Mathematics 581   Problem Set 6   Due Monday, May 20

1. Consider the symmetry group  D5  of the regular pentagon. The pentagon has vertices 1, 2, 3, 4, 5  in counterclockwise order, and the dihedral group has two generators
r = (1 5 4 3 2)  and l = (2 5)(3 4). Show that each of the five symmetries  l,  rl, r2l,  r3l,  r4l   is simply a reflection of the pentagon about a line passing from a vertex to the midpoint of the opposite side.
***
l   =      (2 5)(3 4), reflection through the line passing through vertex 1;
rl =   (1 5 4 3 2)(2 5)(3 4) = (1 2)(3 5), reflection through the line  through vertex 4;
r2l = (1 4 2 5 3)(2 5)(3 4) = (1 3)(4 5), reflection through the line  through vertex 2;
r3l = (1 3 5 2 4)(2 5)(3 4) = (1 4)(2 3), reflection through the line  through vertex 5;
r4l =  (1 2 3 4 5)(2 5)(3 4) = (1 5)(2 4), reflection through the line  through vertex 3.
***
 

2.  (a)   Let be the set of complex numbers of norm  1  under multiplication.  Recall that multiplication of complex numbers is defined as follows.  If  x = a + bi  and  y = c + di  then  xy = (ac - bd) + (ad + bc)i.  The norm of a complex number  a + bi  is the distance of the point  (a, b)  to the origin (0, 0) in the  plane (of complex numbers).   In other words, the norm of  x = a + bi is given by  |x| = (a2 + b2)1/2.   Show that  C  is a group under multiplication.
***
identity element: e = 1 + 0i is the identity.  And,  e  is an element of  C because  |e| = 12 + 02 = 1.  Check it:  ex = x for every  x = a + bi.
inverses: Note that  (a + bi)(a - bi) = a2 + b2.  If  a + bi  is in C, then  a2 + b2 = 1, so that
(a + bi)(a - bi) = 1  and the inverse of  a + bi  is  a - bi.  Note that  |a - bi| =  a2 + b2= 1, so that  a - bi  is in  C.
closure: (needs to be checked, or else it is not clear that the product of two elements in  C  is an element in  C.)  If |x| = |a + bi| = (a2 + b2)1/2 = 1, and  |y| = |c + di| = (c2 + d2)1/2 = 1, then
 |xy| = |(a + bi)(c + di)| = |(ac - bd) + (ad + bc)i| = ((ac - bd)2 + (ad + bc)2)1/2  = (a2c2 - 2acbd + b2d2 + a2d2 +2adbc + b2c2)1/2 = ((a2 + b2)(c2 + d2))1/2 = (a2 + b2)1/2(c2 + d2)1/2 = 1.
associative law: multiplication of complex numbers is associative.  to prove it calculate
[(a + bi)(c + di)](g + hi) = [(ac - bd) + (ad + bc)i](g + hi) =
[(ac -bd)g - (ad + bc)h] - [(ac - bd)h + (ad + bc)g] i = [a(cg - dh) - b(dg + ch)] + [a(ch + dg) + b(cg - dh)] i =  (a + bi)[(c + di)](g + hi)].
***

(b)  Show that for every real number  q,  the complex number   cos q + i sin q, which ought to be written as  cos q + (sin q)i   in strict complex number notation,  is an element of the group C.
***
| cos q + i sin q| =  cos2 q +  sin2 q = 1.
***

(c)  Let  R  be the group of real numbers under addition.  Show that the function  F  from  R  into  defined by  F(r) = cos 2pr + i sin 2pr  is a homomorphism of  R  onto  C.  (Don't be afraid to use some trig identities here.)
***
F(r + s) = cos 2p(r + s) + i sin 2p(r + s) = [(cos 2pr)(cos 2ps) - (sin 2pr)(sin 2ps)] + [(sin 2pr)(cos 2ps) + (sin 2ps)(cos 2pr)] i = (cos 2pr + i sin 2pr)(cos 2ps + i sin 2ps) = F(r)F(s).
***

(d)  What subgroup of  R  is the kernel of  F?
***
 The kernel is all real numbers  r  such that  F(r) = cos 2pr + i sin 2pr = 1 + 0i, that is,  cos 2pr = 1  and sin 2pr = 0.  So the kernel is the group of integers under addition.
***

3.  (Independent of problem  2.)  Describe as best you can the symmetry group of a circle.  (Hint:  the circle behaves in many ways like a polygon with infinitely many sides.  So use rotations and a fixed reflection like we did for n-gons.)
***
Draw a circle.  Fix a point  P  on it.  The subgroup of all symmetries that fix  P  consists of two symmetries, the identity  e  and the reflection    about a diameter through  P. Any rotation is also a symmetry.  Let  R(q) denote a counterclockwise rotation through an angle   q.   Further,
R(q)  =  l R(q)-1.
***