GROUP THEORY
m 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 xy z=(x•y)•z.Show that assuming
x•y=−r−^1 leads to (rx+1)(ry+1) =0.The inverse ofxisx−^1 =−x/(1 +rx);
show that this is not equal to−r−^1.
28.5 (a) Consider bothX=iandX=i. Here,i∼i. (b) In this casei∼i, but the
conclusion cannot be deduced from the other axioms. In both casesiis in a class
by itself and noY, as used in the false proof, can be found.
28.7 †Use|AB|=|A||B|=1×1 = 1 to prove closure. The inverse hasw↔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 IfXY=Z, show thatY=XZandX=ZY,thenformYX. Note that the
elements ofBcan only have orders 1, 2 orp. Suppose they all have order 1 or
2; then using the earlier result, whilst noting that 4 does not divide 2p,leadsto
a contradiction.
28.11 Using the notation indicated in figure 28.3,Rbeing a rotation ofπ/2 about an
axis perpendicular to the square, we have:Ihas order 1;R^2 ,m 1 ,m 2 ,m 3 ,m 4 have
order 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;...], individual
entries in each row being separated by commas.