Automorphism groups and isomorphisms of Cayley digraphs of Abelian groups

Let S be a minimal generating subset of the finite abelian group G. We prove that if the Sylow 2-subgroup of G is cyclic, then Sand S U S-l are CI-subsets and the corresponding Cayley digraph and graph are normal. Let G be a finite group and let S be 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 g E G,s E S}. Obviously we have the following basic facts. Proposition 1 Let X = Cay( G, S) be the Cayley digraph of G with respect to S. Then (1) Aut(X) contains the right regular representation R(G) of G. (2) X is connected if and only if G = (S). (3) X is undirected if and only if S-l = S. 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 So. = T. A Cayley digraph X = Cay(G, S) is called normal if R(G) <l A = Aut(X). Xu [1, Problem 6] asked the following Question (for part (1), see also [2, Problem 8]). Question 2 Let G be a finite group and let S be a minimal generating set of G. *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 16(1997). pp.183-187 (1) Are Sand S U S-l CI-subsets ? (2) Are the corresponding Cayley digraph and graph normal? For cyclic groups, Huang and Meng [3, 4, 5] proved Proposition 3 Let G be a finite cyclic group and let S be a minimal generating set of G. Let X = Cay(G, S) and X = Cay(G, SUS-I). Let 0" be an automorphism of G such that gCT = g-\ \/g E G, and let 2: = (0"). Then Aut(X) = R(G) and Aut(X) = R(G)L.. The answers to both parts (1) and (2) in Question 2 are positive. There is a obvious error in the second assertion of this Proposition. (Let G = Z12 ~ (a) and S = {a3 ,a }. It is easy to check IAut(G,S U S-l)1 = 4 where Aut(G, SUS-I) = {a E Aut(G) 1 sa = S}, so Aut(X) -=I R(G)2:.). However it is true that the answers to both questions (1) and (2) are still positive; we prove this in the Theorem below for a larger family of finite abelian groups than the cyclic groups. For abelian groups, Li [6] gave an example which shows that the answer to question (1) is negative in general. (This is also true for question (2); iflet G = Z4 X Z2 = (a) x (b) and S = {a, ab}, then both Cay(G, S) and Cay(G, SUS-I) are not normal.). However, if the group has odd order, then the answer to (1) is positive. Namely, he proved Proposition 4 (1) Let G = (a) x (x) x (e) ~ Z3 x Z4 X Z2 and let S = {x, xe, ax2} and T = {x, xe, ax2e}. Then S is a minimal generating subset of G and the Cayley digraph Cay(G, S) is isomorphic to Cay(G, T). However, there is no automorphism of G which maps S to T. In other words, S is not a CI-subset. (2) Every minimal generating subset of an abelian group of odd order is a CIsubset. Feng and Xu [7] proved that all generating subsets of an abelian group G with the minimum number of generators are CI, that is, the answer to question (1) for minimum generating sets of a finite abelian group is positive. Actually they proved Proposition 5 Let G be a finite abelian group and let both Sand T be minimal generating subsets of G of minimum size. Suppose that X = Cay(G, S) and Y = Cay(G, T) are isomorphic. Then there exists an a E Aut(G) such that sa = T. Let G be Li's example in Proposition 4. Then the Sylow 2-subgroup of G is not cyclic. The main result of this paper is the following Theorem: Theorem Let G be a finite abelian group such that the Sylow 2-subgroup of G is cyclic. Let S be a minimal generating subset of G. We have (1) Sand S U S-l are CI-subsets. (2) The corresponding Cayley digraph and graph are normal. As a consequence of this, every minimal generating subset of a cyclic group is CI and every minimal generating subset of an abelian group of odd order is CI. (These are Huang and Meng's and Li's results.) Proof of Theorem: To prove the theorem, first we need the following. Fact 1: Let Xl, X2 E S and Xl -=I X2· Then xi -=I x§.