Group theory

More documents

Recommendations

Info

GROUP THEORYm 1 (π)m 2 (π)m 3 (π)m 4 (π)Figure 28.3 The notation for exercise 28.11.28.9 Hints and answers28.1 § (a) Yes, (b) no, there is no inverse for 2, (c) yes, (d) no, 2 × 3 is not in the set,(e) yes, (f) yes, they form a subgroup of order 4, [1, 0; 0, 1] [4, 0; 0, 4] [1, 2; 0, 4][4, 3; 0, 1], (g) yes.28.3 x • (y • z) = x + y + z + r(xy + xz + yz) + r 2 xyz = (x • y) • z. Show that assumingx • y = −r −1 leads to (rx + 1)(ry + 1) = 0. The inverse of x is x −1 = −x/(1 + rx);show that this is not equal to −r −1 .28.5 (a) Consider both X = i and X = i. Here, i ∼ i. (b) In this case i ∼ i, but theconclusion cannot be deduced from the other axioms. In both cases i is in a classby itself and no Y , as used in the false proof, can be found.28.7 † Use |AB| = |A||B| = 1×1 = 1 to prove closure. The inverse has w ↔ z, x ↔ −x,y ↔ −y, giving |A −1 | = 1, i.e. it is in the set. The only element of order 2 is −I;A 2 can be simplified to [−(w + 1), −x; −y, −(z + 1)].28.9 If XY = Z, show that Y = XZ and X = ZY , then form Y X. Note that theelements of B can only have orders 1, 2 or p. Suppose they all have order 1 or2; then using the earlier result, whilst noting that 4 does not divide 2p, leads toa contradiction.28.11 Using the notation indicated in figure 28.3, R being a rotation of π/2 about anaxis perpendicular to the square, we have: I has order 1; R 2 , m 1 , m 2 , m 3 , m 4 haveorder 2; R, R 3 have order 4.subgroup {I, R, R 2 , R 3 } has cosets {I, R, R 2 , R 3 }, {m 1 , m 2 , m 3 , m 4 };subgroup {I, R 2 , m 1 , m 2 } has cosets {I, R 2 , m 1 , m 2 }, {R, R 3 , m 3 , m 4 };subgroup {I, R 2 , m 3 , m 4 } has cosets {I, R 2 , m 3 , m 4 }, {R, R 3 , m 1 , m 2 };subgroup {I, R 2 } has cosets {I, R 2 }, {R, R 3 }, {m 1 , m 2 }, {m 3 , m 4 };subgroup {I, m 1 } has cosets {I, m 1 }, {R, m 3 }, {R 2 , m 2 }, {R 3 , m 4 };subgroup {I, m 2 } has cosets {I, m 2 }, {R, m 4 }, {R 2 , m 1 }, {R 3 , m 3 };subgroup {I, m 3 } has cosets {I, m 3 }, {R, m 2 }, {R 2 , m 4 }, {R 3 , m 1 };subgroup {I, m 4 } has cosets {I, m 4 }, {R, m 1 }, {R 2 , m 3 }, {R 3 , m 2 }.28.13 G = {I, A, B, B 2 , B 3 , AB, AB 2 , AB 3 }. The proper subgroups are as follows:{I, A}, {I, B 2 }, {I, AB 2 }, {I, B, B 2 , B 3 }, {I, B 2 , AB, AB 3 }.28.15 (b) A 3 = {(1), (123), (132)}.(d) For Φ 1 , K = {(1), (123), (132)} is a subgroup.For Φ 2 , K = {(23), (13), (12)} is not a subgroup because it has no identity element.For Φ 3 , K = {(1), (23), (13), (12)} is not a subgroup because it is not closed.§ Where matrix elements are given as a list, the convention used is [row 1; row 2; . . . ], individualentries in each row being separated by commas.1074
28.9 HINTS AND ANSWERSFor Φ 4 , K = {(1), (123), (132)} is a subgroup.Only Φ 1 is a homomorphism; Φ 4 fails because, for example, [(23)(13)] =(23) (13) .28.17 Recall that, for any pair of matrices P and Q, |PQ| = |P||Q|. K is the set of allmatrices with unit determinant. The cosets of K are the sets of matrices whosedeterminants are equal; K itself is the identity in the group of cosets.28.19 (a) No, because the set is not closed, (b) yes, (c) yes, (d) yes.28.21 Each subgroup contains the identity, a rotation by π, and two reflections. Thehomomorphism is ±1 → I, ±i → R 2 , ±j → m x , ±k → m y with kernel {1, −1}.28.23 There are 10 elements in all: I, rotations R i (i = 1, 4) and reflections m j (j = 1, 5).(a) There are five proper subgroups of order 2, {I, m j } and one proper subgroupof order 5, {I, R, R 2 , R 3 , R 4 }.(b) Four conjugacy classes, {I}, {R, R 4 }, {R 2 , R 3 }, {m 1 , m 2 , m 3 , m 4 , m 5 }.1075
Page 2: GROUP THEORYNHMHHH(a)(b)HFigure 28.
Page 5 and 6: 28.1 GROUPSUsing only the first equ
Page 7 and 8: 28.1 GROUPSLMKFigure 28.2 Reflectio
Page 9 and 10: 28.2 FINITE GROUPS28.2 Finite group
Page 11 and 12: 28.2 FINITE GROUPS(a)1 5 7 111 1 5
Page 13 and 14: 28.3 NON-ABELIAN GROUPSAs a first e
Page 15 and 16: 28.3 NON-ABELIAN GROUPSI A B C D EI
Page 17 and 18: 28.4 PERMUTATION GROUPSSuppose that
Page 19 and 20: 28.5 MAPPINGS BETWEEN GROUPS28.5 Ma
Page 21 and 22: 28.6 SUBGROUPS(a)I A B C D EI I A B
Page 23 and 24: 28.7 SUBDIVIDING A GROUP(i) the set
Page 25 and 26: 28.7 SUBDIVIDING A GROUPthis implie
Page 27 and 28: 28.7 SUBDIVIDING A GROUP• Two cos
Page 29 and 30: 28.7 SUBDIVIDING A GROUP(iii) In an
Page 31 and 32: 28.8 EXERCISES28.4 Prove that the r
Page 33: 28.8 EXERCISESSimilarly compute C 2
Page 37 and 38: 29.1 DIPOLE MOMENTS OF MOLECULESAB(
Page 39 and 40: 29.2 CHOOSING AN APPROPRIATE FORMAL
Page 45 and 46: 29.3 EQUIVALENT REPRESENTATIONSresp
Page 47 and 48: 29.4 REDUCIBILITY OF A REPRESENTATI
Page 49 and 50: 29.4 REDUCIBILITY OF A REPRESENTATI
Page 51 and 52: 29.5 THE ORTHOGONALITY THEOREM FOR
Page 53 and 54: 29.6 CHARACTERS3m I A, B C, D, EA 1
Page 55 and 56: 29.7 COUNTING IRREPS USING CHARACTE
Page 61 and 62: 29.8 CONSTRUCTION OF A CHARACTER TA
Page 63 and 64: 29.10 PRODUCT REPRESENTATIONSgive a
Page 65 and 66: 29.11 PHYSICAL APPLICATIONS OF GROU
Page 73 and 74: 29.12 EXERCISESas the sum of two on
Page 75 and 76: 29.12 EXERCISESUse this to show tha
Page 77 and 78: 29.13 HINTS AND ANSWERS(a) Make an

Group theory

Create successful ePaper yourself

Delete template?

Save as template?