The Cayley isomorphism property for Cayley maps

In this paper we study finite groups which have Cayley isomorphism property with respect to Cayley maps, CIM-groups for a brief. We show that the structure of the CIM-groups is very restricted. It is described in Theorem~\ref{111015a} where a short list of possible candidates for CIM-groups is given. Theorem~\ref{111015c} provides concrete examples of infinite series of CIM-groups.


Introduction
We say that a map Cay(H, S, ρ) is connected if the underlying Cayley graph is connected, that is S = H.
Using a less combinatorial approach, a Cayley map is a 2-cell embedding of a Cayley graph into oriented surface with the same cyclic rotation around each vertex. For precise definiton of embedding graphs into orientable surfaces, see [10]. Several different subclasses of Cayley maps have been investigated. The notion of a Cayley map first appeared in the paper of Biggs [2] who investigated balanced Cayley maps. A Cayley map Cay(H, S, ρ) is called balanced if ρ(s −1 ) = ρ(s) −1 and it is called antibalanced if ρ(s −1 ) = ρ −1 (s) −1 . Further, a Cayley map M is called regular if its automorphism group is transitive on the arcs as well. Following Jajcay and Siran [9], we say that for a group H a permutation φ ∈ Sym(H) is a skew-morphism if there exists a mapping π : H → N such that φ(gh) = φ(g)φ π(g) (h) for every g, h ∈ G.
In what follows we say that M 1 and M 2 are Cayley isomorphic if there exists a group isomorphism φ : H 1 → H 2 which is simulteneously a map isomorphism, that is φ(S 1 ) = S 2 and φ(ρ 1 (s)) = ρ 2 (φ(s)) holds for each s ∈ S 1 .
The automorphism group of a Cayley map M = Cay(H, S, ρ) is the set of all isomorphisms from M to M and it will be denoted by Aut(M ). It is clear that H ≤ Aut(M ). Thus Aut(Cay(H, S, ρ)) contains the regular subgroup H. Every group automorphism σ ∈ Aut(H) induces Cayley isomorphism between the maps Cay(H, S, ρ) and Cay(H, σ(S), σ ′ ρ(σ ′ ) −1 ) where σ ′ = σ| S is the restriction of σ on S. Thus a group automorphism σ is an automorphism of a map Cay(H, S, ρ) if and only if σ(S) = S and σ| S ρ = ρσ| S . Since ρ is a full cycle, the latter condition is equivalent to σ| S = ρ k for some integer k.
The so-called CI (Cayley isomorphism) property of groups is well studied with respect to Cayley graphs. A group H is called a CI-group with respect to graphs (CIG-groups, for short) if two Cayley graphs of H are isomorphic if and only if they are isomorphic by a group automorphism as well. A Cayley graph Γ = Cay(H, S) is called a CI-graph if every Cayley graph Cay(H, T ) isomorphic Γ is Cayley isomorphic to Γ. For an old but excellent survey about CI-groups, see [11] and further results can be found in [12]. Similarly to the original definition of the CI property we say that a Cayley map M = Cay(H, S, ρ) is a CI-map of H if every Cayley map M ′ over H isomorphic to M is also Cayley isomorphic to M . We call a group H a CIM-group if for every Cayley map Cay(H, S, ρ) is a CI-map.
A Cayley map M = Cay(H, S, ρ) can also be considered as a ternary relational structure on the vertices of the underlying graph. Three vertices (x, y, z) are in the relation R if and only if x −1 y, x −1 z ∈ S and ρ(x −1 y) = x −1 z. The automorphism group Aut(M ) consists of all those permutations of the vertices which preserves the relation R. In particular, it is a 3-closed permutation group. This observation allows us to use the technique developed by Babai to solve problems concerning CIM-groups. Moreover a theorem of Pálfy [14] shows that the groups which are CI-groups for every m-ary relational structures are the cyclic groups of order n, where (n, φ(n)) = 1 and the Klein group. Pálfy also proved that if a group is not a CI-group with respect to some m-ary relation, then it is not a CI-group with respect to 4-ary relational structures.
CI-groups with respect to ternary relations (CI (3) -groups, for short) were investigated by Dobson [4], [5] and later by Dobson and Spiga [6]. Although the class of CI (3) -groups is rather narrow, its full classification is not finished yet.
The latest results may be found in [5] and [6]. Since map automorphism group is 3-closed, each CI (3) -group is a CIM-group. The converse is not true. For example, every elementary abelian 2-groups of rank at least 6 is a CIM-group but not a CI (3) -group.
As it was also pointed out by Dobson and Spiga [6] every CI (3) -group is also a CI (2) -group, that is a group which has a the CI-property with respect to binary relational structures. However, we will prove that there are CIM-groups which are not CI (2) -groups. The Venn diagram below reflects the relationships between the three classes of CI-groups.

CIM
Our first result formulates necessary conditions for being a CIM-group. Theorem 1.1. Let H be a CIM-group. Then H is isomorphic to one of the following groups where m is an odd square-free number.
The second main result provides several infinite series of CIM-groups.
Theorem 1.2. The following groups are CI-groups with respect to Cayley maps.
where m is an odd square-free number.
As an immediate corollary of the above Theorems we obtain the following criterion. Notice that obtained results do not provide a complete classification of cyclic CIM-group. This is because we do not know which of the groups Z m × Z 8 , m is odd and square-free, are CIM-groups. Proposition 5.7 shows that Z 8 is a CIMgroup. We believe that all groups of the above structure have the CIM-property.
Our paper is organised as follows. In Section 2 we collect a few general results about CI-property which will be used later. In Section 3 we characterize Sylow subgroups of CIM-groups. Section 4 is devoted to the proof of Theorem 1.1. The last section provides proofs of Theorems 1.2 and 1.3.
Most of the group-theoretical notation used in the paper are standard and can be found in [17].

General observations
The original CI property for graphs is inherited by subgroups which gives us a strong tool to determine the list of possible CI-groups. Similar, but a weaker, property holds for CIM-groups as well. Let us call a group H to be a connected CIM-group if it is a CI-group with respect to connected maps.
Proof. Let G be a CIM-group and H ≤ G. Let us assume that Cay(H, S, ρ) and Cay(H, S ′ , ρ ′ ) are isomorphic connected Cayley maps of H. Let f be a map isomorphism from Cay(H, S, ρ) to Cay(H, S ′ , ρ ′ ). Then g 2 f g 1 −1 is an isomorphism between the connected component of Cay(G, S, ρ) on g 1 H and the one Cay(G, S ′ , ρ ′ ) on g 2 H. This shows that the connected components of Cay(G, S, ρ) and Cay(G, S ′ , ρ ′ ) are isomorphic. Therefore Cay(G, S, ρ) and Hence α| H is a Cayley isomorphism between the above maps.
This result suggests that it is worth investigating p-groups which arise as the Sylow p-subgroups of finite groups.
Another important observation is that if Cay(H, S, ρ) is a Cayley map with |S| ≤ 2, then the Cayley graph Cay(H, S,) has to be a CI-graph since there exists only one cyclic ordering on one or two elements. This shows that the automorphism group of a CIM-group H has only one orbit on the elements of order 2 and for every g, h ∈ H with the same order there exists α ∈ Aut(H) with α(g) = h or α(g) = h −1 . Groups having this property were investigated by Li and Praeger [13].
The following lemma is due to Babai [1] and applies for every Cayley relational structures. In what follows we refer to a regular permutation subgroup isomorphic to H as H-regular subgroup.
The statement below describes the structure of the Cayley map automorphism group. Although it was proven by Jajcay [8] we prefer to provide its proof here to make the paper self-contained. Then G e acts faithfully on S and its restriction (G e )| S is contained in ρ . In particular, G e is cyclic.
Proof. Pick an arbitrary φ ∈ G e . Then ∆ e φ = φ| S implying ρφ| S = φ| S ρ. Since ρ is a full cycle on S, any permutation commuting with it belongs to ρ . Therefore (G e )| S ≤ ρ . This inclusion also implies that for each s ∈ S the two-point stabilizer G e,s acts trivially on S. Therefore G h,hs acts trivially on hS for any h ∈ H and s ∈ S. Thus if φ fixes e and s ∈ S, then it fixes pointwise the sets S, S 2 , S 3 etc. Since Cay(H, S) is connected, we conclude that G e,s is trivial, i.e. G e acts faithfully on S.
The above statement shows the full automorphism group G of a connected Cay(H, S, ρ) is a product of H with the cyclic group G e . Moreover the restriction of G e on S is contained in ρ .

Sylow subgroups of CIM-groups
Similarly to the classical case of CI-groups, it follows from Lemma 2.1 that it is important to investigate p-groups. Babai and Frankl proved that if a group H is a CI (2) -group of prime power order, then H is either elementary abelian p-group, the quaternion group of order 8 or a cyclic group of small order. The statement below describes odd order Sylow subgroups of a CIM -group. Proof. It follows from Lemma 2.1 that it is sufficient to show that any subgroup of order p 2 is not a connected CIM-subgroup.
Let K be a group of order p 2 . Then either K ∼ = Z 2 p or K ∼ = Z p 2 . In both cases there exists an automorphism β ∈ Aut(K) of order p (the concrete examples of β are given below). A direct check shows that the bijection α ∈ Sym(K) defined via α(x) = −β(x) is an automorphism of K of order 2p. It follows from α p = −1 that each non-zero α-orbits is symmetric, and, therefore, has even cardinality. This implies that at least one orbit of α contains 2p element. Let us denote this orbit as S. Clearly S = K. Consider a Cayley map M = Cay(K, S, α| K ). The group G := Aut(M ) contains the semidirect product K ⋊ α ≤ Sym(K). Combining this with |Aut(M )| ≤ |K||S| = |K|| α | we conclude that G = K ⋊ α (so, M is a balanced regular map). We claim that M is not a CI-map. According to Lemma 2.2 it is enough to find two K-regular subgroups of G which are not conjugate in G. Since K is normal in G, it is sufficient to find a K-regular subgroup of G distinct from K. To point out such a subgroup we consider the cases of K ∼ = Z p 2 and K ∼ = Z 2 p separately. In both cases we use the fact that β = α p+1 ∈ G.
Case of K ∼ = Z 2 p . In this case we chose β ∈ Aut(Z 2 p ) defined via β((x, y)) = (x + y, y). Then the group G contains the subgroup Z 2 p ⋊ β which consists of all permutations of the form (x, y) → (x + ay + u, y + v) where a, u, v ∈ Z p . A direct check shows that the permutations τ a,b : (x, y) → (x+ay +b, y +a), a, b ∈ Z p form a subgroup, say T , of G isomorphic to Z 2 p . It is easy to check that T acts regularly on Z 2 p .

Sylow 2-subgroups of CIM-groups
Proposition 3.2. For every n ≥ 4 the cyclic group Z 2 n is not a connected CIM-group.
We Straightforward calculation shows that ( 1α) 2 (x) = x + a + 1 = x + 2 n−1 + 2 implying that ( 1α) 2 has order 2 n−1 . Hence the order of 1α is 2 n . Therefore the subgroup A of G contains at least two regular subgroups isomorphic to Z 2 n , both of index two. These subgroups are not conjugate in A, since they are normal in A. Thus it is enough to prove that A = G. The latter is equivalent to showing that the point stabilizer of G 0 has order two. Assume, towards a contradiction, that |G 0 | > 2. The group G 0 is cyclic and acts on S faithfully and semi-regularly. Therefore there exists an element σ ∈ G 0 such that σ 2 = α. In particular, σ has order 4. Since σ| S commutes with ρ, we conclude that σ| S = ρ 2 = (1, 3, a, 3a)(−1, −3a, −a, −3).
Thus P does not contain K 4 . By Burnside's Theorem [3], P is either cyclic or generalized quaternion. If P is cyclic, then by Proposition 3.2 its order is bounded by 8.
Assume now that P is a generalized quaternion group distinct from Q 8 . Then P contains a characteristic cyclic subgroup C = c of index 2. Then it follows from Lemma 2.1 and Proposition 3.2 that |C| ≤ 8. Together with P ∼ = Q 8 we obtain that |C| = 8, and, consequently |P | = 16.
Let a ∈ P denote an element of order 4 outside of C. Then a, c 2 ∼ = Q 8 . Let α be an automorphism of a, c 2 whose action is described by the formulas  and ℓ = m+1 2 . Notice that the condition OO (−1) = K implies that the map is connected.
It follows from the construction that ρ 2 = σ| S∪S (−1) , where σ is the inner automorphism of H mapping x to x c . Therefore σ ∈ Aut(M ) and G := H⋊ σ ≤ Aut(M ).
In order to build a regular subgroup of H ⋊ σ different from H we notice, first, that this group is isomorphic to a direct product H × C where the isomorphism is defined via ψ : hσ i → (hc i , c −i ). Under this isomorphism the point stabilizer G 1 = σ is mapped onto the subgroup ψ( Let π : H → C be a projection homomorphism defined via π(xk) := x for x ∈ C and k ∈ K. Then F := {(h, π(h)) | h ∈ H} is a subgroup of H × C which intersects ψ(G 1 ) trivially. Indeed, By assumption C has odd order. Therefore h = 1.
It follows from F ∩ ψ(G 1 ) = 1 that ψ −1 (F ) is a regular subgroup of G. Thus G contains two regular subgroups isomorphic to H, which are H and ψ −1 (F ). Since H ⊳ G, it is not conjugate to ψ −1 (F ) inside G.
Since G 1 has two orbits on the connection set S ∪ S −1 , either Aut(M ) = G or [Aut(M ) : G] = 2. In the first case we already have two H-regular subgroups of G which are non-conjugate in G. In the second case it follows from ρ(x −1 ) = ρ(x) −1 that M is a regular balanced map over H. It was proved in [15] that H Aut(M ). Since G contains a H-regular subgroup distinct from H, it is not conjugate to H inside Aut(M ).
Remark. The condition OO (−1) = K is always fulfilled if K does not contain a proper non-trivial C-normalized subgroups. For example, if K is of prime order, then OO (−1) = K holds for any non-trivial orbit O. Now we are ready to prove Theorem 1.1.
Proof. Let T denote a Sylow 2-subgroup of H. Our proof is divided into few steps.
Step 1. Any normal subgroup N of H of odd order is cyclic. Since all Sylow subgroups of N have prime order, it is sufficient to prove that any Sylow subgroup of N is normal in H. This would follow if we prove that each Sylow subgroup of N has a normal complement. To show that let us fix a Sylow subgroup P of order p, where p is prime. By Burnside Theorem the existence of a normal complement follows from N N (P ) = C N (P ). Assume towards a contradiction that there exists g ∈ N N (P ) which does not centralize P . We may assume that o(g) is a prime power. By Lemma 3.1 any Sylow subgroup of N has a prime order. Therefore o(g) is prime distinct from p. In this case the group g P satisfies the assumptions of Lemma 4.1 and therefore, is not a connected CIM group. A contradiction.
Step 2. T has a normal complement. By Proposition 3.3, T is isomorphic to one of the groups Z r 2 , Z 4 , Z 8 or Q 8 . If T is cyclic, then the result follows from the Cayley normal 2-complement Theorem.
Assume now that T is not cyclic, i.e. T ∼ = Z r 2 or T ∼ = Q 8 . By Frobenius normal p-complement Theorem it is sufficient to show that N H (T )/C H (T ) is a 2-group. Notice that N H (T )/C H (T ) is embedded into Aut(T ).
If T ∼ = Q 8 , then T ∼ = Z e 2 for some e ≥ 1. Assume, towards a contradiction, that N H (T )/C H (T ) is not a 2-group. Then there exists an element g ∈ N H (T ) of odd order which acts on T nontrivially. Without loss of generality, we may assume that o(g) is a p-power for some odd prime divisor p of |H|. Since |H| p = p, we conclude o(g) = p. Since T is elementary abelian 2-group, it contains a minimal g-invariant subgroup T 1 on which g acts non-trivially. The group g T 1 satisfies the assumptions of Proposition 4.1. Therefore g T 1 is not a connected CIM-group. A contradiction.
If T ∼ = Q 8 and N H (T )/C H (T ) is not a 2-group, then this group contains an element of order 3. Hence N H (T ) contains an element g of order 3 which acts on T non-trivially. Applying Lemma 4.1 once more we get a contradiction.
Step 3. If T is non-cyclic, then H ∼ = N × T . As it was mentioned before, a CIM-group has the property that any two elements of the same order are either conjugate or inverse conjugate by an automorphism of H. In particular, this implies that all involtiuons of H are Aut(H)-conjugate.
If T is non-cyclic, then either it is elementary abelian or Q 8 . Let us assume first that T is an elementary abelian 2-group of order at least 4. Then all nontrivial elements of T are Aut(H)-conjugate. Since N is characteristic in H, the subgroups C N (s) and C N (t) are Aut(H)-conjugate for any s = t ∈ T \ {1}. Since any subgroup of N is characteristic in H, we conclude that K := C N (s) = C N (t) = C N (ts). Let L ≤ N be a unique subgroup complementary to K in N . Then both t and s invert the elements of L, Therefore st acts trivially on L implying L ≤ K, and consequently L = 1. Thus any element of T centralizes N . Therefore H ∼ = N × T .
It remains to settle the case when T ∼ = Q 8 . In this case all cyclic subgroups of order 4 are Aut(H)-conjugate. Since Aut(N ) is abelian the commutator subgroup Z of T acts trivially on N . The quotient groupH = H/Z is isomorphic to N ⋊ Z 2 2 . Moreover all involutions of Z 2 2 are Aut(H)-conjugate. From the previous paragraph we obtain that Z 2 2 acts trivially on N . So, the semi-direct product N ⋊ Z 2 2 is, in fact, the direct one. Therefore H ∼ = N × T .

Proof of Theorem 1.2
We start with introduccing the notation M for the set of groups Z n × Z r 2 , Z n × Z 4 , Z n × Q 8 where n is a square-free odd number. The statement below collects the properties of these groups. We omit the proof because it is straightforward. If H is abelian then by Ito's theorem [7] the group G is metabelian and therefore G is solvable. If H is non-abelian, then it is nilpotent and G is solvable by Kegel-Wielandt theorem.
We will prove Theorem 5.3 by induction on |G| and assume that G is a counterexample of a minimal order. In particular, this implies that the theorem is correct for any proper subgroup X where H ≤ X < G. Since G is a counerexample, there exists an H-regular subgroup F of G which is not conjugate to H inside G. We fix F till the end of the proof. By the minimality of G, we may assume that H, F g = G for each g ∈ G. We write the order of H by 2 r n. Recall that n is an odd square-free number.
Below the following notiation is used. If G is a group acting on a set X, then G X denote the kernel of this action and G X denote the image of G in Sym(X).
, and, therefore, is cyclic. Thus G D ≤ Sym(D) satisfies the assumptions of Theorem 5.3. Since |G D | = |G|/|G D | < |G|, we may apply the induction hypothesis to G D . It yields us that F D and H D are conjugate in G D . Therefore there exists g ∈ G such that (F g ) D = H D implying Proposition 5.5. Let G be a minimal counterexample to Theorem 5.3. Then G admits at most one minimal imprimitivity system.
Proof. Assume, towards a contradiction, that G admits two minimal imprimitivity systems, say D and E. By Proposition 5.4 G ω ≤ G D and G ω ≤ G E . If follows from minimality of E and D that G D ∩ G E = {1}. Therefore G ω = {1} implying G = H contrary to G being a counterexample.
For a set of elements S of a group acting on a set X, we denote by Fix(S), the elements of X fixed by every s ∈ S.
Proposition 5.6. Let G ≤ Sym(Ω) be a transitive permutation group with cyclic point stabilizer. Then for each S ≤ G ω the set Fix(S) is a block of G.
Proof. Assume that Fix(S) ∩ Fix(S g ) is non-empty, and pick an arbitrary δ ∈ Fix(S)∩Fix(S g ). Then S, S g ≤ G δ . Since G δ is cyclic, any two subgroups of G δ of the same order coincide. Therefore S = S g implying that Fix(S g ) = Fix(S).

Proof of Theorem 5.3
Let P be a minimal imprimitivity system of G. Pick an arbitrary block Π ∈ P. Then G Π {Π} is a solvable primitive permutation subgroup of Sym(Π). Therefore |Π| is a power of a prime divisor p of |H|.
We split the proof into few steps.
Step 1. |P| > 1. Assume the contrary, that is |P| = 1, or, equivalently, Π = Ω. In this case H is a p-group. By Proposition 5.6 the set Fix(G α,β ) is a block of G for any pair of points α, β ∈ Ω. Together with primitivity of G this implies that G α,β = 1 whenever α = β. Therefore G is a Frobenius group the kernel of which, K say, has order |H|. Since K is a unique Sylow p-subgroup of G, we conclude H = K = F , a contradiction.
Step 2. G P is a p-group. Assume that there exists a prime divisor q = p of G P . Since G P acts transitively on each block Π ∈ P and G ω ≤ G P , we conclude that |G P | = |G ω | · |Π|. This implies that q divides |G ω |. Thus G ω contains a subgroup Q of order q. By Proposition 5.6 the set Fix(Q) is block of G. It follows from Proposition 5.4 that Q ≤ G ω ≤ G P that Q fixes each block of P setwise. Since blocks of P have a p-power size, the set Fix(Q) intersects each block of P non-trivially. By Proposition 5.5 P is a unique minimal imprimitivity system of G. Therefore non-conjugate regular cyclic subgroups. Thus N P ( Z 8 ) = Z 8 ⋊ α contains nonconjugate regular cyclic subgroups. This yields a unique choice for α ∈ Aut(H), namely: α(x) = 5x, x ∈ Z 8 . Notice that Z 8 ⋊ α contains exactly two regular cyclic subgroups Z 8 and 1α . Each of these subgroups is normal in Z 8 ⋊ α .
Since α ∈ G 0 , it acts semiregularly on S. Combining this with S = Z 8 and S = −S we obtain that the only possibility for S is {1, 5, 3, 7}. It follows from ρ 2 = α| S that either ρ = (1, 3, 5, 7) or ρ = (1, 7, 5, 3). In both cases M is an antibalanced map the full automorphism group of which has order 32 and has a decomposition G = Z 8 ρ where ρ acts trivially on the subgroup 2Z 8 . In both cases all regular cyclic subgroups are conjugate in G.