College Algebra Exam Review 112

College Algebra Exam Review 112 - H in G Proof The distinct...

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122 2. BASIC THEORY OF GROUPS (b). Finally, (d) and (e) are equivalent by taking inverses, and a D bh 2 bH , b ± 1 a D h 2 H , so (a) and (d) are equivalent. n Proposition 2.5.4. Let H be a subgroup of a group G . (a) Let a and b be elements of G . Either aH D bH or aH \ bH D ; . (b) Each left coset aH is nonempty and the union of left cosets is G . Proof. If aH \ bH 6D ; , let c 2 aH \ bH . By the previous proposition cH D aH and cH D bH , so aH D bH . For each a 2 G , a 2 aH ; this implies both assertions of part (b). n Proposition 2.5.5. Let H be a subgroup of a group G and let a and b be elements of G . Then x 7! ba ± 1 x is a bijection between aH and bH . Proof. The map x 7! ba ± 1 x is a bijection of G (with inverse y 7! ab ± 1 y ). Its restriction to aH is a bijection of aH onto bH . n Theorem 2.5.6. (Lagrange’s theorem). Let G be a ﬁnite group and H a subgroup. Then the cardinality of H divides the cardinality of G , and the quotient j G j j H j is the number of left cosets of
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Unformatted text preview: H in G . Proof. The distinct left cosets of H are mutually disjoint by Proposition 2.5.4 and each has the same size (namely j H j D j eH j ) by Proposition 2.5.5 . Since the union of the left cosets is G , the cardinality of G is the cardinality of H times the number of distinct left cosets of H . n Deﬁnition 2.5.7. For a subgroup H of a group G , the index of H in G is the number of left cosets of H in G . The index is denoted ŒG W HŁ . Index also makes sense for inﬁnite groups. For example, take the larger group to be Z and the subgroup to be n Z . Then Œ Z W n Z Ł D n , because there are n cosets of n Z in Z . Every subgroup of Z (other...
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