≥ 2, provided that the transitive
permutation group contains a Sylow p-subgroup that is abelian with rank t (Theorem 6.4)
and p > 2t−1. Note that as p is odd, this condition is always satisfied if t = 2, and so
for abelian groups with rank two, we generalize Jones’ result with no restrictions on p
other than p is odd. We will then explicitly determine all transitive permutation groups
of odd prime-power pk whose Sylow p-subgroup is abelian of rank two (Corollary 6.5),
and finally, we determine all possible automorphism groups of Cayley digraphs of abelian
groups of rank two whose Sylow p-subgroups are abelian (Corollary 7.3).
As the proof of the main result is somewhat long, it will be useful to provide a general
outline of the proof. Let G be a transitive group of odd prime-power degree pk, p an odd
prime, with an abelian Sylow p-subgroup P . IfG is primitive, then one can use the O’Nan-
Scott Theorem together with the Classification of the Finite Simple Groups to analyze
G. This has already been done - Li [16] has determined all primitive groups that contain a
regular abelian subgroup. We are then left with the case whereG is imprimitive. We choose
B to be a complete block system of G such that G admits no nontrivial complete block
system whose blocks are strictly contained within blocks of B. Then the set-wise stabilizer
of a block of B inG, denoted StabG(B), is primitive in its action onB [5, Exercise 1.5.10].
Thus these groups are given by [16], and fall into three classes: almost simple groups, affine
groups, and primitive groups constructed by the product action. We will also need to use
the set-wise stabilizer of every block of B, denoted fixG(B), in our analysis. Combined
with the above possibilities for the action of StabG(B) on B ∈ B, we have the following
possibilities. Both StabG(B) and fixG(B)|B are almost simple, affine, or primitive groups
constructed via the product action, as well as the case where StabG(B) is primitive on B
but fixG(B)|B is imprimitive. In the former three cases, we show that action of fixG(B)
on B is equivalent to the action of fixG(B) on B′ (these actions are considered in Section
3), B,B′ ∈ B, and then use relatively standard techniques to show that G is contained in a
canonical direct product provided that fixG(B) is not semiregular. The case where fixG(B)
is semiregular is considered in Section 4, while the case StabG(B) is primitive on B but
fixG(B)|B is imprimitive is considered in Lemma 6.2. The main results are then proven in
Section 6.
Definition 1.1. Let G be a finite abelian group, so that G ∼= Zpa11 × Zpa22 × · · · × Zparr
in accordance with the Fundamental Theorem of Finitely Generated Abelian Groups. We
define the rank of G to be r.
E. Dobson: On overgroups of regular abelian p-groups 61
The following result is [7, Theorem 33].
Theorem 1.2. Let p ≥ 3 be prime, and G ≤ Spm , m ≥ 1, be transitive such that every
minimal transitive subgroup of G is cyclic. Then either G contains a transitive normal
Sylow p-subgroup, or G is doubly-transitive and
1. G = Apm or Spm , and m = 1,
2. PSL(n, k) ≤ G ≤ PΓL(n, k), for some prime power k and n ≥ 2 with pm =
(kn − 1)/(k − 1),
3. G = PSL(2, 11) or M11 and pm = 11,
4. G = M23 and pm = 23.
Definition 1.3. For a positive integer n, define N(n) = {x→ ax+ b : a ∈ Z∗n, b ∈ Zn}.
Thus N(n) is the normalizer of the left regular representation (Zn)L of Zn in Sn. In
general, for a group G, we define GL = {h → gh : h, g ∈ G}. We remark that for p a
prime, N(p) is usually denoted AGL(1, p).
Definition 1.4. Let G ≤ Sn be imprimitive with complete block system B. We say that
B is normal if B is formed by the orbits of a normal subgroup of G. Also, any g ∈ G
permutes the blocks of B, and we denote this induced permutation by g/B. We set G/B =
{g/B : g ∈ G}, and fixG(B) = {g ∈ G : g(B) = B for all B ∈ B}. If C is another
complete block system of G, we write C  B if every block of B is a union of blocks of C.
If C  B, then G/C admits a complete block system, denoted B/C, induced by B. That is
each block of B/C, say B/C, is the set of those blocks of C whose union is B. For a point
i, we denote the point-wise stabilizer of i in G by StabG(i), and for a subset B of Zn (we
assume that G acts on Zn), we denote the set-wise stabilizer of B in G by StabG(B).
2 Some General Results
The following general results on permutation groups will be useful.
Definition 2.1. For a permutation group H ≤ SΩ, we define the support of H , denoted
supp(H), to be the set of all x ∈ Ω such that there exists h ∈ H such that h(x) 6= x.
Lemma 2.2. Let G ≤ Sn be transitive such that G admits a complete block system B
consisting of m blocks of size k. If T = soc(fixG(B))|B is a transitive nonabelian simple
group, then C = {supp(M) : M is a minimal normal subgroup of fixG(B)} is a com-
plete block system of G, B  C, and soc(fixG(B)) is a direct product of simple groups
isomorphic to T .
Proof. As soc(fixG(B)|B) is a transitive nonabelian simple group, a minimal normal sub-
group M of fixG(B) is nonabelian. In fact, if M |B 6= 1, then M |B contains the so-
cle of fixG(B)|B for B ∈ B, and so if M |B 6= 1, then M |B is a transitive nonabelian
simple group. Let N be another minimal normal subgroup of fixG(B). Note then that
supp(M) and supp(N) are unions of blocks of B. If B1, B2 ∈ B with B1 ⊆ supp(M)
and B2 ⊆ supp(N), then as minimal normal subgroups centralize each other, we have that
B1 ∩B2 = ∅. Define a relation ≈ on Zn by i ≈ j if and only if i, j ∈ C ∈ C. Note that as
G is transitive, ≈ is reflexive. That ≈ is symmetric and transitive are trivial, and so ≈ is an
equivalence relation. Let g ∈ G, and i ≈ j. Then gMg−1 is a minimal normal subgroup
62 Ars Math. Contemp. 2 (2009) 59–76
of fixG(B), and g(i) and g(j) are contained in supp(gMg−1). This last statement follows
as, if i, j ∈ supp(M), there exists hi, hj ∈ M such that hi(i) 6= i and hj(j) 6= j. Then
ghig
−1(g(i)) = g(hi(i)) 6= g(i) as hi(i) 6= i. Similarly, ghjg−1(g(j)) 6= g(j). Thus ≈ is
a G-congruence, and by [5, Exercise 1.5.4], C is a complete block system of G. As C is a
complete block system of G, we have that any two minimal normal subgroups of fixG(B)
are conjugate and so isomorphic, and it is easy to see that a minimal normal subgroup must
be isomorphic to soc(fixG(B)|B).
Lemma 2.3. Let G ≤ Smpk , p a prime, k ≥ 1, be such that G admits a complete block
systemB consisting ofm blocks of size pk such that fixG(B) contains a semiregular element
ρ of order pk. Additionally, assume that fixG(B)|B is doubly-transitive with nonabelian
socle and has a cyclic Sylow p-subgroup 〈ρ|B〉, B ∈ B. Then there exists a complete block
system B  C such that a Sylow p-subgroup of fixG(B) is 〈ρ|C : C ∈ C〉.
Proof. By Lemma 2.2, we have that C = {supp(M) : M is a minimal normal subgroup
of fixG(B)} forms a complete block system of G, consisting of, say, n blocks of size rpk,
and B  C. Let M1, . . . ,Mn be distinct minimal normal subgroups of fixG(B), so that
C = {supp(Mi) : 1 ≤ i ≤ n}. Let P be a Sylow p-subgroup of fixG(B) that contains 〈ρ〉.
As each Mi is transitive on some Bi ∈ B, we have that pk divides |Mi| for 1 ≤ i ≤ n.
Also, if Bi ⊆ supp(Mi), Bi ∈ B, then Mi must act faithfully on Bi or Mi is not a
minimal normal subgroup. Thus a Sylow p-subgroup Pi of Mi has order |Pi|Bi |. As a
Sylow p-subgroup fixG(B)|B is cyclic and regular, Pi|Bi is cyclic and regular, and so a
Sylow p-subgroup of Πni=1Mi has order p
nk. We assume without loss of generality that
each Pi ≤ P .
Now, let ρ′ ∈ P , and for each Mi, 1 ≤ i ≤ n, choose Bi ∈ B such that Bi ⊆
supp(Mi). As ρ′|Bi is contained in a Sylow p-subgroup of fixG(B)|Bi , and all Sylow p-
subgroups of fixG(B)|Bi are conjugate and so contained in Mi|Bi , we have that ρ′|Bi ∈
Mi. Also, as a Sylow p-subgroup of fixG(B)|B is cyclic, ρ|Bi is a Sylow p-subgroup of
Mi|Bi , ρ′ ∈ P , and a transitive abelian group is self-centralizing, we have that ρ′|Bi ∈
〈ρ〉|Bi . We conclude that there exists δi ∈ Mi such that δiρ′|Bi = 1. As δi|Bi ∈ 〈ρ〉|Bi ,
we must have that δi ∈ 〈ρ|supp(Mi)〉 as otherwise 〈δi, ρ〉|supp(Mi) does not act faithfully
on supp(Mi), and so Mi is not a minimal normal subgroup. Let δ = Πni=1δi. Then
δρ′|Bi = 1 for every 1 ≤ i ≤ n. Note that if δρ′ = 1, then the result follows, as then
P = 〈δi : 1 ≤ i ≤ n〉 and δi ∈ 〈ρ|supp(Mi)〉. If δρ′ 6= 1, then 〈δρ′〉fixG(B) is a normal
subgroup of fixG(B), and so contains a minimal normal subgroup M . Furthermore, there
exists B ∈ B such that M |B 6= 1. As C is a complete block system of G, there exists
1 ≤ i ≤ n such that Mi|B is transitive. Also, M and Mi centralize each other as they are
minimal normal subgroups of fixG(B). However, Mi|B and M |B are transitive nonabelian
groups with simple socle, and as they both contain a transitive cyclic subgroup, they are
not regular, contradicting [5, Theorem 4.1A (ii)]. The result then follows.
3 The number of inequivalent actions of certain groups
In order to prove our main results, we shall need to know the number of inequivalent actions
of certain groups of appropriate degrees. These results are all somewhat technical, and so
for convenience are all gathered together in this section.
Definition 3.1. We will have need of the wreath product of two groups in this paper, and
usually, we will need the “permutational” wreath product, which we now define. Let G
E. Dobson: On overgroups of regular abelian p-groups 63
and H be groups acting on X and Y , respectively. We define the wreath product of X
and Y , denoted G o H , to be the permutation group that acts on X × Y consisting of all
permutations of the form (x, y) → (g(x), hx(y)), where g ∈ G and hx ∈ H . We remark
that the order of G and H is backwards from what is usually used in group theory, but is
common in graph theory. We have occasional use of the more abstract definition of the
wreath product with its product action. The reader unfamiliar with this action is referred to
[5, pg. 50] for this definition as well as basic facts regarding the product action.
Lemma 3.2. Let p ≥ 3 be prime, and G a primitive group with nonabelian simple socle
T of degree ps, s ≥ 1. Then every transitive representation of G of degree a power of p is
of degree ps provided that if T = PSL(2, 8) is of degree 9, then G contains a cyclic Sylow
p-subgroup.
Proof. As G is primitive in its action on ps points, the stabilizer of a point in this action
is a subgroup of G of prime-power index ps. By [11, Theorem 1], T acts transitively
on exactly one prime-power degree, which is necessarily ps, and by [11, Corollary 2],
this action is primitive. Now suppose that G has a transitive action of degree pt, t 6= s.
Note that T cannot be transitive in this action, as otherwise the stabilizer of a point in T
is a subgroup of index pt, but by [11, Theorem 1], we must have that t = s. Hence G
admits a nontrivial complete block system C formed by the orbits of T . If t < s, then
T |C / fixG(C)|C and T |C ∼= T is transitive on C ∈ C, and so T |C is a transitive group
acting on C. Then the stabilizer of a point of T |C is subgroup of index pt, and so by [11,
Theorem 1], we must have that t = s, a contradiction. If t > s, then G/C has degree a
power of p, and so p must divide |G/C|. As T / G, we must have that G ≤ NSps (T ).
As G/C is a quotient group of G/T , we must then have that p divides |NSps (T )/T |. By
[11, Theorem 1], T = Aps , T = PSL(n, q) for some prime power q and n a prime with
ps = (qn−1)/(q−1), T = PSL(2, 11) and ps = 11, T = M23 and ps = 23, or T = M11
and ps = 11. Note that if ps is prime, then clearly p cannot divide |NSps (T )/T |, and it
is also clear that T 6= Aps . The only remaining case is that T = PSL(n, q), and by [7,
Lemma 17] or hypothesis, a Sylow p-subgroup of G and T is cyclic, so that p does not
divide |NSps (T )/T |, a contradiction.
Lemma 3.3. Let p ≥ 3 be prime, and let m = pt for some t ≥ 1. Let G be a primitive
nonabelian almost simple group of degree m acting on Ω with socle T , where if T =
PSL(2, 8) is of degree 9, then G contains a cyclic Sylow p-subgroup. Let K ≤ Sr o G be
primitive of degreemr acting on Ωr, where Sr oG has the product action and soc(K) = T r.
Then the number of primitive inequivalent actions of K of degree mr on Ωr is the number
of inequivalent actions of T of degree m acting on Ω, and this number is either 1 or 2.
Proof. Let X be the set of all subgroups H of K of index mr in K such that the action
of K on the left cosets of H is primitive of degree mr, and Y be the set of all subgroups
of T of index m in R. Define f : X → Y by letting f(X) be the projection in the first
coordinate of X ∩T r. We first show that f is well-defined by showing that f(X) is indeed
a subgroup of T of index m.
Note that we may viewX as the stabilizer of a point in a primitive transitive action ofK
of degree mr in which case T r is transitive as T r /K and a normal subgroup of a primitive
group is transitive [19, Theorem 8.8]. By [5, Exercise 1.4.1], we have that XT r = K.
Then by [12, Theorem 1.4.8], we have that [T r : X ∩ T r] = [K : X] = mr. Then the
index of the projection of X ∩T r in the first coordinate, call it L, has index dividing mr in
64 Ars Math. Contemp. 2 (2009) 59–76
T . As m = pt is a prime power, we have that [T : L] = ps for some s ≥ 1. We conclude
that T has a transitive representation of degree ps, and by Lemma 3.2, we have that s = t.
Thus f(X) is a subgroup of T of index m = pt and so f is well defined.
Let Ωr = Πri=1Ωi, where if (ω1, . . . , ωr) ∈ Ωr, then ωi ∈ Ωi. We write T r = Πri=1Ti,
where each Ti is the action of T on Ωi in T r. As K ≤ Sr o G has the product action, the
action of Ti on Ωi is the natural action, and so the stabilizer in Ti of a point in Ωi is the
stabilizer in Tj of a point in Ωj . Thus the action of Ti on Ωi is equivalent to the action
of Tj on Ωj . Thus if Li,ω is the stabilizer in Ti of the point ω, then Li,ω is a conjugate
in T of Lj,ω′ for every 1 ≤ i, j ≤ r, ω ∈ Ωi and ω′ ∈ Ωj . Then the stabilizer of the
point (ω1, . . . , ωr) in T r is Πri=1Li,ωi . As T
r / K, we conclude that for any k ∈ K,
k−1Li,ωk = Lj,ω′ , where again 1 ≤ i, j ≤ r, ω ∈ Ωi, and ω′ ∈ Ωj .
Now let k ∈ K such that kT r 6= T r, and (ω1, . . . , ωr) = ω ∈ Ωr. As T r is transitive,
there exists h ∈ T r such that kh stabilizes ω. Then khT r = kT r and kh normalizes
Πri=1Li,ωi , the stabilizer of ω ∈ T r. Thus |NK(Πri=1Li,ωi)/Πri=1Li,ωi | = |K/T r|. We
conclude that NK(Πri=1Li,ωi) is a subgroup of index m
r in K and so is the stabilizer of a
point of Ωr.
Now suppose that K has another primitive representation, which we will call ρ(K),
of degree mk, that is inequivalent to the natural representation as a subgroup of Sr o G.
Let X be the stabilizer of a point in ρ(K), so that f(X) ≤ T is a subgroup of index
m in T . Set L = L1,ω1 , ω1 ∈ Ω1, where we regard L as a subgroup of T of index
m. If f(X) is a conjugate of L in T , then f(X) = L1,ω′1 for some ω
′
1 ∈ Ω1, and as
k−1Li,ωk = Lj,ω′ for every k ∈ K, 1 ≤ i, j ≤ r, ω ∈ Ωi and some ω′ ∈ Ωj , we have that
Stabρ(K)(ω1, . . . , ωr) = Πri=1Li,ωi . Arguing as in the previous paragraph, we have that
NK(Πri=1Li,ωi) is the stabilizer in ρ(K) of (ω1, . . . , ωr) and so the representation ρ(K) is
the same representation as the natural representation of K. We may thus assume without
loss of generality that f(X) is not a conjugate of L in T . This then implies that T has at
least two inequivalent transitive representations of degree m, and so if T has exactly one
representation up to equivalence, the result follows.
Now, by [11, Corollary 2], we have that T is doubly-transitive. By [3, Table], we have
that T has at most two inequivalent representations of degree m, and so T has exactly
two inequivalent representations of degree m. In ρ(K), let Mi,ωi be the stabilizer of the
point ωi ∈ Ωi in Ti. By arguments above, we have that no Mi,ωi is conjugate to any
Li,ωj , ωj ∈ Ωi. As there are two inequivalent representations of T of degree m, there
are exactly two conjugacy classes of subgroups of index m in T and as for any k ∈ K,
k−1Li,ωk = Lj,ω′ , we have that for any k ∈ K, k−1Mi,ωk = Mj,ω′ . Arguing as above,
we conclude that the stabilizer of the point (ω1, . . . , ωr) ∈ Ωr in ρ(K) isNK(Πri=1Mi,ωi).
Arguing as above, we conclude that K has exactly two inequivalent representations if T
has two inequivalent representations.
Lemma 3.4. Let p ≥ 3, and m = pt. Let T be a doubly-transitive simple group of degree
m acting on Ω. Then the number of inequivalent transitive actions of T r acting coordinate-
wise on Ωr is ar, where a is the number of inequivalent actions of T on Ω. Consequently,
ar = 1 or ar = 2r.
Proof. Similarly to Lemma 3.3, we define X to be the set of all subgroups H of T r of
index mr in T r such that the action of T r on the left cosets of H is transitive of degree
mr, and Y be the set of all subgroups of T of index m in T . Define fi : X → Y by letting
fi(X) be the projection in the ith-coordinate ofX∩T r. Using the argument in Lemma 3.3
E. Dobson: On overgroups of regular abelian p-groups 65
to show that f is well-defined, we have that each fi is well-defined. As there are a choices
for each fi(X) and there are r coordinates, we have that |X | = ar. It then follows by [3,
Table] that a = 1 or 2.
Lemma 3.5. Let G ≤ Spk , p a prime and k ≥ 2, be transitive with abelian Sylow p-
subgroup P and admit a normal complete block system B such that StabG(B)|B is primi-
tive. Suppose that fixG(B) contains a characteristic subgroupK whose orbits form B,K|B
is equivalent to K|B′ for every B,B′ ∈ B, and K|B is imprimitive. Then StabPG(B)|B is
imprimitive.
Proof. Let P1 be a Sylow p-subgroup of fixG(B) that is contained in P . Note that P1 is
necessarily semiregular and abelian. As K is characteristic and Sylow p-subgroups are
conjugate, every Sylow p-subgroup of fixG(B) is contained in K. Let B consist of p`
blocks of size pm. We consider G as acting on Zp` × (P1)L so that B = {{(i, j) : j ∈
P1} : i ∈ Zp`}. Then g(i, j) = (σg(i), ωg,i(j)), for g ∈ G, where σg ∈ Sp` and each
ωg,i ∈ SP1 . We may also assume that if ρ ∈ P1, then ρ(i, j) = (i, j + b), where b ∈ P1.
As K|B is equivalent to K|B′ for every B,B′ ∈ B, we may additionally assume that if
k ∈ K, then k(i, j) = (i, κ(j)), where κ ∈ SP1 . Let g ∈ G. Then g−1P1g ≤ K is a Sylow
p-subgroup of K (and fixG(B)), and so there exists kg ∈ K such that k−1g g−1P1gkg = P1.
Then gkg normalizes P1, and NSZ
p`
×(P1)L
(P1) = {(σ(i), α(j) + bi) : σ ∈ Sp` , α ∈
Aut(P1), bi ∈ P1}. Thus gkg(i, j) = (σ(i), α(j) + bi) for appropriate σ, α, and bi (note
that (gkg)−1(i, j) = (σ−1(i), α−1(j) − α−1(bσ−1(i)))). Also, kg(i, j) = (i, ωkg,i(j)), so
as kg ∈ K, we may set kg(i, j) = (i, ω(j)), ω ∈ SP1 . As P is transitive and abelian, if
β ∈ P , then β(i, j) = (σβ(i), j + ci), ci ∈ P1. Then
g−1βg(i, j) = kg(k−1g g
−1)β(gkg)k−1g (i, j)
= (σ−1σβσ(i), ω(ω−1(j) + α−1(bi) + α−1(cσ(i))− α−1(bσ−1σβσ(i)))).
As each map (i, j) → (i, j + di), di ∈ P1, is contained in fixP1(B)|B ≤ K|B , B ∈
B, and ω, ω−1 ∈ K|B , B ∈ B we conclude that PG ≤ (G/B) o (K|B) for B ∈ B.
Then StabPG(B)|B ≤ K|B which is imprimitive, and the result follows by [5, Exercise
1.5.10].
The following result is ultimately where the requirement that p > 2t−1 originates. A
deeper analysis of the number of inequivalent actions of fixG(B) in the following result
could lead to the removal of the condition that p > 2t−1.
Lemma 3.6. Let p ≥ 3 be prime, and let G ≤ Spk be transitive such that G con-
tains an abelian Sylow p-subgroup P and G admits a complete block system B such that
StabG(B)|B is primitive and not permutation isomorphic to a subgroup of AGL(r, p) for
any r ≥ 1, but fixG(B)|B is imprimitive,B ∈ B. Let t be the rank of a Sylow p-subgroup of
fixG(B). If p > 2t, then PG admits a nontrivial complete block system C such that C ≺ B.
Proof. As StabG(B)|B is primitive, StabG(B)|B is given by the O’Nan-Scott Theorem
(we use the form given by [17]). As StabG(B)|B is a primitive group of odd prime-power
degree, B ∈ B, we have that StabG(B)|B is a subgroup of Sm o K with the product
action, where K is an almost simple group of degree a power of p as StabG(B)|B is
not permutation isomorphic to a subgroup of AGL(r, p). Note that StabG(B)|B cannot
66 Ars Math. Contemp. 2 (2009) 59–76
be an almost simple group as fixG(B)|B is imprimitive and normal in StabG(B)|B . By
[11, Theorem 1], we have that K is doubly-transitive. Then soc(fixG(B)|B) = T r acting
coordinate-wise on Ωr for some doubly-transitive simple group T of degree pk/r acting on
Ω. Also, soc(fixG(B)|B) is characteristic in StabG(B)|B , and so normal, and as a normal
subgroup of a primitive group is transitive [19, Theorem 8.8], we have that T r is transitive
as well.
Observe that T r must act faithfully on each B ∈ B, as otherwise T r contains a normal
subgroup H that is nontrivial (and so has orbits of size dividing a multiple of p) on some
blockB ∈ B whileH acts trivially on someB′ ∈ B. ThenH contains a nontrivial Sylow p-
subgroup which necessarily fixes a point, while P is regular. Define an equivalence relation
≡ on B by B ≡ B′ if and only if the action of T r on B is equivalent to the action of T r on
B′. By [5, Lemma 1.6B], B ≡ B′ if and only if there exists x ∈ B and x′ ∈ B′ such that
StabT r|B (x) = StabT r|B′ (x
′). As conjugation by an element of G maps the stabilizer of
some point of T r|B to the stabilizer of some point of T r|B′ for some B′ ∈ B, we see that
if g ∈ G, then g(B) ≡ g(B′). As G/B permutes the blocks of B, we see that ≡ is a G/B-
congruence, and so by [5, Exercise 1.5.4], the equivalence classes of ≡ are blocks of G/B.
By Lemma 3.4 we have that T r has at most 2r ≤ 2t < p inequivalent representations. We
conclude that there is exactly one equivalence class of≡, and by Lemma 3.5 StabPG(B)|B
is imprimitive. The result then follows by [5, Exercise 1.5.10].
Lemma 3.7. Let H ≤ AGL(k, p) be transitive such that a Sylow p-subgroup of H is
elementary abelian of order pk. Then H has a unique action of degree pk.
Proof. As H ≤ AGL(k, p) is transitive, H contains a regular normal elementary abelian
Sylow p-subgroup P of order pk. As P is solvable, by the Schur-Zassenhaus Theorem, P
contains a complement K of order |H|/pk, and any two subgroups of H of order |H|/pk
are conjugate in H . We conclude by [5, Lemma 1.6B] and the comments following it that
H has a unique action of degree pk.
4 Direct Products
In this section, we have gathered together all of the results that consider when fixG(B) is
not semiregular and conclude with G being contained in a direct product. Our main tool is
the following result.
Lemma 4.1. Let G ≤ Sn be transitive such that G admits a normal complete block system
B consisting of m blocks of size k. If fixG(B) acts faithfully on B ∈ B, fixG(B)|B is
equivalent to fixG(B)|B′ for every B,B′ ∈ B, and the stabilizer of a point in fixG(B)|B
fixes exactly one point forB ∈ B, thenG is permutation isomorphic to subgroup of Sm×Sk
with the canonical action.
Proof. We assume without loss of generality that G acts on Zm×Zk so that B = {{(i, j) :
j ∈ Zpk} : i ∈ Zm}. Define an equivalence relation ≡ on Zm × Zk by (i, j) ≡ (i′, j′)
if and only if StabfixG(B)(i, j) = StabfixG(B)(i
′, j′). It is easy to see that the equivalence
classes of ≡ form a complete block system E of H . If fixG(B)|B has only one action up to
equivalence and the stabilizer of a point in fixG(B)|B fixes only one point, then E consists
of k blocks of size m, and each block of E contains exactly element of each block in B by
[5, Lemma 1.6B]. The result then follows by [6, Lemma 2.2].
E. Dobson: On overgroups of regular abelian p-groups 67
Lemma 4.2. Let G ≤ Sn, n odd, such that G admits a complete block system B consisting
of m blocks of size pk, p a prime and k ≥ 1, and one of the following is true:
1. a Sylow p-subgroup of fixG(B)|B contains a regular cyclic subgroup and fixG(B)|B
is a primitive group with nonabelian socle, or
2. a Sylow p-subgroup of fixG(B) is cyclic and semiregular of order pk.
Then one of the following is true:
i. fixG(B) is cyclic and semiregular of order pk,
ii. G is permutation isomorphic to a subgroup of Sm × Spk with the canonical action.
Furthermore, if G contains a regular abelian group, then there exists H ≤ Sm and
K ≤ Spk such that H ×K / G, or
iii. fixG(B) does not act faithfully on B ∈ B and a Sylow p-subgroup of fixG(B) is not
semiregular.
Proof. We assume without loss of generality thatG acts on Zm×Zpk so that B = {{(i, j) :
j ∈ Zpk} : i ∈ Zm}. Note that if fixG(B) does not act faithfully on B ∈ B, then there
exists some normal subgroup N / fixG(B) and B,B′ ∈ B such that N |B = 1 while
N |B′ 6= 1. As |B′| = pk, the orbits of N |B′ have order a power of p, and so p divides |N |.
Thus N contains an element of order p that fixes a point, and so if fixG(B) does not act
faithfully on B ∈ B, then a Sylow p-subgroup of fixG(B) is not semiregular. We may now
assume without loss of generality that fixG(B) acts faithfully on B ∈ B, as otherwise (iii)
occurs. We may also assume that fixG(B) is not cyclic and semiregular of order pk as then
(i) occurs.
First suppose that fixG(B)|B is primitive with nonabelian socle. If k = 1, then by
Burnside’s Theorem [5, Theorem 3.5B], we have that fixG(B) is doubly-transitive. If
k ≥ 2, then as Zpk is a Burnside group [5, Theorem 3.5A], we also have that fixG(B)|B
is doubly-transitive. Thus in any case, fixG(B)|B is doubly-transitive with nonabelian so-
cle. Define an equivalence relation ≡ on Zm × Zpk by (i, j) ≡ (i′, j′) if and only if
StabfixG(B)((i, j)) = StabfixG(B)((i
′, j′)). It is easy to see that the equivalence classes of
≡ are blocks of G, forming the complete block system E . Now, fixG(B)|B is one of the
groups given in [16, Theorem 1.1], and by [3, Theorem 5.3, Table], each of these groups has
either one or two inequivalent actions. As fixG(B)|B is doubly-transitive, we have that the
stabilizer of a point in fixG(B)|B fixes exactly one point, B ∈ B. The first statement in (ii)
then follows by Lemma 4.1 if fixG(B)|B is equivalent to fixG(B)|B′ for every B,B′ ∈ B.
Otherwise, fixG(B)|B must have exactly two inequivalent actions, and E must consist of
an even number of blocks which is impossible as m is odd. Thus the first statement of (ii)
follows.
If fixG(B)|B is not primitive with nonabelian socle, then fixG(B) contains a cyclic and
semiregular Sylow p-subgroup. Then fixG(B)|B contains a cyclic and transitive Sylow p-
subgroup. By Theorem 1.2 fixG(B)|B contains a normal cyclic Sylow p-subgroup. Thus
fixG(B)|B ≤ Z∗pk · (Zpk)L, and Z
∗
pk has order p
k−1(p− 1), and is solvable. We conclude
that fixG(B)|B has order dividing (p− 1)pk. Let |fixG(B)|B | = a · pk, so that gcd(a, p) =
1. Note that fixG(B) = 〈ρ〉 and is semiregular if a = 1, and the result follows. We
thus assume that a 6= 1. As fixG(B)|B is solvable, fixG(B)|B contains a subgroup A of
order a, and by Hall’s Theorem [12, Proposition II.7.14], any two subgroups of fixG(B)|B
of order a are conjugate in fixG(B)|B . As any action of fixG(B)|B on pk points can be
68 Ars Math. Contemp. 2 (2009) 59–76
viewed as the action of fixG(B)|B on the right cosets of some subgroup A′ of order a,
and all such subgroups are conjugate, by [5, Lemma 1.6A] and the comments following
it, fixG(B)|B has a unique action as well. Note that cx ≡ x (mod pk) is equivalent to
(c − 1)x ≡ 0 (mod pk). This last equation will have a unique solution provided that
c− 1 6≡ 0 (mod p). That is cx ≡ x (mod pk) has a unique solution unless c = 1 + bp for
some b. As 1+bp has multiplicative order a power of p in Zpk for every b and fixH(B)|B ≤
NS
pk
((Zpk)L) = {x → a1x + b1 : a1 ∈ Z∗pk , b1 ∈ Zpk}, we conclude that the stabilizer
of a point in fixG(B)|B fixes exactly one point. Then E ∩B is a singleton for every E ∈ E
and B ∈ B, and the first statement of (ii) follows by Lemma 4.1. Thus in any case, the first
statement of (ii) follows.
If in addition G contains a regular abelian group, set H = fixG(E) 6= 1 and K =
fixG(B). Then the internal direct product H ×K / G and is transitive. Hence (ii) follows.
Lemma 4.3. Let G ≤ Sn, n odd, admit a complete block system B of n/pk blocks of
size pk such that a Sylow p-subgroup of fixG(B) is a semiregular direct product of cyclic
groups of prime-power order. If fixG(B)|B is primitive, then G is permutation isomorphic
to a subgroup of Sn/pk × Spk .
Proof. As fixG(B)|B is primitive and a Sylow p-subgroup of fixG(B) is a semiregular
direct product of cyclic groups of prime-power order, we have a Sylow p-subgroup of
fixG(B)|B is direct product of cyclic groups of prime-power order. It follows by [16, The-
orem 1.1] that fixG(B)|B is permutation isomorphic to a subgroup of AGL(k, p) or Sr oK,
where K is primitive group of degree pk/r with nonabelian socle (note that the possible
socles are given in [16, Corollary 1.2] as K contains a regular cyclic subgroup). Observe
that fixG(B) must act faithfully on B ∈ B, as otherwise fixG(B) contains a normal sub-
group N which is trivial on B but nontrivial on some B′ ∈ B. If this occurs, we have
that pk divides |N |B′ | as a normal subgroup of a primitive group is transitive, and so a
Sylow p-subgroup of fixG(B) has order at least p2k, which does not occur. If fixG(B)|B
is permutation isomorphic to a subgroup of AGL(k, p), then by Lemma 3.7, we have that
fixG(B)|B has a unique action of degree pk, and as fixG(B)|B is primitive, the stabilizer of
a point in fixG(B)|B must fix exactly one point. The result then follows by Lemma 4.1.
If fixG(B)|B is permutation isomorphic to a subgroup of Sr oK, then by Lemma 3.3,
the number of inequivalent actions of fixG(B)|B of degree pk is the number of inequivalent
actions of K of degree pk/r, and this number of inequivalent actions is either 1 or 2. If K
has one action of degree pk/r, then the result follows by Lemma 4.1. If K has two actions
of degree pk/r, then define≡ as in Lemma 4.1. If for someB′ ∈ B we have that fixG(B)|B
is inequivalent to fixG(B)|B′ , then the equivalence classes of ≡ have even order, and so
G admits a complete block system whose blocks have an even number of points. This,
however, is not possible as n is odd. Thus fixG(B)|B is equivalent to fixG(B)|B′ for every
B,B′ ∈ B, and the result follows by Lemma 4.1.
5 fixG(B) is semiregular
In this section, we consider the case when fixG(B) is semiregular.
Lemma 5.1. Let G ≤ Spk , k ≥ 2 such that G contains a regular abelian p-subgroup P
and admits a nontrivial complete block systemB such that StabG(B)|B normalizes P |B for
every B ∈ B, which always occurs if fixG(B) is semiregular. Then fixP (B) is contained in
E. Dobson: On overgroups of regular abelian p-groups 69
the center of PG, the normal closure of P in G, and PG ≤ Spk−j oK, where K ∼= fixP (B)
has order pj .
Proof. As a transitive abelian group is regular [19, Proposition 4.4] and the homomorphic
image of an abelian group is abelian, we have that B is formed by the orbits of a subgroup
K of P of order pj . Let L = P/K, so that L has order pi with i + j = k. We view
G as acting on L × K (even if P 6∼= L × K) in such a way that B is formed by the
orbits of 1L × K. We also assume without loss of generality that K = KL, the left
regular representation of K. Let g ∈ G. Then g(`, k) = (σ(`), β`(k)), where σ ∈ SL
and each β` ∈ SK . As StabG(B)|B normalizes KL, by [5, Corollary 4.2B] we have that
StabG(B)|B is isomorphic to a subgroup of Aut(K) · K. Thus β`(k) = α`(k) + b`,
where α` ∈ Aut(K) and each b` ∈ K. A straightforward computation will show that
g−1(`, k) = (σ−1(`), α−1σ−1(`)(k) − α
−1
σ−1(`)(b`)). Let b ∈ K and ρ(k, `) = (k, ` + b).
An equally straightforward computation will show that g−1ρg(`, k) = (`, k + α−1` (b)).
As g−1ρg ∈ fixG(B), we must have that α−1` (b) = α−1m (b) for every `,m ∈ L. As this
must hold for every b ∈ K, we conclude that α` = αm for every `,m ∈ L. Hence
g(`, k) = (σ(`), α(k) + b`), where σ ∈ SL, α ∈ Aut(K), and each b` ∈ K. As if g ∈ G,
then g(`, k) = (σ(`), α(k) + b`) as above, if h ∈ PG, then h(`, k) = (δ(`), k+ c`), where
δ ∈ SL and each c` ∈ K. As for every ρ ∈ fixG(B), ρ(`, k) = (`, k + b), b ∈ K, it is easy
to see that every element of K commutes with every elements of PG, so that K ≤ Z(PG),
the center of PG. It is equally straightforward to see that PG ≤ Spk−j oK as well.
Definition 5.2. We denote the commutator subgroup of G by G′ and the center of G by
Z(G).
Lemma 5.3. LetG ≤ Spk , p an odd prime and k ≥ 2, be transitive with Sylow p-subgroup
P that is abelian. Suppose that G admits a complete block system B such that fixG(B) ≤
Z(G), and G/B is nonabelian. Then G′ ∩ fixG(B) = 1, G′ 6= 1, and G′/B 6= 1.
Proof. Observe first that fixG(B) is semiregular as fixG(B) ≤ Z(G) and Z(G) ≤ P as
an abelian group is self-centralizing [5, Theorem 4.2A (v)]. Note that P ′, the commutator
subgroup of P , is trivial as P is abelian, and as P is a Sylow p-subgroup of G, P is a
Hall subgroup of G. By [13, Satz IV.2.2], we have that P ∩ G′ ∩ Z(G) = 1. As G/B
is nonabelian, G′ 6= 1 6= G′/B. As fixG(B) ≤ Z(G) and a regular abelian group is self-
centralizing, we have that fixG(B) ≤ P . Also, as fixG(B) ≤ P ∩ Z(G), we have that
G′ ∩ fixG(B) = 1.
Lemma 5.4. Let G ≤ Sn contain a regular abelian subgroup. If B and C are complete
block systems of G such that fixG(B) ∩ fixG(C) = 1, then |B ∩ C| ≤ 1 for every B ∈ B,
C ∈ C, and fixG(C) acts faithfully on B.
Proof. As G contains a regular abelian subgroup, say R, B and C are formed by the orbits
of subgroups of R, say, S and T , respectively. As fixG(B) ∩ fixG(C) = 1, we have that
S ∩ T = 1. As the intersection of blocks is again a block, if B ∈ B and C ∈ C such that
B ∩ C 6= ∅, then B ∩ C is a block of G, and so is an orbit of a subgroup U of R. Then
U ≤ S ∩ T so that U = 1. Thus |B ∩ C| ≤ 1 for every B ∈ B and C ∈ C. Now suppose
that fixG(C) does not act faithfully on B. Then there exists nontrivial H ≤ fixG(C) such
that h(B) = B for every B ∈ B, and so H ≤ fixG(B). As fixG(B) ∩ fixG(C) = 1, we
have that H = 1, a contradiction.
70 Ars Math. Contemp. 2 (2009) 59–76
Lemma 5.5. LetG ≤ Spk , p an odd prime and k ≥ 2, be transitive with Sylow p-subgroup
P that is abelian. Let PG be the normal closure of P in G. If G admits a complete
block system B such that fixPG(B) is semiregular and PG/B is permutation isomorphic
to H ×K (with the canonical action), where H is regular and abelian and K is a direct
product nonabelian simple groups, then PG is permutation isomorphic to L×K (with the
canonical action) where L is abelian.
Proof. As fixG(B) is semiregular, fixG(B) = fixP (B), and as PG/B is permutation iso-
morphic to H ×K, PG/B admits a complete block system C′ formed by the orbits of H .
As the inverse image of a normal subgroup under a homomorphism is a normal subgroup,
we have that if L ≤ PG is maximal such that L/B = H , then L is a normal subgroup of
PG. As fixL(B) = fixG(B) is a p-group and L/B = H is a p-group, we have that L is
a normal p-subgroup of PG. As a Sylow p-subgroup of G is abelian, we have that L is
abelian. Also, PG admits a complete block system C consisting of pk−i blocks of size pi
induced by C′. That is, each block of C is the union of those blocks of B that are contained
in a block of C′. Hence fixPG(C) = L and PG/C = K. By Lemma 5.1, we have that
L ≤ Z(PG). Note also that as C consists of pk−i blocks of size pi, K has degree pk−i.
By Lemma 5.3, (PG)′ ∩ fixPG(C) = 1. Also, as fixPG(C) is a normal p-subgroup of
PG, fixPG(C) is contained in every Sylow p-subgroup of PG. As K is a direct product of
nonabelian simple groups, we have that (PG)′/C ≥ soc(K) = K and is transitive. Let P1
be a Sylow p-subgroup of (PG)′ and P2 a Sylow p-subgroup of PG that contains P1. Then
P2 is abelian, so P1 / P2 and fixPG(C) / P2 as well as a normal p-subgroup is contained in
every Sylow p-subgroup. Of course, P1 ∩ fixPG(C) = 1 as (PG)′ ∩ fixPG(C) = 1, and, as
P1/C is transitive and fixPG(C)|C is transitive, C ∈ C, we have that 〈P1,fixPG(C)〉 = P2,
and so P2 ∼= P1× fixPG(C). Thus P1 has order pk−i, and so no orbit of P1 has order more
than pk−i.
Now, as (PG)′ /PG, the orbits of (PG)′ form a complete block systemD of PG. Then
P1 is transitive on each block of D and as P1 is abelian of order pk−i, each block of D has
order pk−i. Thus P1 is a Sylow p-subgroup of fixPG(D), and fixPG(D) ∩ fixPG(C) = 1.
By Lemma 5.4, |C ∩ D| ≤ 1 for every C ∈ C and D ∈ D. As the number of possible
intersections of blocks of C andD is pk, we conclude that |C∩D| = 1 for every C ∈ C and
D ∈ D. By [6, Lemma 2.2], PG is permutation isomorphic to a subgroup of Spk−i × Spi .
Let h ∈ PG. Then h/C ∈ K ∼= fixPG(D). Thus h ∈ fixPG(C) ·K, and asK∩fixPG(C) =
1, we have that h ∈ K × fixPG(C). The result then follows as L = fixPG(C).
Lemma 5.6. Let G ≤ Spk , k ≥ 2 such that a Sylow p-subgroup of G is a regular abelian
subgroup, andG admits a nontrivial complete block system B such that fixG(B) is semireg-
ular and G/B is primitive. Then there exists H /G such that H is permutation isomorphic
to a subgroup of Spi × Spj , where pj is the size of a block of B, and i+ j = k, or P / G.
Proof. By [16, Theorem 1.1], we have that soc(G/B) is a direct product of nonabelian
simple groups or is elementary abelian. If soc(G/B) is elementary abelian, then G/B
contains a normal p-subgroup. As fixG(B) is a p-group (as it is semiregular), we have that
G contains a transitive normal Sylow p-subgroup Q. As a Sylow p-subgroup of G is a
regular abelian group, Q = P and the result follows. If soc(G/B) is a direct product of
nonabelian simple groups, then first observe that PG/B = (P/B)G/B. Also, soc(G/B)
contains a transitive p-subgroup which we may assume is P/B. Thus every conjugate of
P/B is contained in soc(G)/B. The result then follows by Lemma 5.5 with H = PG.
E. Dobson: On overgroups of regular abelian p-groups 71
6 The main results
In this section, we have the final lemma that we will need to prove the main results, and
then prove the main results.
Definition 6.1. Let G be a group. The Frattini subgroup Φ(G) is the intersection of all
proper maximal subgroups of G. Note that Φ(G) is equal to the set of all nongenerators of
G, where a nongenerator of G is an element g such that if X is a generating set of G, then
X − {g} is also a generating set of G.
Lemma 6.2. Let G ≤ Sp` , p a prime and ` ≥ 2, be transitive such that a Sylow p-
subgroup P of G is regular and abelian. Suppose that G admits a complete block system
B such that StabG(B)|B , B ∈ B, is primitive but fixG(B)|B is imprimitive. If there exists
g ∈ Φ(P ) such that g ∈ fixG(B) but there is no h ∈ fixP (B) such that 〈g〉 < 〈h〉, then
fixG(B)|B ≤ AGL(m, p), where B consists of blocks of size pm.
Proof. As StabG(B)|B is primitive and contains a regular abelian subgroup, StabG(B)|B
is given by [16, Theorem 1.1]. As fixG(B)|B is imprimitive and fixG(B)|B / StabG(B)|B ,
StabG(B)|B cannot have simple socle. We conclude that StabG(B)|B ≤ AGL(m, p)
(where B consists of blocks of size pm), or is contained in Sr o L, where L is a primitive
almost simple group of degree pm/r, and r ≥ 2. In the former case, the result follows with
no additional hypothesis as fixG(B)|B ≤ StabG(B)|B , so we need now only consider the
case where StabG(B)|B ≤ Sr o L.
Let g ∈ Φ(P ) such that g ∈ fixG(B) but there is no h ∈ fixP (B) such that 〈g〉 < 〈h〉.
As P is a finite abelian group, we have that there exists h ∈ P such that 〈g〉 < 〈h〉. Thus
h/B 6= 1. Let H = 〈P,fixG(B)〉. Then H/B is a regular abelian group of degree n/pm,
and so the orbits of 〈h〉/B form a complete block system C′ of H/B. Then C′ induces
a complete block system C of H , where each block of C is the union of the blocks of B
that are a block of C′. As StabG(B)|B ≤ Sr o L, we have that T r ≤ fixG(B)|B ≤ Lr,
where T = soc(L). Also note that as a Sylow p-subgroup of G is regular, we must have
that fixG(B) acts faithfully on each B ∈ B. Hence fixG(B) ∼= fixG(B)|B . Let M be the
smallest normal subgroup of fixG(B) that contains h. Then M = T t for some t ≤ r. As a
Sylow p-subgroup of T t is isomorphic to Zt
pm/r
and 〈h〉 is cyclic, we have that t = 1. We
conclude that M has a normal complement N in T r, and that N is normal in H . Let F be
the complete block system of H formed by the orbits of N . Then A = [fixH(C)/F ]|C/F ,
C ∈ C, is imprimitive, has a cyclic Sylow p-subgroup, but does not have a normal Sylow
p-subgroup, contradicting Theorem 1.2.
The following result generalizes the result of Jones [14] mentioned in the introduction.
Theorem 6.3. Let G ≤ Spk be transitive with an abelian Sylow p-subgroup P . Let t be
the rank of P . If p > 2t−1, then PG is permutation isomorphic to a direct product of cyclic
groups and doubly-transitive nonabelian simple groups with the canonical action, with the
number of factors in the direct product equal to the rank of P .
Proof. If G is primitive, then by [16, Theorem 1.1], we have that G ≤ AGL(k, p) or
G ≤ Sr o N , where N is an almost simple group of degree pk/r. If G ≤ AGL(k, p),
then P / G and the result follows. If G ≤ Sr o N , then soc(G) = Qr with the canonical
action for some simple group Q of degree pk/r. As Qr is transitive, pk | |Qr|. As a
Sylow p-subgroup of G is P and every Sylow p-subgroup of G is conjugate, we have that
72 Ars Math. Contemp. 2 (2009) 59–76
P ≤ Qr, so that PG = Qr and the result then follows as Q is necessarily doubly-transitive
by [11] as Q contains an abelian Sylow p-subgroup. We henceforth assume without loss of
generality that G is imprimitive.
For the imprimitive case, we proceed by induction on k. If k = 2, then the result can be
deduced easily from [14, Proposition A and B] or [9, Theorem 4]. Let k ≥ 3 and assume
that the result holds for all G satisfying the hypothesis of degree pj , j < k.
Let B be a necessarily normal complete block system of G such that there is no non-
trivial complete block system C of G such that C ≺ B. Note that as usual, we must have
that fixG(B) acts faithfully on B ∈ B as a Sylow p-subgroup of G is abelian. Also,
StabG(B)|B is primitive by [5, Exercise 1.5.10]. We also let B consist of pk−m blocks of
size pm.
If fixG(B)|B is primitive then by Lemma 4.3, we have thatG is permutation isomorphic
to a subgroup of Spk−m × Spm . Let H ≤ Spk−m and K ≤ Spm be minimal such that
G ≤ H × K. Let P1 be a Sylow p-subgroup of H and P2 a Sylow p-subgroup of K.
As H and K are contained in appropriate projections of G, P1 and P2 are both regular
abelian groups, and so P = P1 × P2. Thus the sum of the rank of P1 and the rank of
P2 must be the same as the rank of P . By the induction hypothesis both PH1 and P
K
2
can be written as a direct product of cyclic groups and doubly-transitive nonabelian simple
groups, with the number of factors in the direct product equal to rank of a Sylow p-subgroup
of H and K, respectively. By comments earlier in this paragraph, we conclude that the
number of factors in the direct product of PH1 and P
K
2 is the rank of P . Finally, as both
H and K are contained in appropriate projections of G, PG1 = P
H
1 and P
G
2 = P
K
2 . Then
(P1 × P2)G = PH1 × PK2 . The result then follows by the induction hypothesis.
If fixG(B)|B is imprimitive and StabG(B)|B ≤ AGL(m, p), then by Lemma 5.1, we
have that fixP (B) is contained in the center of PG and fixPG(B)|B ≤ fixP (B)|B for every
B ∈ B. As a Sylow p-subgroup of fixG(B) is semiregular, we have that fixPG(B) =
fixP (B). By the induction hypothesis PG/B can be written as a direct product of cyclic
groups or doubly-transitive nonabelian simple groups. If PG/B is abelian, then PG is
abelian and P / G, in which case the result follows. Otherwise, by Lemma 5.5, PG is
permutation isomorphic to a direct product L ×M with the canonical action, where L is
abelian, andM is a direct product of nonabelian simple groups each of which is necessarily
doubly-transitive by [11] as they all have abelian Sylow p-subgroups. It is then easy to see
that the number of factors of L ×M is the rank of P as a Sylow p-subgroup of a doubly-
transitive nonabelian simple group of degree a prime-power is cyclic if it is abelian (this is
implicit in [16, Theorem 1.1]), and the result follows by induction.
Finally, if fixG(B)|B is imprimitive and StabG(B)|B 6≤ AGL(m, p), then let C be a
complete block system of PG such that C  B and there is no nontrivial complete block
system D of PG such that D ≺ C. Note that StabG(B)|B ≤ Sr o K, where K is a
doubly-transitive group with nonabelian socle T of degree pm/r. Then T r / fixG(B), and
T r is transitive on each block of B ∈ B. Let H = 〈T r, P 〉, so that H is transitive, and
admits a complete block system consisting of blocks of size pm/r formed by the orbits of
a factor of T r (note that this is true as StabH(B) = fixH(B) = T r for every B ∈ B). By
arguments above, H is a direct product of cyclic groups and doubly-transitive nonabelian
simple groups, with the number of factors equal to the rank of P . As every doubly-transitive
nonabelian simple group factor of H is contained in fixH(B), we conclude that r < t. By
Lemma 3.6, we conclude that C ≺ B.
Now, PG is the subgroup of G generated by all of the Sylow p-subgroups of G, and
E. Dobson: On overgroups of regular abelian p-groups 73
PP
G
is the subgroup generated by all of the Sylow p-subgroups of PG. As PG contains
every Sylow p-subgroup of G, PG = PP
G
. We conclude by choice of C and Lemma 3.6
that fixPG(C)|C is either primitive or contained in AGL(n, p) for some n. In either case,
the result follows by arguments above and the fact that PP
G
= PG.
The following result is a special case of the previous result.
Corollary 6.4. Let G ≤ Spk , p an odd prime and k ≥ 2, be transitive with Sylow p-
subgroup P that is abelian of rank two. Let PG be the normal closure of P in G. Then
PG / G is permutation isomorphic to L ×M with the canonical action, where L ≤ Sp` ,
M ≤ Spm , are either regular cyclic groups or doubly-transitive simple groups.
If P is of rank two, we can obtain a complete description of the transitive groups of odd
prime-power degree that have P as a Sylow p-subgroup.
Corollary 6.5. Let G ≤ Spk , p an odd prime and k ≥ 2, be transitive with Sylow p-
subgroup P that is abelian of rank two. Then one of the following is true:
1. G ≤ NS
pk
(P ),
2. k is even, G is primitive and G ≤ S2 o T , where T is a doubly-transitive group with
nonabelian socle of degree pk/2 with cyclic Sylow p-subgroup, or
3. there exists H ≤ Sp` , K ≤ Spm , ` + m = k, each of whose Sylow p-subgroups is
cyclic, and H×K ≤ G ≤ NS
p`
(H)×NSpm (K), where H is doubly-transitive and
simple and K is either a doubly-transitive and simple or is cyclic.
Proof. By Theorem 6.4, either the result follows orPG is permutation isomorphic toH×K
with the canonical action, where H and K are transitive permutation groups of degree p`
and pm, respectively, `+m = k, and at least one of H or K, say H , is a doubly-transitive
simple group, and if K is not a doubly-transitive simple group, then it is cyclic. If G is
primitive, then soc(G) must be a direct product of isomorphic nonabelian simple groups
L, and G ≤ Sr o T , r ≥ 2, for some nonabelian almost simple group T by [16, Theorem
1.1]. As a Sylow p-subgroup of G is abelian of rank two, soc(G) must be a direct product
of two isomorphic nonabelian simple groups, and so r = 2. Thus the result follows if G is
primitive.
If G is imprimitive, then either both H and K are doubly-transitive simple groups,
or only H is. In the latter case, PG = H × Zpm , so that Z(PG) = 1H × Zpm and
(PG)′ = H × 1Zpm . As both Z(PG) and (PG)′ are characteristic, we have that Z(PG)
and (PG)′ are normal in G. Thus G admits complete block systems B and C of pm blocks
of size p` and p` blocks of size pm, respectively, formed by the orbits of (PG)′ and Z(PG),
respectively. Clearly the intersection of a block of B and a block of C is a singleton, and so
by [6, Lemma 2.2] we have that G is permutation isomorphic to a subgroup of Sp` × Spm .
It is then not difficult to see that G ≤ NS
p`
(H) × NSpm (K) and the result follows. We
thus assume that both H and K are doubly-transitive simple groups.
Clearly H ×K admits B and C as complete block systems, where B and C are formed
by the orbits of H× 1K and 1H ×K, respectively. Thus B consists of pm blocks of size p`
and C consists of p` blocks of size pm, and the intersection of a block of B with a block of C
is a singleton. Let D be a nontrivial complete block system of H ×K. As H ×K contains
a regular abelian group, fixH×K(D) 6= 1, so that fixH×K(D) = H × 1K or K × 1H , as
these are the only proper nontrivial normal subgroups of H × K. Thus D = B or C. If
74 Ars Math. Contemp. 2 (2009) 59–76
g ∈ G, then g(B) and g(C) are complete block systems of g(H ×K)g−1 = H ×K so that
g(B) = B and g(C) = C or g(B) = C and g(C) = B. Note that if g(B) = C and g(C) = B,
then G has no nontrivial complete block systems, and so G is primitive, a contradiction.
Otherwise, B and C are complete block systems of G and so by [6, Lemma 2.2] we have
thatG is permutation isomorphic to a subgroup of Sp`×Spm as above. Again, we conclude
that G ≤ NS
p`
(H)×NSpm (K) and the result follows.
Remark 6.6. Corollary 6.5 (2) is a special case of [16, Theorem 1.1].
Remark 6.7. If (2) occurs in the previous result, then as a Sylow p-subgroup ofG is abelian
of rank two, a Sylow p-subgroup of T must by cyclic. Similarly, if (3) occurs and either
H or K is doubly-transitive with nonabelian socle, then a Sylow p-subgroup of H or K is
also cyclic. As every minimal transitive subgroup of a group of prime-power degree is a
p-group [19, Theorem 3.4], any Sylow p-subgroup of T andH orK as above are transitive.
Thus T and H or K are given by Theorem 1.2.
We would like to make the following conjecture.
Conjecture 6.8. Let G ≤ Spk be transitive with an abelian Sylow p-subgroup P . Then
PG is permutation isomorphic to a direct product of cyclic groups and doubly-transitive
nonabelian simple groups, with the number of factors in the direct product equal to the
rank of P .
7 2-closed groups
One of the main motivations for this work is to help in determining the full automorphism
groups of Cayley digraphs of abelian p-groups (Cayley digraphs are defined immediately
after this paragraph), which necessarily contain a transitive abelian subgroup. When deter-
mining such full automorphism groups, one of the more difficult cases seems to be when
the Sylow p-subgroup of the full automorphism group is small - and for a Cayley digraph
of an abelian p-group, the smallest possible Sylow p-subgroup is a transitive abelian group.
We now determine the full automorphism groups of Cayley digraphs of abelian groups of
rank two, and do so in the more general context of a 2-closed group. At this time, in order
to determine the full automorphism groups of Cayley digraphs of abelian groups whose
automorphism group contains a regular abelilan subgroup of rank more than two, more
explicit information will be needed about the such digraphs whose automorphism group is
primitive. This problem will be studied in [8].
Definition 7.1. Let G be a group and S ⊂ G. Define a Cayley digraph of G, denoted
Γ(G,S) to be the digraph with V (Γ(G,S)) = G and E(Γ(G,S)) = {(g, gs) : g ∈
G, s ∈ S}. We remark that if Γ(G,S) is a Cayley digraph of G, then GL ≤ Aut(Γ). If
S = S−1, then Γ(G,S) is a Cayley graph of G. A vertex-transitive digraph is a digraph
whose automorphism group acts transitively on the vertices of the graph. Note that GL is
transitive on G, so every Cayley digraph is vertex-transitive.
We now define the 2-closure of a permutation group G.
Definition 7.2. Let Ω be a set and G ≤ SΩ. Let G act on Ω × Ω by g(ω1, ω2) =
(g(ω1), g(ω2)) for every g ∈ G and ω1, ω2 ∈ Ω. We define the 2-closure of G, denoted
G(2), to be the largest subgroup of SΩ whose orbits on Ω × Ω are the same as G’s. Let
E. Dobson: On overgroups of regular abelian p-groups 75
O1, . . . ,Or be the orbits of G acting on Ω×Ω. Define digraphs Γ1, . . . ,Γr by V (Γi) = Ω
and E(Γi) = Oi. Each Γi, 1 ≤ i ≤ r, is an orbital digraph of G, and it is straightforward
to show that G(2) = ∩ri=1Aut(Γi). Clearly the automorphism group of a vertex-transitive
graph or digraph is 2-closed.
Corollary 7.3. Let G ≤ Spk be transitive and 2-closed with Sylow p-subgroup P that is
abelian of rank two. Then one of the following is true:
1. G has a normal Sylow p-subgroup,
2. G is primitive, k = 2, and G is permutation isomorphic to S2 o Sp,
3. k = 2, and G is permutation isomorphic to Sp × Sp, or
4. G is permutation isomorphic to Sp × A, where A ≤ N(pk−1) has order dividing
(p− 1)pk−1.
Proof. Suppose G does not have a normal Sylow p-subgroup. By Corollary 6.5, either k is
even, G is primitive, and G ≤ S2 o T , where T is a nonabelian doubly-transitive group of
degree pk/2 with cyclic Sylow p-subgroup, or there existsH ≤ Sp` , K ≤ Spm , `+m = k,
each of whose Sylow p-subgroups is cyclic, and H ×K ≤ G ≤ NS
p`
(H) × NSpm (K),
and H and K are either doubly-transitive simple groups or are cyclic. By [15] (this result
also appears in [4, Theorem 5.1]), we have that if H or K is doubly-transitive, then, as the
2-closure of a doubly-transitive group is a symmetric group, H = Sp` orK = Spm . By the
same result, we must also have that T is a symmetric group. We conclude that ifH ,K, or T
is doubly-transitive, then as a Sylow p-subgroup ofG is abelian, we must have that ` = 1 or
m = 1. In the primitive case, we then have that G = S2 o Sp and the result follows. If both
H and K are doubly-transitive, then clearly G = Sp×Sp, and if both H and K are cyclic,
then G has a normal Sylow p-subgroup. If exactly one of H and K is doubly-transitive,
say H , then Sp × Zpk−1 ≤ G ≤ NSp(Sp) ×NSpk−1 (Zpk−1) = Sp ×NSpk−1 (Zpk−1), so
that G = Sp × A, where A ≤ NS
pk−1
(Zpk−1) = N(pk−1). Finally, |A| | (p − 1)pk−1 as
|N(pk−1)| = (p− 1)pk−2 · pk−1 and a Sylow p-subgroup of A must have order pk−1 as a
Sylow p-subgroup of G has order pk.
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