By Gregory Karpilovsky
Enable N be an ordinary subgroup of a finite team G and allow F be a box. a major strategy for developing irreducible FG-modules involves the appliance (perhaps repeated) of 3 uncomplicated operations: (i) limit to FN. (ii) extension from FN. (iii) induction from FN. this can be the `Clifford idea' built by way of Clifford in 1937. some time past 20 years, the speculation has loved a interval of lively improvement. the principles were bolstered and reorganized from new issues of view, in particular from the point of view of graded jewelry and crossed items. the aim of this monograph is to tie jointly a number of threads of the advance so as to supply a finished photograph of the present kingdom of the topic. it really is assumed that the reader has had the identical of a customary first-year graduate algebra direction, i.e. familiarity with easy ring-theoretic, number-theoretic and group-theoretic innovations, and an realizing of straight forward houses of modules, tensor items and fields.
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V,. of nonisomorphic irreducible R-modules and V; E R e , for some primitive idempotent ei of R , 1 5 i 5 r . (ii) R R nlVl$. @nrVrfor some positive integers n;, 1 5 i 5 r , where n;K is a direct sum of n; copies of Vi. (iii) R E n;=,M n i ( D ; ) ,where D;= EndR(K)" E e;Re, is a divi- Preliminaries 38 sion ring, 1 5 i 5 r. (iv) The integers ni and r are unique and each Di is determined up to isomorphism. Proof. 2(i), RR is a finite direct sum of irreducible modules. 22, there exist idempotents e l , .
Following conditions are equivalent: T h e n the (i) V is a finite direct sum of irreducible modules. (ii) V is artinian and completely reducible. (iii) V is artinian and J ( V )= 0. Proof. 11. (ii)=+(iii). 11. Hence, by the definition of J ( V ) , J ( V ) = 0. 1, V is finitely cogenerated. ,V, = 0 for some maximal submodules V1,. . ,V, of V. 10(ii). 12. 2. Corollary. Let R be a ring. T h e n the following conditions are equivalent: Preliminaries 26 (i) RR is a finite direct s u m of irreducible modules.
20 for V = R e , we see that the map f H f ( e ) is an isomorphism of the additive group of EndR(Re) onto the additive group of eRe. Given f , g E EndR(Re), write f ( e ) = erle and g ( e ) = erze for some r1,rz E R . Then proving that f H f ( e ) reverses the multiplication. Because e is the identity element of the ring e R e , the above map preserves identity elements. This establishes the first isomorphism. 20 for V = e R . Lemma. Let R R = V1 @ . . ,n, and write 1 = el ... en with ei E V;.