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Let G act on X via A. Show that for x, y EX that belong to the same orbit, there is a bijection between
Let G act on X via A. Show that for x, y EX that belong to the same orbit, there is a bijection between Stabe(a) and Stabc(y). (Hint: Use that x, y being in the same orbit means that there is czy E G with A(Cry, z)= y. Conjugate the stabilizer of a with Cay and show that this is the stabilizer of y. So in fact, Staba(z) and Staba (y) are conjugate subgroups). Note that this implies the following: if |X] and |G| are finite, the stabi- lizers of elements of the same orbit in X all have the same size. We have 2 diamonds, 2 rubies and 2 sapphires that we want to put on a necklace. How many necklaces are possible? (Hint: If we cut open the necklace at some point, we get some sequence of 2 diamonds, 2 rubies, 2 sapphires. Let X be the set of all such sequences, and compute [X]. Then note that different elements of X can give the same necklace; we need to deal with overcount via the Burnside formula. Let De act on X by rotations and reflections; applying a rotation to a string corresponds to changing the place where we cut open the necklace. Reflections turn the necklace upside down while also possibly changing the place where it was cut open. We want X/De, since neither a rotation nor a reflection changes what a particular necklace looks like. Determine the sizes of the fixed point sets of the types of group elements individually of the identity, the of rotations, and the reflections.)
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Part 1 Let G act on X via A and let xy be two elements in X that belong to the same orbit This means that there exists a g in G such that gx y We want ...Get Instant Access to Expert-Tailored Solutions
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