Abstract
For any commutative semigroup S and any positive integer m, the power function f:S→S defined by f(x)=x m is an endomorphism of S. In this paper we characterize finite cyclic semigroups as those finite commutative semigroups whose endomorphisms are power functions. We also prove that if S is a finite commutative semigroup with 1≠0, then every endomorphism of S preserving 1 and 0 is equal to a power function if and only if either S is a finite cyclic group with zero adjoined or S is a cyclic nilsemigroup with identity adjoined. Immediate consequences of the results are, on the one hand, a characterization of commutative rings whose multiplicative endomorphisms are power functions given by Greg Oman in the paper (Semigroup Forum, 86 (2013), 272–278), and on the other hand, a partial solution of Problem 1 posed by Oman in the same paper.
Highlights
We prove that for any finite commutative semigroup S with 1 = 0, every endomorphism of S preserving 1 and 0 is equal to a power function if and only if either S is a finite cyclic group with zero adjoined or S is a cyclic nilsemigroup with identity adjoined
In Theorem 2.2 below we extend the above result to finite commutative semigroups, characterizing finite cyclic semigroups as those finite commutative semigroups whose endomorphisms are power functions
As we have just proved in Theorem 2.2, every finite commutative semigroup S whose endomorphisms are power functions must be cyclic, that is, S is generated by a single element x subject to a defining relation of the form xi = xj with i = j ∈ N
Summary
He proved in [4, Theorem 1] that every endomorphism of the semigroup (R, ·) is a power function if and only if the ring R is a finite field. We prove that for any finite commutative semigroup S with 1 = 0, every endomorphism of S preserving 1 and 0 is equal to a power function if and only if either S is a finite cyclic group with zero adjoined or S is a cyclic nilsemigroup with identity adjoined (see Theorem 3.1).
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