math336_hw5_solns - H of an abelian group G , the cosets of...

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Partial Solutions to Homework 5 for MATH 336 Exercise (11A, E5). We need to show [ d ] generates Z /n Z (under addition) if and only if ( d,n ) = 1. Suppose [ d ] generates Z /n Z . Then, in particular, [1] = a [ d ] = [ ad ] for some (non-negative) integer a . But then ad 1 (mod n ), so ad + bn = 1 for some integer b , i.e. ( d,n ) = 1. Now suppose ( d,n ) = 1. We claim that the equivalence classes [0] , [ d ] , [2 d ] ,..., [( n - 1) d ] are all distinct in Z /n Z , i.e. [ d ] generates the entire group. Suppose not. Then ad bd (mod n ) for some a,b with 0 a,b n - 1. Then n | ( a - b ) d . But ( d,n ) = 1, so n | a - b , which is impossible. ± Exercise (11B, E1(i)). G = U 19 = [1] , [2] ,..., [18] (under multiplication mod 19). An im- portant fact, not stated in the text, is that in the case of any subgroup
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Unformatted text preview: H of an abelian group G , the cosets of H in G actually form a subgroup themselves. The operation is given by ( aH )( bH ) = ( ab ) H . Thus, the easiest way to gure out the cosets of the subgroup H generated by [7] in U 19 is to compute H, [2] H, [4] H, [8] H,. .. (powers of [2] H ) until one gets H back ( H is the identity in the coset group). Lagranges theorem tells us the number of distinct cosets. For H = < [7] > , the six ( | G | | H | = 18 3 = 6) cosets H, [2] H, [4] H, [8] H, [16 H ] , [32] H = [13] H are all distinct....
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