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Unformatted text preview:  H K  = 1, and so H K = { 1 } . 2 . 58 . Prove that an innite group G contains innitely many subgroups. Every group is the union of its cyclic subgroups: in particular, G = [ a G h a i . Suppose that G has only nitely many subgroups: G = h a 1 i h a n i . If every h a i i is nite, then G is nite, a contradiction. Hence, at least one a i is innite. But the subgroups h a n i i , for n = 1 , 2 , 3 ,... are all distinct: if n < k , then a n / h a k i . Therefore, h a i i , and hence G , has innitely many subgroups, a contradiction. We conclude that G has innitely many subgroups....
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This note was uploaded on 11/15/2010 for the course MATH 417 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff
 Algebra

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