College Algebra Exam Review 113

# College Algebra Exam Review 113 - p or 1 If a ¤ e then the...

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2.5. COSETS AND LAGRANGE’S THEOREM 123 than the zero subgroup f 0 g ) has the form n Z for some n , so every nonzero subgroup has ﬁnite index. The zero subgroup has inﬁnite index. It is also possible for a nontrivial subgroup of an inﬁnite group (i.e., a subgroup that is neither f e g nor the entire group) to have inﬁnite index. For example, the cosets of Z in R are in one to one correspondence with the elements of the half-open interval Œ0;1/ ; there are uncountably many cosets! See Exercise 2.5.11 . Corollary 2.5.8. Let p be a prime number and suppose G is a group of order p . Then (a) G has no subgroups other than G and f e g . (b) G is cyclic, and in fact, for any nonidentity element a 2 G , G D h a i . (c) Every homomorphism from G into another group is either trivial (i.e., every element of G is sent to the identity) or injective. Proof. The ﬁrst assertion follows immediately from Lagrange’s theorem, since the size of a subgroup can only be
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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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