College Algebra Exam Review 242

College Algebra Exam Review 242 - 1 through a p 1 can be...

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252 5. ACTIONS OF GROUPS Proof. We prove this by induction on n . If n D 1 , there is nothing to do. So suppose the result holds for all groups of order p n 0 , where n 0 < n . Let N be a proper normal subgroup of G , with the property that every subgroup of N is normal in G . The order of N is p s for some s , 1 ± s < n . Apply the induction hypothesis to N to obtain a sequence f e g D G 0 ² G 1 ² G 2 ² ³³³ ² G s D N with j G k j D p k . Apply the induction hypothesis again to N G D G=N to obtain a sequence of subgroups f e g D N G 0 ² N G 1 ² N G 2 ² ³³³ ² N G n ± s D N G with j N G k j D p k and N G k normal in N G . Then put G s C k D ± ± 1 N G k , for 1 ± k ± n ´ s , where ± W G ´! G=N is the quotient map. Then the sequence .G k / 0 ² k ² n has the desired properties. n We now use similar techniques to investigate the existence of sub- groups of order a power of a prime. The first result in this direction is Cauchy’s theorem: Theorem 5.4.6. (Cauchy’s theorem). Suppose the prime p divides the order of a group G . Then G has an element of order p . The proof given here, due to McKay, 1 is simpler and shorter than other known proofs. Proof. Let X be the set consisting of sequences .a 1 ;a 2 ;:::;a p / of ele- ments of G such that a 1 a 2 :::a p D e . Note that a
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Unformatted text preview: 1 through a p 1 can be chosen arbitrarily, and a p D .a 1 a 2 :::a p 1 / 1 . Thus the cardinality of X is j G j p 1 . Recall that if a;b 2 G and ab D e , then also ba D e . Hence if .a 1 ;a 2 ;:::;a p / 2 X , then .a p ;a 1 ;a 2 ;:::;a p 1 / 2 X as well. Hence, the cyclic group of order p acts on X by cyclic permutations of the sequences. Each element of X is either xed under the action of Z p , or it belongs to an orbit of size p . Thus j X j D n C kp , where n is the number of xed points and k is the number of orbits of size p . Note that n 1 , since .e;e;:::;e/ is a xed point of X . But p divides j X j kp D n , so X has a xed point .a;a;:::;a/ with a e . But then a has order p . n Theorem 5.4.7. (First Sylow theorem). Suppose p is a prime, and p n divides the order of a group G . Then G has a subgroup of order p n . 1 J. H. McKay, Amer. Math. Monthly 66 (1959), p. 119....
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