These homework problems are meant to expand your understanding of what goes on during class. Any you turn in will be graded and returned to you. Answers may or may not be posted on the web, depending on demand.

1. Recall that the center Z(G) of a group G is the set of elements that commute with every other element of G:
   

   Z(G) = { g in G | gh=hg for all h in G }.

   We found in class that the braid group B₃ can be written
   

   B3 = < s1, s2 | s1s2s1 = s2s1s2  >.

   (Here s_k is ``line k over line k+1'' and group multiplication ab is joining the top of braid b to the bottom of braid a.)
   1. Prove that g = s₁ s₂ s₁ = s₂s₁ s₂ (the 180^° twist) is not in the center Z(B₃). Hint: write down the pictures for the products s₁ g and gs₁. Are they the same?

   2. Prove that g² (the ``360^° twist'') is in the center Z(B₃).

2. We will consider moves (as defined in class) as elements of the cube group of transformations of the Rubik's cube. (We will see that this is a group whose order is more than 4.3 ×10¹⁹.) Elements of this group may have small order, as we'll see in this problem.
   1. What is the order of the element BULU^-1L^-1B^-1? Hint: This causes the permutations
      

      uf --> ul --> bu --> uf,        ufr --> rfu --> flu,        ulb --> rub --> bul.

      The first is a 3-cycle (an element of order 3) and the other two are both 6-cycles (elements of order 6). The total move can be thought of as the product of these three (disjoint) permutations on the cubies. What is the order of this product?

   Use the same idea to find the order of the moves...

   2. UR

   3. UR^-1

   4. UR²

   Notice that you're finding the smallest value of k so that, for example, (UR)^k is the identity. If you start with a solved cube, this should be easy enough (though tedious) to check.

3. How big is the cube group? We said in the previous problem that it had more than 4.3 ×10¹⁹ elements, but Anton wants to know how big a number this is. Here we go...
   1. Assume that you can make 10 turns per second. If you started at the birth of the universe, would you have been able to make 4.3×10¹⁹ turns by now? (Assume the age of the universe is 10 billion (10¹⁰) years and that there are 365.25 days per year.)

   2. A Canadian dime is roughly 1.22 mm thick (as compared to 1.35 mm for a US dime). Stack 4.3×10¹⁹ Canadian dimes. How high is this stack? Let's ask this question in a different way: the average distance from the earth to the sun is about 149,597,870 km (this is called 1 AU, or Astronomical Unit). How many AU is our pile of dimes?

   3. Continuing our previous thought: a Canadian dime has a mass of 1.75 g. Leda, the smallest of Jupiter's moons, has mass of 5.68×10¹⁵ kg. How many Ledas would it take to equal the mass of 4.3×10¹⁹ Canadian dimes?

   4. Finally, suppose we actually owned our pile of 4.3× 10¹⁹ Canadian dimes. Bill Gates, according to some web site, is currently worth roughly US$ 59.3 billion. Let us use this as our standard: 1 BG = US$ 5.93 ×10¹⁰. At an exchange rate of $1 Canadian is $0.646 US, how many BG's is our pile of dimes worth?

4. [A repeat assignment!] Participate in the contest! See the course web page. Deadline for entries: Monday, January 6th at midnight.