Hints and answers

GROUP THEORY
m1 (π)
m2 (π)
m4 (π)
m3 (π)
Figure 28.3 The notation for exercise 28.11.
28.9 Hints and answers
28.1
28.3
28.5
28.7
28.9
28.11
28.13
28.15
§
§
(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.
x • (y • z) = x + y + z + r(xy + xz + yz) + r 2 xyz = (x • y) • z. Show that assuming
x • 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 .
(a) Consider both X = i and X = i. Here, i ∼ i. (b) In this case i ∼ i, but the
conclusion cannot be deduced from the other axioms. In both cases i is in a class
by itself and no Y , as used in the false proof, can be found.
†
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;
A2 can be simplified to [−(w + 1), −x; −y, −(z + 1)].
If XY = Z, show that Y = XZ and X = ZY , then form Y X. Note that the
elements of B can only have orders 1, 2 or p. Suppose they all have order 1 or
2; then using the earlier result, whilst noting that 4 does not divide 2p, leads to
a contradiction.
Using the notation indicated in figure 28.3, R being a rotation of π/2 about an
axis perpendicular to the square, we have: I has order 1; R 2 , m1 , m2 , m3 , m4 have
order 2; R, R 3 have order 4.
subgroup {I, R, R 2 , R 3 } has cosets {I, R, R 2 , R 3 }, {m1 , m2 , m3 , m4 };
subgroup {I, R 2 , m1 , m2 } has cosets {I, R 2 , m1 , m2 }, {R, R 3 , m3 , m4 };
subgroup {I, R 2 , m3 , m4 } has cosets {I, R 2 , m3 , m4 }, {R, R 3 , m1 , m2 };
subgroup {I, R 2 } has cosets {I, R 2 }, {R, R 3 }, {m1 , m2 }, {m3 , m4 };
subgroup {I, m1 } has cosets {I, m1 }, {R, m3 }, {R 2 , m2 }, {R 3 , m4 };
subgroup {I, m2 } has cosets {I, m2 }, {R, m4 }, {R 2 , m1 }, {R 3 , m3 };
subgroup {I, m3 } has cosets {I, m3 }, {R, m2 }, {R 2 , m4 }, {R 3 , m1 };
subgroup {I, m4 } has cosets {I, m4 }, {R, m1 }, {R 2 , m3 }, {R 3 , m2 }.
G = {I, A, B, B2 , B3 , AB, AB2 , AB3 }. The proper subgroups are as follows:
{I, A}, {I, B2 }, {I, AB2 }, {I, B, B2 , B3 }, {I, B2 , AB, AB3 }.
(b) A3 = {(1), (123), (132)}.
(d) For Φ1 , K = {(1), (123), (132)} is a subgroup.
For Φ2 , K = {(23), (13), (12)} is not a subgroup because it has no identity element.
For Φ3 , 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; . . . ], individual
entries in each row being separated by commas.
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28.9 HINTS AND ANSWERS
28.17
28.19
28.21
28.23
For Φ4 , K = {(1), (123), (132)} is a subgroup.
Only Φ1 is a homomorphism; Φ4 fails because, for example, [(23)(13)] =
(23) (13) .
Recall that, for any pair of matrices P and Q, |PQ| = |P||Q|. K is the set of all
matrices with unit determinant. The cosets of K are the sets of matrices whose
determinants are equal; K itself is the identity in the group of cosets.
(a) No, because the set is not closed, (b) yes, (c) yes, (d) yes.
Each subgroup contains the identity, a rotation by π, and two reflections. The
homomorphism is ±1 → I, ±i → R 2 , ±j → mx , ±k → my with kernel {1, −1}.
There are 10 elements in all: I, rotations R i (i = 1, 4) and reflections mj (j = 1, 5).
(a) There are five proper subgroups of order 2, {I, mj } and one proper subgroup
of order 5, {I, R, R 2 , R 3 , R 4 }.
(b) Four conjugacy classes, {I}, {R, R 4 }, {R 2 , R 3 }, {m1 , m2 , m3 , m4 , m5 }.
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