College Algebra Exam Review 231

College Algebra Exam Review 231 - of all x 2 X . For...

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5.1. GROUP ACTIONS ON SETS 241 (d) There are eight 3–cycles. The centralizer of the 3-cycle .1;2;3/ is the group generated by .1;2;3/ , and has size 3. Therefore, the number of 3–cycles is 24=3 D 8 . (e) There are six 4–cycles. The computation is similar to that for 3–cycles. Example 5.1.18. How large is the conjugacy class of elements with cycle structure 4 3 in S 12 ? The centralizer in S 12 of the element .1;2;3;4/.5;6;7;8/.9;10;11;12/ is the semidirect product of Z 4 ± Z 4 ± Z 4 and S 3 . Here Z 4 ± Z 4 ± Z 4 is generated by the commuting 4–cycles f .1;2;3;4/; .5;6;7;8/; .9;10;11;12/ g ; while S 3 acts by permutations of the set of three ordered 4-tuples f .1;2;3;4/;.5;6;7;8/;.9;10;11;12/ g : Thus the size of the centralizer is 4 3 ± 6 , and the size of the conjugacy class is 12Š 4 3 6 D 1247400 . The kernel of an action of a group G on a set X is the kernel of the corresponding homomorphism ' W G ! Sym .X/ ; that is, f g 2 G W gx D x for all x 2 X g : According to the general theory of homomorphisms, the kernel is a normal subgroup of G . The kernel is evidently the intersection of the stabilizers
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Unformatted text preview: of all x 2 X . For example, the kernel of the action of G on itself by conjugation is the center of G . Application: Counting Formulas. It is possible to obtain a number of wellknown counting formulas by means of the proposition and its corollary. Example 5.1.19. The number of k-element subsets of a set with n ele-ments is n k D n k.n k/ : Proof. Let X be the family of k element subsets of f 1;2;:::;n g . S n acts transitively on X by f a 1 ;a 2 ;:::;a k g D f .a 1 /;.a 2 /;:::;.a k / g . (Verify!) The stabilizer of x D f 1;2;:::;k g is S k S n k , the group of permutations that leaves invariant the sets f 1;2;:::;k g and f k C 1;:::;n g . Therefore, the number of k-element subsets is the size of the orbit of x , namely, j S n j j S k S n k j D n k.n k/ :...
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