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Journal of Algebraic Combinatorics 8 (1998), 205–215
c 1998 Kluwer Academic Publishers. Manufactured in The Netherlands.
°
On Cayley Graphs of Abelian Groups
CAI HENG LI
li@maths.uwa.edu.au
Department of Mathematics, University of Western Australia, Nedlands, W.A. 6907, Australia
Received March 9, 1995; Revised April 18, 1997
Abstract. Let G be a finite Abelian group and Cay(G, S) the Cayley (di)-graph of G with respect to S, and let
A = Aut Cay(G, S) and A1 the stabilizer of 1 in A. In this paper, we first prove that if A1 is unfaithful on S then S
contains a coset of some nontrivial subgroup of G, and then characterize Cay(G, S) if A1S contains the alternating
group on S. Finally, we precisely determine all m-DCI p-groups for 2 ≤ m ≤ p + 1, where p is a prime.
Keywords: Cayley graph, isomorphism, CI-subset, m-DCI group
1.
Introduction
Let G be a finite group and S a Cayley subset of G, that is, S does not contain the identity
of G. The Cayley (di)-graph Cay(G, S) of G with respect to S has the elements of G
as vertices and the pairs (g, sg), g ∈ G, s ∈ S, as edges. Given a Cayley subset S of
G, if, for any Cayley subset T of G, Cay(G, S) ∼
= Cay(G, T ) implies T = S σ for some
σ ∈ Aut(G), then S is called a CI-subset (CI stands for Cayley Isomorphism). A finite
group G is called an m-DCI group if all of its Cayley subsets of G of size at most m are
CI-subsets; G is called a DCI-group if it is a |G|-DCI group. Similarly, G is called an
m-CI group if all Cayley subsets S of G of size at most m with S = S −1 are CI-subsets, G
is called a CI-group if G is an |G|-CI group. The problem of determining which groups
are m-DCI groups and m-CI groups has been investigated for a long time, see [6, 10, 12]
for references. Recently, all m-DCI groups and all m-CI groups for m ≥ 2 have been
classified in [10] and [9], respectively, in the sense that all the possibilities for such groups
are explicitly listed. However, it is still a difficult question to determine which of them are
really m-DCI (m-CI) groups. Babai and Frankl [2] asked whether the elementary abelian
group Z dp for any p and d was an m-CI group for all m ≤ |G| (in other words, Z dp is a
CI-group). Godsil [6] and Dobson [4] proved this to be true for d = 2, 3, respectively.
However, recently Nowitz [11] gave a negative answer to the question by proving that
Z 26 is not a 31-CI group. It is not known if this the answer of the question is positive
for odd prime p and d ≥ 4. The main aims of this paper are to characterize Cayley
graphs Cay(G, S) of abelian groups by the action of A1 on S, where A1 is the stabilizer
of 1 in Aut Cay(G, S), and to determine precisely m-DCI p-groups for 2 ≤ m ≤ p + 1,
which implies that the answer of Babai and Frankl’s question is positive for any p, d and
m ≤ p + 1.
Notation In this paper, Z n denotes a cyclic group of order n, Q 8 is the quaternion group
of order 8. Recall that a group is called homocylic if it is a direct product of some cyclic
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groups of the same order. For groups G and H , H ≤ G denotes that H is a subgroup
of G, and G o H denotes a semidirect product of G by H . For a positive integer n, Cn
denotes the directed cycle of length n, K n denotes the complete graph on n vertices and
K n,n denotes the complete-bipartite graph on 2n vertices. For a directed graph 0 = (V, E),
its complement 0̄ = (V, Ē) is the directed graph with vertex set V such that (a, b) ∈ Ē if
and only if (a, b) 6∈ E. The direct product 01 × 02 of two directed graphs 01 = (V1 , E 1 )
and 02 = (V2 , E 2 ) is the directed graph with vertex set V1 × V2 such that ((a1 , a2 ), (b1 , b2 ))
is an edge if and only if either (a1 , b1 ) ∈ E 1 and a2 = b2 , or (a2 , b2 ) ∈ E 2 and a1 = b1 .
The lexicographic product 01 [02 ] of two directed graphs 01 = (V1 , E 1 ) and 02 = (V2 , E 2 )
is the graph with vertex set V1 × V2 such that ((a1 , a2 ), (b1 , b2 )) is an edge if and only if
either (a1 , b1 ) ∈ E 1 or a1 = b1 and (a2 , b2 ) ∈ E 2 . For any vertex x of graph Cay(G, S),
the neighborhood 0(x) of x in Cay(G, S) equals x S = {xai | 1 ≤ i ≤ m}. Let 0i (x) =
{y ∈ G | d(x, y) = i}, where d(x, y) denotes the distance from x to y in Cay(G, S). Note
that 0(x) = 01 (x).
In Section 2, we quote some results which are used in the following sections. Section 3
characterizes some Cayley graphs on Abelian groups, and Section 4 precisely determines
m-DCI p-groups for certain values of m.
2.
Preliminaries
In this section, we quote some results which we need in the following sections. Let G be a
finite group, S a Cayley subset of G and let A = Aut Cay(G, S). Babai [1] gave a criterion
for a subset of G to be a CI-subset.
Theorem 2.1 ([1]) For a given group G and a Cayley subset S of G, S is a CI-subset if
and only if for any τ ∈ Sym(G) with τ Gτ −1 ≤ A, there exists α ∈ A such that αGα −1 =
τ Gτ −1 , where Sym(G) is the symmetric group on G.
The normalizer of G in A is often useful for characterizing Cay(G, S).
Lemma 2.2 ([5]) Let A = Aut Cay(G, S) and Aut(G, S) = {α ∈ Aut(G) | S α = S}. Then
N A (G) equals a semidirect product of G by Aut(G, S), that is, N A (G) = G o Aut(G, S).
All finite m-DCI groups for m ≥ 2 have been explicitly listed in [10], in particular, we
have
Lemma 2.3 ([10, Proposition 3.1]) Let G be a finite m-DCI p-group, where m ≥ 2 and
p is a prime.
(1) If p is odd and 2 ≤ m ≤ p − 1, then G is homocyclic.
(2) If m = p, then either G is elementary Abelian, cyclic, or G = Q 8 .
(3) If m = p + 1, then either G is elementary Abelian, or G = Z 4 or Q 8 .
Lemma 2.4 ([16])
The quaternion group Q 8 is a DCI-group.
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CAYLEY GRAPHS OF ABELIAN GROUPS
3.
Cayley graphs of Abelian groups
In this section, we characterize some properties of Cayley graphs of Abelian groups. Let
G be a finite group, S = {a1 , a2 , . . . , am } be a Cayley subset of G and 0 = Cay(G, S).
Let A be the full automorphism group of 0 and A1 the stabilizer of 1 in A. For h distinct
elements ai1 , ai2 , . . . , aih ∈ S and y ∈ G, let
( ¡
¢
¡
¢
¡
¢
0 yai1 , . . . , yaih = 0 yai1 ∩ · · · ∩ 0 yaih ,
¡
¢
¡
¢² S
0 ∗ yai1 , . . . , yaih = 0 yai1 , . . . , yaih
x∈R 0(yx),
where R = S\{ai1 , . . . , aih }, that is, 0 ∗ (yai1 , . . . , yaih ) is the set of all vertices of 0 which
are joined to every element of {yai1 , . . . , yaih } and to no element of y R. Let
0i∗ = max{|0 ∗ (u 1 , . . . , u i )| | u 1 , . . . , u i ∈ S}.
If R = {u 1 , . . . , u i ) ⊆ S, then denote 0 ∗ (u 1 , . . . , u i ) by 0 ∗ (R) sometimes.
Lemma 3.1 Suppose that G is an Abelian group. Then
(i) 1 ∈ 0 ∗ (W ) for W ⊆ S if and only if W = W −1 and (S\W ) ∩ (S\W )−1 = ∅;
(ii) 0 ∗ (x x1 , . . . , x xk ) = x0 ∗ (x1 , . . . , xk ) for any x ∈ G and any x1 , . . . , xk ∈ S;
(iii) 0k∗ ≤ k for every k ≥ 1;
(iv) every element of 02 (1) lies in 0 ∗ (x1 , . . . , xk ) for some x1 , . . . , xk ∈ S.
By the definition of 0 ∗ (x1 , . . . , xk ), part (i) is clear. Again by definition, we have
²[
0(x z)
y ∈ 0 ∗ (x x1 , . . . , x xk ) ⇔ y ∈ 0 ∗ (x x1 ) ∩ · · · ∩ 0 ∗ (x xk )
Proof:
z∈R
²[
⇔ x −1 y ∈ 0(x1 ) ∩ · · · ∩ 0(xk )
0(z)
z∈R
¡
²[
¢
⇔ y ∈ x 0(x1 ) ∩ · · · ∩ 0(xk )
0(z)
z∈R
∗
= x0 (x1 , . . . , xk ),
where R = S\{x1 , . . . , xk }. Thus part (ii) is true. Now suppose that 0k∗ = |0 ∗ (x1 , . . . , xk )|
for some x 1 , . . . , xk ∈ S. By definition, x1 x 6∈ 0 ∗ (x1 , . . . , xk ) for any x ∈ S\{x1 , . . . , xk ),
so 0 ∗ (x1 , . . . , xk ) ⊆ {x1 x1 , . . . , x1 xk }. Hence 0k∗ = |0 ∗ (x1 , . . . , xk )| ≤ k as in (iii).
Finally, for any y ∈ 02 (1), let {x1 , . . . , xk } = {x ∈ S | y ∈ 0(x)}. Then y ∈ 0 ∗ (x1 , . . . , xk )
is as in (iv).
2
It is clear that if Cay(G, S) ∼
= Cl [ K̄ m ] for m > 1 then A1 is not faithful on S. Conversely,
the following theorem shows that if A1 is not faithful on S then Cay(G, S) contains such a
subgraph.
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Theorem 3.2 Let G be an Abelian group and 0 = Cay(G, S) for some S ⊂ G such that
G = hSi. Let A = Aut 0 and A1 the stabilizer of 1 in A. Then either A1 is faithful on S,
or S contains a coset of some nontrivial subgroup of G and 0 has a subgraph isomorphic
to Cl [ K̄ n ] for some integers l and n.
Proof: Let S = {a1 , a2 , . . . , am }. Assume first that for any integer h ≥ 1 and any h
elements x1 , . . . , x h ∈ S, |0 ∗ (x1 , . . . , x h )| ≤ 1. We claim that A1 is faithful on S. For
any y ∈ 02 (1), let {ai1 , . . . , aih } = {x ∈ S | y ∈ 0(x)}. Then y is the unique element of
0 ∗ (ai1 , . . . , aih ). If α ∈ A1 such that x α = x for all x ∈ S, then α fixes ai1 , . . . , aih . Thus
α fixes 0 ∗ (ai1 , . . . , aih ), and so α fixes y. Hence x α = x for all x ∈ 02 (1). Since hSi = G,
Cay(G, S) is connected, and it follows that x α = x for all x ∈ V 0. Hence α = 1 and A1
is faithful on S.
Assume now that there are some h vertices ai1 , . . . , aih such that |0 ∗ (ai1 , . . . , aih )| ≥ 2.
Let w, y ∈ 0 ∗ (ai1 , . . . , aih ). Without loss of generality, we may assume that {i 1 , . . . , i h } =
{1, . . . , h}. By the definition of 0 ∗ (a1 , . . . , ah ), there exist u 1 , . . . , u h , v1 , . . . , vh ∈
{a1 , . . . , ah } such that
(
a1 u 1 = a2 u 2 = · · · = ah u h = w,
a1 v1 = a2 v2 = · · · = ah vh = y,
where u i 6= vi and {u 1 , . . . , u h } = {v1 , . . . , vh } = {a1 , . . . , ah }. Since {u 1 , . . . , u h } =
{v1 , . . . , vh }, there exist i 1 6= 1, i 2 6= i 1 , . . . , i k 6= i k−1 for some k ≤ h such that
v1 = u i1 , vi1 = u i2 , . . . , vik−1 = u ik and vik = u 1 . Thus
(
a1 u 1 = ai1 u i1 = · · · = aik u ik ,
a1 u i1 = ai1 u i2 = · · · = aik u 1 .
For convenience, without loss of generality, we may assume that i 1 = 2, i 2 = 3, . . . , i k =
k + 1. Then we have
(
a1 u 1 = a2 u 2 = · · · = ak+1 u k+1 ,
a1 u 2 = a2 u 3 = · · · = ak+1 u 1 .
Thus a1 u 1 ai u i+1 = a1 u 2 ai u i for i ≤ k and a1 u 1 ak+1 u 1 = a1 u 2 ak+1 u k+1 . Therefore,
u 1 u i+1 = u 2 u i for i ≤ k and u 21 = u 2 u k+1 . Let U = {u 1 , . . . , u k+1 }. Then u 1 U = u 2 U .
Similarly, we have u 1 U = · · · = u k+1 U . We claim that a1−1 U is a subgroup of G. In fact,
for any i, j with 1 ≤ i, j ≤ k + 1, there exists an integer l such that u 1 u i = u j u l because
−1
u 1 U = u j U . Thus u i u −1
j = u 1 u l and so
¡ −1 ¢−1
−1
−1
= u i u −1
u −1
j = u 1 u l ∈ u 1 U.
1 ui · u1 u j
−1
Therefore, u −1
1 U is a subgroup of G and U is a coset of the subgroup u 1 U . Now
∼
Cay(hU i, U ) = Cl [ K̄ |U | ] is a subgraph of Cay(G, S) as in the theorem. This completes the
proof of the theorem.
2
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209
Next we are going to characterize Cayley graphs Cay(G, S) for which A1S is the alternating group or the symmetric group of degree |S|. To do this, we first prove the following
lemma.
Lemma 3.3 Let G be an Abelian group, and let S, T be two Cayley subsets of G such
that G = hSi and Cay(G, S) ∼
= Cay(G, T ). If 0 ∗ (x, y) = {x y} for all x, y ∈ S and
∗
0 (u, v) = {uv} for all u, v ∈ T, then every isomorphism preserving 1 between Cay(G, S)
and Cay(G, T ) induces an automorphism of G.
Proof: Let S = {a1 , a2 , . . . , am } and T = {b1 , b2 , . . . , bm }. Without loss of generality,
assume that ρ is an isomorphism from Cay(G, S) to Cay(G, T ) such that 1 → 1, ai → bi
for i = 1, 2, . . . , m. Then for any i 6= j,
ρ : {ai a j } = 0 ∗ (ai , a j ) 7→ 0 ∗ (bi , b j ) = {bi b j }.
We claim that ρ is an automorphism of G. To prove this, we need only verify that for all
integers n 1 , n 2 , . . . , n m ≥ 0,
¢ρ
¡ n1 n2
(1)
a1 a2 · · · amn m = b1n 1 b2n 2 · · · bmn m ,
by induction on n 1 + n 2 + · · · + n m . Since
½
ρ:
ai → bi ,
ai a j → bi b j ,
for 1 ≤ i ≤ m,
for i 6= j,
we have ρ:
© ª
© 2ª
ai = 0(ai )\{ai a j | j 6= i} 7→ 0(bi )\{bi b j | j 6= i} = bi2
for all i = 1, 2, . . . , m. In other words, (1) holds for n 1 + n 2 + · · · + n m ≤ 2. Now assume
inductively that the equality (1) holds for n 1 + n 2 + · · · + n m ≤ N , where N ≥ 2. Let
a=
m
Y
n0
aj j,
where
j=1
m
X
n 0j = N − 1.
j=1
By the induction assumption, we have
(
ρ:
a→b=
Qm
aai → bbi ,
j=1
n0
bj j,
for 1 ≤ i ≤ m.
Since G is Abelian, for any x ∈ hSi, y ∈ hT i and any i 6= j, we have 0 ∗ (xai , xa j ) =
{xai a j } and 0 ∗ (ybi , yb j ) = {ybi b j }. Hence ρ:
(
{aai a j } = 0 ∗ (aai , aa j ) 7→ 0 ∗ (bbi , bb j ) = {bbi b j }, for 1 ≤ i 6= j ≤ m,
© ª
© 2ª
aai = 0(aai )\{aai a j | j 6= i} 7→ 0(bbi )\{bbi b j | j 6= i} = bbi2 .
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Therefore, the equality (1) holds for n 1 + n 2 + · · · + n m = N + 1. By induction, the
equality (1) holds for all n 1 , n 2 , . . . , n m ≥ 0. Hence ρ is an automorphism of G sending S
to T .
2
To prove our next theorem, we need some notation. If a, b ∈ S and b 6= a −1 , then the
product ab (in G) is said to be a word of length 2 on S. Let w(ab) be the number of all words
of length 2 on S which are equal to ab, that is, w(ab) = |{uv | uv = ab and u, v ∈ S}|.
For Y ⊆ 02 (1), let w(Y ) be thePnumber of all words of length 2 on S which are equal to an
element of Y , that is, w(Y ) = y∈Y w(y).
Lemma 3.4 Using the notation defined above, we have
(i) if 1 6= ab ∈ 0 ∗ (u 1 , u 2 , . . . , u i ), then w(ab) = i for any u 1 , u 2 , . . . , u i ∈ S;
(ii) if |0 ∗ (u 1 , . . . , u i )| = j then
(
ij
if 1 6∈ 0 ∗ (u 1 , . . . , u i ),
w(0 ∗ (u 1 , . . . , u i )) =
i( j − 1) if 1 ∈ 0 ∗ (u 1 , . . . , u i );
(iii) |02 (1)| ≤ w(02 (1)) and if 1 ∈ 0 ∗ (R) then w(02 (1)) = m 2 − |R| for any R ⊆ S;
(iv) if A1S ≥ Alt(|S|), then 1 ∈ 0 ∗ (R) for R ⊆ S implies R = S.
Proof: By definition, part (i) is clear. It follows that part (ii) holds. Now let S =
{a1 , . . . , am }. Then 02 (1) = {ai a j | 1 ≤ i, j ≤ m}\{1}. It follows that part (iii) is true.
Noting that A1S is (m − 2)-transitive on S, in particular, transitive and 2-set-transitive on S,
part (iv) is clearly true.
2
Now we can prove our next result.
Theorem 3.5 Let G be an Abelian group, and let S be a generating subset of G of size
m. Let 0 = Cay(G, S), and let A = Aut 0 and A1 the stabilizer of 1 in A. If A1S ≥ Alt(m),
the alternating group of degree m, then one of the following holds:
(i) S = G\{1} and 0 ∼
= K m+1 ;
(ii) S = a H for some H ≤ G, and 0 ∼
= K m,m or C|G|/m [ K̄ m ];
|G|
(iii) S = bH \{b} for some H ≤ G, 0 ∼
= C|G|/(m+1) [ K̄ m+1 ] − o(b) Co(b) ;
L
(iv) S = a for some a ∈ S and some L ≤ Aut(G, S), and GCA;
(v) either G is cyclic,
S or G = Z n × B, where n is odd and B is a 2-group of exponent 4,
and 02 (1) = u,v∈S 0 ∗ (u, v) ∪ 0 ∗ (S)\{1}.
Proof: First assume that m = 2 and S = {a, b}. If b = a −1 then G = hai is cyclic and 0
is a cycle of length n := o(a). Thus A ∼
= D2n , and so part (iv) holds in this case. Suppose
that b 6= a −1 . If |0 ∗ (a, b)| = 1, then a 2 6= b2 and so 0 ∗ (a, b) = {ab}. It follows from
Lemma 3.3 that part (iv) holds. If |0 ∗ (a, b)| = 2 then 0 ∗ (a, b) = {ab = ba, a 2 = b2 }.
Thus {1, a −1 b} is a subgroup of G of order 2, and S = a{1, a −1 b} as in part (ii). Hence
0i (1) = a i {1, a −1 b} for all i ≥ 1. Hence |0i (1)| = 2, and it follows that Cay(G, S) ∼
=
C|G|/2 [ K̄ 2 ].
In the following, assume that m ≥ 3 and S = {a1 , a2 , . . . , am }. Since A1S ≥ Alt(m), A1S
is (m − 2)-transitive on S, in particular, A1S is 2-set-transitive on S. By Lemma 3.1(iv),
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any element of 02 (1) belongs to 0 ∗ (R) for some R ⊆ S. Since either 0 ∗ (ai ) = ∅ or
0 ∗ (ai ) = {ai2 } and ai2 6= a j ak for any j, k 6= i, there is at least one n ∈ {2, . . . , m} such
that 0n∗ ≥ 1.
(1) Assume that there exists an integer n with 3 ≤ n ≤ m −2 such that 0n∗ = r ≥ 1. Then
there are n vertices c1 , . . . , cn ∈ S such that |0 ∗ (c1 , . . . , cn )| = r . Thus 0 ∗ (c1 , . . . , cn ) contains exactly r elements of 02 (1). By Lemma 3.4(ii) and (iv), w(0 ∗ (c1 , c2 , . . . , cn )) = r n.
Since A1 is (m −2)-transitive on S, for any n elements x1 , . . . , xn of S, w(0 ∗ (x1 , . . . , xn )) =
r n. Hence
µ ¶
X
m
2
∗
m ≥ w(02 (1)) ≥
w(0 (x1 , . . . , xn )) = r n
.
n
x1 ,...,xn ∈S
However, it is easy to see that r n( mn ) > m 2 since 3 ≤ n ≤ m − 2, a contradiction. Thus
0n∗ = 0 for 3 ≤ n ≤ m − 2.
∗
(2) Assume that 02∗ = 0m−1
= 0. Then 02 (1) = (0 ∗ (S)\{1}) ∪ 0 ∗ (a1 ) ∪ · · · ∪
2
∗
∗
0 (am ) ⊆ (0 (S)\{1}) ∪ {a1 , . . . , am2 }. Thus ai a j ∈ 0 ∗ (S) for any ai 6= a j . Since no two of
a1 a2 , . . . , a1 am are equal, |0 ∗ (S)| ≥ m − 1 ≥ 2. Thus for any i, j 6= 1, there are integers
h, k such that a1 ai = a j ah and a1 a j = ai ak . It follows that a12 = ah ak and so 0 ∗ (a1 ) = ∅.
Thus 0 ∗ (S)\{1} = 02 (1). Hence every vertex in 02 (1) is joined to all vertices in 0(1) = S.
Thus if 1 ∈ 0 ∗ (S) then Cay(G, S) ∼
[ K̄ m ]
= K m,m ; if 1 6∈ 0 ∗ (S) then Cay(G, S) ∼
= C |G|
m
−1
where |G| > 2m. It follows that ai S = a j S for any ai , a j ∈ S. Thus H = a1 S is a subgroup
of G and S = a1 H . This case is as in part (ii).
(3) Suppose that 02∗ = r ≥ 1. By Lemma 3.1(iii), r ≤ 2. If r = 2, then since A1 is 2-settransitive on S, for any u, v ∈ S, |0 ∗ (u, v)| = 2 and so 0 ∗ (u, v) = {uv = vu, u 2 = v 2 }. It
follows that a12 = a22 and a22 = a32 , a contradiction. Thus r = 1. Since A1 is 2-set-transitive
on S, |0 ∗ (u, v)| = 1 for any u, v ∈ S. Hence 0 ∗ (u, v) = {uv = vu} or {u 2 = v 2 }. By
Lemma 3.4(iv), 1 6∈ 0 ∗ (u, v) and so w(0 ∗ (u, v)) = 2.
First assume that there are two elements a, b ∈ S such that 0 ∗ (a, b) = {a 2 = b2 }.
Then 0 ∗ (a) = ∅ and ab 6∈ 0 ∗ (a, b), so ab = cd for some c, d ∈ S\{a, b}. Thus
ab ∈ 0 ∗ (x1 , . . . , xi ) for some x1 , . . . , xi ∈ S where i > 2. Since 0n∗ = 0 for 3 ≤ n ≤ m −2
∗
6= 0 or 0m∗ 6= 0. Since
shown in (1), i ≥ m − 1 and so w(ab) ≥ m P
− 1. Thus 0m−1
w(0 ∗ (u, v)) = 2 for all u, v ∈ S where u 6= v, u,v∈S w(0 ∗ (u, v)) = 2( m2 ) = m(m − 1).
∗
= s 6= 0 then since A1S P
is transitive on S, |0 ∗ (S\{u})| = s for all u ∈ S. Thus
If 0m−1
∗
w(0 (S\{u})) = s(m−1) and so u∈S w(0 ∗ (S\{u})) = ms(m−1). Since 1 6∈ 0 ∗ (S\{u}),
we have
w(02 (1)) ≥
X
u,v∈S
w(0 ∗ (u, v)) +
X
w(0 ∗ (S\{u}))
u∈S
= (s + 1)m(m − 1) > m 2 ≥ w(02 (1)),
S
∗
a contradiction. Thus 0m−1
= 0, so 0m∗ = s 6= 0 and 02 (1) = u,v∈S 0 ∗ (u, v) ∪ 0 ∗ (S)\ {1}.
Without loss of generality, suppose that a = a1 and b = a2 , and let ai = oi ei such that
oi ∈ G 20 and ei ∈G 2 , where G 2 is a Sylow 2-subgroup and G 20 is a Hall 20 -subgroup of G.
Since a12 = a22 , o1 = o2 =: o and e12 = e22 . For any ai ∈ S with i 6= 1, 2, since a1 a2 ∈ 0 ∗ (S),
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there is an a j such that a1 a2 = ai a j . If j = i then oi2 = o1 o2 = o2 and ei2 = e1 e2 , so
oi = o. If j 6= i then since ai a j = a1 a2 , ai a j 6∈ 0 ∗ (ai , a j ). Since 0 ∗ (ai , a j ) 6= ∅, we have
0 ∗ (ai , a j ) = {ai2 = a 2j }. It follows that oi = o j and ei2 = e2j . Since ai a j = a1 a2 = o2 e1 e2 ,
we have oi = o and ei e j = e1 e2 . Thus, whether j = i or not, we have oi = o and
ei4 = (ei e j )2 = (e1 e2 )2 = e14 . Hence o1 = o2 = · · · = om and e14 = e24 = · · · = em4 . Note
that G = hSi, so G 20 = hoi and G 2 = he1 , e2 , . . . , em i. If e14 6= 1 then G 2 has only one
subgroup of order 2. By [14, p. 59], G 2 is cyclic; if e14 = 1 then G 2 is of exponent 4. This
case is as in part (v).
Now assume that 0 ∗ (u, v) = {uv} for any u, v ∈ S and that G is not as in part (v).
For any T ⊆ S\{1}, by the previous paragraph, Cay(G, S) ∼
= Cay(G, T ) implies that
0 ∗ (u 0 , v 0 ) = {u 0 v 0 } for any u 0 , v 0 ∈ T with u 0 6= v 0 . By Lemma 3.3, S is conjugate in
ρ
Aut(G) to T and so S is a CI-subset. For any ρ ∈ A1 , let bi = ai and T = {b1 , . . . , bm }.
Then Cay(G, S) = Cay(G, T ). By Lemma 3.3, ρ induces an automorphism of G. Thus
A1 ≤ Aut(G), so A1 = Aut(G, S) and A = G A1 = G o Aut(G, S), which is as in part (iv).
∗
(4) Assume that 02∗ = 0 and 0m−1
= r ≥ 1. Then m ≥ 4. Since A1 is transitive on S, we
∗
have |0 (S\{x})| = r for every x ∈ S. If r ≥ 2, then since 1 6∈ 0 ∗ (S\{x}) for any x ∈ S,
X
w(0 ∗ (S\{x})) = r m(m − 1) > m 2 ≥ w(02 (1)),
w(02 (1)) ≥
x∈S
a contradiction. Thus r = 1. Let v(x) be the unique element of 0 ∗ (S\{x}). If v(a1 ) = a1 ,
then for any ai , we have v(ai ) = ai because A1 is transitive on S. Thus Cay(G, S) ∼
= K m+1
as in part (i). Now suppose that v(a1 ) 6= a1 . Then v(a1 ) ∈ 0(x) for all x ∈ S\{a1 }. Let
b = a1−1 v(a1 ) and S ∗ = S ∪ {b}. We shall prove that b−1 S ∗ is a subgroup of G. To do
this, we need to prove that b−1 ai · (b−1 a j )−1 ∈ b−1 S ∗ for any i 6= j. Since i 6= j, we may
assume that j 6= 1. Then v(a1 ) ∈ 0(a j ) and so v(a1 ) = a j ak for some ak ∈ S. Thus
b−1 ai · (b−1 a j )−1 = b−1 · b · ai a −1
j
= b−1 · a1−1 v(a1 ) · ai a −1
j
= b−1 · a1−1 a j ak · ai a −1
j
= b−1 a1−1 ai ak ·
If ai ak ∈ 0(a1 ), that is, ai ak = a1 ak 0 for some ak 0 ∈ S, then b−1 ai (b−1 a j )−1 = b−1 a1−1 ai ak
= b−1 ak 0 ∈ b−1 S ∗ . Hence H := b−1 S ∗ is a subgroup of G and S ∗ is a coset of H . Thus S =
|G|
|G| [ K̄ m+1 ] −
Ck
S ∗ \{b} = bH \{b}, and Cay(G, S) = Cay(G, S ∗ ) − Cay(G, {b}) ∼
= C m+1
k
where k = o(b), which are as in part (iii) of the theorem. Thus, in the following, we
only need to prove that ai ak ∈ 0(a1 ). Since i 6= j, ai ak 6= a j ak = v(a1 ). If i 6= k,
then since 0n∗ = 0 for 2 ≤ n ≤ m − 2, ai ak ∈ 0 ∗ (S) ∪ 0 ∗ (S\{x}) for some x ∈ S\{a1 }
and so ai ak ∈ 0(a1 ). Thus we may assume that i = k, so ai ak = ak2 . If ak2 = a12 then
ak2 ∈ 0(a1 ). Hence suppose that ak2 6= a12 . Since a j ak = v(a1 ) ∈ 0 ∗ (S\{a1 }), there exist
ah , al ∈ S\{a j , ak } such that a j ak = ah al . If l = h then ah2 = a j ak and so 0 ∗ (ah ) = ∅.
Since A1 is transitive on S, 0 ∗ (ak ) = ∅ and so ak2 ∈ 0 ∗ (S) ∪ 0 ∗ (S\{x}) for some x ∈ S.
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Since ak2 6= v(a1 ), we have ak2 ∈ 0(a1 ). If l 6= h, then at least one of ak ah and ak al does not
belong to 0 ∗ (S\{a j }), say, ak ah 6∈ 0 ∗ (S\{a j }). Thus ak ah ∈ 0 ∗ (S) ∪ 0 ∗ (S\{x}) for some
x ∈ S\{a j }, and so ak ah = a j al 0 for some l 0 , which, together with a j ak = ah al , implies
ak2 = al al 0 . Thus 0 ∗ (ak ) = ∅ and so ak2 ∈ 0 ∗ (S) ∪ 0 ∗ (S\{x}) for some x ∈ S. Since
ak2 6= v(a1 ), ak2 6∈ 0 ∗ (S\{a1 }) and so ak2 ∈ 0(a1 ). This completes the proof of the theorem.
2
Theorem 3.5 gives an application to Babai and Frankl’s question.
Corollary 3.6 Let G be an elementary Abelian p-group, p a prime, and S a Cayley
subset. Let A = Aut Cay(G, S). If A1S ≥ Alt(S), then S is a CI-subset of G.
Proof: Since any subgroup of G is still elementary Abelian group and each isomorphism
between any two subgroups can be extended as an automorphism of G, we may assume
that hSi = G. By Theorem 3.5, Cay(G, S) satisfies parts (i)–(iv). It is easy to check that S
is a CI-subset of G.
2
Remark By Theorem 3.5, the graphs in parts (i)–(iii) have been completely characterized. The graphs Cay(G, S) in part (iv) satisfies a very strong condition Aut Cay(G, S) ≤
G o Aut(G)’s.
4.
Finite m-DCI p-groups, p a prime
By definition, a finite group G is a 1-DCI group if and only if all elements of G of the same
order are conjugate in Aut(G). Suppose that G is a 1-DCI p-group. If p is an odd prime
then G is homocyclic by the result of Shult [13]; if p = 2 then by [7], G is a homocyclic
group or the quaternion group Q 8 , or G satisfies the following conditions:
(i)
(ii)
(iii)
(iv)
G 0 = 8(G) is homocyclic of rank n;
G/G 0 is of order 2n or 22n ;
the centre Z(G) of G consists of the identity and all the involutions of G;
either Z(G) = G 0 , or CG (G 0 ) = G 0 with Z(G) = 8(G 0 ).
It is easy to see that homocyclic groups and Q 8 are 1-DCI groups, however, it is still difficult
to characterize precisely 1-DCI 2-groups, see [7]. For m ≥ 2, the problem of determining
m-DCI groups is very different from the case m = 1. By Lemma 2.4, we need to consider
mainly Abelian p-groups. We first prove a property of Cayley graphs of arbitrary Abelian
p-groups.
Proposition 4.1 Let G be an Abelian p-group, S a Cayley subset of G such that hSi = G
and A = Aut Cay(G, S). If p 2 6 | |A1 | then either S is a CI-subset, or p k |A1 | and S contains
a coset of some subgroup of G, where A1 is the stabilizer of 1 in A.
Proof: Suppose that |G| = p d . If p 6 | |A1 |, then G is a Sylow p-subgroup of A. By
Sylow Theorem and Theorem 2.1, S is a CI-subset. Thus assume that p k |A1 |. Let P be
a Sylow p-subgroup of A containing G. Then |P : G| = p and P1 ∼
= Z p where P1 is the
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stabilizer of 1 in P, and so P is non-Abelian, see [15, 4.4]. Assume that S is not a CI-subset
of G. By Theorem 2.1, there is a τ ∈ Sym(G) such that G τ < A and G τ is not conjugate
to G. Let g ∈ A such that (G τ )g < P. Then G τ g 6= G and P ≥ hG τ g , Gi > G. Hence
P = hG τ g , Gi = G τ g G as |P : G| = p. Since any element in G τ g ∩ G commutes with all
elements of G τ g and G, we have G τ g ∩ G ≤ Z(hG τ g , Gi) = Z(P). Further
|G τ g ∩ G| =
|G τ g ||G|
pd · pd
= p d−1 .
=
τ
g
|G G|
p d+1
Since P is non-Abelian, G τ g ∩ G = Z(P). For any a ∈ Z(P), Pa = P1a = P1a = P1 , so P1
fixes all vertices in Z(P). Now hZ(P), P1 i is an Abelian subgroup of index p in P. Hence
hZ(P), P1 i C P and hZ(P), P1 i has orbits {xZ(P) | x ∈ G} on V 0 = G. Thus P1 fixes
every xZ(P) setwise. Moreover, P1 = hαi has an orbit O on S of length p. If a ∈ O ⊆ S,
then since P1 fixes xZ(P) setwise for each x ∈ G, a α ∈ aZ(P), so a α = az for some
z ∈ Z(P). Thus O = a hαi = {a, az, az 2 , . . . , az p−1 } = ahzi. Thus the proposition holds.
2
This result has been generalized in [8] to general abelian groups under certain conditions.
The following lemma enables us to focus our attention on connected graphs.
Lemma 4.2 Assume that G is a homocyclic p-group and that S is a Cayley subset of G.
If S is a CI-subset of hSi and for any subset T of G, Cay(hT i, T ) ∼
= Cay(hSi, S) implies
hT i ∼
= hSi, then S is a CI-subset of G.
Proof: Assume that S is a CI-subset of hSi and that T is a Cayley subset of G such that
Cay(hT i, T ) ∼
= Cay(hSi, S). Then hT i ∼
=σ hSi for some isomorphism σ from hT i to hSi.
0
σ
0 ∼
Let T = T . Then Cay(hSi, T ) = Cay(hT i, T ) ∼
= Cay(hSi, S). Since S is a CI-subset of
hSi, there is α ∈ Aut(hSi) such that T 0σ = S. Thus β = σ α is an isomorphism from hT i to
hSi such that T β = (T σ )α = T 0α = S. Since G is a homocyclic p-group, it is easy to show
that every isomorphism between any two isomorphic subgroups of G can be extended as
an automorphism of G. Let ρ ∈ Aut(G) be an extension of β. Then T ρ = T β = S, so S
is a CI-subset of G.
2
Now we can determine m-DCI p-groups for 2 ≤ m ≤ p + 1.
Theorem 4.3 Let G be a finite p-group, where p is prime. Then
(1) G is an m-DCI group for 2 ≤ m ≤ p − 1 if and only if p ≥ 3 and G is homocyclic;
(2) G is a p-DCI group if and only if G is elementary Abelian, cyclic, or G = Q 8 ;
(3) G is a ( p + 1)-DCI group if and only if G is elementary Abelian, or G = Z 4 , Q 8 .
Proof:
(1) By Lemmas 2.3 and 2.4, we only need to prove that homocyclic p-groups are m-DCI
groups. Let S be a Cayley subset of G of size m. By [8, Theorem 1.1], S is a CI-subset
of hSi. Thus by Lemma 4.2, S is a CI-subset of G and G is an m-DCI group.
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(2) By Lemmas 2.4 and 4.2, we only need to prove that elementary Abelian p-groups and
cyclic p-groups are p-DCI groups. By [8, Theorem 1.1], S is a CI-subset of hSi. Thus
by Lemma 4.2, S is a CI-subset of G and G is a p-DCI group.
(3) By Lemmas 2.4 and 4.2, we only need to prove that elementary Abelian p-groups are
( p + 1)-DCI groups. Let G = Z dp and let S be a Cayley subset of G such that |S| ≤
p + 1. By parts (1) and (2), we only need to consider the case where |S| = p + 1. Since
G is elementary Abelian, any two subgroups of G of the same order are isomorphic.
Thus, by Lemma 4.2, we may assume that hSi = G.
If p = 2, then by [3, Theorem 1], G is a 3-DCI group. Thus assume p ≥ 3 in the
following. Suppose first that S contains a coset a H of some subgroup H of G for some
a ∈ S. Since |S| = p + 1, we have |H | = p and S = a H ∪ {b} for some b ∈ S. If
b ∈ ha H i then G = ha, H i is of order p 2 , and thus by [6], S is a CI-subset. If b 6∈ ha H i
then G = ha H i × hbi ∼
= Z 3p , and thus by [4], again S is a CI-subset. Suppose now that S
does not contain any coset of subgroups of G. By Theorem 3.2, A1 is faithful on S. Since
|S| = p + 1, it follows that p 2 6 | |A1 |. By Proposition 4.1, S is a CI-subset and so G is a
( p + 1)-DCI group. This completes the proof of the theorem.
2
Acknowledgments
The author gratefully acknowledges the support of an Overseas Postgraduate Research
Scholarship of the Australian Department of Education, Employment and Training and a
University Postgraduate Award from the University of Western Australia.
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