math336_hw5_solns

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

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

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 ﬁgure 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). Lagrange’s 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. ±...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online