Modern Algebra

MTH 461: Survey of Modern Algebra, Spring 2021 Homework 8 Homework 8 1. Determine which of the following are rings, with the usual operations of addition and multiplication. If it is a ring you do not need to write out all the axioms, but if it is not a ring, explain why. If the set is a ring, is it a field? If not, explain why. ? ? (a) Qp 2q “ ta ` b 2 ∶ a, b P Qu (b) Zris “ ta ` bi ∶ a, b P Zu, where i “ ? ´1, so that i2 “ ´1 (c) The set of following matrices with entries in the ring Z2 : “ˆ ˙ ˆ ˙ ˆ ˙ ˆ ˙* 0 0 1 0 1 1 0 1 , , , 0 0 0 1 1 0 1 1 (d) The set of continuous functions f ∶ R Ñ R. Addition and multiplication is defined pointwise. That is, for two such functions f, g, we have f ` g and f g defined by pf ` gqpxq “ f pxq ` gpxq and pf gqpxq “ f pxqgpxq for x P R. 2. For each ring R given, determine the set of units Rˆ . (a) The integers Z (b) The ring M2 pZq, the 2 ˆ 2 matrices with integer entries (c) Each of the rings in problem #1 1 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 20 The Rubik’s cube group Today we will study the Rubik’s cube using some of the group theory we have developed. During lecture, I’ll hand out Rubik’s cubes and you should play along. If you don’t know how to solve the cube, however, be careful not to scramble it! Let’s begin with a top (on the left) and bottom (on the right) view of the Rubik’s cube: We label the 6 faces of the cube F (front), B (back), U (up), D (down), L (left), R (right). To each face we associate a symmetry of the cube which is a 90˝ clockwise rotation of that face, if you are facing the face. For example, we have the moves R and U ´1 : And R2 , for example, is 180˝ rotation of the right face. We write G for the Rubik’s cube group, which is the group of symmetries generated by the moves F, B, U, D, L, R. Our convention is that the expression LU ´1 RD means: first do D, then do R, then U ´1 , and then L. Warning: This is opposite the order in which Rubik’s cube enthusiasts write moves, but it goes better with our conventions in group theory. Let us try to understand the group G. First, compare the following two sequences of moves: 1 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 20 This shows that U ´1 R ‰ RU ´1 . In particular G is non-abelian. However we can find non-trivial abelian subgroups. Here’s an example. Consider the subgroup generated by R. This is clearly xRy “ te, R, R2 , R3 u as R is a 90˝ rotation of a face. So we get a cyclic subgroup of order 4, which is isomorphic to Z4 . Similarly, xLy – Z4 . Note also that L and R commute as these two faces are not sharing any part of the cube. From this we get a subgroup xR, Ly Ă G isomorphic to Z4 ˆ Z4 . Let’s look at the orders of some elements in G. Check with your cube the following: ordpR´1 U Rq “ 4, ordpU R´1 U Rq “ 5 As another example, you can check (after a few minutes!) that ordpU Rq “ 105. For a more extravagant example, one can check that ordpB ´1 U B ´1 F 2 Rq “ 1260, although I suggest you don’t waste your time doing it! This order, 1260, is in fact the highest possible order of any element in G. How should we better understand G? One way to is map it into a symmetric group, just as we’ve done for the triangle and the tetrahedron. Label the “sticker positions” of the cube: In the picture we labelled one face, but continuing in this fashion we will get p9qp6q “ 54 labels. Each move a P G is determined by how these sticker positions are permuted, and to each such a we get permutation in S54 . This gives a 1-1 homomorphism G Ñ S54 . 2 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 20 But we can do better! Note each a P G fixes the middle sticker position on each face. So forget about these. We only label the other sticker positions, of which there are 48. This gives a homomorphism G Ñ S48 . Note that 48! « 1061 . However it turns out |G| “ 8!12!210 37 « 4 ˆ 1019 , which is very large but much smaller than |S48 | “ 48!. How can you solve the Rubik’s cube? The basic strategy is as follows: (1) first, come up with some basic moves that permute only a few parts of the cube; (2) use these basic moves to progressively solve parts of the cube. (In reality it is best to just learn one of the algorithms that solves it.) We won’t solve the cube but rather aim to understand G better. But let’s focus on (1) for a moment. Here is a “basic move” called an edge 3-cycle: Both top and bottom views are shown. It is called an edge 3-cycle because exactly 3 pieces (subcubes) are permuted. A representation of a P G is given by a “ F 2 U L´1 RF 2 LR´1 U F 2 . Having shown that we can permute 3 edge cubes as above, we might suspect we can permute any given set of subcubes as we like. Not true! For example, consider the configuration: 3 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 20 This configuration has exactly 2 corner cubes swapped. To show this is impossible, we introduce another homomorphism. Consider subcube positions which are not in the middles of faces. You can count there are 20 such non-middle subcube positions in the cube. Next, label these subcube positions 1, . . . , 20 in any fashion. By seeing how a symmetry a P G permutes these 20 subcube positions, we obtain a homomorphism φ ∶ G ÐÑ S20 For example, take the subcubes starting on the right face at a corner and label them 1, . . . , 8 in a clockwise fashion (and label the others arbitrarily) then φpRq “ p1357qp2468q. Note this permutation is even. A similar computation shows φpaq is even for F, B, U, D, L. But every element in G can be written as a composition of these moves, so we have shown: § The homomorphism φ ∶ G Ñ S20 has impφq Ă A20. In other words, for any symmetry a P G, the permutation φpaq is even. Now the proof of why the configuration drawn previously is impossible is easy! Just note that it swaps exactly 2 subcubes, so the picture corresponds to a transposition in S20 , which is odd, and this contradicts what we have just found. We can also prove that the following configuration is impossible: Note that this configuration does not swap any subcubes; it only changes the “orientation” of exactly one subcube. It is in fact in the kernel of the homomorphism φ, so that homomorphism will not help us here. We can define another homomorphism as follows. Notice that each edge piece (not a corner) has 2 stickers. Collecting all such pairs of stickers, there are 24. Labelling these 1, . . . , 24 and seeing how each a P G permutes these, we obtain a homomorphism ψ ∶ G Ñ S24 . Again, it is not hard to show that ψpRq is even, and the same is true for F, B, U, D, L, and thus for all a P G. On the other hand, the above drawn configuration corresponds to a transposition which swaps exactly 2 of these edge stickers, which is impossible! 4 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 21 Definition of a ring A ring R is a set, with elements 0, 1 P R, together with two binary operations ` (addition) and ˆ (multiplication, often written ¨ or omitted) such that the following properties hold: (Associativity for `) a ` pb ` cq “ pa ` bq ` c (Associativity for ˆ) apbcq “ pabqc (Identity for `) a`0“0`a“a (Identity for ˆ) 1¨a“a¨1“a (Inverses for `) a ` p´aq “ p´aq ` a “ 0 (Distributivity) apb ` cq “ ab ` ac, (Commutativity for `) a`b“b`a pa ` bqc “ ac ` bc These properties are for all a, b, c P R. The “Inverses for `” property should be understood as: for each a P R, there is an element called ´a P R such that a ` p´aq “ p´aq ` a “ 0. A ring is commutative if ab “ ba for all a, b P R. Examples of rings include Z, Q, R, C and Zn , all with the usual operations of addition and multiplication. Note these are all commutative rings. A “shorter” definition of a ring is as follows: a ring R is a set with binary operations ` and ˆ such that pR, `q is an abelian group with identity 0 P R; and ˆ is associative and has an identity 1 P R; and the distributivity properties hold. We remark that the axiom that “`” is commutative is implied by the other axioms. To see this we compute p1 ` 1qpa ` bq in two different ways: p1 ` 1qpa ` bq “ 1pa ` bq ` 1pa ` bq “ a ` b ` a ` b p1 ` 1qpa ` bq “ p1 ` 1qa ` p1 ` 1qb “ a ` a ` b ` b We have used distributivity and also that 1 is a multiplicative identity. Then we can add ´a to the left sides and ´b to the right sides of these equations to get b ` a “ a ` b. § Let R be a ring and a, b P R. Then: (i) a0 “ 0a “ 0 (ii) ap´bq “ ´ab “ p´aqb (iii) p´aqp´bq “ ab Proof. To prove (i), we compute a0 “ ap0 ´ 0q “ a0 ´ a0 “ 0, and similarly 0a “ 0. For (ii), use distributivity: ab ` ap´bq “ apb ` p´bqq “ a0 “ 0; this implies ap´bq “ ´ab. Similarly p´aqb “ ´ab. Finally, for (iii): from (ii), p´aqp´bq “ ´pap´bqq “ ´p´pabqq “ ab. 1 MTH 461: Survey of Modern Algebra, Spring 2021 § Suppose 0 “ 1 in a ring R. Lecture 21 Then R “ t0u. To see this, we compute for any a P R: a “ a1 “ a0 “ 0. Thus every elemnent in R is equal to 0, and this proves the claim. We call R “ t0u, with the operations ` and ˆ defined in the only possible way, the zero ring. From now on we assume R ‰ t0u. For a ring R, pR, ˆq is not a group. To see this, recall we are assuming R is not the zero ring, so 0 ‰ 1. Then a0 “ 0 for all a P R. Thus there is no a P R such that a0 “ 1. This means 0 does not have a multiplicative inverse. So pR, ˆq is not a group. § Let R be a ring. We make the following definitions: (i) a P R is a unit if there is a b P R such that ab “ 1. We write b “ a´1 . (ii) a P R is a zero divisor if there is a non-zero b P R such that ab “ 0. The following is a straightforward verification from the definitions. § Let R be a ring and define Rˆ “ ta P R ∶ a is a unitu. Then pRˆ , ˆq is a group. Note this notation agrees with our earlier notations for Cˆ , Qˆ , Rˆ and Zˆ n. Here is an example of a non-commutative ring. Consider the set of 2 ˆ 2 real matrices: ” ˆ ˙ * a b M2 pRq “ A “ ∶ a, b, c, d P R c d Then we define ` to be addition of matrices and ˆ to be multiplication of matrices. Then, not surprisingly, M2 pRq is a ring with additive and multiplicative identities given by: ˆ ˙ ˆ ˙ 0 0 1 0 “0” “ “1” “ 0 0 0 1 An example of a zero divisor in M2 pRq (which is not 0) is the following matrix: ˆ ˙ 0 1 A“ 0 0 In fact AA “ A2 “ 0. Finally, we note that A P M2 pRq is a unit if and only if A P GL2 pRq. We conclude M2 pRqˆ “ GL2 pRq. § Let R be a ring. We make the following definitions: (i) R is an integral domain if it is commutative and ab “ 0 implies a “ 0 or b “ 0. (ii) R is a division ring if every non-zero a P R is a unit. (iii) R is a field if it is a commutative division ring. 2 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 21 Note a commutative ring R is an integral domain if and only if the only zero divisor in R is 0. Also, every field is an integral domain. Note also that R is a division ring if and only if the group of units Rˆ is exactly Rzt0u. Examples of fields are Q, R and C. The ring Z is an integral domain but not a division ring or a field. The ring M2 pRq is not commutative (hence not an integral domain or a field) and also not a division ring, because it has non-zero zero divisors. Let’s look at some other examples. Consider Z3 “ t0, 1, 2u. Note 1 ¨ 1 “ 1 and 2 ¨ 2 “ 1. Thus Z3 is an integral domain, and even a field. For another example, consider Z4 . Note that 2 ¨ 2 “ 0, so this ring has a non-zero zero divisor, and is not an integral domain. Now consider Zn for n a general positive integer n. We know that Zˆ n “ tunits in Zn u “ tapmod nq ∶ gcdpa, nq “ 1u The only case in which Zˆ n “ Zn zt0u is when n is a prime. Thus the commutative ring Zn is a field if and only if n is prime. Furthermore, suppose n is not prime, and write n “ ab for some positive integers a, b less than n. Then a, b pmod nq are non-zero, but ab ” n ” 0 pmod nq. Thus a, b pmod nq are zero divisors in Zn . Thus Zn is not even an integral domain when n is not prime. Thus: § If n is prime, the ring Zn is a field. If n is not prime, Zn has (non-zero) zero divisors and so is not an integral domain. 3
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