Unformatted text preview:  H ∩ K  = 1, and so H ∩ K = { 1 } . 2 . 58 . Prove that an inﬁnite group G contains inﬁnitely many subgroups. Every group is the union of its cyclic subgroups: in particular, G = [ a ∈ G h a i . Suppose that G has only ﬁnitely many subgroups: G = h a 1 i ∪ · · · ∪ h a n i . If every h a i i is ﬁnite, then G is ﬁnite, a contradiction. Hence, at least one a i is inﬁnite. 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 inﬁnitely many subgroups, a contradiction. We conclude that G has inﬁnitely many subgroups....
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 Spring '08
 Staff
 Algebra, Abelian group, Subgroup, Cyclic group, inﬁnitely many subgroups

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