On the full automorphism group of a Hamiltonian cycle system of odd order

0.35 MB·English

by  Marco Buratti

On the full automorphism group of a
Hamiltonian cycle system of odd order
Marco Buratti ∗ Graham J. Lovegrove † Tommaso Traetta ‡
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Abstract
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2 It is shown that a necessary condition for an abstract group G to be the
n fullautomorphismgroupofaHamiltoniancyclesystemisthatGhasodd
a order or it is either binary, or the affine linear group AGL(1,p) with p
J prime. We show that this condition is also sufficient except possibly for

8 the class of non–solvable binary groups.
2
Keywords: Hamiltonian cycle system; automorphism group.

]
O ∗Dipartimento di Matematica e Informatica, Universita` degli Studi di Perugia, via Van-
vitelli1-06123Italy,email: [email protected]
C †Department of Mathematics and Statistics, The Open University, Walton Hall, Milton
. KeynesMK76AA,U.K.,email: [email protected]
h ‡Dipartimento di Matematica e Informatica, Universita` degli Studi di Perugia, via Van-
t
a vitelli1-06123Italy,email: [email protected]
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1 Introduction
DenoteasusualwithK thecompletegraphonv vertices. AHamiltoniancycle
v
systemoforderv(briefly,anHCS(v))isasetofHamiltoniancyclesofK whose
v
edges partition the edge-set of K . It is very well known [20] that an HCS(v)
v
exists if and only if v is odd and v ≥3.
Two HCSs are isomorphic if there exists a bijection (isomorphism) between
theirsetofverticesturningoneintotheother. AnautomorphismofaHCS(v)is
anisomorphismofitwithitself,i.e.,apermutationoftheverticesofK leaving
v
it invariant.
HCSs possessing a non-trivial automorphism group have drawn a certain
attention (see [10] for a short recent survey on this topic). Detailed results can
befoundin: [11,17]forthecyclic groups;[12]forthedihedral groups;[6]forthe
doubly transitive groups; [9] for the regular HCS; [1, 8] for the symmetric HCS;
[13] for those being both cyclic and symmetric; [4, 15, 3] for the 1-rotational
HCS and 2–pyramidal HCS.
Given a particular class of combinatorial designs, to establish whether any
abstract finite group is the full automorphism group of an element in the class
is in general a quite hard task. Some results in this direction can be found in
[21] for the class of Steiner triple and quadruple systems, in [22] for the class of
finiteprojectiveplanes, in[7]fortheclassofnon–Hamiltonian2−factorizations
of the complete graph, and very recently in [16, 18, 19] for the class of cycle
systems.
This paper deals with the following problem:
Determining the class G of finite groups that can be seen as the full
automorphism group of a Hamiltonian cycle system of odd order.
As a matter of fact, some partial answers are known. In [15] it is proven that
anysymmetricallysequenceablegroupliesinG (see[3]forHCSsofevenorder).
In particular, any solvable binary group (i.e. with a unique element of order 2)
except for the quaternion group Q , is symmetrically sequenceable [2], hence it
8
can be seen as the full automorphism group of an HCS. In [6] it is shown that
the affine linear group AGL(1,p), p prime, is the full automorphism group of
the unique doubly transitive HCS(p).
HereweprovethatanyfinitegroupGofoddorderisthefullautomorphism
group of an HCS. This result will be achieved in Section 3 by means of a new
doubling construction described in Section 2. On the other hand, in Section 3
we also prove that if G∈G has even order, then G is necessarily binary or the
affine linear group AGL(1,p) with p prime; still, we show that Q lies in G. We
8
obtain, in this way, the major result of this paper:
Theorem 1.1. If a finite group G is the full automorphism group of a Hamil-
tonian cycle system of odd order then G has odd order or it is either binary,
or the affine linear group AGL(1,p) with p prime. The converse is true except
possibly in the case of G binary non-solvable.
We therefore leave open the problem only for non–solvable binary groups.
2
2 A new doubling construction
We describe a new doubling construction that will allow us to constuct an
HCS(4n+1) starting from three HCS(2n+1) not necessarily distinct.
For any even integer n≥1, take three HCS(2n+1), say H ,H ,H , on the
1 2 3
set {∞} ∪ [2n], where [2n]={1,2,...,2n}. Denote by A , B , C , for i∈[n],
i i i
the cycles composing H , H , H , respectively, let
1 2 3
A =(∞,α ,...,α ) B =(∞,β ,...,β ) C =(∞,γ ,...,γ ).
i i,1 i,2n i i,1 i,2n i i,1 i,2n
We need these HCS to satisfy the following property:
α =β =γ , and α =β =γ , for i∈[n]. (2.1)
i1 i1 i1 i,2n i,2n i,2n
We construct an HCS(4n + 1) T on the set {∞} ∪ ([2n] × {1,−1}). For
convenience,ifz =(x,y)∈[2n]×{1,−1},thenthepoint(x,−y)willbedenoted
by z(cid:48).
Let T = {T ,T | i ∈ [n]} be the set of 2n cycles of length 4n+1 and
i1 i2
vertex-set {∞} ∪ ([2n]×{1,−1}) obtained from the cycles of H ,H , H as
1 2 3
follows: set a = (α ,1), b = (β ,1), and c = (γ ,1), for i ∈ [n] and
ij ij ij ij ij ij
j ∈ [2n] and define the cycles T , T , of first and second type respectively, as
i1 i2
follows:
T =(∞,a ,a ,...,a ,b(cid:48) ,b(cid:48) ,...,b(cid:48) ),
i,1 i,1 i,2 i,2n i,2n i,2n−1 i1
T =(∞,c ,c(cid:48) ,c ,c(cid:48) ,...,c(cid:48) ,c ,c(cid:48) ,c ,c(cid:48) ,...,c(cid:48) ).
i,2 i,2n i,2n−1 i,2n−2 i,2n−3 i1 i,1 i,2 i,3 i,4 i,2n
Remark 2.1. Notethat,byconstruction,theneighborsof∞andthemiddleedge
ofanyofthecyclesofT arebothpairsoftheform(z,z(cid:48))forz ∈[2n]×{−1,1}.
Also, if T = (∞,z,...,w,w(cid:48),...,z(cid:48)) is a cycle of T, then there is a cycle
1
T =(∞,w,...,z(cid:48),z,...,w(cid:48)) of T of alternate type.
2
We first show that our doubling construction yields a Hamiltonian cycle
system.
Lemma 2.1. T is an HCS(4n+1).
Proof. Every unordered pair of form (x ,x ) or (∞,x ), x ,x ∈ [2n],x (cid:54)= x
1 2 1 1 2 1 2
iscontainedinauniquecycleofeachofH ,H ,H . Thefirsttypeofcycleabove
1 2 3
containsallpairsofform((x ,y),(x ,y)),x ,x ∈[2n],x (cid:54)=x ,y =−1,1,and
1 2 1 2 1 2
the second cycle type contains all pairs of form ((x ,1),(x ,−1)), x ,x ∈[2n],
1 2 1 2
x (cid:54)= x . We are left to show that T contains the following edges: (∞,z) and
1 2
(z,z(cid:48)) for z ∈[2n]×{−1,1}.
In view of Property (2.1), we have that a =b =c for j =1,2n. There-
ij ij ij
fore, the middle edges of the cycles in T are exactly (a ,a(cid:48) ) and (a ,a(cid:48) )
i1 i1 i,2n i,2n
for i∈[n]. Considering that {a ,a |i∈[n]}=[2n]×{1}, we conclude that
i1 i,2n
T covers the edges (z,z(cid:48)) for z ∈[2n]×{−1,1}.
Finally, the edges incident with ∞ and covered by T are the following:
(∞,a ), (∞,b(cid:48) ),(∞,c ),(∞,c(cid:48) ), for i ∈ [n]. With a reasoning similar to
i1 i1 i,2n i,2n
theformerweeasilyseethatalledges(∞,z)withz ∈[2n]×{−1,1}arecovered
by T.
3
Example 2.2. Here we show how to construct an HCS(13) by applying the
doubling construction to three HCSs or order 7. Let G=(cid:104)g(cid:105) be the cyclic group
of order 6 generated by g, and let H ,H , and H denote the three HCSs of
1 2 3
order 7 defined as follows:
1. H = {A ,A ,A } with A = (∞,1,g,g5,g2,g4,g3) and A = A ·gi−1
1 1 2 3 1 i 1
for i=2,3,
2. H = {B ,B ,B } with B = (∞,1,g4,g5,g2,g,g3) and B = B ·gi−1
2 1 2 3 1 i 1
for i=2,3,
3. H ={C ,C ,C } and C =B for i=1,2,3 (hence, H =H ),
3 1 2 3 i i 3 2
where A ·gi−1 (B ·gi−1) is the cycle that we obtain by replacing each vertex
1 1
of A (B ) different from ∞, say x, with x·gi−1. It is easy to check that H
1 1 1
and H are HCS(7); also, property 2.1 is satisfied (see Figure 1).
2
Figure 1: Two HCSs of order 7.
The cycles T and T for i=1,2,3 are shown in Figure 2. Each cycle T
i1 i2 i1
is basically constructed from the paths that we obtain from A (continous path)
i
and B (dashed path) after removing ∞, by joining two of their ends with ∞
i
and joining to each other the other two ends (zigzag edges). The construction
of each cycle T is only based on C =B . Consider two copies of the path we
i2 i i
obtain from B (dashed path) after removing ∞ and replace the horizontal edges
i
with the diagonal ones. At the end, we add the zigzag edges.
It is easy to check that the set T ={T ,T |i=1,2,3} is an HCS(13).
i1 i2
Remark 2.2. This construction can be used to construct many different HCS
fromthesamebasesystemsbyrelabellingtheverticesofoneormoreofH ,H ,H
1 2 3
in a way that Property (2.1) is still satisfied.
Lemma 2.3. All automorphisms of T fix the ∞ point for n>1.
4
Figure 2: An HCS(13) resulting from the doubling condstruction.
Proof. Suppose there is an automorphism φ of T which does not fix the ∞
point, and let u = φ(∞) and φ−1(∞) = v. Also denote φ(v(cid:48)) by w. The
cycle T containing (∞,v) as an edge has form: (∞,v,...,z,z(cid:48),...,v(cid:48)). This is
1
mapped by φ to the cycle φ(T )=(u,∞,...,φ(z),φ(z(cid:48)),...,w).
1
Now consider the cycle T that is of alternate type to T and that has the
2 1
middle edge (v,v(cid:48)), i.e. T = (∞,z,...,v(cid:48),v,...,z(cid:48)). Then T is mapped to
2 2
the cycle φ(T ) = (u,φ(z),...,w,∞,...,φ(z(cid:48))). The middle edge of φ(C ) is
2 2
(u,φ(z)), hence there exists a cycle where the neighbors of ∞ are u and φ(z).
Since u is adjacent to ∞ in φ(T ), then φ(C ) = (u,∞,φ(z),φ(z(cid:48)),w) and
1 1
n=1.
3 HCSs with a prescribed full automorphism
group
In this section we prove that any group of odd order lies in the class G of finite
groupsthancanbeseenasthefullautomorphismgroupofanHCSofoddorder.
After that, we prove that whenever a group G ∈ G has even order then it is
either binary or the affine linear group AGL(1,p), with p prime; also, we show
that the quaternion group Q lies in G. As mentioned in the introduction, it is
8
knownthatanybinarysolvablegroup(cid:54)=Q [15]andAGL(1,p)(pprime)[6]lie
8
inG. Therefore,weleaveopentheproblemofdeterminingwhethernon–solvable
binary groups lie in G as well.
5
Inordertoshowthatanygroupofoddorderisthefullautomorphismgroup
ofasuitableHCS(4n+1),wewillneedsomepreliminarieson1–rotationalHCSs.
We will use multiplicative notation to denote any abstract group; as usual,
the unit will be denoted by 1.
An HCS(2n+1) H is 1-rotational over a group Γ of order 2n if Γ is an
automorphism group of H acting sharply transitively on all but one vertex. In
this case, it is natural to identify the vertex-set with {∞} ∪ Γ where ∞ is the
vertexfixedbyanyg ∈ΓandviewtheactionofΓonthevertex-setastheright
multiplication where ∞·g =∞ for g ∈Γ.
Itisknownfrom[14](asaspecialcaseofamoregeneralresulton1-rotational
2-factorizationsofthecompletegraph)thatGisbinary,namely,ithasonlyone
element λ of order 2. As usual, we denote by Λ(Γ) = {1,λ} the subgroup of Γ
of order 2.
Inthesamepaper,theauthorsalsoprovethattheexistenceofa1-rotational
HCS(2n+1) H is equivalent to the existence of a cycle A = (∞,α ,...,α )
1 2n
with vertex-set {∞} ∪ Γ such that
A·λ=A and {α α−1 ,α α−1 |i∈[n−1]}=Γ\{1,λ}, (3.1)
i i+1 i+1 i
In this case, α = α ·λ and H = {A·x | x ∈ X}, where X is a complete
n+1−i i
system of representatives for the cosets of Λ(Γ) in Γ. This means that H is the
set of distinct translates of any of its cycles.
We are now ready to prove the following result.
Theorem 3.1. Any group G of odd order n is the full automorphism group of
a suitable HCS(4n+1).
Proof. Consider the set H of the following 13–cycles:
(∞,0 ,1 ,5 ,2 ,4 ,3 ,3 ,1 ,2 ,5 ,4 ,0 )
0 0 0 0 0 0 1 1 1 1 1 1
(∞,1 ,2 ,0 ,3 ,5 ,4 ,4 ,2 ,3 ,0 ,5 ,1 )
0 0 0 0 0 0 1 1 1 1 1 1
(∞,2 ,3 ,1 ,4 ,0 ,5 ,5 ,3 ,4 ,1 ,0 ,2 )
0 0 0 0 0 0 1 1 1 1 1 1
(0 ,4 ,5 ,2 ,1 ,3 ,∞,3 ,1 ,2 ,5 ,4 ,0 )
0 1 0 1 0 1 0 1 0 1 0 1
(1 ,5 ,0 ,3 ,2 ,4 ,∞,4 ,2 ,3 ,0 ,5 ,1 )
0 1 0 1 0 1 0 1 0 1 0 1
(2 ,0 ,1 ,4 ,3 ,5 ,∞,5 ,3 ,4 ,1 ,0 ,2 )
0 1 0 1 0 1 0 1 0 1 0 1
It is not difficult to check that H is an HCS(13) with no non-trivial automor-
phism.
Now, given a group G of odd order n ≥ 3, we show that there exists an
HCS(4n+1)whosefullautomorphismgroupisisomorphictoG. LetZ ={1,λ},
2
andsetΓ=Z ×G. SinceΓisasolvablebinarygroup,itisknownfrom[2]that
2
thereexistsanHCS(2n+1)Hwithvertices{∞} ∪ Γthatis1-rotationalunder
Γ. Let G = {g = 1,g ,...,g } and for a given cycle C = (∞,x ,...,x )
1 2 n 1 2n
of H set C = C · g for i ∈ [n]; therefore, H = {C | i ∈ [n]}. We have
i i i
previously pointed out that C satisfies (3.1). Therefore, we can write C =
(∞,x ,...,x ,x ,...,x ), where x =x ·λ. Also, there exists k ∈[n−1]\{1}
1 n n 1 i i
such that x x−1 =x x−1λ or x x−1 =x x−1λ. We define the (2n+1)-cycle
k k+1 1 2 k k+1 2 1
C∗ as follows:
C∗ =(∞,x ,x ,...,x ,x ,...,x ,x ,...,x ,x ,...,x ,x ).
1 2 k k+1 n n k+1 k 2 1
6
Of course, C∗·λ=C∗. Also, we have that
{x x−1,x x−1}={x x−1λ,x x−1λ}={x x−1 ,x x−1}
1 2 2 1 1 2 2 1 k k+1 k+1 k
{x x−1 ,x x−1}={x x−1 λ,x x−1λ}={x x−1,x x−1}, and
k k+1 k+1 k k k+1 k+1 k 1 2 2 1
{x x−1 |i∈[n−1]\{1,k}}={x x−1 |i∈[n−1]\{1,k}}.
i i+1 i i+1
Therefore, C∗, as well as C, satisfies both conditions in (3.1). It follows that
H∗ = {C∗,...,C∗}, with C∗ = C∗ ·g , (namely, the G-orbit H∗ of C∗) is an
1 n i i
HCS(2n+1).
Now we apply the doubling construction defined above with H = H∗ and
1
H = H = H and let T denote the resulting HCS(4n+1) with vertex set
2 3
{∞}∪Γ×{1,−1}. Note that Property 2.1 is satisfied, as the vertices adjacent
with∞inC andC∗ coincide. ThestartercyclesT , T ofT havethefollowing
1 2
form:
T =(∞,a =b ,a ,...,a ,a =b ,b(cid:48) ,...,b(cid:48)),
1 1 1 2 2n−1 2n 2n 2n 1
T =(∞,b ,b(cid:48) ,...,b ,b(cid:48),b ,b(cid:48),...,b ,b(cid:48) ),
2 2n 2n−1 2 1 1 2 2n−1 2n
where b =(x ,1), b(cid:48) =(x ,−1), a =(α ,1), and (α ,...,α ) is the ordered
j j j j j j 1 2n
sequence (x ,x ,x ,...) of the vertices of C∗.
1 2 3
The cycles of T are generated from T , T by the action (x,i) (cid:55)→ (xg,i),
1 2
x∈Γ, i=1,−1, g ∈G.
For any g ∈ G, this action is an automorphism of T, and we denote the
group of such automorphisms by τ .
G
We are going to show that τ = Aut(T). Note that b = b ·(λ,1) and
G 2n 1
that b ,b(cid:48),b ,b(cid:48) form a complete system of representatives of the G-orbit on
1 1 2n 2n
Γ×{1,−1}. Therefore, given an automorphism ϕ of T, then there exists z ∈G
such that φ = τ ϕ maps b to one of b ,b(cid:48),b ,b(cid:48) . It is enough to prove that
z 1 1 1 2n 2n
it is always b . It will follow that φ is the identity, because, since the ∞ point
1
is always fixed, all the vertices of T are fixed. Hence, ϕ=τ−1 ∈τ . Since τ
1 z G G
is isomorphic to G we get the assertion.
Suppose that φ(b ) = b(cid:48). It follows that the edge (∞,b(cid:48)) of T is also an
1 1 1 1
edge of φ(T ), therefore T =φ(T ). In other words, φ is the reflection of T in
1 1 1 1
the axis through ∞. In particular, φ swaps b and b(cid:48), and also a and b(cid:48). This
1 1 2 2
meansthatφfixestheedge(b ,b(cid:48))andhenceitfixesthecycleT containingthis
1 1 2
edge. Therefore, φ is the reflection of T in the axis through the edge (b ,b(cid:48)).
2 1 1
It follows that φ swaps b and b(cid:48), but this contradicts the previous conclusion
2 2
as b =(x ,1)(cid:54)=(x ,1)=a .
2 2 2 2
Nowsupposethatφ(b )=b . ThenφmapsT toT . Therefore,φ(a )=
1 2n 1 2 2n−1
b = (x ,1) = a , that is, φ fixes a . However this implies that φ is the
2 2 2n−1 2n−1
identity, since then the edge (∞,a ) and all the points in the cycle that
2n−1
contains it are also fixed. A similar argument applies to the possibility that
φ(b )=b(cid:48) . Thus φ fixes b .
1 2n 1
We point out to the reader that the HCS(13) T of Example 2.2 has been
constructed following the proof of Theorem 3.1. In fact, H and H are two
1 2
1-rotational HCS(7) and if we set C = B , then C∗ = A . It then follows that
1 1
Aut(T)=Z .
3
We finally consider the case where a group G∈G has even order and prove
the following:
7
Theorem 3.2. If H is an HCS(2n+1) whose full isomorphism group has even
order, then either Aut(H) is binary or 2n+1 is a prime and Aut(H) is the
affine linear group AGL(1,2n+1).
Proof. Let H be an HCS(2n+1) and assume that Aut(H) has even order. We
first show that any involution of Aut(H) has exactly one fixed point and fixes
eachcycleofH. Thismeansthataninvolutionisuniquelydeterminedoncethe
pointitfixesisknown. Therefore,distinctinvolutoryautomorphismsofHhave
distinct fixed points.
Suppose α is an involutory automorphism of H and let x be any point not
fixed by α. Then the edge [x,α(x)] occurs in some cycle C, and α fixes this
edge. ItthenfollowsthattheentirecycleC isfixedbyαwhichthereforeactson
C as the reflection in the axis of [x,α(x)]. The point a opposite to this edge is
then the only fixed point of α. Note that there are n edges of the form [x,α(x)]
with x(cid:54)=a, and that they are partitioned among the n cycles of H. Therefore,
reasoning as before, we get that α fixes all cycles of H
Now, assume that Aut(H) is not binary and let β be a second involutory
automorphismwithfixedpointb. WedenotebyS thesubgroupofAut(H)gen-
eratedbyαandβ. SincebothinvolutionsfixallcyclesofH,allautomorphisms
in S fix them. If C(cid:48) is the cycle of H containing the edge [a,b], then the map
βα is the rotation of C(cid:48) with step 2. Since C(cid:48) has odd order, this means that S
also contains all rotations of C(cid:48). Hence, S is the dihedral group of order4n+2.
We also have that 2n+1 is prime. In fact, let ϕ ∈ S be an automorphism
of prime order p; given a point x not fixed by ϕ we denote by C(cid:48)(cid:48) the cycle
of H containing [x,ϕ(x)]. Of course, all edges of the form [ϕi(x),ϕi+1(x)]
with i = 0,...,p − 1 lies in C(cid:48)(cid:48). It follows that C(cid:48)(cid:48) is the cycle (x,ϕ(x),
ϕ2(x),...,ϕp−2(x),ϕp−1(x)), hence 2n+1=p.
One can easily see that H is then the unique 2-transitive HCS(p) whose full
automorphism group is AGL(1,p) [6] and this completes the proof.
The following theorem provides a sufficient condition for a binary group H
of order 4m to be the full automorphism group of infinitely many HCSs of odd
order.
Theorem 3.3. Let H be a binary group of order 4m and let d ≥ 3 be an odd
integer. If there exists a 1-rotational HCS(4md+1) under H ×Z , then there
d
exists an HCS(4md+1) whose full automorphism group is H.
Proof. Let Z =(cid:104)z(cid:105) denote the cyclic group of order d generated by z and let λ
d
be the unique element of order 2 in H. It is straightforward that G = H ×Z
d
is a binary group and its element of order 2 is λ.
Now, let H be a 1–rotational HCS(4md+1) under G, and let A = (∞,a ,
1
a2,...,a2md,a2mdλ,...,a2λ,a1λ)denoteitsstartercycle. Also,letOrbZd(A)=
{A ,A ,...,A }betheZ –orbitofAwhereA =A·zi fori=0,1,...,d−1.
0 1 d−1 d i
We can then see H as the union of the H-orbits of each cycle in OrbZ (A), that
d
is, H=∪d−1(Orb (A )).
i=0 H i
We now construct a new HCS(4md+1) H∗ through a slight modification of
the cycle A :
0
1. Since the binary group H has order 4m, there is at least an element
x ∈ H of order 4. Also ∆A covers all non-zero elements of H ×Z . It
d
follows that there exists j ∈ [2md − 1] such that x = a a−1. Now,
j+1 j
8
let A∗ denote the graph we get from A by replacing the edges in E =
0 0
{[a ,a ],[a λ,a λ]} with those in E∗ ={[a ,a λ],[a λ,a ]}.
j j+1 j j+1 j j+1 j j+1
2. Set H∗ =∪d−1Orb (A ) ∪ Orb (A∗).
i=1 H i H 0
ItiseasytoseethatA∗ isa(4md+1)–cycle. Also,E(A )\E =E(A∗)\E∗. We
0 0 0
show that E∗ = E ·y, where y = a−1xa . We first point out that x2 = λ, since
j j
x2 has order 2 and H is binary. Also, note that a = xa and recall that λ
j+1 j
commutes with every element in G. Therefore,
[a ,a ]y =[xa ,x2a ]=[a ,a λ], and
j j+1 j j j+1 j
[a λ,a λ]y =[a ,a ]yλ=[a ,a λ]λ=[a λ,a ].
j j+1 j j+1 j+1 j j+1 j
Sincey ∈H,wehavethatOrb (E)=Orb (E∗). ItfollowsthatOrb (A )and
H H H 0
Orb (A∗) cover the same set of edges. Since H\Orb (A )=H∗\Orb (A∗),
H 0 H 0 H 0
we have that H∗ covers the same set of edges covered by H, that is, H∗ is an
HCS(4md+1).
We now show that the full automorphism groups of H∗ is isomorphic to
H. We set A = Aut(H∗) and denote by A the A-stabilizer of ∞. Also, let
∞
τ denote the translation by the element g ∈ G, that is, the permutation on
g
G ∪ {∞} fixing ∞ and mapping x ∈ G to xg for any x,g ∈ G. Finally, let
τ and τ denote the group of all translations by the elements of G and H,
G H
respectively. It is easy to check that, by construction, τ is an automorphism
H
group of H∗ fixing ∞, that is, τ ⊆ A ; on the other hand, the replacement
H ∞
of Orb (A ) with Orb (A∗) ensures that τ is not an automorphism of H∗
H 0 H 0 g
whenever g ∈G\H. In other words, for any g ∈G we have that
τ ∈A if and only if g ∈H. (3.2)
g
WearegoingtoshowthatA =τ . Letϕ∈A andnotethat|Orb (A∗)|<
∞ H ∞ H 0
|H∗ \Orb (A∗)|. Therefore, there exists a cycle C = (∞,g ,g ,...,g ) ∈
H 0 1 2 4mp
H∗ \ Orb (A∗) such that ϕ(C) = H∗ \ Orb (A∗). Since ϕ fixes ∞, then
H 0 H 0
ϕ(C) = (∞,ϕ(g ),ϕ(g ),...,ϕ(g )). Note that H∗\Orb (A∗) ⊆ H, there-
1 2 4mp H 0
foreϕ(C)isatranslateofC,namely,thereexistsx∈Gsuchthatϕ(g )=g ·x.
i i
This means that ϕ = τ , and in view of (3.2), x ∈ H. It follows that ϕ ∈ τ ,
x H
hence A =τ .
∞ H
Sinceτ isisomorphictoH,wearelefttoshowA=A . Assumethatthere
H ∞
exists w ∈ OrbA(∞)\{∞}. From a classic result on permutations groups, we
have that A and A are conjugate hence, in particular, they are isomorphic.
w ∞
ThismeansthatA containsaninvolutionψ. SinceA ∩A ={id}(otherwise
w ∞ w
therewouldbeanon-trivialautomorphismfixingtwovertices),wehavethatτ
λ
and ψ are distinct. Therefore, in view of Theorem 3.2, we have that 4md+1 is
aprimeandA=AGL(1,4md+1); inparticular,|A|=4md(4md+1). Butthis
leads to a contradiction since |A|=|A∞||OrbA(∞)| and |OrbA(∞)|≤4md+1,
that is, |A|≤4m(4md+1) where d≥3. Therefore, OrbA(∞)={∞}, namely,
A=A .
∞
As a consequence we obtain the following:
Theorem 3.4. The quaternion group Q is the full automorphism group of an
8
HCS(8d+1), for any d≥3.
9
Proof. It is enough to observe that Q ×Z , d ≥ 3, is a binary solvable group
8 d
(cid:54)=Q . As mentioned earlier in this paper, any binary solvable group is the full
8
automorphismgroupofanHCSofoddorder. Then,theconclusionimmediately
follows by Theorem 3.3
Collecting the above results, we can prove Theorem 1.1.
proof of Theorem 1.1. ThefirstpartofthestatementisproveninTheorem3.2.
Now,letGbeafinitegroup. IfGhasoddorderoritisQ ,thenbyTheorems
8
3.1 and 3.3 we have that G is the full automorphism group of a suitable HCS
of odd order. If G is a binary solvable group (cid:54)= Q or AGL(1,p) with p prime,
8
it is known [15, 6] that G∈G.
Remark 3.1. Note that Theorem 3.3 would allow us to solve completely the
problemunderinvestigationinthispaper,ifonecouldshowthatanysufficiently
large binary group has a 1–rotational action on an HCS of odd order. In this
case, given a binary non-solvable group H of order 4m, we could consider the
group G=H×Z with d sufficiently large to ensure that G has a 1–rotational
d
action on an HCS(4md+1). By Theorem 3.3 we would have that H is the full
automorphism group of an HCS of odd order.
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