On the isomorphism problem for Cayley graphs of abelian groups whose Sylow subgroups are elementary abelian or cyclic

We show that if certain arithmetic conditions hold, then the Cayley isomorphism problem for abelian groups, all of whose Sylow subgroups are elementary abelian or cyclic, reduces to the Cayley isomorphism problem for its Sylow subgroups. This yields a large number of results concerning the Cayley isomorphism problem, perhaps the most interesting of which is the following: if p1, . . . , pr are distinct primes satisfying certain arithmetic conditions, then two Cayley digraphs of Za1 p1×· · ·×Z ar pr , ai 6 5, are isomorphic if and only if they are isomorphic by a group automorphism of Za1 p1 ×· · ·×Z ar pr . That is, that such groups are CI-groups with respect to digraphs. Mathematics Subject Classifications: 05E18


Introduction
The history of the modern Cayley isomorphism problem begins in 1967 when Ádám [1] conjectured that any two Cayley graphs of the cyclic group Z n of order n are isomorphic if and only if they are isomorphic by a group automorphism of Z n .While Ádám's conjecture was quickly shown to be false [19], the conjecture nonetheless generated much interest in the following obvious generalization: Are two Cayley graphs of a group G isomorphic if and only if they are isomorphic by a group automorphism of G? If so, we say that G is a CI-group with respect to graphs.This problem naturally generalizes to any class of combinatorial objects (see [33] for several equivalent formulations of the precise definition of a combinatorial object), and in fact was considered much earlier for designs [2,27].Our question is then: For a group G is it true that any two Cayley objects of G in some class K of combinatorial objects are isomorphic if and only if they are isomorphic by a group automorphism of G? If the answer to this question is yes, we say that G is a CI-group with respect to K. If G is a CI-group with respect to every class of combinatorial objects, we say that G is a CI-group.In 1987, Pálfy [34] proved the following remarkable result: Theorem 1.A group G of order n is a CI-group if and only if gcd(n, ϕ(n)) = 1 or n = 4, where ϕ is Euler's phi function.
Definition 2. Given a group G and g ∈ G, define g L : G → G by g L (h) = gh, and G L = {g L : g ∈ G}.Then G L ∼ = G is a group, the left regular representation of G.A Cayley object of G is a combinatorial object X with G L Aut(X), the automorphism group of X.
We remark that a classical result of Sabidussi [35] gives that the definition of a Cayley object above is consistent with the usual definition of a Cayley digraph when the object is a digraph.
An essential tool in proving Pálfy's Theorem is the following result of Babai, which characterizes the CI-property: Lemma 3.For a group G and a class K of combinatorial objects the following are equivalent: 1. G is a CI-group with respect to K, 2. whenever X is a Cayley object of G in K and δ ∈ Sym(G) such that δ −1 G L δ Aut(X), then G L and δ −1 G L δ are conjugate in Aut(X).
Babai's Lemma has been generalized to give a similar characterization of the solution to the isomorphism problem for Cayley objects X of G in classes K when G is not a CI-group with respect to K (see [33,Lemma 1.1], [8, Lemma 13], and [13,Lemma 20]).All such results basically reduce to determining the conjugacy classes of G L in Aut(X).In fact, in the positive direction, Pálfy showed that if gcd(n, ϕ(n)) = 1, then there is always one conjugacy class of (Z n ) L in (Z n ) L , δ −1 (Z n ) L δ (we remark that every group of order n is cyclic if and only if gcd(n, ϕ(n)) = 1 [36,Theorem 9.2.7]).To place the work in this paper in its proper context, it will be useful to discuss the structure of Pálfy's proof in more detail, and to do this, we will need the following additional definition: Definition 4. Let G be a transitive group acting on Ω.Let Y be the set of all complete block systems of G. Define a partial order on Y by B C if and only if every block of C is a union of blocks of B. Let n = Π r i=1 p a i i be the prime factorization of n and define Ω : N → N by Ω(n) = Σ r i=1 a i .Let m = Ω(n).A transitive group G of degree n is m-step imprimitive if there exists a sequence of complete block systems B 0 ≺ B 1 ≺ • • • ≺ B m .A complete block system B will be said to be normal if B is formed by the orbits of a normal subgroup.We will say that G is normally m-step imprimitive if each B i , 0 i m, is formed by the orbits of a normal subgroup of G.
G is a CI-group with respect to color digraphs (Corollary 39).Finally, additional open problems are discussed in Section 6.
An additional comment is in order.Our main motivation in considering the Cayley isomorphism problem in general is to investigate the Cayley isomorphism problem for Cayley digraphs.This is what has guided our restricting Sylow p-subgroups to be either elementary abelian or cyclic, as the Cayley isomorphism problem for digraphs has been has been studied for these groups (many of the corresponding results are mentioned in Section 5), while for other abelian p-groups the isomorphism problem has been solved only for the group Z p × Z p 2 [6].The following result of C. H. Li [28, Theorem 1.1] will be crucial.Theorem 6.Let X be a primitive permutation group of degree n.Then X contains an abelian regular subgroup A if and only if either 1. X AGL(d, p), where p is prime, d 1, and n = p d , or Ultimately, we will show by induction on m = Ω(n) that, under certain arithmetic conditions, if G Sym(n) is transitive, abelian and has Sylow subgroups elementary abelian or cyclic, then for Gδ with no nontrivial block system B ≺ C, then either C consists of blocks of prime size, in which case we are finished using the induction hypothesis, or the blocks of C are of composite size and Stab G,δ −1 Gδ (B)| B is primitive in its action on B ∈ B by [4,Exercise 1.5.10].We next show that we need only consider when the blocks of C are of prime power size.We then analyze the various cases for Stab G,δ −1 Gδ (B)| B given by Theorem 6 in Section 3.
Finally, one of our intentions in this paper is begin to determine when the isomorphism problem for a direct product Π r i=1 G i of groups reduces to the isomorphism problem for the G i , 1 i r (as opposed to specifically studying abelian groups whose Sylow subgroups are elementary abelian or cyclic).Induction seems to be the best way to attack this problem, but there is an inherent difficulty.Namely, if B is a complete block system of G, δ −1 Gδ where G is a transitive abelian group and δ ∈ S G , it need not be the case that , and fix δ −1 Gδ (B) are all abelian groups.So if possible we will use G 1 and G 2 instead of G and δ −1 Gδ, and only assume that G 1 and G 2 are abelian and not assume that G 1 ∼ = G 2 .In fact, we will often only need that fix G 1 (B) ∼ = fix G 2 (B), and then certain properties of the quotient.
Throughout this paper, all groups are finite.For group theoretic terms not defined in this paper, see [4].
2 We may assume G 1 , G 2 is imprimitive with blocks of primepower order We shall have need of the following elementary result whose straightforward proof is left to the reader.
Sym(n) be transitive such that both G 1 and G 2 admit B as a complete block system.Then G 1 , G 2 admits B as a complete block system.
The following result is proven in more generality than is needed here.Lemma 8. Let n = kp a , where p is an odd prime, and p a 5. Let G 1 , G 2 Sym(n) be transitive such that G 1 , G 2 admits a complete block system C of n/p a blocks of size p a such that Alt(p a ) Stab G 1 ,G 2 (C)| C and fix G i (C) is abelian and semiregular while fix G i (C)| C is transitive for every C ∈ C and i = 1, 2. Then there exists γ ∈ G 1 , G 2 such that G 1 , γ −1 G 2 γ admits a complete block system B with n/p blocks of size p.
Proof.If a = 1, the result is trivial, so we assume that a 2. As We conclude that the p elements of C 0 not in any O 2 i are mapped by γ −1 0 to the p elements of C 0 not in any O 1 i as p > 2. Hence the orbits of Let E 0 be the equivalence class of ≡ 0 that contains C 0 .Then the orbits of , then for every C ∈ E 0 , we have that ω| C = 1 as well.Then L 1 = 1.We conclude that Alt(p a ) L 1 | C 1 for every C 1 ≡ 0 C 0 as Alt(p a ) is simple.Also observe that as p a is odd, By arguments analogous to those above, there exists γ 1 ∈ L 1 such that the orbits of As before, if the action of L 1 on C 1 is equivalent to the action of L 1 on C, then we also have that the orbits of Continuing inductively, we find γ ∈ G 1 , G 2 such that the orbits of J 1 and γ −1 J 2 γ are identical.Then G 1 admits a complete block system B formed by the orbits of J 1 G 1 and γ −1 G 2 γ admits B as a complete block system formed by the orbits of γ −1 J 2 γ γ −1 G 2 γ, and so by Lemma 7, G 1 , γ −1 G 2 γ admits B as a complete block system of n/p blocks of size p.
Sym(n) be transitive abelian groups, with n odd and com- Proof.Assume G 1 , G 2 is primitive.If Theorem 6 (1), (2a) with ( Ti , G i ) = (M 12 , Z 2 2 ×Z 3 ) or (2c) occurs, then n is a power of some prime p, and both G 1 and G 2 are p-groups.Then there exists δ ∈ G 1 , G 2 such that G 1 , δ −1 G 2 δ is a p-group.As a p-group contains a nontrivial center, G 1 , δ −1 G 2 δ contains a normal subgroup of order p, and so has blocks of size p.If Theorem 6 (2) holds and 2, then If Theorem 6 (2b) holds, then both G 1 and G 2 are cyclic, and the result follows by [32,Theorem 4.9].Note that Theorem 6 (2a) with ( Ti , G i ) = (M 12 , Z 2 2 × Z 3 ) cannot hold as n is odd and composite.Finally, if Theorem 6 (2d) holds, then the result follows by Lemma 8 with C as in Lemma 8 the trivial complete block system consisting of one block of size n.
Lemma 10.Let n be a positive integer such that if k is a proper divisor of n and H 1 , H 2 Sym(k) are transitive abelian groups, then there exists Proof.Let C be a nontrivial complete block system of G 1 , G 2 such that there exists no nontrivial block system D with D ≺ C. As both G 1 /C and G 2 /C are transitive abelian groups, and a transitive abelian group is regular, fix G i (C) = 1, i = 1, 2 and acts transitively on C ∈ C.
Let C 0 ∈ C. By hypothesis there exists We thus may assume without loss of generality that fix product of nilpotent groups is nilpotent and a subgroup of a nilpotent group is nilpotent.
Let p| |C|, be prime, p a be the largest power of p that divides |C|, C ∈ C, and Clearly after an appropriate conjugation of G 2 , if necessary, we may assume that P i P , i = 1, 2, where P is a Sylow p-subgroup of fix G 1 (C), fix G 2 (C) that contains P 1 .Then P i , i = 1, 2, as well as P have orbits of size p a by [8, Lemma 10] Hence the orbits of P 1 , P 2 , are the same because P 1 , P 2 P .Then the orbits of P i form a complete block system B of n/p a blocks of size p a of G i as P i G i .Then B is a complete block system of G 1 , G 2 by Lemma 7.

The possibilities for Stab
In this section, we will consider the various possibilities for Stab G 1 ,G 2 (C) in its action on C ∈ C. We begin with T i = Alt(p a ) and 2, with = 1 in fact considered in the previous section.The results for the case where T i = Alt(p a ) will not require any arithmetic conditions on n.In the proofs (but not the statements) of results in this section, we let G = G 1 , G 2 .

T
We will ultimately reduce this case to the case where Stab G 1 ,G 2 (C)| C AGL(a, p).We begin with a preliminary result.Proof.First observe that if H 1 , H 2 Sym(p a ) are p-groups and p is odd, then H 1 , H 2 Alt(p a ).Furthermore, there exists δ ∈ Alt(p a ) such that H 1 , δ −1 H 2 δ is contained in a Sylow p-subgroup P of Alt(p a ) (which, as p is odd, is also a Sylow p-subgroup of Sym(p a )).We thus now need only show that any two regular elementary abelian subgroups of P are conjugate in P .
We proceed by induction on a.If a = 1, then the result is trivial as a Sylow p-subgroup of Sym(p) has order p. Assume that the result is true for all a − 1 1, and let P be a Sylow p-subgroup of Sym(p a ).Then P = Z p (Z p (• • • Z p )) (a times), which we view as acting canonically on Z a p .For 1 i a, define τ i : Z a p → Z a p by Clearly then τ i : 1 i a is a regular elementary abelian subgroup of P .Let R be any other regular elementary abelian subgroup of P .Note that P admits a complete block system B consisting of p blocks of size p a−1 formed by the orbits of 1 By the induction hypothesis and the fact that fix Let The result then follows by induction.
We remark that it is not the case that the second half of the previous lemma holds for every regular abelian p-group.See [15,Example 6.4].
Definition 12. Let G be a group and H G. We define the normal closure of ) and the result follows.Let , each consisting of blocks of size m = p a/ .
For C ∈ C and 1 i , let π C,i : K → Alt(m) be the projection map from K to the i th copy of Alt(m . As Alt(m) has a unique representation [3,Table] as m = 6, we have that π C ,i (ω But this contradicts our choice of γ, and hence π C,i (γ AGL(a, p) for every C ∈ C as required.

A Common Hypothesis
We now consider when T i = Alt(p a ).All of the results in the rest of this section share some common hypothesis, which we will call Hypothesis 1.

Let π = {p
We prove the results needed to deal with the case where Stab G 1 ,G 2 (C)| C AGL(a, p).
Lemma 14.In addition to Hypothesis 1, assume that Let K i be S π -subgroups of G that contain H i , i = 1, 2. Then there exists γ 1 ∈ G such that γ −1 1 K 2 γ 1 = K 1 .We may thus assume without loss of generality that K 1 = K 2 and H 1 , H 2 are contained in the same S π -subgroup K of G. Assume G acts in the natural fashion on Z n/p ar r × Z p ar r .Then h 1 ∈ H 1 can be written as h 1 (i, j) = (σ(i), ω i (j)) and h 2 ∈ H 2 can be written as h 2 (i, j) = (ι(i), ξ i (j)), where σ, ι ∈ Sym(n/p ar r ) and each ω i , ξ i ∈ Sym(p ar r ).As H i commutes with Π i , we see that each ω i centralizes Π 1 and each ξ i centralizes Π 2 .We now assume that h 1 , h 2 are chosen so that for some b ∈ Z n/p ar r we have that As G/C = H × P , every element of G/C of order relatively prime to p r commutes with every element of G/C of order a power of p r .Let Q be the transitive permutation group obtained by the action of P on the blocks of C whose union is the block B ∈ B where B = {(b, x) : x ∈ P }, and ξ σ(b) ω b be the permutation obtained by the action of ξ σ(b) ω b on the blocks of C whose union is B. Then ξ σ(b) ω b commutes with every element of Q so that ξ σ(b) ω b , Q admits a complete block system formed by the orbits of ξ σ(b) ω b .As the the electronic journal of combinatorics 25(2) (2018), #P2.49degree of Q is a power of p r , we conclude that ξ σ(b) ω b fixes each block of C contained in B set-wise.That is, h 2 h 1 ∈ Stab G (C ) for some C ∈ C. As As ω b centralizes Π 1 , we see that ξ σ(b) centralizes Π 1 .As b was arbitrary, we see that ξ i centralizes Π 1 for every i ∈ Z n/p ar r .This implies that H 2 centralizes Π 1 , and similarly, Lemma 15.In addition to Hypothesis 1, assume that Let Pi be a Sylow p r -subgroup of P 1 , P 2 G that contains P i .By a Sylow Theorem, we may assume without loss of generality that P1 = P2 .Then P1 and the result follows by Lemma 14.

3.4
The case where Ti = Alt(p a ) We will need the following result [10, Lemma 3.3].
Lemma 16.Let PGL(d, q) X PΓL(d, q) be primitive of degree n = (q d − 1)/(q − 1) and contain a regular cyclic subgroup, where q is a prime-power and d 2. If n = p k for some odd prime p and (d, q) = (2, 8), then a Sylow p-subgroup of X is regular and cyclic.Definition 17.A group is homocyclic if it is the direct product of isomorphic cyclic groups.
In the following result, we will use the terminology of Theorem 6.
Theorem 18.In addition to Hypothesis 1, set m = Ω(n), and assume that the following conditions hold: 1. p i does not divide p j − 1 for any distinct primes p i , p j |n, the electronic journal of combinatorics 25(2) (2018), #P2.49

for any δ ∈ Sym(n) and nontrivial complete block system
and fix G 2 (C) are semiregular and elementary abelian or cyclic, and (2,8) for some z (here we are using the notation of Theorem 6).
Proof.By (2) we may, after an appropriate conjugation of G 2 , assume that G/C is nilpotent.As X is primitive and contains a regular abelian subgroup, by Theorem 6 we have that one of the following is true: , Q is a transitive permutation group of degree , and one of the following holds: (a) Tj = PGL(d, q), and P i = Z (q d −1)/(q−1) is a Singer subgroup; (b) Tj = PΓL(2, 8) and Z 9 = P i PSL(2, 8); (c) Tj = Sym(z) or Alt(z), and P i is abelian of order z.
If X AGL(a, p), then X normalizes Z a p and the result follows by Lemma 15.If Ti = PGL(d, q), and A i = Z (q d −1)/(q−1) is a Singer subgroup then by Lemma 16, a Sylow p-subgroup of T1 × • • • × T is regular and homocyclic.Then P i | C , i = 1, 2, are regular homocyclic subgroups and 1 G 2 γ 1 is nilpotent, and the result follows.If Ti = Alt(z), Sym(z), or PΓL (2,8), then by hypothesis we must have = 1.If Ti = Alt(z) or Sym(z), then by Lemma 8, there exists 1 G 2 γ 1 admits a complete block system B of n/p blocks of size p.After a suitable conjugation, we may assume that Lemma 9], and the result follows by [8,Theorem 12].
If T1 = PΓL (2,8), then p a = 9 and a Sylow p-subgroup of T1 has order 27.Also, a Sylow 3-subgroup of fix G 1 (C) is cyclic of order 9, say fix G 1 (C) = ρ .As PΓL(2, 8) contains a regular cyclic subgroup of order 9, we have by [17,Theorem 9] and [17, Lemma 9], that a Sylow 3-subgroup of PΓL (2,8) is the same as a Sylow 3-subgroup P of {x → ax+b : We may assume, after an appropriate conjugation, that a Sylow 3-subgroup of fix Theorem 22.Let n be odd with prime-power decomposition n = p a 1 1 • • • p ar r , and G be a transitive abelian group of order n such that every Sylow p i -subgroup P i of G is elementary abelian or cyclic.If no prime divisor of |Aut(P i )| other than p i divides n for every 1 i r, then whenever δ ∈ Sym Proof.The second statement follows from the first and [9, Lemma 9], so it suffices to show the first statement.First, if r = 1 then by a Sylow Theorem there exists γ ∈ G, δ −1 Gδ such that G, γ −1 δ −1 Gδγ is a p 1 -group, and so nilpotent.We thus assume that r 2. We now proceed by induction on m = Ω(n).If m = 1, then the result follows by the immediately preceding argument, so we assume the result holds for all n with 1 Ω(n) m−1.Let n be such that Ω(n) = m, G Sym(n) satisfy the hypothesis, and δ ∈ Sym(n).By Corollary 9, we may assume without loss of generality that G, δ −1 Gδ is imprimitive, and by Lemma 10, that G, δ −1 Gδ admits a complete block system B consisting of n/p a blocks of size p a , for some prime p|n and a 1.
Choose B in such a way that a is as small as possible.By [4, Exercise 1.5.10] Note that p i − 1 divides |Aut(P i )| for every 1 i r as P i is abelian and hence a direct product of cyclic groups.By the induction hypothesis, there exists Lemma 9].The result then follows by [8,Theorem 12].If a 2, then as H contains a regular abelian subgroup A and is primitive, by Theorem 6 one of the following is true: , Q is a transitive permutation group of degree , and one of the following holds: (a) Ti = PGL(d, q), and A i = Z (q d −1)/(q−1) is a Singer subgroup; (b) Ti = PΓL(2, 8) and Z 9 = A i PSL(2, 8); (c) Ti = Sym(z) or Alt(z), and A i is abelian of order z.

Suppose that if
2 then Ti = Sym(z), Alt(z), or PΓL (2,8).Let P be a Sylow p-subgroup of H.By [36, 3.If Ti = Alt(z) or Sym(z) and 2, then a regular abelian subgroup of H cannot be cyclic, and so by hypothesis is elementary abelian.By Lemma 13 there exists γ 2 ∈ the electronic journal of combinatorics 25(2) (2018), #P2.49 AGL(a, p) for every B ∈ B, and this case reduces to one considered above.Finally, if Ti = PΓL(2, 8) and 2, then a regular abelian subgroup of H cannot be cyclic or elementary abelian, a contradiction.

The Main Results
We begin with the basic definitions and results concerning Cayley objects and the Cayley isomorphism problem, some of which we have seen before.Definition 23.We define a Cayley object of G to be a combinatorial object X such that G L Aut(X), where Aut(X) is the automorphism group of X.If X is a Cayley object of G in some class K of combinatorial objects with the property that whenever Y is another Cayley object of G in K, then X and Y are isomorphic if and only if they are isomorphic by a group automorphism of G, then we say that X is a CI-object of G in K.If every Cayley object of G in K is a CI-object of G in K, then we say that G is a CI-group with respect to K. If G is a CI-group with respect to every class of combinatorial objects, then G is a CI-group.
We will also have need of the notion of solving sets.Definition 24.Let G be a finite group.We say that S ⊆ Sym(G) is a solving set for a Cayley object X in a class of Cayley objects K if for every Cayley object X ∈ K such that X ∼ = X , there exists s ∈ S such that s(X) = X .We say that S ⊆ Sym(G) is a solving set for a class K of Cayley objects of G if whenever X, X ∈ K are Cayley objects of G and X ∼ = X , then s(X) = X for some s ∈ S. Finally, a set S is a solving set for G if whenever X, X are isomorphic Cayley objects in any class K of combinatorial objects, then s(X) = X for some s ∈ S.
Note that X is a CI-object of G if and only if Aut(G) is a solving set for X.The following characterization of a solving set for an abelian group is [8,Lemma 15] and generalizes [33,Lemma 1.1].
Lemma 25.Let G be a finite abelian group, and S ⊆ Sym(G) a set of permutations.Then the following conditions are equivalent: 1. S is a solving set for a Cayley object X in a class K of Cayley objects of G, Aut(X), there exists s ∈ S and v ∈ Aut(X) such that v −1 δ −1 g L δv = s −1 g L s for every g ∈ G.
The following result [8,Theorems 16] will be needed to apply our main result.We remark that in [8] this result is stated for an arbitrary Cayley object X in an arbitrary class of combinatorial objects).
the electronic journal of combinatorics 25(2) (2018), #P2.49Theorem 26.Let k be a positive integer with gcd(k, ϕ(k)) = 1 and k = p 1 p 2 • • • p r be the prime-power decomposition of k.Let n = p a 1 1 p a 2 2 • • • p ar r , where a i 1 is a positive integer.Let G = Π r i=1 P i be an abelian group where P i is a Sylow p i -subgroup of G, and let S(i) be a solving set for P i .If, whenever δ ∈ Sym(G) there exists ω ∈ G L , δ −1 G L δ such that G L , ω −1 δ −1 G L δω is normally m-step imprimitive, then a solving set for G is contained in Π r i=1 S(i).
Let k and n be as above and P i a Sylow p i -subgroup of G. Suppose that for 1 i r, no prime divisor of n other than p i divides |Aut(P i )|.As G is abelian, P i is abelian, and p i − 1 divides |Aut(P i )| as p i − 1 divides |H| for every cyclic group of order p a i , a 1.Thus in this case gcd(k, ϕ(k)) = 1.Combining Theorem 26 with Theorem 22, we have the following result.
Corollary 27.Let n = p a 1 1 p a 2 2 • • • p ar r be the prime-power decomposition of n.Let G be an abelian group of order n such that every Sylow p i -subgroup P i of G is elementary abelian or cyclic.Let S(i) be the solving set for P i .If no prime divisor of n other than p i divides |Aut(P i )|, then a solving set for G is Π r i=1 S(i).
As solving sets are known for Z 2 p [17, Corollary 2] and Z p 2 [24] or [17,Corollary 1], and as all groups of order p 2 are elementary abelian or cyclic, we have the following result.
Corollary 28.Let n = p a 1 1 p a 2 2 • • • p ar r be the prime-power decomposition of n and n be cube-free.Let G be an abelian group of order n with Sylow p i -subgroup P i , 1 i r.If no prime divisor of n other than p i divides |Aut(P i )|, then a solving set for G is known.
1 p a 2 2 • • • p ar r be the prime-power decomposition of n.Define g : Z + → Z + by g(n) = Π r i=1 f (p a i i ) (recall that f was defined in Definition 20).
Let G be an abelian group of order n = p a 1 1 • • • p ar r such that every Sylow p i -subgroup P i of G is elementary abelian or cyclic.As |Aut(Z p a )| = p a − p a−1 and Aut(Z a p ) = AGL(a, p) so that |Aut(Z a p )| = Π a−1 i=0 (p a − p i ), if gcd(n, g(n)) = 1, then no prime divisor of n other than p i divides |Aut(P i )|.Hence we have the following results.
Corollary 30.Let n be a positive integer such that gcd(n, g(n)) = 1, and G an abelian group of order n such that every Sylow p i -subgroup P i of G is elementary abelian or cyclic.Let S(i) be the solving set for P i .Then a solving set for G is Π r i=1 S(i).
Corollary 31.Let n be a cube-free positive integer such that gcd(n, g(n)) = 1, and G be an abelian group of order n.Then a solving set for G is known.
It may be interesting to observe that if gcd(n, ϕ(n)) = 1, then every group of order n is cyclic, while if gcd(n, g(n)) = 1 then every group of order n is nilpotent, and if in addition to gcd(n, g(n)) = 1 n is also cube-free, then every group of order n is abelian [36,Theorem 9.2.7].
While the previous results definitely give the flavor of the consequences of the work in this paper, from a strictly computational point of view they can be quite inefficient.
A solving set for P i will contain the solving sets of every Cayley combinatorial object in any class K of combinatorial objects, and so in practice can be too large to be useful for efficient isomorphism testing.In practice, for a Cayley object X of G of prime-power order p a a solving set for X is determined by the Sylow p-subgroup P of Aut(X) that contains G L (see also [17,Lemma 15]).Lemma 32.Let X be a Cayley object of the abelian p-group G and P a Sylow p-subgroup of Aut(X) that contains G L .Then there exists a Cayley object W in some class of combinatorial objects such that Aut(W ) = P and any solving set for W is a solving set for X.
Proof.By [40, Theorem 5.12], there exists a Cayley object W of G such that Aut(W ) = P .Let Y be a Cayley object of G in the same class of combinatorial objects as X and δ ∈ Sym(G) such that δ(X) = Y .Then δ −1 G L δ Aut(X).Hence there exists v ∈ Aut(X) such that v −1 δ −1 G L δv P , so we assume without loss of generality that δ −1 G L δ P .As δ −1 G L δ Aut(W ), δ(W ) is a Cayley combinatorial object of G isomorphic to W .Let S be a solving set for W .By Lemma 25, there exists s ∈ S and v ∈ Aut(W ) such that v Thus S is a solving set for X by Lemma 25.
Corollary 33.Let n = p a 1 1 p a 2 2 • • • p ar r be the prime-power decomposition of n.Let G be an abelian group of order n such that every Sylow p i -subgroup P i of G is elementary abelian or cyclic, and X a Cayley object of G.If no prime divisor of n other than p i divides |Aut(P i )|, then there exists Cayley objects X i of P i such that if S(i) is a solving set for X i , then a solving set for X is Π r i=1 S(i).Many of our main results hold for so-called -closed groups, introduced by Wielandt [40].
Definition 34.Let Ω be a set.An -ary relational structure on Ω is an ordered pair (Ω, U ), where U ⊆ Ω = Π i=1 Ω.A group G Sym(Ω) is called -closed if G is the intersection of the automorphism groups of some set of -ary relational structures.The -closure of G, denoted G ( ) , is the intersection of all -closed subgroups of Sym(Ω) that contain G.
The following result is [8,Theorem 20], and is the analogue of Theorem 26 for -ary relational structures (again in [8], this result is stated for an arbitrary -ary relational structure X).
Theorem 35.Let k be a positive integer with gcd(k, ϕ(k)) = 1 and k = p 1 p 2 • • • p r be the prime-power decomposition of k.Let n = p a 1 1 p a 2 2 • • • p ar r , where a i 1 is a positive integer.Let G = Π r i=1 P i be an abelian group where P i is a Sylow p i -subgroup of G, and S (i) be a solving set for P i in the class K of -ary relational structures.If, whenever δ ∈ Sym(G) the electronic journal of combinatorics 25(2) (2018), #P2.49

Problems
The isomorphism problem for Cayley digraphs of abelian group seems to be quite difficult in general.There are now at least three obstacles to a group being CI with respect to digraphs that will depend at least to some extent on the structure of the full automorphism group of the digraph.The first is that there may not be an appropriate conjugate such that G L , δ −1 Gδ is normally m-step imprimitive [16].Second, isomorphic regular abelian subgroups of a p-subgroup P of the symmetric group need not be conjugate in P .The third is that in general the direct product of two CI-groups of relatively prime order need not be a CI-group by Theorem 1.All of these obstacles can occur in the automorphism group of a ternary relational structure ( [16] in the first instance, and [9] for the latter two).It thus seems wise to begin to break these large difficult problems into more manageable pieces.For example, the following problem seems natural in this context: Which finite groups G have the property that whenever δ ∈ Sym(G), there exists γ ∈ G L , δ −1 G L δ ( ) such that G L , γ −1 δ −1 G L δγ is (normally) mstep imprimitive?
For the previous problem, particular attention should be paid to the groups G which may be CI-groups with respect to digraphs (see [29,Theorem 1.2]) or graphs (see [18,Corollary 21]), and most especially, to groups which may be CI-groups with respect to ternary relational structures (see [16,Theorem 4.2]).Abelian groups also seem worthy of consideration.Of course, p-groups and cyclic groups have the above property, as well as groups given in Theorem 22. Also observe that for any group G, G (1) = G, so a special case of the previous problem is to find all groups G such that whenever δ ∈ Sym(G) there exists γ ∈ G L , δ −1 G L δ such that G L , γ −1 δ −1 G L δγ is (normally) m-step imprimitive.
The problem corresponding to the second obstacle given above is simply to solve the isomorphism problem for abelian p-groups, so we make no more mention of this well-known problem.
Finally, for the third obstacle given above, the following conjecture has been made: Conjecture 43.Let G and H be CI-groups with respect to digraphs of relatively prime order.Then G × H is a CI-group with respect to digraphs.
Virtually no progress has been made on this general conjecture, other than showing specific groups are CI-groups with respect to digraphs.We propose a much simpler problem (which still seems quite difficult, and would also be happy for partial solutions for particular G): Problem 44.Let G be a cube-free abelian group.Assume that whenever δ ∈ Sym(G) such that G L , δ −1 G L δ is normally m-step imprimitive.Does the isomorphism problem for Cayley digraphs of G reduce to the Cayley isomorphism problem for digraphs of its Sylow subgroups?Finally, we believe that a much more general result than the positive part of Pálfy's Theorem is true.Note that if m and n are positive integers such that gcd(m, ϕ(m)) = 1, gcd(n, ϕ(n)) = 1 and gcd(ϕ(n) • n, m) = 1 = gcd(ϕ(m) • m, n), then gcd(mn, ϕ(mn)) = 1, and so Pálfy's Theorem implies the conjecture is true for cyclic groups of order m and n respectively.
Let k = p 1 • • • p r be such that gcd(k, ϕ(k)) = 1, and m = p a 1 1 • • • p ar r , = q 1 • • • q s and gcd( , ϕ( )) = 1 and n = q a 1 1 • • • q as s , where the p i 's are distinct primes and the q j 's are distinct primes.If in addition, gcd(ϕ(n) • n, m) = 1 = gcd(ϕ(m) • m, n), then p 1 , . . ., p r , q 1 , . . ., q s are distinct primes, and if a = p 1 • • • p r q 1 • • • q s , then gcd(a, ϕ(a)) = 1.Thus the conjecture holds for cyclic groups of order m and n by Muzychuk's result [33,Theorem 1.9].More generally, combining [13,Corollary 4.4] and [33,Theorem 4.9], the conjecture is true for all cyclic groups satisfying the hypothesis (this also follows from the results in this paper).Of course, the results in this paper verify the result for abelian groups G and H such that every Sylow subgroup is either elementary abelian or cyclic.

Definition 5 .
Let G be a transitive group acting on Ω and B a complete block system of G.By fix G (B) we mean the subgroup of G which fixes each block of B set-wise.That is, fix G (B) = {g ∈ G : g(B) = B for all B ∈ B}.We denote by Stab G (x) the stabilizer of x ∈ Ω, and for B ∈ B, Stab G (B) is the set-wise stabilizer of the block B. That is, Stab G (x) = {g ∈ G : g(x) = x} and Stab G (B) = {g ∈ G : g(B) = B}.Finally, for g ∈ G we denote by g/B permutation of B induced by g, and G/B = {g/B : g ∈ G}.
then we also have that the orbits of J 1 | C and γ −1 0 J 2 γ 0 | C are identical.Define an equivalence relation ≡ 0 on C by C ≡ 0 C if and only the action of fix G 1 ,G 2 (C) on C is equivalent to the action of fix G 1 ,G 2 (C) on C .As Alt(p a ) has only one representation as p a = 6 [3, Theorem 5.3], and two transitive actions of a group are equivalent if and only if the stabilizer of a point in one action is the same as the stabilizer of a point in the other [4, Lemma 1.6B], if C ≡ 0 C , then the action of fix G 1 ,G 2 (C)| C∪C on C cannot be a faithful action.Thus if C ≡ 0 C , then there exists

Lemma 11 .
Let p be an odd prime.Then any two regular elementary abelian subgroups of Sym(p a ) are conjugate in Alt(p a ), a 1, and any two regular elementary abelian subgroups of Sym(p a ) contained in a Sylow p-subgroup P of Sym(p a ) are conjugate in P .
the electronic journal of combinatorics 25(2) (2018), #P2.49Lemma 13.Let p be an odd prime, and G 1 , G 2 Sym(n) such that G = G 1 , G 2 admits a complete block system C of n/p a blocks of size p a , and both fix G 1 (C)| C and fix G 2 (C)| C are regular and elementary abelian for every C ∈ C. Further, assume that soc(Stab G (C))| C = Alt(m) .Then there exists γ

Hypothesis 1 .
Let n = p a 1 1 . . .p ar r be the prime power factorization of n.Let G 1 , G 2 be transitive groups of order n such that G = G 1 , G 2 satisfies the following conditions:1.G admits a complete block system C of n/p a r blocks of size p a r , a 1, andStab G (C)| C is primitive, C ∈ C. Additionally, fix G i (C) is a p r -group transitive on each C ∈ C, the electronic journal of combinatorics 25(2) (2018), #P2.49 2. no prime divisor of |N Sym(p a r ) (P 1 )| other than p r divides n, where P 1 is a Sylow p r -subgroup of fix G 1 (C) (we denote the Sylow p r -subgroup of fix G 2 (C) by P 2 ),

G 2 (
C) is contained in the same Sylow 3-subgroup P of fix G (C) as fix G 1 (C).As a transitive abelian group is self-centralizing [4, Theorem 4.2A (v)], we must have that Z(P | C ) ρ | C for every C ∈ C, and as Z( P ) = {x → x + 3b : b ∈ Z 3 } we have that Z(P | C ) = ρ 3 | C for every C ∈ C. Also, as fix G 2 (C)| C is a regular cyclic subgroup for every C ∈ C, we must the electronic journal of combinatorics 25(2) (2018), #P2.49 also have that ρ 3 | C fix G 2 (C)| C for every C ∈ C. We conclude that every element of both G 1 and G 2 normalizes ρ 3 | C : C ∈ C , and so M = G ∩ ρ p | C : C ∈ C G. Then G admits a complete block system with blocks of size 3, and the result follows by arguments above.
and that a Sylow p-subgroup of fix P (B)| B is permutation isomorphic to a Sylow psubgroup of Sym(p a−1 ) for every B ∈ B. As P/B ∼ = Z p , we have that fix R (B) is semiregular of order p a−1 .Without loss of generality, we assume that τ i : 2 i a fix P (B).Let As a normal subgroup of a primitive group is transitive [41, Theorem 8.8], we have that K| C is transitive for every C ∈ C. It is then easy to see that Alt(m) K| C for some and every C ∈ C as Alt(m) is simple.As H i | C , i = 1, 2, are transitive abelian subgroups (acting on C), we have by Theorem 6 that H i | C Sym(m) for every C ∈ C and i = 1, 2. As p is odd, if H K| C and is regular and elementary abelian, then H Alt(m) .In particular, K| C Alt(m) for every C ∈ C so that K| C = Alt(m) for every C ∈ C. Then K| C admits complete block systems B C,i formed by the orbits of one of the copies of Alt(m), 1 i and 2. G/C = H × P where H is a solvable π-subgroup, and P is the unique Sylow p rsubgroup of G/C.Sym(p a r ) (P 1 ) by the Embedding Theorem [30, Theorem 1.2.6] we have G (G/C) N Sym(p a r ) (P 1 ).We conclude that |G| divides |G/C|•|N Sym(p a By the Schur-Zassenhaus Theorem [22, Theorem 6.2.1], G possesses an S π -subgroup K, which is a complement to fix G (B), and any two S π -subgroups of G are conjugate in G as G/B = H is solvable.
Then there exists γ ∈ G such that G 1 , γ −1 G 2 γ = H × Π, where Π is a p r -subgroup andH = H 1 , H 2 .Proof.Let |N Sym(p a r ) (P 1 )| = p v r • c, where gcd(p r , c) = 1.As Stab G (C)| C N r ) (P 1 )| n/p a r .As |G/C| = |H|•p dr for some d 0, and no prime divisor of n other than p r divides |N Sym(p a r ) (P 1 )|, we have that H is a π-subgroup of G.As P G/C, G admits a (possibly trivial) complete block system B consisting of n/p ar r blocks of size p ar−a r .Hence G admits a complete block system B of n/p ar r blocks of size p ar r and C B. Then G/B = H is a π-group, and fix G (B) is a normal π -subgroup of G.