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View Full DocumentNotes: c circlecopyrt F.P. Greenleaf 2003 - 2010 v43-f10products.tex, version 12/1/10 Algebra I: Chapter 6. The structure of groups. 6.1 Direct products of groups. We begin with a basic product construction. 6.1.1 Definition (External Direct Product). Given groups A 1 ,...,A n we define their external direct product to be the Cartesian product set G = A 1 ... A n equipped with component-by-component multiplication of n-tuples. If a = ( a 1 ,...,a n ) , b = ( b 1 ,...,b n ) in the Cartesian product set G , their product is (1) a b = ( a 1 ,...,a n ) ( b 1 ,...,b n ) = ( a 1 b 1 ,...,a n b n ) for all a i ,b i A i The identity element is e = ( e 1 ,...,e n ) where e i is the identity element in A i ; the inverse of an element is a 1 = ( a 1 1 ,...,a 1 n ) . There is a natural isomorphism between A i and the subgroup A i = ( e 1 ) ... A i ... ( e n ) , the n-tuples whose entries are trivial except for a i . From (1) it is clear that (a) Each A i is a subgroup in G . (b) The bijective map J i ( a i ) = ( e 1 ,...,a i ,...,e n ) defines an isomorphism from A i to A i . (c) The A i commute with each other in the sense that xy = yx if x A i , y A j and i negationslash = j . (d) Each A i is a normal subgroup in G . (e) The product set A 1 ... A n = { x 1 ...x n : x i A i , 1 i n } is all of G . Note carefully what (c) does not say: the subgroup A i need not commute with itself (the case when i = j ) unless the group A i happens to be abelian. The subsets H i = A 1 ... A i 1 ( e i ) A i +1 ... A n G are also normal subgroups, and in a group-theoretic sense the H i are complementary to the A i . We have the following properties. (2) (i) H i A i = ( e ) (ii) G = A 1 A 2 ... A n (product of subsets in G ) (iii) Each complement H i is a normal subgroup in G (iv) A i = G/ H i via the bijection f i : a i mapsto J i ( a i ) H i 6.1.2 Exercise. Verify the claims (a) (e) regarding the subgroups A i in a direct product G = A 1 ... A n . square 6.1.3 Exercise. Verify the relations (2) between the subgroups A i = A i and their complementary subgroups H i . square 6.1.4 Exercise. Verify that the map f i : A i G/ H i defined in (iii) above is actually a 1 bijection, and that it is a homomorphism from A i to the quotient group G/ H i , so that G/ H i = A i . square The order of entries in an n-tuple makes a difference; therefore the Cartesian product sets A 1 A 2 and A 2 A 1 are not the same thing (unless A 1 = A 2 ). For instance, are the direct product groups Z 3 Z 5 and Z 5 Z 3 the same ? What do elements in these groups look like? However, in dealing with groups we only care whether they are isomorphic. It happens that Z 3 Z 5 = Z 5 Z 3 even though these groups are not identical. 6.1.5 Exercise. Let A 1 ,A 2 ,... be groups. Prove that the following product groups are isomorphic.... View Full Document
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