On the Heyde Theorem for Finite Abelian Groups

On a characterization theorem on finite Abelian groups

“…On the other hand, since the subgroup L contains no elements of order 2 and b −1 ∈ Aut(L), we find that 2 m b −1 l 0 = 0 in a contradiction with (12). We have thus proved that the characteristic function g(h, l) may be represented in the form…”

“…Since X satisfies condition (α) of Proposition A, there exists a direct summand G in X isomorphic to one of the groups (Z(3 r )) 2 , (Z(3 r )) 3 , (Z(5 r )) 2 , (Z(5 r )) 3 , and Z(p r ), where p is a prime and p ≥ 7, so that the subgroup K, with X = G + K, satisfies also condition (α) of Proposition A and K does not satisfy condition (i). As proved in [12], there are independent random variables ξ j , j = 1, 2, 3, with values in G and distributions µ j , and there are automorphisms α, β ∈ Aut(G) satisfying (15) such that the conditional distribution of L 2 = ξ 1 + αξ 2 + βξ 3 given L 1 = ξ 1 + ξ 2 + ξ 3 is symmetric while the distributions µ j are not idempotent. The automorphisms α and β can be extended to automorphisms of the whole group X so that the extended automorphisms satisfy (15) as well.…”

“…Problem 1 was solved in [12] in the class of finite Abelian groups (see Theorem A below). In this article we solve Problem 2 in the class of finite Abelian groups as well as address some related questions.…”

“…To show this, it suffices to verify that there are automorphisms α, β ∈ Aut(K p ) satisfying (15) on each p-prime component K p . Since K satisfies condition (α) of Proposition A and fails to (i), such automorphisms exist (see [12]) for each p-prime component K p for p > 2. Construct these automorphisms for K 2 = X 2 .…”

“…In [12] the corresponding idempotent distributions µ j were considered under the condition µ 1 = µ 2 . It is also obvious that if σ(µ j ) ⊂ X (2) then the distributions µ 1 and µ 2 solve this problem.…”