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
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 C
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
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 C 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
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 l 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 = l R(q)-1.