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College Algebra Exam Review 231

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

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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
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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 well–known 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 ² D nŠ kŠ.n ² k/Š : 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 nŠ kŠ.n ² k/Š :...
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