# Permutation Groups. Tom Davis April 2, 2003

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Permutation Groups Tom Davis April 2, 2003 Abstract This paper describes permutations (rearrangements of objects): how to combine them, and how to construct complex permutations from simpler ones. We ll talk a bit about groups of permutations as well. Some interesting examples here are related to solving the Rubik s Cube puzzle. It may help have a Rubik s Cube with you as you read along (and a screwdriver to take it apart if you don t know how to solve it). 1 Permutations A permutation is a rearrangement of objects. Here we will only consider permutations of a finite number of objects, and since the object names don t really matter, we will often simply consider permutations of the numbers 1, 2, 3,..., n. When we work with Rubik s Cube, however, there are better names for the faces than integers see Section 3. Of course we ll learn about permutations first by looking at permutations of small numbers of items, but if you think of the 54 colored faces of the little cubelets ( cubies ) on Rubik s Cube, you can see that every time you twist a side of the cube, you are rearranging those little faces. There are plenty of other examples of permutations, many of which are extremely important and practical. For example, when you have a list of items to sort, either by hand or with a computer program, you are essentially faced with the problem of finding a permutation of the objects that will put them in order after the permutation. If we consider permutations of n objects, there are n! of them. To see this, first consider where object number 1 winds up. There are n possibilities for that. After the fate of object 1 is determined, there are only n 1 possible fates for object number 2. Thus there are n(n 1)(n 2) = n! permutations of a set of n objects. For example, if we consider all possible rearrangements of the set {1, 2, 3}, there are 3! = = 6 of them, listed in Table Table 1: Permutations of 3 objects A good way to think of permutations is this (using permutations of three objects as an example): Imagine that there are three boxes labeled 1, 2, and 3, and initially, each contains a ball labeled with the same number box 1 contains ball 1, and so on. A permutation is a rearrangement of the balls but in such a way that when you re done there is still only a single ball in each box. In the table above, the notation a b indicates that whatever was in box a moves to the box labeled b, so to apply permutation number 3 above means to take whatever ball is in box 1 and move it to box 3, to leave the contents of box 2 alone, and to take the ball from box 3 and put it into box 1. In other words, permutation number 3 above tells us to swap the contents of boxes 1 and 3. 1

2 The notation above is pretty clumsy. Here are a couple of other possibilities: 1.1 Two Row Notation Write the permutation like this: ( where the example above indicates that the contents of box 1 moves to box 4, box 2 is unchanged, the ball in box 3 moves to box 1, and the ball in box 4 moves to box 3. The advantage to this notation is that it is very easy to figure out where everything goes. The disadvantage is that it requires writing down each number twice. Since the top row can always be put in order, however, there is no real need to write it, so simply listing the second row is sufficient (assuming there is an obvious way to put the boxes in order). But there is another way that is often far more useful. ) 1.2 Cycle Notation Write the example above like this: (143) This indicates that the contents of box 1 moves to box 4, the contents of box 4 to box 3, and the contents of box 3 moves back into box 1. The system is called cycle notation since the contents of the boxes in parentheses move in a cycle: 1 to 4, 4 to 3, and 3 back to 1. Some permutations have more than one cycle. For example, the cycle notation for the permutation corresponding to: is ( (13)(24). There are two cycles. 1 moves to 3 and 3 moves back to 1. At the same time, 2 moves to 4, and 4 back to 2. In other words, the contents of boxes 1 and 3 are cycled, and at the same time, the contents of boxes 2 and 4 are cycled. In cycle notation, there cannot be any duplicate elements in the various cycles that make up the permutation, so something like (13)(12) is not a valid form. As we will see in the next section, something like (13)(12) can be reduced to a valid form in this particular case to (132). As a final example, consider this permutation of {1, 2, 3, 4, 5, 6, 7, 8}: (135)(2768). It moves the ball in box 1 to box 3, 3 to 5, and 5 back to 1. At the same time, it moves 2 to 7, 7 to 6, 6 to 8, and 8 back to 1. Notice that 4 is not involved, so it stays fixed. If you want to make it clear that 4 is a member of the set of items under consideration, but that in this particular permutation it is not moved, you can write: (135)(2768)(4). In fact, the special permutation that does not move anything is often written as: (1). Note also that the ordering doesn t matter as long as each item to be permuted appears only once, and that you can list a cycle beginning with any member of it. All of the following indicate exactly the same permutation: (135)(2768) (2768)(135) (7682)(135) (351)(6827) (8276)(513) (6827)(513) ) 2

3 Since there are so many possible ways to label a particular permutation, if there is no convention about the labeling, it may be difficult to tell if two permutations are the same or different as the example above illustrates. If it is easy to assign an order to the elements being permuted as in the example above, then it is best to choose the representation of a permutation as follows, where we assume they elements are 1, 2, 3,..., n: 1. Begin with the smallest element. If it is moved, list the permutation beginning with that smallest element as the first entry in the cycle. 2. After listing the complete cycle, check to see if there are other elements moved by the permutation that do not appear in the cycle. If so, choose the smallest remaining, and repeat the process until all moved elements are listed. In the example above, this canonical representation would be the first one listed: (135)(2768). In the rest of this document, we ll use the cycle notation and unless there is a good reason to do otherwise, we will list permutations using their canonical representations. 2 Combining Permutations Of course it s nice to have a method to write down a permutation, but things begin to get interesting when we combine them. If you twist one face of Rubik s Cube and then twist another one, each twist jumbles the faces, and the combination of two twists usually causes a jumbling that is more complicated than either of the two individual twists. Rather than begin with Rubik s Cube, let s begin by looking at permutations of just 3 objects. We listed them in Table 1, but there we used a very clumsy notation. Here are the six possible permutations of three items listed in the same order as in Table 1: (1), (12), (13), (23), (123), (132). What happens if we begin with ball 1 in box 1, ball 2 in box 2, and ball 3 in box 3, and then we apply (12) followed by (13)? A good way to think about this is to follow the contents of the boxes one at a time. For example, ball 1 begins in box 1, but after (12) it has moved to box 2. The second permutation, (13), does not move the contents of box 2, so after both permutations have been applied, ball 1 will have moved to box 2. So the final result will look like this: (12 where we re not sure what comes next. We don t know if 2 will go back to one and the cycle will close, or whether it will continue to another box. So since the fate of 2 is in question, let s see where it goes. The first permutation, (12) moves box 2 to box 1, and then (13) will move box 1 to box 3, so now we know the combination of permutations looks like this: (123 Since there are only three objects, we know that 3 will go back to 1 and close the cycle, but (especially when you re beginning), it s good to trace each ball, including ball 3 in this case. The first permutation, (12), does not move the contents of box 3, but the second, (13) moves it to box 1, so the combination of (12) followed by (13) is equivalent to the single permutation (123). Combining permutations as above is written just like a multiplication in algebra, and we can write our result as follows 1 : (12)(13) = (123). 1 In other places, sometimes this multiplication of permutations is written in the opposite order: (13)(12) = (123). There are good reasons to choose either ordering, but here we ll write them in the order they occur from left to right, so (12)(13) means that first (12) is applied, followed by (13). 3

4 Beware, however. This is not the same as multiplication that you re used to for real numbers. By doing the same analysis as above, convince yourself that: (13)(12) = (132) (123) = (12)(13). In other words, the order of multiplication makes a difference. If P 1 and P 2 are two different permutations, it may not be true that P 1 P 2 = P 2 P 1. Multiplication of permutations is not commutative. Test your understanding of multiplication of permutations by verifying all of the entries in the multiplication table for the permutations of three objects in Table 2. (1) (12) (13) (23) (123) (132) (1) (1) (12) (13) (23) (123) (132) (12) (12) (1) (132) (123) (23) (13) (13) (13) (123) (1) (132) (12) (23) (23) (23) (132) (123) (1) (13) (12) (123) (123) (13) (23) (12) (132) (1) (132) (132) (23) (12) (13) (1) (123) Table 2: Multiplication table of permutations Remember that the order of multiplication is important. In Table 2, if you are trying to look up the product of (12)(13), find the column labeled (12) and the row labeled (13). If you use the row labeled (12) and the column labeled (13) you will be looking up the product (13)(12) which may be different. As a final check on your understanding of multiplication of permutations, verify the following multiplications of permutations: (1342)(3645)(1623) = (126435) (12)(23)(34)(45) = (15432) (135)(32)(54321)(413)(13) = (1) Here are some general properties of multiplication of permutations. They hold for the sets of permutations of any number of elements, but you should check to see that they do hold in the particular case of the three-element permutations in Table 2. Closure: If P and Q are two permutations, then the products P Q and QP both make sense in that the result is also a permutation. This is obviously true since a rearrangement of a rearrangement is itself a rearrangement. Identity: The permutation (1) that leaves everything fixed is an identity under multiplication. If P is any permutation, then P (1) = (1)P = P. In other words, the permutation (1) behaves for permutation multiplication just like the number 1 behaves for multiplication of real numbers. Sometimes the identity is written as e. It is not hard to prove that the identity is unique. Inverses: Every permutation has an inverse that undoes the operation. In other words, if you apply a permutation to a set and then apply its inverse, the result is that the final order is unchanged from the original. If P is a permutation and P 1 is its inverse, we can write P P 1 = P 1 P = (1) = e. If you have a permutation written in cycle notation and you want to find its inverse, simply reverse all the cycles. For example, [(134)(256)] 1 = (652)(431). To see why this works, multiply: (134)(256)(652)(431). The result will be (1) 2. 2 Notice that we have reversed not only the contents of each cycle, but also the order of the cycles. For cycles in the canonical form, reversing the order of cycles is not important, but if the permutation is not in canonical form, reversing both orders will produce the inverse, no matter what. 4

5 Associativity: If P, Q, and R are any three permutations, then P [QR] = [P Q]R. In other words, if you have to multiply 3 or more permutations together, it doesn t matter how you group them to do the multiplications. We use braces [ and ] to indicate the grouping since we ve used parentheses to indicate the cycles of the permutations. For example, let s work out (1345)(243)(163) two different ways. First we ll multiply (1345) by (243) and then take that result and multiply it by (163). Then we ll do the multiplication beginning with the last two permutations (check these yourself): [(1345)(243)](163) = (1245)(163) = (124563) (1345)[(243)(163)] = (1345)(16324) = (124563) Not (necessarily) Commutativity: This is really just a reminder that the commutative law does not hold in general. If you swap the order of a product, the result may change, so P Q and QP are not necessarily the same. There are some cases, however, where things do commute. For example, if your permutations are two cycles that share no elements in common, the order in which they occur does not matter. So (123)(45) = (45)(123). This is obviously true since each cycle rearranges a different subset of elements, so their operations are completely independent and can be reversed in order with no effect on the final outcome. 2.1 Powers of Cycles Because the associative law holds, it makes sense to write something like P n where P is a permutation and n is a positive integer. P 4 = P P P P, and because the operation of permutation multiplication is associative, you get the same answer no matter how you choose to multiply them together. For example, let s compute P 3, where P = (134)(25). P 3 = P P P = (134)(25)(134)(25)(134)(25) = (25). In fact, it s easy to see how powers work on cycles. Let s look at P = (123456), for example. Here are the various powers of P : P 1 = (123456) P 2 = (135)(246) P 3 = (14)(25)(36) P 4 = (153)(264) P 5 = (165432) P 6 = (1) When raising a cycle to a power k, each elements steps forward by k steps, cycling back to the beginning, if necessary. It s just like modular (clock) arithmetic. Clearly if the cycle P is n items long, then P n = (1) = e. It s a great exercise to calculate P k for all powers of k, where P is a cycle whose length is a prime number. Try it with P = ( ) and calculate P 1, P 2, P 3, P 4, P 5, P 6, and P 7. If a permutation is written in proper cycle form where there is possibly more than one cycle, but there are no items that appear in more than one cycle, then taking powers of such a permutation is easy just raise the individual cycles to the power and combine the results. This is because individual cycles that do not share items do commute, so, for example, [(123)(45)] 3 = (123)(45)(123)(45)(123)(45), but the (123) and the (45) cycles commute, so the right hand side can be rearranged to be: [(123)(45)] 3 = (123)(123)(123)(45)(45)(45) = (123) 3 (45) 3. Clearly, if P is a cycle of length n, then P n = e because each application of the cycle moves all the elements in the cycle one step forward. For any permutation, we say that the order of the permutation is the smallest power of that 5

6 permutation that is the identity. Thus if P is a cycle of 17 elements, it will have order 17, since 17 applications of it will return every ball to its original box. If P is not a cycle, but is written in proper cycle form, then the order of P is the least common multiple of the cycle lengths. This is pretty obvious consider the permutation P = (12345)(678). If we consider that P n = (12345) n (678) n, then to make P n = e, we must have that both (12345) n = e and (678) n = e. The first will be true if n is a multiple of 5; the second if n is a multiple of 3. For both to be true, n must be a multiple of both 5 and 3, and the smallest number that is both is the least common multiple of the two: 15 in this case. 3 The Befuddler Notation From now we will use Rubik s Cube for some of our examples of permutations. For that reason, we need a reasonable notation to describe the moves that can be made. Here we are talking only about the standard cube, although much of what we do can easily be applied to other versions. The cube has six faces, each of a different color, but different cubes have different coloring patterns, so it is useful to have a notation that is independent of the particular coloring of a cube. Here is a good method to describe a general move. Imagine that you hold the cube in front of you looking directly at the center of one face, and with the top and bottom faces parallel to the ground. There are six faces the front and back, the up and down, and the left and right. Conveniently, the first letters of these words are all different: F, B, U, D, L, R. Rearrange them as BF UDLR, and it reminds you of the English word, befuddler, which is also appropriate for describing the general difficulty of the problems presented by the cube. There are six primitive moves that can be made any of the six faces can be turned 1/4 turn clockwise. Obviously, if you want to turn a face counter-clockwise, that s what you would do, but to keep the description mathematically simple using the minimum number of primitive operations, remember that a single twist counter-clockwise is the same as three clockwise twists. By clockwise is meant that if the face in question is grasped in the right hand, it is turned in the direction pointed to by the right thumb. We will use the befuddler letters as names for these primitive moves. Thus the move F means that the front face is turned 1/4 turn clockwise, et cetera. We can combine letters as well. F UB means first twist the front face clockwise, then twist the up face, then the back face. All twists are 1/4 turn clockwise. To turn the front face by 1/2 turn or 3/4 turn (3/4 turn clockwise = 1/4 turn counter-clockwise), use the notation F 2 or F 3. Note that F 4 = B 4 = U 4 = D 4 = L 4 = R 4 = e, so we could write F 3 as F 1 equally well. We will tend to use the F 1 form here. As a final example, F 2 U 3 T BD means to turn the front face a half turn, then twist the up face 1/4 turn counterclockwise, followed by a 1/4 clockwise twist of the top, then back, and then down faces. If you think of the entire cube as being composed of a bunch of smaller cubies, the befuddler notation gives a good method to name the individual cubies. The cubies in the corners are identified by the three faces they share. The cube on the up right front can be called URF, and so on. The edge cubies are identified by the two faces it lies on, so the one on the up and front faces would be called UF. But in order to distinguish between the cubie UF and the transformation that is a rotation about the up face followed by a rotation about the front face, we will put boxes around the cubie names: URF and UF for the cubes just mentioned. With this cubie notation, we can describe (using our permutation cycle notation) certain results that transformations may achieve. For example, ( LD F D RD ) refers to an operation that cycles the left-down, front-down, and right-down cubies. The left-down cubie moves into front-down postion, et cetera. The notation still isn t perfect. You ll find that when you solve Rubik s cube that sometimes a cubie will be in the right place in the cube, but rotated (if it s a corner cubie) or flipped (if it s an edge cubie). But we can describe it as follows. Suppose there is an operation that leaves everything fixed, but flips U B and U L in place. We can write this as: UB, UL BU, LU. 6

7 4 Groups and Subgroups A group is a system consisting of a set of objects and a binary operation that produces from any two objects another object that satisfies the conditions listed in Section 2 (having an identity, an inverse, and with an associative operation). The set of all possible permutations of a set of elements is a special group called the symmetric group. The symmetric group on n objects is the group consisting of all permutations of n elements, so it contains n! elements in other words, the symmetric groups get big pretty fast as n gets larger. In this paper we will denote the symmetric group on n elements by S n. Most practical applications use only a subset of the possible permutations. In Rubik s Cube, for example, although there are 54 little colored faces, it is clear that the ones in the corners will always be in some corner, the ones on the edges remain on the edges, and the ones in the centers of the faces remain centers of faces. Thus in the collection of permutations reachable from a solved cube, there are none that move, say, a corner to an edge. We will be interested in special subsets of groups that are themselves groups in other words, a non-empty subset of the permutations so that any product of permutations in the subset is another permutation in the subset. In our earlier example of S n (the symmetric group on three elements), there are the following subgroups (including the group that contains only the identity and the entire symmetric group): {(1)}, {(1), (12)}, {(1), (13)}, {(1), (23)}, {(1), (123), (132)}, {(1), (12), (13), (23), (123), (132)}. There aren t any others. If you try to construct some, you ll see what happens. As an example, suppose we try to make one that contains (12) and (123). It will have to contain (12) 2 = (1) and (123) 2 = (132). It will also have to contain (123)(12) = (23) and (132)(12) = (13). But now we ve shown that it must contain all the permutations in the symmetric group, so S 3 is the group generated by (12) and (123). If you are a beginner with Rubik s Cube and you want to practice with some operations that jumble the cube but do not jumble it into a nightmare, consider restricting yourself to a subgroup of all the allowable moves. Here are a couple of good examples: Only allow moves that consist of 180 turns of two opposite faces at the same time. Basically, there are only 3 moves: R 2 L 2, U 2 D 2, and F 2 B 2. These generate some nice patterns as well. This is a very simple subgroup. This one is more complicated, but still not too bad. It s basically the same as the one above, except that you re allowed to do single turns of the opposite faces, such as LR 1, UD 1, and F B 1. By repeating these moves you can, of course, get to any position in the subgroup above, but there are many more possibilities. 5 Even and Odd Permutations Begin with the following exercise: verify the following products of permutations: (12) = (12) (12)(13) = (123) (12)(13)(14) = (1234) (12)(13)(14)(15) = (12345) Although the expressions on the left are not in proper cycle notation, this does show that any cycle can be expressed as a product of 2-cycles, or exchanges. This example shows that a cycle of n objects can be written as a product of (n 1) 2-cycles. 7

8 In fact, there are clearly an infinite number of ways to express any permutation as a product of 2-cycles: (123) = (12)(13) = (12)(13)(12)(12) = (12)(13)(12)(12)(12)(12) = But it is true that if a permutation can be written as an even number of cycles, any representation will contain an even number of cycles. In the example above, (123) was expressed as 2, 4, 6, cycles. Similarly, if a permutation allows a representation as an odd number of cycles, all its 2-cycle representations will contain an odd number of 2-cycles. All permutations can be divided into these even and odd permutations. It requires proof, of course, that it is impossible to represent a permutation with both an even and an odd number of 2-cycles, and that will be shown in Section 5.1. The identity is an even permutation (zero 2-cycles), and clearly if you multiply any even permutation by another even permutation, you will get an even permutation. Thus the set of all permutations that are even form a subset of the full symmetric group. This is called the alternating group, and the alternating group on n objects is called A n. Table 3 is the multiplication table for the alternating group A 4. It is a great example of a group that is complicated, but not too complicated. See what subgroups of it you can find. (1) (123) (124) (134) (234) (132) (142) (143) (243) (12)(34) (13)(24) (14)(23) (1) (1) (123) (124) (134) (234) (132) (142) (143) (243) (12)(34) (13)(24) (14)(23) (123) (123) (132) (13)(24) (234) (12)(34) (1) (143) (14)(23) (124) (134) (243) (142) (124) (124) (14)(23) (142) (13)(24) (123) (134) (1) (243) (12)(34) (143) (132) (234) (134) (134) (124) (12)(34) (143) (13)(24) (14)(23) (234) (1) (132) (123) (142) (243) (234) (234) (13)(24) (134) (14)(23) (243) (142) (12)(34) (123) (1) (132) (143) (124) (132) (132) (1) (243) (12)(34) (134) (123) (14)(23) (142) (13)(24) (234) (124) (143) (142) (142) (234) (1) (132) (14)(23) (13)(24) (124) (12)(34) (143) (243) (134) (123) (143) (143) (12)(34) (123) (1) (142) (243) (13)(24) (134) (14)(23) (124) (234) (132) (243) (243) (143) (14)(23) (124) (1) (12)(34) (132) (13)(24) (234) (142) (123) (134) (12)(34) (12)(34) (243) (234) (142) (124) (143) (134) (132) (123) (1) (14)(23) (13)(24) (13)(24) (13)(24) (142) (143) (243) (132) (234) (123) (124) (134) (14)(23) (1) (12)(34) (14)(23) (14)(23) (134) (132) (123) (143) (124) (243) (234) (142) (13)(24) (12)(34) (1) Table 3: The Alternating Group A Parity Preservation Earlier in this section we showed that it is possible to represent any permutation as a product of 2-cycles, and we stated without proof that any such representation of a given permutation will always contain an even number of cycles or it will always contain an odd number of cycles. We will prove that fact by showing that it is possible to assign a parity (even or odd) to any permutation in a unique way and that multiplication of an even permutation by a 2-cycle produces an odd permutation and vice-versa. When written in canonical form, a permutation will have a certain number c 1 of 1-cycles, c 2 of 2-cycles, c 3 of 3-cycles and so on. We define the parity of such a permutation as: c n (n + 1) mod 2. (1) n This makes sense from the observation that (123 n) = (12)(13) (1n) so every cycle of length n can be written as a product of n 1 different 2-cycles. What we will show is that if this permutation is multiplied by another 2-cycle that the parity of the sum in equation 1 switches. Assume that every element is listed in the canonical permutation representation including those elements that do not move which are listed as 1-cycles. If this permutation is multiplied by the 2-cycle (km) then there are two possibilities: either the two elements k and m are in the same cycle or they appear in different cycles (where the cycles in which they appear may have length 1). 8

9 First consider the case where they lie in the same cycle. That cycle will look like this: (kk 2 k 3 k i mm 2 m j ). This cycle contains i + j elements and will thus add i + j + 1 to the sum in equation 1. If this cycle is multiplied on the right by (km), we obtain: (kk 2 k i )(mm 2 m j ). The two resulting cycles of lengths i and j will add i j + 1 to the sum, so the parity will switch. Finally, if k and m appear in different cycles: (kk 2 k 3 k i )(mm 2 m j ), then if this is multiplied on the right by (km) we obtain: (kk 2 k i mm 2 m j ). In this case, the original cycle form contributed i j + 1 to the sum in equation 1 and after multiplication the single cycle result contributes i + j + 1, so again, the parity changes. 5.2 The 4 4 Sliding Block Puzzle You have probably seen the sliding block puzzle with 4 4 spaces and 15 blocks numbered 1 through 15, and the object is to try to slide them until they are in order. If you begin will all of them in order except that 14 and 15 are reversed, there is no solution. In other words, there is no way to convert the situation shown below on the left to the solved condition on the right This can be proved by showing that the sliding operation is like a permutation group, and that the swapping of two blocks amounts to an odd permutation in that group, but the operation of sliding a block is an even permutation. No matter how many even permutations you put together, it will never be odd. A B C D E F G H I J K L M N O A B C D E F G K H I J O L M N A B C D E I F G H J K O L M N The basic idea is illustrated in the diagrams above. Suppose that initially the situation is as shown in the left-most diagram where the variables A through N represent the numbers 1 through 15 in some order. For any such situation where the open square is not in the lower-right corner, we agree to convert it to that form first by moving all possible blocks to the left and then moving all possible blocks up. The result of this when applied to the left diagram is the diagram in the center. After these moves are made, the situation is a pure permutation of the numbers 1 through 15 which has either even or odd parity. On the left, there are four possible moves. If E or F is moved, then the situation is unchanged after moving the blank square to the lower-right as described above. But if block I is moved up then after the movement of the blank square to the lower-right corner is shown in the diagram on the right. It is easy to check that this is obtained from the one in the middle by the permutation (F IJKG) an even permutation. Thus in both cases, the result of a movement of a single block results in an even permutation of the blocks. Obviously a few other cases need to be considered, but the results will be similar. Hence, it is impossible to reverse just one pair of blocks such as swapping the 14 and 15 pair. 9

10 6 Generators Suppose you pick some complete symmetry group and choose some number of permutations from it. Then you construct the smallest subgroup that contains all of them. This is the subgroup generated by the initial set of permutations you chose. Using again S 3 as our example, what is the subgroup generated by {(123)}? Well, it has to contain (123) itself, (123) 2 = (132), (123) 3 = (1) and nothing else since higher powers of (123) start repeating: (123) 4 = (123) 3 (123) = (1)(123) = (123). In general, if n >= 3, (123) n = (123) n 3. We know that {(1), (123), (132)} is a subgroup of S 3 so it is the subgroup generated by (123). Using Rubik s Cube as an example, if you ve played with it, you know that there are billions (acutally, there are a lot more!) of permutations in the Rubik s Cube group, but if we consider as a generator a 1/4-twist of the front (in other words, the F move), you can see that if that is the only operation you re allowed to do, there are only four possible rearrangments of the cube you can achieve. So that particular generator will generate a subgroup of size 4. If you have a cube handy, here s an exercise. Begin with a solved cube. You are only allowed to make two sorts of move: F 2 and R 2, in other words, only 180 rotations about the front and right faces are allowed. These moves certainly are permutations of the cube s faces. Show that the order of the subgroup they generate is 6. In other words, show that you can only get the cube into 6 different patterns (including the solved pattern) if F 2 and R 2 are the only allowed moves. If we have a finite group (and that s the only sort we consider in this paper), every element has some finite order. The proof is easy: Let P be some permutation, and consider P 1, P 2, P 3,.... eventually, since there are only a finite number of elements in the group, there has to be some i and j such that P i = P j. Assume j > i. Then P j i P i = P j = P i. Thus P j i = e, the identity. We can also see that it s true since every permutation expressed in proper cycle form will have an order equal to the least common multiple of the lengths of its cycles as we stated previously. The advantage of the proof in the previous paragraph is that it applies to all finite groups not just permutation groups. On the other hand, it is sometimes quite difficult to guess what the order of an element of the permutation group might be. For example, consider the following permutation of Rubik s Cube: F R. Suppose that it is one indivisible move. it is obvious that both the F and R moves by themselves have order 4 4 such turns return the cube to the original condition. But if you consider the pair of turns to be a single move, what is the order of that? The answer turns out to be 105 not an obvious result! My initial (and very painful) solution to the cube was based on the above concept. I knew that any operation, if repeated enough times, would return an initially solved cube back to the solved condition. But by experimentation, I found that if my operation required, say, 24 repeats to get from solved to solved, very often the condition after 12 or 8 moves (these are divisors of 24) would leave most of the cubies fixed. It is pretty obvious why this works. Imagine a permutation with a cycle structure like this: P = (123)(4567)(89). We know that P 12 = e, but what does P 6 or P 4 look like? Work it out: P 6 = (46)(57), P 4 = (123). Do you see what s going on? Here is an interesting exercise. Show that if you have a subgroup of S n that contains (12) and (123 n), then the subgroup is the entire group S n. In other words, any possible permutation of n objects can be expressed as some product of a 2-cycle and an n-cycle. (Hint: If P = (123 n), consider the permutations P (12)P 1, P 2 (12)P 2,..., P n (12)P n.) 7 Conjugates If P and Q are any permutations, then the permutation P QP 1 is called a conjugate of P. In group theory, a conjugate operation is very much like a change in coordinate system. 10

11 Here s a concrete example from Rubik s Cube. Suppose that you know how to swap two corner pieces that are on the same edge (see Section 10.1; let s call this operation Q), but you re faced with a cube where the corners you would like to swap are not on the same edge. No problem find a simple operation (call it P ) that brings the two corners of interest to be on the same edge. If you perform P, then Q, and then undo P (in other words, perform P 1, the net effect will be to move the corners to the same edge, swap the corners on that edge, and then move the corners back to where they began. Doing P, then Q, then P 1 is the same as doing P QP 1 a conjugate of Q. 8 Commutators If P and Q are any permutations, then the commutator of P and Q is P QP 1 Q 1. It s just a conjugate with one additional operation of Q 1 tagged onto the end. Here s an example of a commutator in action. Suppose that you want to find an operation that flips two edge cubies on the same face in place without affecting any of the other cubies. It s not hard to find a series of moves that leaves one face completely fixed except for flipping a single cubie on it but perhaps hopelessly jumbles the rest of the cube. Call the operation that does this P. Now let Q be a single twist of that face that puts another cubie in the same slot where the flipped cubie was. What does P QP 1 Q 1 do? P flips the cubie (but trashes the rest of the cube that s not on the face). Q moves a different cubie to that slot. P 1 then undoes all of the damage caused by P on the rest of the cube, but flips the new cubie. Q 1 just rotates the face in question back to its original condition. The operation in Section 10.4 is just such a commutator. 9 Representations of Arbitrary Groups In Section 4 we stated that groups are usually defined without reference to permutations. The mathematical definition of a group is simply a set of elements and a binary operation on that set that has an identity, inverses of each element, and is associative. In fact, the multiplication table listed below represents a valid binary operation on the elements E, A, B, and C where E is the identity: E A B C E E A B C A A E C B B B C E A C C B A E It is easy to prove, however, that for any such group it is possible to define a permutation group that behaves in exactly the same way. Here is how to do it for the group above: E (e)(a)(b)(c), A (ea)(bc), B (eb)(ac) and C (ec)(ab). Do you see how this is related to the group multiplication table? Check to see that the operations work the same. For example, show that if AB = C, then the product of the permutations corresponding to A and B yield the permutation that corresponds to C, et cetera. Thus, in a sense, all groups can be considered to be permutation groups. 10 Interesting Rubik Permutations (Spoiler!) This section contains enough information for you to solve your cube without much thinking. Don t look at it if you like to solve puzzles by yourself. Here are some operations that may prove to be useful in solving the cube puzzle. The first three move the cubies as indicated, but some of them also may flip the edges or rotate the corners. The final two operations leave all the cubies in place, but flip edge cubies or rotate corner cubies as indicated. 11

12 There are almost certainly better methods available these are just the ones I found myself. For the initial stages of solving a cube, they are also too powerful. If you re just trying to get the top face correct from a completely jumbled cube, you don t really care what you do to the other cubies, but all the examples below are very restrictive only the indicated cubies move; the others are left fixed by the operations. Warning: If you re starting with a pure cube, be careful to follow the instructions below exactly one error and your cube will be trashed. Take it slowly and remember that B is back, not bottom. Also remember that to undo an operation, you can reverse the steps starting from the back. For example, to reverse F L 1 D 2 RUR 1, perform RU 1 R 1 D 2 LF 1. Also, be sure to keep the top cube on top and the right cube on the right as you do these operations. For example, if the top cube is white when you begin, make sure it stays white through all the operations. Another way to avoid problems when you are a beginner is always to twist the faces with your right hand. Then if it is one of F, B, L, R, U, or D, you will twist in the direction of your thumb. If the operation is among F 1, B 1, L 1, R 1, U 1, or D 1, you ll twist away from your thumb. If you re left-handed, think different. Finally, it is much easier to study movements if you can begin with a solved cube, but it s pretty easy to make an error and to wind up with a cube that s totally jumbled. If you have no idea how to solve it, this situation can be pretty depressing. But there is a way to cheat just take the cube apart, and put it back together in the solved configuration. To take the cube apart, the easiest way is to take a screwdriver and to put it between the center cubie of a face and one of the edge cubies, and then to pry out the edge cubie. Once it s out, it is easy to remove all the rest of the cubies, leaving a central skeleton. From the skeleton, put the cubies back one at a time into their correct positions Swap Two Corners This operation swaps two corners, but also jumbles some edge cubies. Use it if you re planning to get the corners in place first, and then work on the edges. In addition to the permutation specified, it also twists LF D. ( RBD LBD )( RB F D BD LD ): RD 1 R 1 D 1 B 1 DB Cycle Three Edge Cubies ( LD F D RD ): R 1 LBRL 1 D 2 R 1 LBRL 1. ( UF UB DB ): R 1 LU 2 RL 1 B Cycle Three Corner Cubies ( LUF RUB LUB ): URU 1 L 1 UR 1 U 1 L Flip UB and UL In Place UB, UL BU, LU : R 1 LB 2 RL 1 D 1 R 1 LBRL 1 ULR 1 B 1 L 1 RDLR 1 B 2 L 1 RU Rotate U RB and U RF In Place URB, URF RBU, RF U : F D 2 F 1 R 1 D 2 RUR 1 D 2 RF D 2 F 1 U 1. 12

### Permutation Groups. Rubik s Cube

Permutation Groups and Rubik s Cube Tom Davis tomrdavis@earthlink.net May 6, 2000 Abstract In this paper we ll discuss permutations (rearrangements of objects), how to combine them, and how to construct

### Group Theory via Rubik s Cube

Group Theory via Rubik s Cube Tom Davis tomrdavis@earthlink.net http://www.geometer.org ROUGH DRAFT!!! December 6, 2006 Abstract A group is a mathematical object of great importance, but the usual study

### A Solution of Rubik s Cube

A Solution of Rubik s Cube Peter M. Garfield January 8, 2003 The basic outline of our solution is as follows. (This is adapted from the solution that I was taught over 20 years ago, and I do not claim

### An Interesting Way to Combine Numbers

An Interesting Way to Combine Numbers Joshua Zucker and Tom Davis November 28, 2007 Abstract This exercise can be used for middle school students and older. The original problem seems almost impossibly

### Kenken For Teachers. Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles June 27, 2010. Abstract

Kenken For Teachers Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles June 7, 00 Abstract Kenken is a puzzle whose solution requires a combination of logic and simple arithmetic skills.

### Comments on the Rubik s Cube Puzzle

Comments on the Rubik s Cube Puzzle Dan Frohardt February 15, 2013 Introduction The Rubik s Cube is a mechanical puzzle that became a very popular toy in the early 1980s, acquiring fad status. Unlike some

### Extended Essay Mathematics: Finding the total number of legal permutations of the Rubik s Cube

Extended Essay Mathematics: Finding the total number of legal permutations of the Rubik s Cube Rafid Hoda Term: May 2010 Session number: 000504 008 Trondheim Katedralskole Word Count: 3635 Abstract In

### CHAPTER 3 Numbers and Numeral Systems

CHAPTER 3 Numbers and Numeral Systems Numbers play an important role in almost all areas of mathematics, not least in calculus. Virtually all calculus books contain a thorough description of the natural,

### Fractions and Decimals

Fractions and Decimals Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles December 1, 2005 1 Introduction If you divide 1 by 81, you will find that 1/81 =.012345679012345679... The first

### S on n elements. A good way to think about permutations is the following. Consider the A = 1,2,3, 4 whose elements we permute with the P =

Section 6. 1 Section 6. Groups of Permutations: : The Symmetric Group Purpose of Section: To introduce the idea of a permutation and show how the set of all permutations of a set of n elements, equipped

### Group Theory in Rubik's Cube. Pritish Kamath special thanks to Nishant Totla

Group Theory in Rubik's Cube Pritish Kamath special thanks to Nishant Totla Overview Notations Counting Rubik's cube configurations... Quick review of Group Theory. The Rubik's Cube Group. More Group Theory

### Introduction to Diophantine Equations

Introduction to Diophantine Equations Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles September, 2006 Abstract In this article we will only touch on a few tiny parts of the field

### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

### ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

### Graph Theory Problems and Solutions

raph Theory Problems and Solutions Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles November, 005 Problems. Prove that the sum of the degrees of the vertices of any finite graph is

### Pigeonhole Principle Solutions

Pigeonhole Principle Solutions 1. Show that if we take n + 1 numbers from the set {1, 2,..., 2n}, then some pair of numbers will have no factors in common. Solution: Note that consecutive numbers (such

### Course notes on Number Theory

Course notes on Number Theory In Number Theory, we make the decision to work entirely with whole numbers. There are many reasons for this besides just mathematical interest, not the least of which is that

### Introduction to Matrices

Introduction to Matrices Tom Davis tomrdavis@earthlinknet 1 Definitions A matrix (plural: matrices) is simply a rectangular array of things For now, we ll assume the things are numbers, but as you go on

### Group Theory and the Rubik s Cube. Janet Chen

Group Theory and the Rubik s Cube Janet Chen A Note to the Reader These notes are based on a 2-week course that I taught for high school students at the Texas State Honors Summer Math Camp. All of the

### THE 15-PUZZLE (AND RUBIK S CUBE)

THE 15-PUZZLE (AND RUBIK S CUBE) KEITH CONRAD 1. Introduction This handout is on a fun topic: permutation puzzles. There are a couple of famous puzzles which are related to permutations and their signs.

### Induction Problems. Tom Davis November 7, 2005

Induction Problems Tom Davis tomrdavis@earthlin.net http://www.geometer.org/mathcircles November 7, 2005 All of the following problems should be proved by mathematical induction. The problems are not necessarily

### Homework 5 Solutions

Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

### Congruences. Robert Friedman

Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition

### Mathematics of Cryptography

CHAPTER 2 Mathematics of Cryptography Part I: Modular Arithmetic, Congruence, and Matrices Objectives This chapter is intended to prepare the reader for the next few chapters in cryptography. The chapter

### The Mathematics of the Rubik s Cube

Introduction to Group Theory and Permutation Puzzles March 17, 2009 Introduction Almost everyone has tried to solve a Rubik s cube. The first attempt often ends in vain with only a jumbled mess of colored

### Induction. Margaret M. Fleck. 10 October These notes cover mathematical induction and recursive definition

Induction Margaret M. Fleck 10 October 011 These notes cover mathematical induction and recursive definition 1 Introduction to induction At the start of the term, we saw the following formula for computing

### Algebra of the 2x2x2 Rubik s Cube

Algebra of the 2x2x2 Rubik s Cube Under the direction of Dr. John S. Caughman William Brad Benjamin. Introduction As children, many of us spent countless hours playing with Rubiks Cube. At the time it

### A Little Set Theory (Never Hurt Anybody)

A Little Set Theory (Never Hurt Anybody) Matthew Saltzman Department of Mathematical Sciences Clemson University Draft: August 21, 2013 1 Introduction The fundamental ideas of set theory and the algebra

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

### Introduction to the Mathematics of Zome

Introduction to the Mathematics of Zome Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles July 1, 008 Abstract This article provides a gentle introduction to the Mathematics of Zome,

### Math 016. Materials With Exercises

Math 06 Materials With Exercises June 00, nd version TABLE OF CONTENTS Lesson Natural numbers; Operations on natural numbers: Multiplication by powers of 0; Opposite operations; Commutative Property of

### If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?

Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question

### A Mathematical Introduction to the Rubik s Cube

Sept 29th, 2015 Outline Initial Observations Some Notation Socks and Shoes The Cross First 2 Layer - F2L Last Layer Commutators Conjugation 6 faces, 12 edges, 8 corners 54 stickers Scrambling faces is

### 13 Solutions for Section 6

13 Solutions for Section 6 Exercise 6.2 Draw up the group table for S 3. List, giving each as a product of disjoint cycles, all the permutations in S 4. Determine the order of each element of S 4. Solution

### 3. BINARY NUMBERS AND ARITHMETIC

3. BINARY NUMBERS AND ARITHMETIC 3.1. Binary Numbers The only place in a computer where you will find the number 9 is on the keyboard. Inside the computer you will only find 0 s and 1 s. All computer memory

### AN ANALYSIS OF THE 15-PUZZLE

AN ANALYSIS OF THE 15-PUZZLE ANDREW CHAPPLE, ALFONSO CROEZE, MHEL LAZO, AND HUNTER MERRILL Abstract. The 15-puzzle has been an object of great mathematical interest since its invention in the 1860s. The

### Ryan s Guide to Solving the 2x2 By Ryan Goessl

From This... To This! Ryan s Guide to Solving the 2x2 By Ryan Goessl Before you start learning to solve, there are a few things you should have, and a few things you should know: First of all, you ll need

### 1 Symmetries of regular polyhedra

1230, notes 5 1 Symmetries of regular polyhedra Symmetry groups Recall: Group axioms: Suppose that (G, ) is a group and a, b, c are elements of G. Then (i) a b G (ii) (a b) c = a (b c) (iii) There is an

### Solution to Homework 2

Solution to Homework 2 Olena Bormashenko September 23, 2011 Section 1.4: 1(a)(b)(i)(k), 4, 5, 14; Section 1.5: 1(a)(b)(c)(d)(e)(n), 2(a)(c), 13, 16, 17, 18, 27 Section 1.4 1. Compute the following, if

### Steps to Proving a Theorem

Steps to Proving a Theorem Step 1: Understand Goal What am I looking for? What should the last sentence in my proof say? What are some equivalent ways to state what I want? Are there any complicated definitions

### MAT2400 Analysis I. A brief introduction to proofs, sets, and functions

MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take

### 1.16 Factors, Multiples, Prime Numbers and Divisibility

1.16 Factors, Multiples, Prime Numbers and Divisibility Factor an integer that goes into another integer exactly without any remainder. Need to be able to find them all for a particular integer it s usually

### 8 Primes and Modular Arithmetic

8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.

### Pick s Theorem. Tom Davis Oct 27, 2003

Part I Examples Pick s Theorem Tom Davis tomrdavis@earthlink.net Oct 27, 2003 Pick s Theorem provides a method to calculate the area of simple polygons whose vertices lie on lattice points points with

### List of Standard Counting Problems

List of Standard Counting Problems Advanced Problem Solving This list is stolen from Daniel Marcus Combinatorics a Problem Oriented Approach (MAA, 998). This is not a mathematical standard list, just a

### Notes on polyhedra and 3-dimensional geometry

Notes on polyhedra and 3-dimensional geometry Judith Roitman / Jeremy Martin April 23, 2013 1 Polyhedra Three-dimensional geometry is a very rich field; this is just a little taste of it. Our main protagonist

### MATH10212 Linear Algebra. Systems of Linear Equations. Definition. An n-dimensional vector is a row or a column of n numbers (or letters): a 1.

MATH10212 Linear Algebra Textbook: D. Poole, Linear Algebra: A Modern Introduction. Thompson, 2006. ISBN 0-534-40596-7. Systems of Linear Equations Definition. An n-dimensional vector is a row or a column

### Chapter 11 Number Theory

Chapter 11 Number Theory Number theory is one of the oldest branches of mathematics. For many years people who studied number theory delighted in its pure nature because there were few practical applications

### MATHEMATICAL INDUCTION

MATHEMATICAL INDUCTION MATH 5A SECTION HANDOUT BY GERARDO CON DIAZ Imagine a bunch of dominoes on a table. They are set up in a straight line, and you are about to push the first piece to set off the chain

### Introduction to mathematical arguments

Introduction to mathematical arguments (background handout for courses requiring proofs) by Michael Hutchings A mathematical proof is an argument which convinces other people that something is true. Math

### Students in their first advanced mathematics classes are often surprised

CHAPTER 8 Proofs Involving Sets Students in their first advanced mathematics classes are often surprised by the extensive role that sets play and by the fact that most of the proofs they encounter are

### Session 7 Fractions and Decimals

Key Terms in This Session Session 7 Fractions and Decimals Previously Introduced prime number rational numbers New in This Session period repeating decimal terminating decimal Introduction In this session,

### Chapter 7: Products and quotients

Chapter 7: Products and quotients Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 7: Products

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### L1-2. Special Matrix Operations: Permutations, Transpose, Inverse, Augmentation 12 Aug 2014

L1-2. Special Matrix Operations: Permutations, Transpose, Inverse, Augmentation 12 Aug 2014 Unfortunately, no one can be told what the Matrix is. You have to see it for yourself. -- Morpheus Primary concepts:

### Solving for Voltage and Current

Chapter 3 Solving for Voltage and Current Nodal Analysis If you know Ohm s Law, you can solve for all the voltages and currents in simple resistor circuits, like the one shown below. In this chapter, we

### Relations Graphical View

Relations Slides by Christopher M. Bourke Instructor: Berthe Y. Choueiry Introduction Recall that a relation between elements of two sets is a subset of their Cartesian product (of ordered pairs). A binary

### Math 4310 Handout - Quotient Vector Spaces

Math 4310 Handout - Quotient Vector Spaces Dan Collins The textbook defines a subspace of a vector space in Chapter 4, but it avoids ever discussing the notion of a quotient space. This is understandable

### RUBIK'S CUBE BY: MOSES TROYER & NATE VARDELL ALGORITHMS & PARALLELIZATION

RUBIK'S CUBE ALGORITHMS & PARALLELIZATION BY: MOSES TROYER 1 & NATE VARDELL WHAT IN THE WORLD IS A RUBIK S CUBE??? Originally, a 3x3x3 Cube Shaped Puzzle. There are now 2x2x2 all the way up to 10x10x10.

### A Mathematical Analysis of The Generalized Oval Track Puzzle

Rose- Hulman Undergraduate Mathematics Journal A Mathematical Analysis of The Generalized Oval Track Puzzle Samuel Kaufmann a Volume 12, no. 1, Spring 2011 Sponsored by Rose-Hulman Institute of Technology

### Lecture 13 - Basic Number Theory.

Lecture 13 - Basic Number Theory. Boaz Barak March 22, 2010 Divisibility and primes Unless mentioned otherwise throughout this lecture all numbers are non-negative integers. We say that A divides B, denoted

### Squares and Square Roots

SQUARES AND SQUARE ROOTS 89 Squares and Square Roots CHAPTER 6 6.1 Introduction You know that the area of a square = side side (where side means the length of a side ). Study the following table. Side

### diffy boxes (iterations of the ducci four number game) 1

diffy boxes (iterations of the ducci four number game) 1 Peter Trapa September 27, 2006 Begin at the beginning and go on till you come to the end: then stop. Lewis Carroll Consider the following game.

### Square roots by subtraction

Square roots by subtraction Frazer Jarvis When I was at school, my mathematics teacher showed me the following very strange method to work out square roots, using only subtraction, which is apparently

### 3. Mathematical Induction

3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)

### It is time to prove some theorems. There are various strategies for doing

CHAPTER 4 Direct Proof It is time to prove some theorems. There are various strategies for doing this; we now examine the most straightforward approach, a technique called direct proof. As we begin, it

### The Mathematics of Origami

The Mathematics of Origami Sheri Yin June 3, 2009 1 Contents 1 Introduction 3 2 Some Basics in Abstract Algebra 4 2.1 Groups................................. 4 2.2 Ring..................................

### Applications of Methods of Proof

CHAPTER 4 Applications of Methods of Proof 1. Set Operations 1.1. Set Operations. The set-theoretic operations, intersection, union, and complementation, defined in Chapter 1.1 Introduction to Sets are

### 5544 = 2 2772 = 2 2 1386 = 2 2 2 693. Now we have to find a divisor of 693. We can try 3, and 693 = 3 231,and we keep dividing by 3 to get: 1

MATH 13150: Freshman Seminar Unit 8 1. Prime numbers 1.1. Primes. A number bigger than 1 is called prime if its only divisors are 1 and itself. For example, 3 is prime because the only numbers dividing

### Arithmetic Circuits Addition, Subtraction, & Multiplication

Arithmetic Circuits Addition, Subtraction, & Multiplication The adder is another classic design example which we are obliged look at. Simple decimal arithmetic is something which we rarely give a second

### LAMC Beginners Circle: Parity of a Permutation Problems from Handout by Oleg Gleizer Solutions by James Newton

LAMC Beginners Circle: Parity of a Permutation Problems from Handout by Oleg Gleizer Solutions by James Newton 1. Take a two-digit number and write it down three times to form a six-digit number. For example,

### Chapter 10 Expanding Our Number System

Chapter 10 Expanding Our Number System Thus far we have dealt only with positive numbers, and, of course, zero. Yet we use negative numbers to describe such different phenomena as cold temperatures and

### Sets and functions. {x R : x > 0}.

Sets and functions 1 Sets The language of sets and functions pervades mathematics, and most of the important operations in mathematics turn out to be functions or to be expressible in terms of functions.

### Clock Arithmetic and Modular Systems Clock Arithmetic The introduction to Chapter 4 described a mathematical system

CHAPTER Number Theory FIGURE FIGURE FIGURE Plus hours Plus hours Plus hours + = + = + = FIGURE. Clock Arithmetic and Modular Systems Clock Arithmetic The introduction to Chapter described a mathematical

### a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

### (Refer Slide Time: 1:41)

Discrete Mathematical Structures Dr. Kamala Krithivasan Department of Computer Science and Engineering Indian Institute of Technology, Madras Lecture # 10 Sets Today we shall learn about sets. You must

### 5. GRAPHS ON SURFACES

. GRPHS ON SURCS.. Graphs graph, by itself, is a combinatorial object rather than a topological one. But when we relate a graph to a surface through the process of embedding we move into the realm of topology.

### Let s just do some examples to get the feel of congruence arithmetic.

Basic Congruence Arithmetic Let s just do some examples to get the feel of congruence arithmetic. Arithmetic Mod 7 Just write the multiplication table. 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0

### Binary Operations. Discussion. Power Set Operation. Binary Operation Properties. Whole Number Subsets. Let be a binary operation defined on the set A.

Binary Operations Let S be any given set. A binary operation on S is a correspondence that associates with each ordered pair (a, b) of elements of S a uniquely determined element a b = c where c S Discussion

3. QUADRATIC CONGRUENCES 3.1. Quadratics Over a Finite Field We re all familiar with the quadratic equation in the context of real or complex numbers. The formula for the solutions to ax + bx + c = 0 (where

### An Innocent Investigation

An Innocent Investigation D. Joyce, Clark University January 2006 The beginning. Have you ever wondered why every number is either even or odd? I don t mean to ask if you ever wondered whether every number

### [[SOlVE THE CUBE]] This is the second of four installments: Beginner, Intermediate, Advanced, and the full Fridrich Method.

[[SOlVE THE CUBE]] Intermediate Method by Tim Pow Welcome to the Intermediate Method, a method specifically designed to follow the Beginner Method I wrote. This solution is proven to be a lot faster than

### Chapter 3. Distribution Problems. 3.1 The idea of a distribution. 3.1.1 The twenty-fold way

Chapter 3 Distribution Problems 3.1 The idea of a distribution Many of the problems we solved in Chapter 1 may be thought of as problems of distributing objects (such as pieces of fruit or ping-pong balls)

### Math Workshop October 2010 Fractions and Repeating Decimals

Math Workshop October 2010 Fractions and Repeating Decimals This evening we will investigate the patterns that arise when converting fractions to decimals. As an example of what we will be looking at,

### Practice Problems for First Test

Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate.-

### Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00

18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list

### A fairly quick tempo of solutions discussions can be kept during the arithmetic problems.

Distributivity and related number tricks Notes: No calculators are to be used Each group of exercises is preceded by a short discussion of the concepts involved and one or two examples to be worked out

### Equations and Inequalities

Rational Equations Overview of Objectives, students should be able to: 1. Solve rational equations with variables in the denominators.. Recognize identities, conditional equations, and inconsistent equations.

### Unit 1 Number Sense. In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions.

Unit 1 Number Sense In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions. BLM Three Types of Percent Problems (p L-34) is a summary BLM for the material

### Chapter 3. Cartesian Products and Relations. 3.1 Cartesian Products

Chapter 3 Cartesian Products and Relations The material in this chapter is the first real encounter with abstraction. Relations are very general thing they are a special type of subset. After introducing

### Writing Algorithms / Intro to Speed Solving

Writing Algorithms / Intro to Speed Solving Common Core: Solve word problems leading to equations of the form px + q= r and p(x + q) = r, where p, q, and r are specific rational numbers. Solve equations

### Elementary Algebra. Section 0.4 Factors

Section 0.4 Contents: Definitions: Multiplication Primes and Composites Rules of Composite Prime Factorization Answers Focus Exercises THE MULTIPLICATION TABLE x 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5

### Combinatorial Concepts With Sudoku I: Symmetry

ombinatorial oncepts With Sudoku I: Symmetry Gordon Royle March 29, 2006 (Version 0.2) bstract The immensely popular logic game Sudoku incorporates a significant number of combinatorial concepts. In this

### Mathematical Induction

Mathematical Induction (Handout March 8, 01) The Principle of Mathematical Induction provides a means to prove infinitely many statements all at once The principle is logical rather than strictly mathematical,

### Solving the Rubik's Revenge (4x4x4) Home Pre-Solution Stuff Step 1 Step 2 Step 3 Solution Moves Lists

Solving your Rubik's Revenge (4x4x4) 07/16/2007 12:59 AM Solving the Rubik's Revenge (4x4x4) Home Pre-Solution Stuff Step 1 Step 2 Step 3 Solution Moves Lists Turn this... Into THIS! To solve the Rubik's

### 26 Integers: Multiplication, Division, and Order

26 Integers: Multiplication, Division, and Order Integer multiplication and division are extensions of whole number multiplication and division. In multiplying and dividing integers, the one new issue