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College Algebra Exam Review 90

College Algebra Exam Review 90 - of cardinality q for each...

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100 2. BASIC THEORY OF GROUPS it follows that r D 0 . Thus n is divisible by d . It now follows from the previous lemma that j H j D n=d . n Corollary 2.2.25. Fix a natural number n 2 . (a) Any subgroup of Z n is cyclic. (b) Any subgroup of Z n has cardinality dividing n . Proof. Both parts are immediate from the proposition. n Corollary 2.2.26. Fix a natural number n 2 . (a) For any positive divisor q of n , there is a unique subgroup of Z n of cardinality q , namely h OEn=qŁ i . (b) For any two subgroups H and H 0 of Z n , we have H H 0 , j H j divides j H 0 j . Proof. If q is a positive divisor of n , then the cardinality of h OEn=qŁ i is q , by Lemma 2.2.23 . On the other hand, if H is a subgroup of cardinality q , then by Proposition 2.2.24 (b), n=q is the least of positive integers s such that s 2 H , and H D h OEn=qŁ i . So h OEn=qŁ i is the unique subgroup of Z n of cardinality q . Part (b) is left as an exercise for the reader. n Example 2.2.27. Determine all subgroups of Z 12 and all containments between these subgroups. We know that Z 12 has exactly one subgroup
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Unformatted text preview: of cardinality q for each positive divisor of 12. The sizes of subgroups are 1;2;3;4;6 , and 12 . The canonical generators of these subgroups are Œ0Ł;Œ6Ł;Œ4Ł;Œ3Ł;Œ2Ł , and Œ1Ł , respectively. The inclusion relations among the subgroups of Z 12 is pictured in Figure 2.2.2 on the next page . Corollary 2.2.28. Let b 2 Z , b ¤ . (a) The cyclic subgroup h ŒbŁ i of Z n generated by ŒbŁ is equal to the cyclic subgroup generated by ŒdŁ , where d D g : c : d :.b;n/ . (b) The order of ŒbŁ in Z n is n= g : c : d :.b;n/ . (c) In particular, h ŒbŁ i D Z n if, and only if, b is relatively prime to n . Proof. One characterization of d D g : c : d :.b;n/ is as the smallest positive integer in f ˇb C ±n W ˇ;± 2 Z g . But then d is also the smallest of positive...
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