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is surjective, ReSy^g/ V(^(9P\ is an indecomposable ^-module fulfilling the same condition: all the simple factors of a Jordan-Holder sequence of /(X)ReSç',^' ( V0OP\ are isomorphic to /(X)Res^„^ (F); moreover, it is clear

that the restriction of Res^,^' lv(^OP]to&P through the structural homomor- phism P-^B"^ is a projective (9P-modulQ. Hence, by Lemma 7.3 below, Resç„^' I F0 б^Р j is a projective ^-module too and by Lemma 7.4 below, we have

^ ( B ) = {0} and rank^{B) = m^\P\ rank^(V) = rank^,(S)

( since B(ô)^/^(^S and therefore, m^ — rank^(F)). о Finally, by Corollary 6.9, any automorphism of S as interior P-algebra is an inner one.

7 . 3 . Lemma. With the notation above, let N be a B-module and assume that the restriction of N through the structural homomorphism P-^B* is a projective ßP-module. Then N is a projective B-module.

Proof Set M = Indp(A^) and A = Indf (^); clearly M is an ^-module and the restriction of M through the structural homomorphism G -> Л * is a projective (9 G- module. But it follows from [9], Th. 3.4 that there is a unique embedding c: ФОЬ-^А such that 3^{B) = Res?(c) о ê (cf. (7.2.1)). Consequently, c(b) • M is a projective (PGb-module (since it is a direct summand of M as (PG-module) and therefore, ReSg(c(è) • M) is a projective 5-module; but, as с о e is a representative for 3p{B), we have ReSg(c(è) • M) = Res^^ ДМ) ^ iV as J?-modules.

7 . 4 . Lemma. With the notation above, let N be a projective indecomposable B-module and assume that all the simple factors of a Jordan-Holder sequence of Â(^N are isomorphic. Then ^(B) = {3} and rank^(^) = w^rank^(7V).

Proof Let Ô be the point of В such that N^Bi where ieS; if у is a primitive idempotent of B, we have7Ä*4= {0} if and only if Bj^Bi as 5-modules (since jBi ^ Hom^ (Bj, Bi))\ hence, if/is an idempotent of ^ with multiphcity m^ on ô and zero everywhere else (cf. (2.2.1)), we have (1 ~f)Bf= {0}; but identifying В with its image through e, we get (1 —f)&Gf= (1 —f)Bf= {0} and therefore, we have also {0} =/(PG(l —/)=/5(l —/) since the Cartan matrix is symmetric (cf (2.15.1) and (2.15.2)); consequently,/belongs to ZB and as ZB a B^, we get/= 1 (cf. 2.20). So, ^{B) = {0} and rank^^) = m^rank^iV) (cf. (2.2.1)).

7 . 5 . Henceforth we prove the main theorem. Assume first that (1.6.2) holds. Then, as rankc,(^)/|P| = \Е^{Ру)\тоа{р) (cf. [12], Prop. 14.6), jç? does not divide rankg,(S) (cf. [12], 14.5) and the uniqueness of S follows now from Corollary 6.9. As P stabilizes by conjugation an (P-basis of (P G and 5 is a direct summand of 6^^ as OP-modulQ, P stabilizes by conjugation an й^-basis of В (cf. [12], (2.8.5)); but we have В ^ SP = @ Su and therefore S is still a direct summand of В as & P-module;

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