2.2.4 Prove that the angular re¬‚ector made of three pairwise perpendicular

mirrors in R3 sends a ray of light back in the direction exactly opposite to the

one it came from, (Figure 2.6).

A. & A. Borovik • Mirrors and Re¬‚ections • Version 01 • 25.02.00 39

h

hh h

y

hh

T

±

©

@@

X

@@@

‚@

Figure 2.5: For Exercise 2.2.3.

Reflections and Linear Algebra.

2.2.5 We say that a subpace U of the real Euclidean space V is perpendicular to

the subspace W and write U ⊥ W if U = (U © W ) • U where U is orthogonal

to W , i.e. (u, w) = 0 for all u ∈ U and w ∈ W . Prove that this relation is

symmetric; U ⊥ W if and only if W ⊥ U .

2.2.6 Prove that if a re¬‚ection s leaves a subspace U < V invariant then U is

perpendicular to the mirror Hs of the re¬‚ection s.

2.2.7 Prove that two re¬‚ections s and t commute, that is, st = ts, if and only

if their mirrors are perpendicular to each other.

Planar geometry.

2.2.8 Prove that the product of two re¬‚ections in AR2 with parallel mirrors is

a parallel translation. What is the translation vector?

2.2.9 If a bounded ¬gure in the Euclidean plane AR2 has a center of symmetry

and a mirror of symmetry then it has two perpendicular mirrors of symmetry.

Is the same true in AR3 ?

2.3 Dihedral groups

In this section we shall study ¬nite groups generated by two involutions.

Theorem 2.3.1 There is a unique, up to isomorphism, group W gener-

ated by two involutions s and t such that their product st has order n.

Furthermore,

40

)

x

0

q

Figure 2.6: Angular re¬‚ector (for Exercise 2.2.4).

(1) W is ¬nite and |W | = 2n.

(2) If r = st then the cyclic group R = r generated by r is a normal

subgroup of W of index 2.

(3) Every element in W R is an involution.

We shall denote the group W as D2n , call it the dihedral group of order

2n and write

D2n = s, t | s2 = t2 = (st)n = 1 .

This standard group-theoretical notation means that the group D2n is gen-

erated by two elements s and t such that any identity relating them to each

other is a consequence of the generating relations

s2 = 1, t2 = 1, (st)n = 1.

The words ˜consequence of the generating relations™ are given precise mean-

ing in the theory of groups given by generators and relations, a very well

developed chapter of the general theory of groups. We prefer to use them

in a informal way which will be always clear from context.

Proof of the Theorem 2.3.1. First of all, notice that, since s2 = t2 = 1,

s’1 = s and t’1 = t.

A. & A. Borovik • Mirrors and Re¬‚ections • Version 01 • 25.02.00 41

In particular, (st)’1 = t’1 s’1 = ts. Set r = st. Then

rt = trt = t · st · t = ts = r’1

and analogously rs = r’1 .

Since s = rt, the group W is generated by r and t and every element

w in W has the form

w = rm1 tk1 · · · rml tkl ,

where mi takes the values 0, 1, . . . , n ’ 1 and ki is 0 or 1. But one can check

that, since trt = r’1 ,

tr = r’1 t,

trm = r’m t

and

km

tk rm = r(’1) tk .

Hence

(rm1 tk1 )(rm2 tk2 ) = rm tk (2.1)

where

k = k1 + k2 , m = m1 + (’1)k1 m2 .

Therefore every element in W can be written in the form

w = rm tk , m = 0, . . . , n ’ 1, k = 0, 1.

Furthemore, this presentation is unique. Indeed, assume that

rm1 tk1 = rm2 tk2

where m1 , m2 ∈ { 0, . . . , n ’ 1 } and k1 , k2 ∈ { 0, 1 }. If k1 = k2 then

rm1 = rm2 and m1 = m2 . But if k1 = k2 then

rm1 ’m2 = t.

Denote m = m1 ’ m2 . Then m < n and rm = (st)m = t. If m = 0 then

t = 1, which contradicts to our assumption that |t| = 2. Now we can easily

get a ¬nal contradiction:

st · st · · · st = t

implies

s · ts · · · ts · · · ts = 1.

The word on the left contains an odd number of elements s and t. Consider

the element r in the very center of the word; r is either s or t. Hence the

previous equation can be rewritten as

sts · · · r · · · sts = [sts · · ·]r[sts · · ·]’1 = 1,

42

C B

ts The group of symmetries of the

“

¨t

t

t ¨”

t regular n-gon ∆ is generated

¨

t

¨¨

t

r by two re¬‚ections s and t in

t

¨tA

rr

t ¨¨ T

r t

t the mirrors passing through the

¨

¨rr

t

t

rr

¨ c midpoint and a vertex of a side

t¨ t r

¨¨ t of ∆.

t t

t

t t

Figure 2.7: For the proof of Theorem 2.3.2.

which implies r = 1, a contradiction.

Since elements of w can be represented by expressions rm tk , and in a

unique way, we conclude that |W | = 2n and

W = { rm tk | m = 0, 1, . . . , n ’ 1, k = 0, 1 },

with the multiplication de¬ned by Equation 2.1. This proves existence and

uniqueness of D2n .

Since |r| = n, the subgroup R = r has index 2 in W and hence is

R then w = rm t for some m, and a direct

normal in W . If w ∈ W

computation shows

w2 = rm t · rm t = r’m+m t2 = 1.

Since w = 1, w is an involution.

Theorem 2.3.2 The group of symmetries Sym ∆ of the regular n-gon ∆

is isomorphic to the dihedral group D2n .

Proof. Denote W = Sym ∆. The mirrors of symmetry of the polygon ∆

cut it into 2n triangle slices4 , see Figure 2.7. Notice that any two adjacent

slices are interchanged by the re¬‚ection in their common side. Therefore

W acts transitively on the set S of all slices. Also, observe that only the

identity symmetry of ∆ maps a slice onto itself. By the well-known formula

for the length of a group orbit

«

order of the

number

· stabiliser = 2n · 1 = 2n.

|W | =

of slices

of a slice

4

Later we shall use for them the more terms fundamental regions or chambers.

A. & A. Borovik • Mirrors and Re¬‚ections • Version 01 • 25.02.00 43

Next, if s and t are re¬‚ections in the side mirrors of a slice, then their

product st is a rotation through the angle 2π/n, which can be immediately

seen from the picture: st maps5 the vertex A to B and B to C. By

Theorem 2.3.1, | s, t | = 2n; hence W = s, t is the dihedral group of

order 2n.

Exercises

2.3.1 Prove that the dihedral group D6 is isomorphic to the symmetric group

Sym3 .

2.3.2 The centre of a dihedral group. If n > 2 then

{1} if n is odd,

Z(D2n ) = n n

{ 1, r 2 } = r 2 if n is even.

2.3.3 Klein™s Four Group. Prove that D4 is an abelian group,

D4 = { 1, s, t, st }.

(It is traditionally called Klein™s Four Group.)

2.3.4 Prove that the dihedral group D2n , n > 2, has one class of conjugate

involutions, in n is odd, and three classes, if n is even. In the latter case, one of

the classes contains just one involution z and Z(D2n ) = { 1, z }.

2.3.5 Prove that a ¬nite group of orthogonal transformations of R2 is either

cyclic, or a dihedral group D2n .

2.3.6 If W = D2n is a dihedral group of orthogonal transformations of R2 ,

then W has one conjugacy class of re¬‚ections, if n is odd, and two conjugacy

classes of re¬‚ections, if n is even.