Unformatted text preview: p or 1 . If a ¤ e , then the subgroup h a i is not f e g , so must be G . The last assertion also follows from the ﬁrst, since the kernel of a homomorphism is a subgroup. n Any two groups of prime order p are isomorphic, since each is cyclic. This generalizes (substantially) the results that we obtained before on the uniqueness of the groups of orders 2, 3, and 5. Corollary 2.5.9. Let G be any ﬁnite group, and let a 2 G . Then the order o.a/ divides the order of G . Proof. The order of a is the cardinality of the subgroup h a i . n The index of subgroups satisﬁes a multiplicativity property: Proposition 2.5.10. Suppose K ± H ± G are subgroups. Then ŒG W KŁ D ŒG W HŁŒH W KŁ: Proof. If the groups are ﬁnite, then by Lagrange’s theorem, ŒG W KŁ D j G j j K j D j G j j H j j H j j K j D ŒG W HŁŒH W KŁ:...
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 Fall '08
 EVERAGE
 Algebra, Sets, Normal subgroup, Subgroup, Cyclic group, HAI

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