A note on isomorphisms of Cayley digraphs of Abelian groups

Let Sand T be two minimal generating s~bsets of a finite abelian group G. We prove that if Cay(G, S) ~ Cay(G, T) then there exists an a E Aut(G) such that SOl = T. Let G be a finite group and S a subset of G not containing the identity element 1. The Cayley digraph X = Cay( G, S) of G with respect to S is defined by V(X) = G, E(X) = {(g,sg) I 9 E G,s E S}. It is easy to check that if a is an automorphism of G then Cay( G, S) ~ Cay( G, SOl). Conversely, we call a subset S of G a CI-subset, if for any subset T of G with Cay( G, S) ~ Cay( G, T), there is an automorphism a of G such that SOl = T. The concept of a CI-subset is important for the study of the isomorphism problem for Cayley digraphs. However, there are not many results in the literature about this concept. The second author conjectured at the third China-US conference on graph theory in 1993 that every minimal generating subset of a finite group is a CI-su bset, see also [3, Problem 8]. Huang and Meng [1] verified it for cyclic groups, and Li [2] verified it for abelian groups of odd order. However, in the same paper Li found a counterexample for an abelian group of even order. Namely, Li proved Proposition (1) Every minimal generating subset of an abelian group of odd order is CI. (2) Let G = (a) X (x) X (e) ~ Z3 X Z4 X Z2 and let S = {x,xe,ax 2 } and T = {x, xe, ax 2e}. Then S is a minimal generating subset of G and the Cayley digraph *The work for this paper was supported by the National Natural Science Foundation of China and the Doctoral Program Foundation of Institutions of Higher Education of China. Australasian Journal of Combinatorics 15(1997), pp.87-90 Cay( G, S) is isomorphic to Cay( G, T). However, there is no automorphism of G which maps S to T. In other words, neither S nor Tare CI. In Li's paper there is a small mistake. Li claimed that both the sets Sand T in above proposition are minimal generating subsets of G. However, it is easy to check that S is, but T is not. The purpose of this note is to prove the following theorem. Theorem Let G be a finite abelian group and let both Sand T be minimal generating subsets of G. Let X = Cay( G, S) and Y = Cay( G, T) be isomorphic. Then there exists an a E Aut( G) such that Sex. = T. As a consequence of this, for abelian groups, all generating subsets with the minimum number of generators are CI. That is the above-mentioned conjecture is true for minimum generating subsets of abelian groups. Proof of Theorem: If lSI = 1, then G is cyclic and the theorem holds. So we may assume that lSI> 1 in what follows. Take an isomorphism (]' from Cay( G, S) to Cay( G, T) with 10" = 1. Then T = SO". Now we define an equivalence relation "rv" on G by Then the restriction of "rv" to S (or to T) is also an equivalence relation on S (or on T). Let Sll ... ,Sk be the equivalence classes of Sunder (the restriction of) the relation "rv" on S. Then S = U7=1 Si. Let Ti = Sf, for i = 1",', k. We have Fact 1: T l ,"', Tk are also the equivalence classes of T under the relation "rv". To see this, first we have the following. Observation 2: For any S1, 82 E S, 81 :f 82, we have the intersection of the outneighborhoods X l (81) and X 1 (82) of 81 and 82 in the graph X is when 81 rf 82, when 81 rv 82. (1) Proof of Observation 2: Obviously, X1(sdnXl (82) contains {8l82} or {8l82,Si} depending on 81 rf 82 or 81 rv 82, respectively. To prove the reverse inclusion, we assume that there is another common out-adjacent vertex of 81 and S2, say x = 8lS = 82S' where 8,8' E S. Then at least one of 8 and 8', say 8, is not in {81, 82}. It follows that 8 = 81"1 828, and G can be generated by S \ {8}, contradicting the minimality of S. 0 By symmetry, for any tl, t2 E T, tl :f t 2, we also have 88 when tl rf t2, when t1 rv t 2. (2) Now we are ready to prove Fact 1. Given S1,S2 E S, set ti = sf for i = 1,2. By (1) and (2), S1 rv S2 {:::::} IX1 (Sl) n X1(S2)1 = 2 {:::::} I(Xl(Sl) n Xl(S2)YI = IYl(t l ) n Y1 (t2)1 = 2 {:::::} tl rv t2, so Fact 1 holds. Next, we consider the image of a product SlS2'" Sn under 0', where Si E S for i = 1,2"" ,n. We shall prove the following. Fact 3: (SlS2'" snY = YlY2'" Yn for some Yi E T, i = 1,2"" ,n, with Yi rv sf Vi. Proof Note that when n = 1, the conclusion is trivially true. So, by induction, it suffices to prove that for any x E G and Sll S2 E S, if (SiX yr = YiXu, i = 1,2, where Yi rv sf, then (Sls2xt = ZlZ2Xu, where Zi rv sf for i = 1,2. To prove this, we distinguish two cases: (i) SI rf sz, and (ii) SI rv sz. In the first case, we also have Yl rf Yz (by Fact 1). Since, by the same argument as in the proof of Observation 2, SlSZX is the only common out-adjacent vertex of SIX and SZX, YIYZX u is also the only common out-adjacent vertex of Yl XU and yzxU, so (SI SZX t YIYZX u and the conclusion holds. In the second case, we also have Yl rv Yz (by Fact 1). In this case SIX and SZX, (resp. YIXu and YZXU ), have two common out-adjacent vertices SISZX and six, (resp. YIYZXu and yrxU), so (Slszxt = YIYZXu or yrxu. The conclusion also holds. 0 Now we are in the position to prove the main theorem. Let Si {Sill SiZ,' .. ,SinJ for i = 1,2"", k. Let sfj = tij for i = 1,,", k, and j = 1"", ni. Then Ti = Sf = {til, tiZ,"', tin.}, and S = Uf=1 Si, T = Uf=l Ti. Now we define a mapping a : G -+ G by k ni k ni II II s~? f--?IT II <? where eij are arbitrary non-negative integers. i=lj=l i=lj=l We shall prove that a is an automorphism of G and hence complete the proof of the theorem. Note that Sand T are generating subsets of G. It suffices to prove that a is well-defined. To do so, we need to prove "i:' t 4' If rrk rrn , eij 1 th rrk rrn ; t 1 rae. i=l j=lSij , en i=1 j=1 ij . Proof Assume that k ni IT IT eij Sij = l. (3) i=l j=l First we observe that if Si has at least two elements for some i, then the exponent eij in above product is even for any j = 1"", ni. (For if eij is odd for some j, then eij eij -1 f ,'../.. " Z Z R l' eij b eij -1, (3) Sij = SijSij' or any J I J smce Sij = sij" ep acmg Sij y SijSij' m , we can express Sij as a product of elements in S \ {Sij}, contradicting the minimality of S.) Again since Srj = Srj' for any i,j,j', we have k ni k IT IT s:? = IT sii = 1, (4) i=l j=l i=l