By Cohn P.M.

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We say that A is s e m i l o c a l for U, or r e l a t i v e if there exists a U-submodule A1 of A such that A is equal direct sum A = O c-lAl" r We then say that the U-module A1 is the local c o m p o n e n t . It is clear that A is uniquely determined by its local component, up to an isomorphism. 9. Let A1,A t E Mod(U). (i) Let fl : A1 "~ A~ be a U-isomorphism, and let A , A ' be G-modules, which are semilocal for U, with local components A1 and A~ respectively. Then there exists a unique G-isomorphism f : A ---* A' which extends f l .

Let 0 ---* A I ---* A --~ A " ---, 0 be a short exact sequence in Mod(G). H-I(A '') HI(A ') = H-I(A ') H~ ') H~ H~ Suppose given an exact hexagon of finite abelian groups as shown: H2 H6 \ / Hs 9 Ha H4 and let hi be the order of Hi, that is hi = (Hi : 0). Let fi : Hi ---* Hi+l with i mod 6 be the corresponding homomorphism in the diagram. Then hi = (Hi : f i - l H i - 1 ) ( f i - l H i - 1 : O) = m i m i - 1 . 1. Assume that each group Hi(A) is finite, and let h i ( A ) = order of Hi(G, A). 5 35 Now let A E Mod(G) be arbitrary.

12. Let G be finite with subgroup U. If A is semilocal for U with local component (U, A1) and if A1 is Uregular, then A is G-regular. Proof. If one can write 1A1 = S u ( f ) with some Z-morphism f, then 1A = ScU(~rSu(f)) = SGU(Su(~f)) = Sa(zrf), which proves that A is G-regular. From the present view point, we recover a result already found previously. 13. Let G be a -finite group, with subgroup U and B in Mod(U). If B is U-regular then M g ( B ) is G-regular. 14. Let U be a subgroup of finite index in a group G.

### Algebraic Numbers and Algebraic Functions by Cohn P.M.

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