Our second application of formula (9.1) is to find the finite subgroups ofSO(3). We already know the subgroupsT,O,I. The dihedral groups also can be realized
9.4. FINITE SUBGROUPS OFSO(3) 143
as groups of rotations of geometric objects. Take a regularn-gon in the plane. Construct a pyramid above it and one of the same height below it. A rotation of the n-gon in the plane can be extended to a rotation of the solid about the line joining the peaks of the two pyramids. A reflection can be extended to a rotation through an angle ofπabout the axis of the reflection. ThusDncan be
embedded inSO(3) as a group of symmetries of this solid. SinceDn contains
a cyclic subgroup of order n, it too is a subgroup ofSO(3). We shall see that these are essentially all the finite subgroups. The way we shall demonstrate this is to consider the fixed points of a group of rotations acting on the unit sphere
S2,
S2 :={v ∈R3 | ∥v∥= 1}.
Remark9.10. If α ∈ O(3) then for anyv ∈ R3, ∥αv∥ = ∥v∥, in particular if
∥v∥ = 1, then ∥αv∥ = 1. So O(3) acts onS2. Any subgroup of O(3), for
exampleT, also acts onS2.
Definition 9.11. If a groupGacts on a setX, the set offixed points ofGis {x|αx=xfor some α ̸= 1}={x|Gx ̸={1}}.
For example, takeG = T, acting onS2. Each non-trivial element in Tis a
rotation about an axis. The axis meetsS2 in a pair of antipodal points, which are fixed by the rotation. These two points belong to the set of fixed points. The rotations about a line through a vertex and the centre of the opposite face give
4pairs of fixed points. The rotations about an axis joining the midpoints of a pair of opposite edges give another3pairs. In the picture below the arcs on the sphere are the edges of an inscribed tetrahedron projected onto the sphere. The centre of one face is shown.
Theorem 9.12. Let G < SO(3) be a finite subgroup. ThenGis conjugate to a cyclic group, toDn,n ≥2, toT, toO, or toI.
Proof. As a subgroup ofSO(3), G acts on S2. Each non-trivial rotation in G
fixes the two points where its axis meets the sphere. The set of all such pairs of antipodal points is the set of fixed points ofG, which we shall denote byX. Now supposex∈ Xis fixed byα ∈G. Take anyβ ∈ G. Thenβxis fixed by
βαβ−1. Soβx∈X. ThusGacts onX.
LetO1, . . . , Osbe the orbits ofGinX. The stabilizer of a point in an orbit
Oj has order
nj :=|G|/|Oj|, (9.1)
for 1 ≤ j ≤ s, by equation (9.1). Since all the points in X have non-trivial stabilizers,nj ≥2.
We now count fixed points as in the proof of Burnside's lemma. Let
Y ={(α, x)|αx=x, α∈G\ {1}, x∈X}.
For fixedx,(α, x)∈Y if and only ifα∈Gx\{1}. Ifx∈Oj, then the number
9.4. FINITE SUBGROUPS OFSO(3) 145
elementsα. Summing overj, we then get |Y|=
s
∑
j=1
|Oj|(nj −1).
On the other hand, if we fixα ∈ G\ {1}, then(α, x) ∈ Y if and only ifxis a fixed point ofα. As we already noted, each rotationα has2fixed points. So summing overα∈G\ {1}, we obtain
|Y|= ∑ α∈G\{1} 2 = 2(|G| −1). Thus 2(|G| −1) = s ∑ j=1 (nj −1)|Oj|.
Substitute the value of|Oj|from equation (9.1):
2(|G| −1) = s ∑ j=1 nj −1 nj |G|=|G|s− |G| s ∑ j=1 1 nj .
Now divide through by|G|and rearrange terms:
s ∑ j=1 1 nj =s−2 + 2 |G| . (9.2)
This is the equation we must analyze. First, notice that since allnj ≥ 2, each
term on the left is at most1/2. So we have the inequality
s
2 ≥s−2 + 2
|G| > s−2,
which implies that
2> s
2 .
Thuss≤3. This leaves us with three cases to discuss. (i) s = 1.
Equation (9.2) becomes:
1
n1
= 2
|G| −1≤0,
since|G| ≥2. Thenn1 ≤0which is impossible.
(ii) s = 2. Equation (9.2) becomes: 1 n1 + 1 n2 = 2 |G| .
Multiplying by|G|, and inserting (9.1), we have |O1|+|O2|= 2.
Therefore|O1| = |O2| = 1 andn1 = n2 = |G|. Now if Ghas only 2
fixed points, they must be antipodal. And the line passing through them must be the axis of rotation of the elements of G. These are then rota- tions in the plane perpendicular to this axis. So Gcan be regarded as a subgroup ofSO(2). We saw earlier that finite subgroups of SO(2) are cyclic. ThereforeGis cyclic.
(iii) s = 3. Equation (9.2) becomes: 1 n1 + 1 n2 + 1 n2 = 1 + 2 |G| . (9.3) In particular 1 n1 + 1 n2 + 1 n3 >1.
9.4. FINITE SUBGROUPS OFSO(3) 147 n1 n2 n3 |G| 2 2 n 2n 2 3 3 12 2 3 4 24 2 3 5 60
Proof of lemma. Ifn1, n2, n3 ≥3, then
1 n1 + 1 n2 + 1 n3 ≤1.
So at least one is2, sayn1 = 2. Ifn2, n3 ≥4, then
1 2 + 1 n2 + 1 n3 ≤ 1.
Therefore we can assume thatn2 <4. The first possible solution isn1 = 2, n2 =
2, n3 =n, wheren ≥2is arbitrary. Now supposen2 = 3. Then the inequality
becomes 1 2+ 1 3+ 1 n3 >1.
Thus we must have thatn3 <6. This gives the other three solutions in the table.
To compute|G|substitute the values ofn1,n2andn3 in equation (9.3).
We return to the proof of the theorem. The entries in the table correspond toDn,T,OandIrespectively. The four cases are similar. We will do the second
one.
So suppose thatn1 = 2, n2 = 3, n3 = 3, and|G| = 12. Then|O1| = 6,
|O2| = 4 and |O3| = 4. We want to convince ourselves that these orbits are
the set of midpoints of the edges of a regular tetrahedron, the set of its vertices and the set of centres of its faces. Begin with a point inP1 ∈ O2. Letl1 be the
line throughP1and the origin. It is the axis of the rotations inGP1, which have
angles of rotation2π/3and4π/3. Pick a pointP2 ∈ O2 different fromP1. Its
orbit underGP1 is{P2, P3, P4}. They all lie in a plane perpendicular tol1 and
P1 P2 P3 P4 l1 l2
Take the line joining any of these to the origin, say the linel2 joiningP2 to
the origin. It is the axis for the rotations ofGP2 which permute {P1, P3, P4}.
So these all lie in a plane perpendicular to l2 and form an equilateral triangle.
Thus the4points inO2are equidistant from one another and are the vertices of
a regular tetrahedron. The groupG is the group of proper symmetries of this tetrahedron. The orbitsO3andO1are the centres of its faces and the midpoints
of its edges respectively. The symmetry groups of any two regular tetrahedra are conjugate inSO(3). This proves the second case.
This result can also be proved using only spherical geometry: see [8],§3.8.
9.5
Exercises
1. Verify the class formula ofS5.
2. Determine the conjugacy classes ofDn, forn ≥4.
3. Calculate the terms in the class equation forSL(2,F5).
9.5. EXERCISES 149
5. Use the Mathematica function ConjugacyClass to compute the conjugacy classes ofA6.
6. What is the centre of the permutation group of degree8generated by {(1 2 3 4 5 6 7 8),(1 3 5 7)}?
7. How many different necklaces with 6 beads can be made from beads of 3 colours?
8. How many ways can the faces of a cube be coloured black and white? 9. Complete the proof of theorem9.12in the case of the cube.
10
Cosets
10.1
Lagrange's Theorem
At the beginning of the previous chapter when we looked at the evaluation map
e:G→Ox, we came upon subsets ofGof the formαGx, wherex∈X, α ∈
G. Such subsets are called cosets of the stabilizerGx in G. In general, given a
subgroupK < G, we call a subset
αK :={ακ|κ∈K}
aleft coset of K in G. For example, a left coset of nZin Zis a set of the form
m+nZ,m∈Z. This is just the congruence class ofmmodulon. A left coset is not a subgroup ofGexcept for the one coset1·K = K, because this is the only coset containing1.
Examples 10.1. (i) What are the left cosets ofAninSn? Well, ifα ∈ Snis
even, thenαAn =An. Ifαis odd, thenαAn is the set of odd permuta-
tions. So there are 2cosets: the set of even permutations and the set of odd permutations.
(ii) Suppose we takeG=S3 andK =⟨(1 2)⟩. Then (1)K =K (1 2)K =K (1 3)K ={(1 3), (1 3 2)} (2 3)K ={(2 3), (1 2 3)} (1 2 3)K ={(2 3), (1 2 3)} (1 3 2)K ={(2 3), (1 2 3)}.
Thus there are 3 left cosets: K = {(1), (1 2)}, {(1 3), (1 3 2)}, and {(2 3), (1 2 3)}.
(iii) Here are the left cosets ofV inA4:
V ={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} (1 2 3)V ={(1 2 3), (2 4 3),(1 4 2), (1 3 4)} (1 3 2)V ={(1 3 2), (1 4 3),(2 3 4), (1 2 4)}.
(Check this calculation yourself!)
(iv) LetG=R2 with vector addition, and letK be a line through(0,0). The
left cosets ofK inGare the translatesv+K,v ∈R2, ofK.
v+K K
The last section of this chapter explainsMathematicafunctions which compute cosets. One can also look at theright cosetsofKinG: they are subsets of the form
10.1. LAGRANGE'S THEOREM 153
used the notation Z/nZ for the set of congruence classes mod n, otherwise known as the integers modn. IfG/K is finite, then the number of elements in it is called theindex ofK inG, written[G:K].
Notice that in the examples we have just looked at, every left coset has the same number of elements, namely|K|, and distinct cosets are disjoint. As a result of this,[G:K]|K|=|G|. For example, there are3left cosets ofV inA4. Each
has4elements, and|A4|= 12.
Theorem 10.2(Lagrange's Theorem). IfGis a finite group andK a subgroup ofG, then
|G|= [G:K]|K|.
Proof. The theorem can be proved in the way suggested above. This is done in exercise1. It also follows from formula9.1, as we going to see now. The group
Gacts on the setG/Kby left multiplication: define
α·(βK) := (αβ)K ,
forα, β ∈ G. This is a variant of the action ofGon itself by multiplication on the left and you show that it is an action in the same way. It too is transitive: given two cosets,βK, γK∈G/K, the group elementα=γβ−1satisfies
α·βK =γK .
So there is only the one orbit, with[G : K]points in it. The stabilizer of the cosetK is the subgroupK. We can now apply our formula relating the number of points in an orbit to the order of the stabilizer:
|G|= [G:K]|K|.
For example,S3can have subgroups of order1,2and3, but not of order4or
5. It is not hard to write down these subgroups. First there is the trivial subgroup of order1. Subgroups of order 2are those generated by a transposition. There are3of these. There is exactly one subgroup of order3, namelyA3 =⟨(123)⟩.
The graph below shows how these subgroups fit together. It is called the lattice of subgroups ofS3.
S3
A3
⟨(12)⟩ ⟨(13)⟩ ⟨(23)⟩
{(1)}
Corollary 10.4. The order of an element divides the order of the group. A consequence of this is that if|G|=n, then for anyα ∈G
αn= 1.
Corollary 10.5. A group of prime order is cyclic.
Proof. LetGbe a group of orderp, wherepis prime. The order of any element inGmust dividep. Therefore it must be either1orp. So ifα∈G, α̸= 1, then ⟨α⟩=G.
Thus groups of order2,3,5and7are all cyclic. On the other hand we know of a group of order4which is not cyclic, namelyV, and one of order6which is not cyclic,S3.
10.1. LAGRANGE'S THEOREM 155
Let's apply corollary 10.4 to the group F×p. Since it has order p −1 the corollary tells us thatap−1 = 1for anya∈F×
p, or equivalently
ap−1 ≡1 (modp)
for anya ∈ Z,(a, p) = 1. But we can extend this to all integers if we multiply the congruence bya:
Theorem 10.6(Fermat's Little Theorem).
ap ≡a (mod p)
for alla∈Z.
The converse to Lagrange's theorem is not true. IfGis a finite group andd
divides|G|, there need not be a subgroup of orderd. Here is an example. Example 10.7. Look atA4. It consists of eight3-cycles,3products of disjoint
transpositions and the identity. Each3-cycle generates a cyclic subgroup of order
3. Any two of them generate the whole group (see page46). Each element of type {2,2} generates a cyclic subgroup of order 2. Two of them generate the subgroupV of order4. A3-cycle and a product of two transpositions generate the whole group. So there is no subgroup of order 6. Here is the lattice of subgroups ofA4.
A4 V ⟨(123)⟩ ⟨(124)⟩ ⟨(134)⟩ ⟨(234)⟩ ⟨(12)(34)⟩ ⟨(13)(24)⟩ ⟨(14)(23)⟩ {(1)}
10.2
Normal Subgroups
Another situation where cosets naturally arise is the following. Supposef :G→ His a group homomorphism. When do two elements inGhave the same image inH? Well, letα, α′ ∈G, α¯ ∈Hwith
f(α) = f(α′) = ¯α . Then f(α−1α′) = 1H so that α−1α′ ∈ker(f) or α′ ∈αker(f).
Conversely, iff(α) = ¯αandα′ ∈αker(f) thenf(α′) = ¯α. So
10.2. NORMAL SUBGROUPS 157
The kernel and its cosets have a special property. The most convenient way to express it is this: for anyα ∈G,
αker(f)α−1 =ker(f).
(Forβ ∈kerf, f(αβα−1) =f(α)f(α−1) = 1
H). We give a name to subgroups
with this property.
Definition 10.8. A subgroupKofGis called anormal subgroup, writtenK ▹ G, if
αKα−1 =K ,
for allα ∈G.
This property can be expressed in terms of the cosets ofK. Theorem 10.9. A subgroupK of a groupGis normal if and only if
(i) for anyα ∈G,αK =Kα, or equivalently,
(ii) for anyα, β ∈G,(αK)(βK) = (αβ)K, or equivalently, (iii) for anyα ∈G,αKα−1 ⊂K.
Proof. Suppose thatKis a normal subgroup. Take anα∈G. We have that
αKα−1 =K . (10.1)
Multiplying on the right byαwe get
αK =Kα .
Thus every left coset coincides with the corresponding right coset. In terms of elements ofK this means that forκ∈K, there exists aλ∈K, such that
and vice versa. Now assume that equation (i) holds for allα. Then givenα, β ∈ G,
(αK)(βK) = α(Kβ)K =α(βK)K = (αβ)K
So the product of the cosets ofαand ofβis the coset ofαβ. If equation (ii) holds, then takingβ=α−1, we have
(αK)(α−1K) = (αα−1)K = 1·K =K , (10.2) which implies that
αKα−1 ⊂K (10.3)
(Why? Well, lying in the set on the left-hand side of (10.2) are all elements of the form(ακ)(α−11), withκ ∈ K). Finally, since equation (10.3) holds for all
elements ofG, it holds forα−1:
α−1Kα⊂K .
Conjugating both sides byαgives
K ⊂αKα−1 ,
and combining this with (10.3) gives equation (10.1).
In practice, to check whetherK ▹ G, you need only verify whetherαKα−1 ⊂
K, whenαruns through a set of generators ofG. In fact it is sufficient to check if
ακα−1 ∈K, whereαruns through a set of generators ofGandκruns through
a set of generators ofK. In example (10.1), we know that{(1 2 3), (1 2)(3 4)}
generatesA4. Since(1 2)(3 4)∈V, it is enough to check that(1 2 3)V(1 3 2) =
V. We see then thatV is a normal subgroup ofA4.
Remark10.10. Any subgroupKof index2is normal. There are2left cosets: K
andαK, whereα̸∈K. And there are2right cosets: KandKα. The left cosets are disjoint from one another and so are the right cosets. ThereforeαK =Kα, andKis normal. For example,Anis a normal subgroup ofSn, for alln.
10.3. QUOTIENT GROUPS 159
IfGis abelian, then every subgroup is normal since αβα−1 = β for anyα
andβ. The centre of any group is a normal subgroup because the same equation holds for anyα, withβin the centre.
We should also see an example of a subgroup which is not normal. LetG=
S3 and K = ⟨(1 2)⟩. Then (1 2 3)(1 2)(1 3 2) = (2 3) ̸∈ K. So K is not
normal.
10.3
Quotient Groups
Condition (ii) in theorem10.9 tells us that we can makeG/K into a group by multiplying cosets whenK is a normal subgroup (and only then). Let's do this carefully. First, we have a binary operation onG/K:
G/K×G/K →G/K ,
given by
(αK, βK)7→(αK)(βK) = αβK ,
forα, β ∈G. This operation is associative:
(αKβK)γK = (αβ)KγK = (αβ)γK
=α(βγ)K =αK(βγ)K =αK(βKγK).
Secondly there is an identity element, namely1·K =K:
(αK)K =αK =K(αK).
Thirdly, the inverse of a cosetαK isα−1K:
αKα−1K = (αα−1)K =K =α−1KαK .
G/Kwith this operation is called thequotient group ofGmodK.
You have already seen a quotient group: Z/nZ, the integers modn, which is the quotient group of the subgroupnZ. In fact, historically this is the example
which lead to the general construction. If we look again at example10.1(iii), we see that ( (1 2 3)V)2 = (1 2 3)V(1 2 3)V = (1 2 3)2V = (1 3 2)V ( (1 2 3)V)3 = (1 2 3)V(1 3 2)V =V .
This shows thatA4/V is a cyclic group of order3.
Using Mathematica to generate left cosets and then multiply them together makes it easy to see the multiplication inG/K in examples where[G : K] is larger.