Groups of Symmetries
J. K. Verma
Department of Mathematics, IIT Bombay [email protected]
Contents
1 Group actions 2
2 Groups of symmetries of figures 7
3 Orthogonal maps of Rn : The Cartan-Dieudonn´e Theorem 8
4 Isometries of Rn 12
5 Isometries of the plane 14
6 Finite subgroups of M 15
7 Symmetries of the Platonic solids 17
8 Finite subgroups of SO3 19
9 Exercises 25
Acknowledgements: These are notes of six lectures compiled from the books mentioned in the list of references and delivered in the Instructional School in Algebra, a school organised
under the aegis of the National Centre of Mathematics at Himachal Pradesh University, Shimla during 5-9 June 2017. Thanks are due to Professors R. P. Sharma (HP University) and Dinesh
1
Group actions
Examples of group actions
Definition 1.1. Let Gbe a group and X be a set. We say G operates on X or G acts on X if there is a map
G×X −→X, (g, x)→gx,
satisfying the conditions (a) 1x=x and(b) (gh)x=g(hx) for all g, h∈G and x∈X.
Let SX denote the group of permutations of X. For eachg∈Gdefine the map
mg :X −→X, mg(x) =gx.
Note thatmgmg−1(x) =g(g−1x) = 1x=x for all g∈G andx∈X. Similarlymg−1mg(x) =x
for all g ∈G and x ∈ X. Hence mg is a bijection ofX. Define F :G −→ SX by F(g) =mg.
The condition (b) for group action implies thatF is a group homomorphism.
Conversely let F :G −→ SX be a group homomorphism. Define the map G×X −→ X
by (g, x) −→ F(g)(x). Check that this defines an action of G on X. In fact there is a 1−1
correspondence between
{actions of Gon X} ←→ {group homomorphismsG−→SX}.
Example 1.2. Let G be a group. The map G×G −→ G given by (g, h) = gh is an action of G on itself. This gives a group homomorphism F : G −→ SG given by F(g) = mg where mg(h) =gh.In fact F is injective. Indeed,
mg =Id ⇐⇒ mg(x) =gx=x for all x∈G
⇐⇒ g= 1
By the first homomorphism theorem, we see thatGis isomorphic to a subgroup of permutations of G.
Theorem 1.3 (Cayley’s Theorem). Let Gbe a group. Then Gis isomorphic to a subgroup of permutations of G.
Example 1.4. Any groupGacts on Gby conjugation:
Indeed, (1, h) = h and (f g, h) = (f g)h(f g)−1 = f ghg−1f−1 = f(gh) for f, g, h ∈ G. This action induces a group homomorphism γ :G−→G given byγ(g) =γg where γg(h) =ghg−1.
Let us find kerγ :
γg =Id ⇐⇒ gxg−1 =x∀x∈G
⇐⇒ gx=xg∀x∈G
⇐⇒ g∈Z(G)
Therefore we have an injective map G/Z(G)−→ Aut (G) whose image is the group I(G)
of inner automorphisms of G. Thus G/Z(G)'I(G).Hence for a finite group G, there are as many inner automorphisms as the order of G/Z(G).
Example 1.5. Let G=GL(n, F) denote the group of n×ninvertible matrices with entries in a fieldF.The groupGacts on the vector spaceV =Fnby the rule (g, v)−→gv.Forn= 2
and F =F2,the order o(G) ofGis 6.The set of nonzero vectors inV is X={e1, e2, e1+e2}.
Since Gacts on X, we have a group homomorphismφ:G−→S3.
kerφ={A∈G|Ae1 =e1, Ae2 =e2}={I}.
Since o(G) = 6,we conclude that GL(2,F2)'S3.
Orbits and stabilizers
Let a groupG act on a setX and a∈X. The orbit ofa, denoted byO(a),is the set
Ga=O(a) ={ga|g∈G}.
The stabilizer ofa∈X is the subgroup of G,
Ga={g∈G|ga=a}.
Theorem 1.6. Let G act on a set X and x∈X. Then
(a) There is a bijectionφ:G/Gx−→O(x) given by φ(gGx) =gx.
(b) |O(x)|= [G:Gx].
(c) X is a disjoint union of theG-orbits of X.
(d) |X|=P
x∈X[G:Gx].Here the sum is taken over one element from each orbit.
Proof. (a) and (b): The map φis well-defined and 1−1.Indeed, aGx=bGx fora, b∈G⇐⇒
a−1bx=x⇐⇒ax=bx. The map is clearly onto. Henceφ is a bijection which implies (b).
(c) Let x, y ∈ X and z ∈ Gx∩Gy. Then z =gx =hy for some g, h ∈G. Therefore x ∈Gy
and y∈Gx.This proves thatGx=Gy. ThereforeX is a disjoint union of the G-orbits.
(d) This follows from (b) and (c).
(e) h∈Ggx⇐⇒h(gx) =gx⇐⇒g−1hgx=x⇐⇒g−1hg∈Gx⇐⇒h∈gGxg−1.
Cauchy’s Theorem
We shall prove Cauchy’s theorem as a simple consequence of group action.
Theorem 1.7 (Cauchy’s Theorem). If p is a prime divisor of the order of a finite group G, then Ghas an element of oder p.
Proof. Consider the set
X ={(x1, x2, . . . , xp)∈Gp |x1x2. . . xp = 1}.
We will find an element (x1.x2, . . . , xp) ∈ X such that x1 = x2 = · · · = xp = x 6= 1 so that x is an element of order p. The number of elements in X is np−1 where n= o(G). Indeed, if
we choose any x1, x2, . . . , xp−1 inG then xp is uniquely determined since x1x2. . . xp = 1.Let
σ = (12. . . p) and H be the cyclic group of order p generated by σ. The group H acts on X
by permuting x1, x2. . . , xp.The set X splits as a disjoint union of orbits say O1, O2, . . . , Or.
The orbit of a= (1,1, . . . ,1) is O1 ={a}. If we can find another orbit consisting of only one
element say (x1, x2. . . , xp) then x1 =x2 =· · ·=xp =x which means that x is an element of
order p.Note that
np−1=|O1|+|O2|+· · ·+|Or|.
The cardinality of any orbit is either 1 or p since |H| = p. If O1 is the only orbit with one
element thenpdoes not divide the right hand side of this equation while it dividesnp−1.Hence
there is one more orbit with only one element. Thus Ghas an element of order p.
The class equation
Any group Gacts on itself by conjugation:G×G−→G, (g, x)−→gxg−1.Under this action
the orbit of x∈G isO(x) ={gxg−1 |g∈G}.The orbit O(x) is called the conjugacy class of x.It is denoted by C(x).The stablizer ofx is the subgroup
called the centralizer of x.It is clear that Z(G)⊂Z(x) for allx∈G. If Gis finite then by the orbit-stablizer formula, we have|G|=|C(x)||Z(x)|.In other words the number of elements
in a conjugacy class C(x) is the index of the centralizerZ(x).Note thatx∈Z(G) if and only if the conjugacy class of x consists of x alone. Let C(x1), C(x2), . . . , C(xr) be all the disjoint
orbits of the noncentral elements x1, x2, . . . , xr.SinceG is the disjoint union of the orbits, we
have the class equation of G:
|G|=|Z(G)|+
r
X
i=1
[G:Z(xi)].
Example 1.8. There are three conjugacy classes inS3.
C(1) ={1}, C((12)) ={(12),(13),(23)} C((123)) ={(123),(132)}.
Example 1.9. Conjugacy classes in Sn. Let σ = (a1a2. . . ar) be an r-cycle in Sn. Let τ ∈Sn.Then
τ στ−1 = (τ(a1)τ(a2). . . τ(ar)).
Hence the conjugacy class of an r-cycle is the set of all r-cycles in Sn. Any r-cycle can be written in r different ways. For example (123) = (231) = (312).Hence the number of r-cycles
is n(n−1)(n−2). . .(n−(r−1))/r. Thus
|Z(σ)|=n!r/n(n−1)(n−2). . .(n−r+ 1) = (n−r)!r.
This count helps us in determination ofZ(σ).If τ is a permutation on then−r symbols not
appearing in σ then τ σ = στ. Hence τ ∈ Z(σ). There are (n−r)! such permutations. The powers (1), σ, σ2, . . . , σr−1,commute with σ.Hence
T ={τ σi|τ ∈Sn−r, i= 1,2, . . . , r} ⊂Z(σ).
Since T and Z(σ) have the same cardinality,T =Z(σ).
Definition 1.10. A sequence of positive integers m1 ≤m2 ≤ · · · ≤mr is called a partition
of an integer nif n=m1+m2+· · ·+mr.The number of partitions of nis denoted byp(n).
Theorem 1.11. The number of conjugacy classes in Sn is p(n).
Proof. Recall that any permutation is a product of disjoint cycles in a unique way except
for the order in which we write the cycles. Let σ be a product of disjoint cycles of length
k1, k2, . . . , ks where 1 ≤k1 ≤ k2 ≤ · · · ≤ ks ≤n and k1+k2 +· · ·+ks = n. The conjugacy
Example 1.12. Let us find the class equation ofG=SL(2,F3).Its order is 12(32−3)(32−1) =
24.One can show thatZ(G) ={I,−I}.We find the order of the conjugacy class of the matrix
A =
"
0 −1
1 0
#
.This is the index of its centralizer Z(A) = {B ∈G | AB =BA}. One can
show that there are four matrices in Z(A).Hence |C(A)|= 6. The characteristic polynomial
CA(x) =x2+ 1.All the matrices in C(A) have the same characteristic polynomial. Therefore, to find another conjugacy class, we find a matrix whose characteristic polynomial is notx2+ 1.
Check that for the matrix B =
"
1 1
0 1
#
, CB(x) = (x−1)2. Moreover |Z(B)| = 6. Thus
|C(B)| = 4. One can find other conjugacy classes in a similar way and show that the class
equation of Gis 1 + 1 + 4 + 4 + 4 + 4 + 6.
Burnside’s formula for the number of orbits
LetGbe a finite group acting on a finite setX.Then there are finitely many orbits. Burnside’s
formula gives the number of orbits in terms of fixed points of permutations. Let g∈G. Let
F ix(g) ={x∈X|gx=x}.
Let mg denote the permutation induced on X by g. Then F ix(g) is the subset of X fixed
pointwise bymg.
Proposition 1.13. For an h∈G, the bijectionx−→hx of X induces a bijection F ix(g)←→F ix(hgh−1).
Proof. Note thatx∈F ix(g)⇐⇒gx=x⇐⇒hgh−1hx=hx.
Theorem 1.14 (Burnside’s Theorem). Let a finite group G act on a finite set X. Then the number of orbits is
1
|G|
X
g∈G
|F ix(g)|.
In other words, the number of orbits is the average number of fixed points.
Proof. Let F(g) denote the number of fixed points of g.Consider the set
Then for a fixed g∈ G, there are F(g) elements in S. For a fixed x there are Gx such pairs. Hence
|S|=X
g∈G
F(g) = X
x∈X
|Gx|.
Let X1, X2, . . . , Xk be the orbits of this action. Using the fact that Ggx = gGxg−1 and the
orbit-stabilizer formula, we have|S|=Pk
i=1|Xi||Gxi|=k|G|where we choose onexi from Xi.
Thus
k= |S|
|G| =
1
|G|
X
g∈G F(g).
2
Groups of symmetries of figures
In this section we introduce the mathematical concept of symmetry. Let d(u, v) denote the distance between two points u, v ∈ Rn. Hence d(u, v) = ||u−v|| where ||u|| =
√
u.u. Recall
that u.v=utv where At denotes the transpose of a matrixA.
Definition 2.1. By a figure we mean a set of X points in Rn.A map T :Rn−→Rn is called
an isometry ofRn ifd(u, v) =d(T(u), T(v))for allu, v∈Rn.The mapT :Rn−→Rn is called
a symmetry of X if T(X) =X and T is an isometry of Rn.
Example 2.2. The matrix of rotation of R2 by an angle θ in counterclockwise direction is
given by:
A=
"
cosθ −sinθ
sinθ cosθ
#
Note that the column vectors as well as the row vectors form an basis ofR2andAtA=I. Such
a matrix is called an orthogonal matrix. Let T :R2 −→ R2 denote the linear transformation
induced byA.ThenT is one-to-one as Ais invertible. Moreover T preserves distance between
points. Indeed, Au.Av= (Au)tAv=utAtAv=utv. Hence
d(u, v) =p(u−v).(u−v) =p(T u−T v).(T u−T v) =d(T u, T v).
Example 2.4. Rotations by any angle θ are symmetries of the unit circle centered at the origin. Moreover reflections with respect to any line passing through the origin are also its
symmetries.
Proposition 2.5. The symmetries of a figure X in Rn form a group under composition of
maps. This group is called the group of symmetries of X.
Proof. Note that composition of distance preserving maps is again distance preserving. More-over, distance preserving maps are one-to-one maps. Hence symmetries of X are distance
preserving bijections of X. The identity map is clearly distance preserving. The inverse of a symmetry ofX is again a symmetry ofX. Thus the symmetries of X constitute a group.
3
Orthogonal maps of
R
n: The Cartan-Dieudonn´
e Theorem
Definition 3.1. An n×n real matrix A is said to be orthogonal if the column vectors of A form an orthonormal basis of Rn.Equivalently AtA=I.
In this section we shall prove the Cartan-Dieudonn´e Theorem which shows that all
orthog-onal matrices are product of matrices which induce reflections in hyperplanes.
Example 3.2. Let u be a nozero vector in Rn.Consider the Householder matrix of u,
H=I −2 uu
t
kuk2.
Then Hu=u−2u(u
tu)
kuk2 =−u. If w ⊥ u then Hw=w−2
uutw
kuk2 =w. Let L(u) denote the
linear span of the vector u.For any subspace V of Rn, define:
V⊥={u∈Rn|u.v= 0 ∀ v∈V}
Proposition 3.3. For any subspaceV of Rn,we have Rn=V ⊕V⊥.
Proof. It is clear thatV ⊕V⊥ ⊂Rn.Let{v
1, v2, . . . , vm}is an orthonormal basis of V. Define T :Rn−→V by:
T u=
m
X
i=1
(u.vi)vi
The set {v1, v2, . . . , vm} can be extended to an orthonormal basis {v1, v2, . . . , vm, . . . , vn} for
Rn.T is clearly onto and
ker(T) ={u∈Rn|u.v
Hence by the rank-nullity theorem, n= dimV + dimV⊥.Note that u =T u+ (u−T u). We show thatu−T u∈V⊥.
(u−T u).T u =u.T u−T u.T u
=
n
X
i=1
(u.vi)vi
!
. m
X
i=1
(u.vi)vi
!
−
m
X
i=1
(u.vi)2
= 0
Hence (u−T u)⊥T u.Hence Rn⊂V ⊕V⊥. ThereforeRn=V ⊕V⊥.
By this proposition Rn=L(u)⊕L(u)⊥. Hence L(u)⊥ is an (n−1)- dimensional subspace
of Rnwhich is perpendicular toL(u).
Definition 3.4. A linear transformation T :Rn→ Rn is called a reflection with respect to a
hyperplane H of dimension(n−1)if T u=−u where u⊥H and T u=u for all u∈H.
Thus H =I −2kuuukt2 is a reflection with respect to the hyperplane L(u)⊥. Moreover H is
an orthogonal symmetric matrix with H2=I. Indeed
HtH =
I−2 uu
t
kuk2 I−2
uut
kuk2
=I−2 uu
t
kuk2 −2
uut
kuk2 + 4
uutuut
kuk4
=I
Hence H is symmetric and orthogonal.
Definition 3.5(Orthogonal Transformation). Let V be a vector space with an inner prod-uct. A linear transformation T :V →V is called orthogonal ifkT uk=kuk for all u∈V.
Theorem 3.6. Let V be a finite dimensional inner product space and T :V →V be a linear transformation. Then the following are equivalent:
(1) T is an orthogonal transformation.
(2) (T u, T v) = (u, v) for all u, v∈V where (u, v) represents the inner product of u and v.
(3)If{u1, u2, . . . , un}is an orthonormal basis ofV then{T u1, T u2, . . . , T un}is an orthonormal
basis of V.
Proof. (1)⇒(2). Let T be an orthogonal transformation. Since kT(u+v)k =ku+vk for all
u, v∈V, we have
(T u+T v, T u+T v) = (T u, T u) + 2(T u, T v) + (T v, T v)
= (u, u) + 2(u, v) + (v, v)
Hence (T u, T v) = (u, v).
(2)⇒(3). Let{u1, u2, . . . , un}be an orthonormal basis ofV. Then {T u1, T u2, . . . , T un}is also
an orthonormal basis ofV sinceT preserves inner product.
(3)⇒(1). Suppose for some orthonormal basis{u1, u2, . . . , un},{T u1, T u2, . . . , T un}is also an
orthonormal basis. Let u=x1u1+· · ·+xnun for somex1, . . . , xn∈R. Then
T u =x1T u1+· · ·+xnT un
(T u, T u) =x21+· · ·+x2n=kuk2
(3) ⇒ (4). Let B = {u1, . . . , un} be an orthonormal basis of V and T uj = Pni=1aijui for j = 1,2, . . . , n. We show that A = (aij) is an orthogonal matrix. Since T u1, . . . , T un is an
orthonormal basis, we get
1 = (T uj, T uj) =
n
X
i=1
aijui, n
X
i=1
aijui
!
=Xa2ij.
Forj 6=kwe have
(T uj, T uk) = n
X
i=1
aijui, n
X
i=1
aikui
!
=
n
X
i=1
aijaik= 0.
(4)⇒(3). IfA is orthogonal then the above equations show thatT maps an orthonormal basis to an orthonormal basis.
Theorem 3.7 (Cartan-Dieudonn´e Theorem). Every orthogonal transformation of a real inner product space V of dimension n is a product of at most n reflections.
Proof. Apply induction on n= dimV. Suppose n= 1. ThenV =L(u) for some unit vector u.
Therefore kT uk = 1. HenceT u =±u. Now let n ≥2 and assume the theorem holds true for orthogonal transformations on (n−1)- dimensional real inner product spaces.
Hence T u ∈ L(x)⊥. Hence T : W → W is an orthogonal transformation of an (n − 1)-dimensional inner product space W.By the induction hypothesis, there exists reflections ofW
says1, s2, . . . , sr such that r≤n−1 and
T|W =s1s2· · ·sr
Define Ti :V → V by Ti(x) = x and Ti|W =si. Let Hi be the hyperplane in W fixed by si.
ThenVi=Hi⊕L(x) is fixed byTi. LetHi⊥⊂W andHi⊥=L(xi) fori= 1,2, . . . , r. Note that
xi ⊥x andxi ⊥Hi. Hencexi ∈Vi⊥. ThusV =Vi⊕L(xi). We show that eachTi is a reflection in V. Since Ti(xi) = si(xi) = −xi and Ti(u) = u for all u ∈ Vi, therefore Ti is a reflection.
From the definition of Ti, it follows that T =T1T2· · ·Tr.
O
x
x+T x
T x
Case 2 :SupposeT x6=xfor anyx∈V. Thenu=T x−x6= 0 and let H =L(u)⊥ and U be a reflection with respect H. Then
(T x+x, T x−x) = (T x, T x) + (x, T x)−(T x, x)−(x, x) = 0
So T x+x ⊥u⇒ T x+x∈H. Hence U(T x+x) = T x+x
and U(T x−x) =x−T x
Note that
U(T x+x) =U T x+U x =T x+x and
U(T x−x) =U T x−U x =x−T x
Therefore U T x =x and by Case 1, U T =T1T2· · ·Ts for some reflections T1, T2, . . . , Ts of V
and s≤n−1. Hence
U U T =T =U T1T2· · ·Ts
is a product of at mostn reflections.
Corollary 3.8. Let A∈SO(3). Then1 is an eigenvalue of A.
Proof. If A=I then the result is clear. So letA6=I.LetT be the orthogonal transformation ofR3 induced byA.Since detA= 1,by the Cartan-Dieudonn´e Theorem, T =R1R2 whereR1
and R2 are distinct reflections with respect to planesH1 and H2 respectively. Then
dim(H1+H2) = 3 = dimH1+ dimH2−dim(H1∩H2) = 4−dim(H1∩H2).
It follows that dim(H1 ∩H2) = 1. Since the vectors in H1 and H2 are fixed by R1 and R2
respectively, any u∈H1∩H2 is fixed byT.ThusT u=u.Hence uis an eigenvector of Awith
4
Isometries of
R
nThroughout this section V denotes the vector space Rn.
Definition 4.1. A map m:Rn−→Rn is called an isometry ‘ofRn if ||m(u)−m(v)||=||u−v||.
It is clear that an isometry of Rn is a one-to-one map and composition of isometries is an
isometry. We shall show that isometries of Rn form a group.
Example 4.2. (1) LetT :Rn−→Rn be an orthogonal linear transformation. Then ||T(u)−T(v)||=||T(u−v)||=||u−v||
for all u, v∈Rn.Hence T is an isometry.
(2) Let w ∈ Rn and tw : Rn −→ Rn be the translation map: tw(u) = w+u for all u ∈ Rn.
Then tw is an isometry.
We wish to show that all isometries are composed of orthogonal transformations and trans-lations. First we show that any isometry fixing the origin is an orthogonal transformation.
Lemma 4.3. Let f :V −→V be a function which preserves dot product and f(ei) =ei for all i= 1,2, . . . , n. Thenf =Id.
Proof. Note that (f(u), ei) = (f(u), f(ei)) = (u, ei) for all i= 1,2, . . . , n. Hence f(u) = u for all u∈V. Hencef is the identity map of V.
Theorem 4.4. The following are equivalent for a map m:V −→V.
(1) m:V −→V is an isometry which fixes the origin, i.em(0) = 0.
(2) m preserves dot product of vectors, i.e. m(u).m(v) =u.v for all u, v∈V.
(3) There exists an n×n orthogonal matrixA so that m(u) =Au for allu∈V.
Proof. (1) ⇒ (2) Sincemis an isometry fixing the origin, for any u∈V we have
||m(u)−m(0)||=||m(u)||=||u||.
Hence for any u, v∈V we have
||m(u)−m(v)||2 = (m(u)−m(v)).(m(u)−m(v))
= m(u).m(u)−2m(u).m(v) +m(v).m(v)
= (u−v).(u−v)
Hence it follows thatm(u).m(v) =u.v for allu, v∈V.
(2)⇒(3) Sincempreserves dot product of vectors, the matrixA= [m(e1)m(e2) . . . m(en)] is
orthogonal. Let Lbe the linear transformation induced by the orthogonal matrix A−1.Hence
L andm both preserve dot product of vectors. Hence for alli= 1,2, . . . , nwe have
L◦m(ei) =A−1A(ei) =ei.
By the previous lemma L◦m =Id. Hence m =L−1. Since L−1 is the linear transformation
induced by the orthogonal matrix A,it follows that m(u) =Aufor all u∈V.
(3)⇒ (1) We have proved this in Theorem (3.6).
Theorem 4.5 (Structure of isometries of Rn). Let f :Rn−→Rn be an isometry. Then
f(u) =Au+w for allu∈Rn
for some orthogonal matrix A and a vector w∈Rn.
Proof. Let f(0) =w.Then
(t−w◦f)(0) =f(0)−w= 0.
Thereforet−w◦f is an isometry fixing the origin. Hence it is a linear transformation induced
by an orthogonal matrix A∈Rn×n.Thusf(u) =Au+wfor all u∈Rn.
Theorem 4.6. (a)The set Mn of isometries of Rn is a group under composition of maps.
(b) Let On and Tn denote the group of orthogonal transformations of Rn and the group of
translations of Rn. Then Tn is a normal subgroup of Mn and
Mn Tn 'On.
(c) Mn=TnOn.
Proof. Let tvTAand twTB be inMn where A, B∈On and v, w∈Rn.Then
tvTAtwTB(u) =tvTA(Bu+w) =ABu+Aw+v=tAw+vTAB(u).
ThusMnis closed under composition. Bothtv andTAare bijections ofRn.Hence all isometries
are bijections. The inverse of twTA is TBt−w whereB =A−1.Note thatTBt−w(u) =B(−w+ u) = Bu−Bw = t−BwTB(u). Therefore inverse of every isometry is again an isometry. It is
clear that Tn is a group.
Define φ:Mn−→On byφ(twTA) =TA.Then φis a group homomorphism. The Kernel is
Tn.Hence Tn is a normal subgroup ofMn and Mn/Tn'On.
5
Isometries of the plane
In this section we classify the isometries of the plane algebraically and geometrically. Let M
denote the group of isometries of the plane. We shall show that M is a non-abelian group generated by the following isometries:
1. Translations :tw(u) =w+ufor all u∈R2 and w∈ R2.
2. Rotation of R2 in anticlockwise direction by an angle θ:
ρθ
"
u1
u2
#
=
"
cosθ −sinθ
sinθ cosθ
# "
u1
u2
#
.
3. Reflection with respect to thex-axis:
r
"
u1
u2
#
=
"
u1
−u2
#
.
Theorem 5.1. Any isometrym of the plane if of the formm=tvρθ orm=tvρθr for uniquely deteminedv and θ which may be zero.
Proof. We know that m=tvφwhere v∈R2 and φis an orthogonal map. A 2×2 orthogonal
matrix is either a rotationρθ or a reflectionr0 with respect to a line passing through origin. If
this line is having an angle ofθ/2 with respect to the x-axis thenr0 =ρθr.
Proposition 5.2. The Matrices inSO2 represent rotations of R2.
Proof. LetA∈SO2 andAe1 =v1. There is a rotation of R2, sayRsuch thatRe1=v1. Then
B =R−1Afixese1 andB ∈SO2. Hence the second column ofB ise2 or−e2. Since detB = 1,
it must be e2. Therefore B = R−1A = I and hence A =R which is a rotation of R2 in the
anti-clockwise direction by the angle between e1 and v1.
Formulas for computations in M :
ρθtv=tρθ(v)ρθ rtv =tr(v)r rρθ=ρ−θr
tvtw =tv+w ραρβ =ρα+β rr= 1
Theorem 5.3. Every isometry of the plane is one of the following:
(1) translation: tw(u) =u+w,
(2) rotation of the plane about a point through an angle θ, tvρθ,
(3) reflection about a line,
(4) glide reflection, i.e. reflection about a line followed by a translation by a vector parallel to the line.
Proof. (1) and (2) : Any m∈M is uniquely written as tvφwhere φis an orthogonal transfor-mation. If m is not a translation then φ 6= Id and it is either a rotation or a reflection with
respect to a line`passing through the origin. Ifm=tvρθ then it is a rotation through a point
p in the plane. Let us find this fixed pointpof m.Lettvρθ(p) =p.Thenv+ρθ(p) =p.Hence
(Id−ρθ)(p) =v.Since θ6= 0,det(Id−ρθ)6= 0.Hence (Id−ρθ) is an invertible map ofR2.Thus
p is unique. Hence m has a unique fixed point. Any q ∈ R2 can be written as q = s+p for
somes∈R2.Thus
m(s+p) = tvρθ(p+s)
= v+ρθ(s) +ρθ(p)
= p+ρθ(s).
Thereforem is a rotation aroundp by θ.
(3) and (4): Now letm=tvφwhereφis a reflection about the line`:y=xtan(θ/2).We may change the co-ordinates so that φ=r is the reflection about thex-axis. Ifv = 0 thenm is a
reflection. If v6= 0 then we show thatm is a glide reflection. Letv= (v1, v2)t.Then
m
"
x
y
#
=
"
x+v1 −y+v2
#
Hencem(x, v2/2)t= (x+v1, v2/2)t.Hencemis a glide motion parallel to the liney=v2/2.A
vector (x, y)t is first mapped to (x+v1, y)tand then reflected with respect to the liney=v2/2
to yield (x+v1,−y+v2)t.
6
Finite subgroups of
M
In this section we find all finite subgroups of M. We shall show that any such group is a conjugate of a finite subgroup of the groupO2.We shall then show that any finite subgroup of
group of rotations and reflections. A key observation for characterisation of all finite subgroups of M is the following:
Theorem 6.1 (A fixed point theorem). Let G be a finite subgroup of Mn.Then there is a common fixed point of all the isometries in G.
Proof. Step 1: First we show that the centre of gravity of {s1, s2, . . . , sn} is mapped by any
isometry m ∈ Mn to the centre of gravity of {m(s1), m(s2), . . . , m(sn)}. For this we use the
fact that m=tvT wherev∈Rn and T is an orthogonal transformation ofRn.Then
tvT
Pn
i=1si
n
= v+
Pn
i=1T(si)
n
= nv+
Pn
i=1T(si) n
=
Pn
i=1T(si) +v n
=
Pn
i=1m(si) n
Step 2: There is a fixed point of all isometries inG. Indeed, for any s∈R2 consider the
orbit of sunder the action ofG:
O(s) ={m(s)|m∈G}.
Let O(s) = {s1, s2, . . . , sn}. If m ∈ G then m(s1), m(s2), . . . , m(sn) is a permutation of
s1, s2, . . . , sn.Hence the center of gravity of s1, s2, . . . , sn is a fixed point of every m∈G.
We now show that all finite subgroups of Mn are conjugates of finite subgroups of On.
Theorem 6.2. Let G be a finite subgroup of Mn and s be a fixed point of all m ∈ G. Then t−sGts is a finite subgroup ofOn.
Proof. Letm∈G. Thent−smts(0) =t−sm(s) =s−s= 0.Hencet−smtsis an isometry which
fixes the origin. Hence it is an orthogonal transformation. Hencet−sGtsis a finite subgroup of
On.
Finite subgroups of O2
Theorem 6.3. A finite subgroup ofO2 is one of the following groups:
(aThe cyclic group Cn generated by the rotation ρ2π/n for a positive integer n.
Proof. Step 1: Let G be a finite subgroup of SO2 having order n ≥ 2. Then G consists of
finitely many rotations about the origin. Let θ be smallest positive angle so that ρθ ∈G. We
show thatθ= 2π/n for some positive integernandG=< ρθ > .Letρα∈Gandα6= 0.Then
α =mθ+β for some positive integer m and an angle β < θ. Hence ρβ =ρα−mθ ∈G. Hence
β = 0.Since |G|=n, ρnθ= Id.Thus G=< ρθ>and θ= 2π/n.
Step 2: Now let G be a finite subgroup of O2 containing a reflection r0 about a line `.
By change of coordinates, we may assume that r0 = r, the reflection about the x-axis. Let
H = G∩SO2.Then H =< ρθ >where θ = 2π/n for some n∈N. If g∈ G\H thengr is a
rotation. Hence gr∈H =G∩SO2.Thusg∈rH. Therefore G=Dn=H∪rH.
7
Symmetries of the Platonic solids
In this section we find the groups of symmetries of the Platonic solids. If we join the centers of adjacent faces of the cube, we obtain the octahedron. Hence they have same symmetry
groups of rotations which is denoted as O. Similarly, the rotational symmetry groups of the dodecahedron and the icosahedron are equal which is denoted by I.The group of rotational symmetries of the tetrahedron is denoted byT.
the faces of the tetrahedron. Hence the four faces form an orbit. The stablizer group of each face is cyclic of order 3.Henceo(T) = 12.Consider the axisLjoining a vertex with the center
of the opposite face. The rotations by 1200 and 2400 about L are in T. This is true for each of the four vertices. These rotations are represented by 3-cycles. Hence the group has all the 8
3-cycles inS4.Rotations through the angleπ about the axis joining the midpoints of opposite
edges give rise to 3 symmetries represented by product of two disjoint transpositions. Along with the identity map we have total of 12 rotations which are all even permutations on the
four vertices. Hence T 'A4.
Example 7.2. The group of rotational symmetries of the cube: The group O acts transitively on the six faces of the cube. The stabilizer of each face is a cyclic group of order 4.
Hence the order ofO is 24.The group acts on the set of four body diagonals of the cube. This
action gives rise to a group homomorphism φ : O −→ S4. The Kernel of φ consists of those
rotations which fix all the four body diagonals. But only identity rotation has this property.
Hence O 'S4.
Example 7.3. The group of rotational symmetries of the icosahedron:The icosahedral groupI has 60 rotations. The 20 faces form a single orbit. The stabilizer of each triangle has 3 elements and there are 20 triangles. Hence the orbit-stabiliser formula gives|I|= 60. There
are 12 vertices and 6 axes joining opposite vertices. There are four nontrivial rotations with each of these axes as an axis of rotation. These give 24 rotations different from identity. There
are ten axes joining the centres of the opposite faces as there are 20 faces. These give us 20 nontrivial rotations. There are 30 edges which give 15 axes joining centres of opposite edges and the identity. Hene we have
|I|= 6·4
|{z}
vertices
+
f aces
z }| {
10·2 + 15·1
| {z }
edges
+1 = 60
Rotations of angle 2π/5 about the center of each pentagon in a dodecahedron are conju-gates. The rotations of 4π/5 around these axes are conjugates. These two conjugacy classes
have 12 elements each. There are 15 axes joining 30 edges. Rotations of angleπ constitute one conjugacy class. Thus the class equation of I is
60 = 1 + 20 + 12 + 12 + 15
Theorem 7.4. The groupI is a simple group of order 60.
Theorem 7.5. I 'A5. Hence A5 is a simple group.
Proof. There are five cubes in an icosahedron. I operates on them transitively. Hence we get
a homomorphism ψ : I → S5. Since I acts non trivially, kerψ 6= I. Since I is simple,
ψ is injective. Let σ : S5 → {±1} be the sign homomorphism. Then σψ : I → {±1} is a
homomorphism. But Im(σψ)6={±1}since otherwise ker(σψ) is a nontrivial normal subgroup
ofI. ButI is simple. Henceσψ(I) ={1}. Thus Imψ consists of only even permutations in
S5. So I ∼=A5. SinceI is simple,A5 is also simple.
8
Finite subgroups of
SO
3In this section we find all the finite subgroups of SO3. We shall prove the following important
theorem:
Theorem 8.1. Let Gbe a finite subgroup of rotations of R3.Then G is isomorphic to one of
the groups in the list:
(1) The cyclic group Cn of rotations of the plane of a regular n-gon.
(2) The group of symmetries of the dihedral groupDn of order 2n.
(3) The groups of symmetries of rotations of the five Platonic solids:T,O and I.
The set of poles of rotations: The axis of a nontrivial rotation ρ(u,α) of R3 intersects the
rotation. LetP denote the set of all the poles of nonidentity rotations inG. This is clearly a finite set. We show thatGacts on P and there can be atmost three orbits of this action.
Theorem 8.2. The group G acts on the set P of poles of nontrivial rotations in G. There are either two or three orbits of this action.
Proof. Let g ∈ G\ {e} and p be a pole of g. We show that hp is a pole of hgh−1. Indeed,
hgh−1(hp) =hgp=hp. Since h is a rotation, ||hp||=||p||= 1.Since hgh−1 6=e, we conclude that hpis a pole ofhgh−1.Hence the map
G×P −→P, (h, p)7→hp
defines an action ofG on the set of poles.
Now we show that there are at most three orbits of this action. For a pole p ∈ P, the stabiliser of p is the subgroup
Gp={g∈G|gp=p}.
Notice that if gp =p then g(xp) = xp for any real number x. Hence L(p) is the axis of the rotationg.Thusginduces a rotation of the plane perpendicular toL(p).Since the order ofGp
is finite, it is a cyclic group generated by, say, a rotation of an angle 2π/rpwhererp = orderGp.
Let O(p) denote the orbit of p under the action of G and np be the number of poles in the orbitO(p).Hence the orbit-stabiliser formula givesN :=rpnp =|G|.
Each g 6= e has two poles. Each p ∈ P, is a pole of rp−1 rotations in Gp. We say that (g, p) is a spin ifg6=eand pis a pole of g.Ifpis fixed then there arerp−1 spins (g, p) and if
g is fixed then there are two spins (g, p).Thus the number of spins equals
X
p∈P
(rp−1) = 2(N−1).
LetO1, O2, . . . Okbe the distinct orbits of the action ofGonP.Let|Oi|=nifori= 1,2, . . . , k.
The stablizers of any two poles in an orbit are conjugates of each other. Hence they have equal
order. Therefore
k
X
i=1
ni(ri−1) = 2N −2.
Divide by N to get
k
X
i=1
1− 1
ri
= 2− 2
N (1)
Since N ≥2,the right hand side of the equation (1) is at least 1 and at most 2.Hence k≤3.
Theorem 8.3. (1) If there are two orbits for the action of GonP thenGis a cyclic group.
(2) If there are three orbits of ordersn1 ≥n2 ≥n3 then there are four possibilities forG:
r1 r2 r3 n1=E n2 =F n3=V N Group
2 2 N/2 N/2 1 N/2 N DN/2
2 3 3 6 4 4 12 T
2 3 4 12 8 6 24 O
2 3 5 30 20 12 60 I
Here V, E, F denote the number of vertices, edges and faces respectively of the Platonic solid.
Proof. (1) Suppose that there are two orbits. Then the equation (1) is reduced to:
1
r1
+ 1
r2
= 2
N.
Hence n1+n2= 2.Thusn1 =n2 = 1.Thus there are two poles p1 andp2 both fixed by each
element of G. The axis L joining p1 and p2 is the axis of all rotations in G. The rotations in
G map the plane W perpendicular to L to itself. Hence each g ∈G induces a rotation of W.
HenceGis cyclic of order N.The regular polygon is located along the equator of the 2-sphere and the axis is L.
(2) Now assume that there are three orbits. Then the equation is
1
r1
+ 1
r2
+ 1
r3
= 1 + 2
N.
Since the right-hand-side of the above equation is bigger than 1, it follows that r1 = 2 after
arranging the orders asr1≤r2 ≤r3.
z
−z x
−x
−y
Case 1: Let r1 = r2 = r3 = 2. Then N = 4. Let x, y, z be the poles so that |Gx| = |Gy| = |Gz| = 2. This implies that G is the Klein’s 4-group. Suppose that (g) = Gz. Since
g preserves distance, the poles x and g(x) are equidistant from z. Same is true for y and
g(y). Since any 1 6= g ∈ G restricted to the plane perpendicular to the axis of the rotation
g induces a rotation of 1800,it follows that the three orbits are {x,−x},{y,−y}and {z,−z}
and g(x) =−x, g(y) =−y.The axes joiningx,−xand y,−y are perpendicular to each other.
z
−z
x
g(x) g2(x)
g3(x) g4(x)
g5(x)
Plane W
Case 2: Let r1 = r2 = 2. Then r3 = k ≥ 3 and N = 2k. Let O1 = O(x), O2 = O(y) and
O3 =O(z).Then Gx, Gy have order 2 and o(Gz) =k.The axis L through z is fixed by every
rotation inGz.These rotations map the planeW perpendicular to the axisLtoW.Hence Gz
is cyclic of order k.LetGz = (g).
Observe that the poles x, g(x), . . . , gk−1(x) are all distinct. Suppose to the contrary that 1 ≤ r < s≤ k−1 and gr(x) = gs(x). Then gs−r(x) = x. Since gs−r 6= 1,and it fixes z and
−z, x cannot be fixed by gs−r as any rotation has only two poles. Moreover x 6= −z since
otherwiseGx =G−z=Gz.ButGxand Gz have different order. Hencex, g(x), . . . , gk−1(x) are kdistinct poles. SinceO(x) haskpoles in it,O(x) ={x, g(x), . . . , gk−1(x)}.Sincegis distance
preserving and g(z) =z,we have
||x−g(x)||=||g(x)−g2(x)||=· · ·=||gk−1(x)−x||.
Hence x, g(x), . . . , gk−1(x) are the vertices of a regulark-gon P. Any rotation in G induces a
permutation of the vertices ofP,we conclude that P is mapped to itself by anyG∈G.Since the group of symmetries ofP has orderk, G must be the dihedral group of symmetries ofP.
Case 3:Now let r1= 2< r2≤r3.If r2 ≥4 then
1
r2
+ 1
r3 ≤ 1
2 < 1 2 +
2
which is a contradiction. Hence r2 = 3.Then the equation for r3 is
1
r3
= 1 + 2
N −
1 2−
1 3 =
1 6 +
2
N.
Hencer3 ≤5.In factr3 = 3,4,5 are possible cases. These cases lead to the groups of rotational
symmetries of the Platonic solids.
z
−z
u g(u)
g2(u)
Case 3a : Letr3= 3.ThenN = 12 andn1 = 6, n2 = 4, n3 = 4.LetO1=O(x), O2 =O(y)
and O3 =O(z).The orbit of zhas 4 points. Let u∈O(z) so that ||u−z||<2.Let (g) =Gz.
Thenu, g(u), g2(u) are all distinct poles and they are equidistant fromz.Hencez, u, g(u), g2(u) are vertices of a regular tetrahedron T. Every rotation on G is a symmetry of T. Since the
group of rotational symmetries ofT and Ghave same order, it follows thatG=T.
u
g(u)
g2(u)
g3(u)
z
Case 3b: Let r3 = 4. Then N = 24 and n1 = 12, n2 = 8, n3 = 6. The poles in the third
orbit are the vertices of a regular octohedron. The group G is the group O of its rotational
symmetries. There are six points in the third orbit of z. Let g be a generator of Gz. Let
u 6= z,−z. Then the four points u, g(u), g2(u), g3(u) are equidistant from z and they are the
vertices of a square. SinceO(z) has six points, it follows that
O(z) ={z,−z, u, g(u), g2(u), g3(u)}.
Observe that−u∈O(z).Certainly −u=6 z,−z and it cannot beg(u) or g3(u) since
||g(u)−u||=||g3(u)−u||<2.
Therefore −u = g2(u). It follows that z, u, g(u), g2(u), g3(u),−z are the vertices of a regular octahedron. ThusG=O.
Case 3c: Let r3 = 5. Then N = 60 and
n1 = 30, n2 = 20, n3 = 12. Let us show that
the poles in the third orbit are vertices of a
regular icosahedron and the group G is the groupI of rotational symmetries of an icosa-hedron. Chooseu, v∈O(z) =O3 so that
0<||z−u||<||z−v||<2.
Letg be a generator of the cyclic groupGz.
Then u, g(u), g2(u), g3(u), g4(u) are all distinct and equidistant from z. These five points are vertices of a regular pentagon. Similarly v, g(v), g2(v), g3(v), g4(v) are vertices of a
reg-ular pentagon and they are equidistant from z. The twelfth point in O(z) is −z. There-fore −u ∈ O(u) = O(z). Since the distance between u and −u is 2, −u is one of the points v, g(v), g2(v), g3(v), g4(v). Without loss of generality, we may set −u = v and thus −gr(u) =gr(v) for all r = 1,2,3,4. The five points closest to u must be equidistant from u.
These points are z, g(u), g3(v), g2(v), g4(u). Therefore ||u−z|| = ||u−g(u)|| = ||u−g2(v)||.
9
Exercises
Group Actions
1. LetG=GLn(R) operate on the set S=Rn by left multiplication. Describe the
decom-position ofS into orbits. What is the stabilizer of e1?
2. LetF :=F3.There are 4 one-dimensional subspaces of the two-dimensional vector space
F2.Describe them. Left multiplication by an invertible matrix in F2×2 permutes these
subspaces. Prove that this operation defines a homomorphism φ:GL2(F)→S4.
Deter-mine the kernel and the image of this homomorphism.
3. Determine the order of the conjugacy class of the diagonal matrix diag(1,2) in the group
GL2(F5).
4. Determine the class equation for each of the following groups (a) the quaternion group
(b)D6 (e) D2n (f) the group of upper triangular matrices inGL2(F3) (g)SL2(F3).
5. LetN be a normal subgroup of a group G. Suppose|N|= 5 and|G|is odd. Prove that
N ⊆Z(G).
6. LetGbe a finite group and let H a proper subgroup ofGwith indexnso that|G|does
not divide n!.Show that Gis not simple.
7. Let G be a finite group with a proper subgroup H whose index is the smallest prime number pdividing|G|.Show that H is a normal subgroup of G.
8. LetGbe a group of orderpn wherepis a prime number andp > n. Show that ifH is a subgroup of orderp then it is a normal subgroup of G.
Rotations and reflections in R2 and R3
1. Show that a reflection is not a rotation inR2.
2. Show that a matrix in O2 induces either a rotation of a reflection.
3. Find the matrix of refection with respect to the line ` which is inclined at the angle θ
from the positivex-axis.
4. Prove that a linear operator on R2 is a reflection if and only if its eigenvalues are 1 and −1, and the eigenvectors with these eigenvalues are orthogonal.
Isometries of the plane
1. Letm be an orientation-reversing isometry. Prove algebraically thatm2 is a translation.
2. Prove that a conjugate of a glide reflection inM is a glide reflection, and that the glide vectors have the same length.
3. Write formula’s for the isometriesρθ, r andtv in terms of a complex variablez=x+iy.
4. Lets be the rotation of the plane with angle π2 about about the point (1,1)t.Write the formula forsas a producttaρθ.
5. Let s denote reflection of the plane about the vertical axis x = 1. Find an isometry g
such thatgrg−1=s, and write sin the form taρθr.
6. Letl1 and l2 be lines through the origin inR2 that intersect in an angle πn, and letri be
the reflection about li.Prove that r1 and r2 generate the dihedral group Dn.
7. Letf andgbe rotations of the plane about distinct points, with arbitrary non-zero angles of rotationθ and φ. Prove that group generated byf and g contain a translation.
8. Letmbe a glide reflection along a linel. Prove thatxlies onlif and only ifx, m(x), m2(x) are collinear. Conversely, prove that if m is an orientation-reversing motion and x is a
point such thatx, m(x), m2(x) are distinct points on a linel, then mis a glide reflection along l.
9. Let SM denote the subset of orientation-preserving motions of the plane. Prove SM is
a normal subgroup of M and determine its index inM.
Groups of isometries
1. Prove that On is not a normal subgroup ofMn.
2. Identify the group of all symmetries of a regular tetrahedron.
3. Determine the order of the group of symmetries of a dodecahedron, when orientation
reversing symmetries such as reflections in the plane are allowed.
4. Show that I has a subgroup of order 10.
5. Let G = D4 be the group of symmetries of a square. Find stabilizers of vertices and
References
[1] M. A. Armstrong, Groups and Symmetry, Undergraduate Texts in Mathematics,
Springer-Verlag, 1988.
[2] M. Artin, Algebra, Second Ed. Prentice Hall of India, 2011.
[3] H. S. M. Coxeter, Introduction to Geometry, Second Ed. Wiley, 1961.
[4] C. W. Curtis, Linear Algebra, An Introductory Approach, Springer-Verlag, 1984.
[5] J. Stillwell, Naive Lie Theory, Springer-Verlag, 2008
