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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


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X
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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.

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