SupposeGandHare groups andf : G→H is a homomorphism. Thekernel off, written ker(f), is defined by
ker(f) := {α∈G|f(α) = 1H} ⊂G .
It measures how closef is to being injective, as we shall see below. Theimageof
f, written im(f), is the set of images of the elements ofGinH, i.e. im(f) := {f(α)|α∈G} ⊂H .
6.4. KERNEL AND IMAGE OF A HOMOMORPHISM 91
Theorem 6.7. The kernel of a homomorphismf is a subgroup of G, and the image a subgroup ofH. A homomorphism is injective if and only if its kernel is trivial.
We just need to verify that ker(f)and im(f)fulfill the definition of a sub- group. Well, supposeα, β ∈Gandf(α) = 1andf(β) = 1. Then
f(αβ) = f(α)f(β) = 1
and
f(α−1) =f(α)−1 = 1.
Therefore ker(f) is a subgroup of G. And for any α, β ∈ G, f(α)f(β) =
f(αβ)∈ im(f), andf(α)−1 =f(α−1)∈ im(f). Thus im(f)is a subgroup of
H.
Lastly, suppose f is injective. Then in particular, if f(α) = 1 for some
α ∈ G, sincef(1) = 1, we have α = 1. Thus ker(f) = {1}. On the other hand, assume ker(f) ={1}. Now iff(α) = f(β)for someα, β ∈G, we have
1 = f(α)f(β)−1 =f(α)f(β−1) =f(αβ−1).
Soαβ−1 ∈ker(f), and thereforeαβ−1 = 1, i.e. α=β. Thusf is injective.
Examples 6.8. (i) Recall that det : GL(n, F) → F×is a homomorphism for any fieldF. By the definition ofSL(n, F)
ker(det) =SL(n, F).
(ii) The kernel of the canonical map: Z→Z/nZis the subgroupnZ⊂Z. (iii) Letf :S →Sbe given byf(z) :=zn, for somen∈N. The kernel off
is
{z ∈C|zn= 1}.
A complex numberzsatisfying this equation is called annth root of unity. We shall denote the group ofnth roots of unity byµn.
(iv) IfGandHare two groups, then the kernel of the projectionp1 :G×H→
GisH.
(v) In chapter 5, we saw that there is a natural homomorphism
br :Bn→Sn, wherebr(ζ) =αζ.
The permutationαζ mapsito the endpoint of theith strand ofζ. So
αζi = (i i+ 1)
Since
{(1 2),(2 3), . . . , (n−1n)}
generatesSn, the homomorphismbris surjective.
6.5
Exercises
1. What are the elements of finite order inSO(2)?
2. •Show thatµnis a cyclic subgroup ofS. Notice that every element of finite
order ofS is annth root of unity for somen.
3. In example6.5, verify thatτ2 =−I and thatτ has order4. 4. Use the algorithm in example6.5to write
(
7 31 2 9
)
as a word inσandτ.
5. Write a Mathematica function which expresses an element of SL(2,Z) as a word inσandτ.
6. Leta, b ∈Zbe relatively prime. The Euclidean algorithm produces integers
sandtsuch that
6.5. EXERCISES 93 Set α= ( a b t s ) .
Prove that if one applies the algorithm in example6.5toαthen the integerk
there is0(cf. exercise 1.13).
7. Show thatF×5, F×7 andF×11are cyclic groups.
8. LetGbe a finite group and suppose thatH ⊂G, H ̸=∅and for allα, β ∈ H,
αβ ∈H .
Prove that H is a subgroup of G. Suppose that|G| = ∞. Does this still hold? If not give a counter-example.
9. Find all generators of Z/60Z.
10. •Letpbe prime. Suppose that a subgroupH ⊂ Sp contains a transposition
and an element of orderp. Prove thatH=Sp.
11. •Show that every subgroupHofZis either trivial or of the formnZ, where nis the least positive integer inH.
12. Suppose that H andK are subgroups of a groupG. Show thatH∪K is a subgroup if and only ifH ⊂K orK ⊂H.
13. LetGbe a group and let
H ={(α, α)∈G×G|α∈G}.
Verify thatH is a subgroup ofG×Gand thatH ∼=G.
14. Can you find two matrices which generateSL(2,F3)?SL(2,F5)? (You may
15. Prove thatO(3)is isomorphic toSO(3)× {±1}.
16. • Let G be a subgroup of O(3). Suppose that −I ∈ G. Prove that G is isomorphic toG+× {±1}, whereG+=G∩SO(3).
17. Show that det:GL(n, F)→F×is surjective. 18. Let
sgn:Sn→ {±1}
be given by the sign of a permutation. Show that sgn is a homomorphism. Is it surjective? What is its kernel?
19. Let the homomorphism
f : (Z/27Z)× →(Z/3Z)×
be reduction modulo3. Write down the elements of the kernel off. Show that it is cyclic.
20. •LetF be a field and letf :F× →F×be the map
f :a7→a2 ,
fora∈F×. Verify thatf is a homomorphism and determine its kernel. 21. Show that inBn,
a) forj ̸=i±1, ζiζj =ζjζi ;
b) for1< i < n, ζiζi+1ζi =ζi+1ζiζi+1 .
22. Prove that inBn,
6.5. EXERCISES 95
b) ζi+1 =ρiζ1ρ−i ,
and that therefore{ζ1, ρ}generatesBn.
23. Show that the image of Sn under the homomorphism pin example 5.7lies
inO(n), and that the image ofAn lies inSO(n). Show that the image of a
7
Symmetry Groups
Intuitively we know when an object has symmetry and when not. Symmetry is closely related to our sense of what is beautiful. A rose window in a cathedral with its rotational symmetry is beautiful, as is a face with strong bilateral symmetry (see [5]). The octagon on the left below has no symmetry, while the one on the right is highly symmetric.
We would like a mathematically precise definition of symmetry which fits with our intuitive sense. The regular octagon above can be reflected in the dotted line. We can also rotate it about its centre through an angle ofπ/4. The rotation and reflection are both isometries of the Euclidean plane. This suggests a general definition.
SupposeXis an object in Euclideann-space. A symmetry ofXis an isom- etry which mapsX to itself. We denote bySym(X)the set of all symmetries of
X. Not surprisingly,Sym(X)is a subgroup ofIso(n). For the identity map is 97
always a symmetry ofX so thatSym(X)is not empty. The inverse of a sym- metry is again a symmetry and so is the product of two symmetries. To say that
Xis highly symmetric is to say thatSym(X)is large. IfXhas no symmetry at all, thenSym(X)is the trivial group As references for Euclidean geometry, [6], [4], and [7] are recommended.
7.1
Symmetries of Regular Polygons
Let's begin by looking at the symmetry groups of regular polygons in the plane. We will place the centre of the polygon at the origin, so that the symmetries are all elements of the orthogonal group (see [4], p.11).
1
2 3
Figure 7.1: Symmetries of an equilateral triangle
First we consider an equilateral triangleP3. We can rotateP3about its centre
through2π/3,4π/3, and2π. (All angles are measured counterclockwise.) And we can reflectP3in the line joining a vertex to the midpoint of the opposite side.
ThusSym(P3)has6elements,3rotations and 3reflections. Notice that if we
label the vertices1,2,3(counterclockwise), then the symmetries permute the3
7.1. SYMMETRIES OF REGULAR POLYGONS 99
1
2
3
4
Figure 7.2: Symmetries of a square
reflections leave a vertex fixed and switch the other two. So they each correspond to a transposition. This mapping
Sym(P3)→S3
is in fact an isomorphism of groups since it maps the product of two symmetries to the product of the corresponding permutations.
Next look at a square P4 (see introduction of chapter 3). We can rotateP4
about its centre throughπ/2,π,3π/2, and2π. There are two types of reflections: reflections in the diagonals, and reflections in the lines joining the midpoints of opposite sides. Of each type there are two. So in all, Sym(P4) consists of 4
rotations and4reflections. Again we can label the vertices ofP4: 1,2,3,4. Each
symmetry gives a permutation of the vertices and the mapping
Sym(P4)→S4
defined this way is a homomorphism: the product of two symmetries is mapped to the product of the corresponding permutations. The rotations correspond to
(1 2 3 4),(1 3)(2 4),(1 4 3 2)and(1)respectively. Reflections about a diagonal correspond to the transpositions (1 3) and (2 4). The other two reflections
correspond to(1 2)(3 4)and(1 4)(2 3). As we saw in chapter 3, these are just the elements of the permutation groupD4. So we have an isomorphism
Sym(P4)
∼
=
−−−→ D4.
These examples generalize. LetPnbe a regularn-gon,n≥3. We can rotate
Pn about its centre through angles 2π/n,4π/n, . . . ,(2n − 2)π/n,2π. The
reflectional symmetry ofPndepends on whethernis even or odd.
Figure 7.3: Symmetries of a regular n-gon
First suppose n is odd. Pn is symmetric about the line joining a vertex to
the midpoint of the opposite side. There aren reflections of this type. Now supposenis even. Then there are two types of reflections: those in a line joining the midpoints of opposite sides, and those in a diagonal joining opposite vertices. There aren/2reflections of each type. In either case,Sym(Pn)hasnrotations
and n reflections. It is called the dihedral group of order 2n We shall denote it by Dn. As in the cases n = 3,4, we can label the vertices 1,2, . . . , n. The
symmetries ofPnpermute the vertices and thus each correspond to an element
of Sn. The rotations correspond to the powers of (1 2 · · · n). Since each
symmetry is determined by its action on the vertices ofPn, this mapping from
Dn intoSn is injective and gives us an isomorphism ofDn with a permutation