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College Algebra Exam Review 234

# College Algebra Exam Review 234 - 9Š 4Š3Š2Š...

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244 5. ACTIONS OF GROUPS 5.1.19. Give the details of the proof of Example 5.1.20 . In particular, define an action of S n on the set of ordered sequences of k -elements from f 1; 2; : : : ; n g , and verify that it is indeed an action. Show that the action is transitive. Calculate the size of the stabilizer of a particular k -element sequence. 5.1.20. Show that the transitive subgroups of S 4 are S 4 A 4 , which is normal D 4 D h .1 2 3 4/; .1 2/.3 4/ i , and two conjugate subgroups V D f e; .12/.34/; .13/.24/; .14/.23/ g , which is normal Z 4 D h .1 2 3 4/; .1 2/.3 4/ i , and two conjugate subgroups 5.2. Group Actions—Counting Orbits How many different necklaces can we make from four red beads, three white beads and two yellow beads? Two arrangements of beads on a circu- lar wire must be counted as the same necklace if one can be obtained from the other by sliding the beads around the wire or turning the wire over. So what we actually need to count is orbits of the action of the dihedral group D 9 (symmetries of the nonagon) on the
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Unformatted text preview: 9Š 4Š3Š2Š arrangements of the beads. Let’s consider a simpler example that we can work out by inspection: Example 5.2.1. Consider necklaces made of two blue and two white beads. There are six arrangements of the beads at the vertices of the square, but only two orbits under the action of the dihedral group D 4 , namely, that with two blue beads adjacent and that with the two blue beads at opposite corners. One orbit contains four arrangements and the other two arrange-ments. We see from this example that the orbits will have different sizes, so we cannot expect the answer to the problem simply to be some divisor of the number of arrangements of beads. In order to count orbits for the action of a ﬁnite group G on a ﬁnite set X , consider the set F D f .g;x/ 2 G ² X W gx D x g . For g 2 G , let Fix .g/ D f x 2 X W gx D x g , and let 1 F .g;x/ D ( 1 if .g;x/ 2 F otherwise :...
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