Morais, Roberto J. F. de

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Czechoslovak Mathematical JournalRoberto J. F. de MoraisOn some commutativity theorems for finite rings and finite groupsCzechoslovak Mathematical Journal, Vol. 50 (2000), No. 2, 245–247Persistent URL: http://dml.cz/dmlcz/127566Terms of use:© Institute of Mathematics AS CR, 2000Institute of Mathematics of the Czech Academy of Sciences provides access to digitized documentsstrictly for personal use. Each copy of any part of this document must contain these Terms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.czCzechoslovak Mathematical Journal, 50 (125) (2000), 245–247ON SOME COMMUTATIVITY THEOREMS FOR FINITE RINGSAND FINITE GROUPSRoberto J. F. de Morais, Paraíba(Received October 25, 1996)In this note we investigate the relation between nilpotent elements in a ring andcommutative conditions in both finite rings and finite groups.Theorem 1. Let R be a finite ring without nilpotent elements. Then R iscommutative. . Since R is finite and has no nilpotent element then for any a ∈ R wehave(1) an(a) = am(a)where n and m are integers which depend on a. First, we will restrict ourselves tothe case in which R is a division ring. In fact our condition (1) then implies thatan(a)−m(a)+1 = a,that isak(a) = aand this condition for our division ring clearly leads to commutativity by Jacobson’sCommutativity Theorem.Since our ring has no nilpotent element, Jacobson’s radical is {0} and the ringis semi-simple. Hence we can apply Wedderburn-Artin decomposition theorem forsemi-simple artinian rings. Now suppose that we have a matrix ring n × n in thisdecomposition; the matrix 0 0 . . . 00 0 . . . 0. . . . . . . . . . . .1 0 . . . 0245is clearly nilpotent. Therefore no matrix rings over division rings can occur in thisdecomposition; that is, all components of R are division rings, which by the first partof our proof are commutative. Hence R is commutative. Theorem 2. Suppose R is a finite ring of characteristic p (a prime). The R iscommutative if, and only if, the group of invertible elements of R is commutative. . The theorem is an immediate consequence of the following one. De Morais Basis Theorem. Let A be a finite dimensional algebra with unitover a field F . Then A has a base of invertible elements.Now we pass to a different category of results, namely to theorems about com-mutativity of groups. These theorems, however, are based on ideas which dependessentially on the assumptions and statements of the previous part of our paper.Theorem 3. Let G be a finite group of order n. Then G is non-commutative if,and only if, for any finite field F such that the characteristic of F does not dividethe order of G, there exist m and a family {ki : ki ∈ F} not all zero such that ifa =∑kigi, then am = 0. . In order to prove sufficiency and necessity of our condition we willconsider the group algebra F (G). Under our hypothesis F (G) is semi-simple, henceif F (G) has nilpotent elements then in the decomposition of F (G) by Wedderburn’sStructure Theorem some matrix algebra will occur and F (G) is non-commutativeand the same holds for G. Necessity follows easily by proceeding in the oppositedirection. Indeed, if G and consequently F (G) are non-commutative this impliesthat some matrix algebra occurs in the decomposition of F (G) and hence we havenilpotent elements in F (G) as we have show in the proof of Theorem 1. Theorem 4. Let G be a finite group of order n. A necessary and sufficientcondition for G to be commutative is that given any field F with charF ×n, for anya with a =∑kigi, ki ∈ F we have(1) am(a) = al(a)with am(a) = 0, where m and 1 are integers which depend on a. . From (1) supposing m > 1 we have(2) al(1− am−l) = 0.246Now if a is nilpotent then 1− am−l is invertible and equation (2) is absurd. HenceF (G) has no nilpotent element and the commutativity of both F (G) and G followsfrom Theorem 1.To prove necessity, we remark that F (G) is semi-simple and the commutativityof G implies the commutativity of F (G), therefore in the decomposition of F (G) nomatrix algebra and hence no nilpotent element occurs. Now F (G) is finite and thusfor all a =∑kigi, ki ∈ F , we haveam(a) = al(a).References[1] Artin, Emil: Collected Papers. Springer Verlag, 1965.[2] De Morais, Roberto J. F.: On a Representation Theorem for Finite Dimensional Alge-bras.[3] Jacoboson, Nathan: Collected Mathematical Papers. Birkhäuser, 1989.[4] Wedderburn, J.H.M.: On hypercomplex numbers. Proc. London Math. Soc. 6 (1908).Author’s address: Av. D. Pedro II, no 2581, Apto 104, Tôrre, Joâo Pessoa, Paraíba,Brazil.247

On some commutativity theorems for finite rings and finite groups

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On some commutativity theorems for finite rings and finite groups

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Institute of Mathematics AS CR, v. v. i.