In example (xii) there is a natural mapping fromBntoSn. Each braidζ defines
a permutationαζwith
5.2. HOMOMORPHISMS 73
For anyα ∈Snthere are many braidsζ ∈Bnwithαζ =α. So this mapping is
surjective but not injective. It is well-behaved with respect to the multiplications inBnandSn: from the diagrams we see that
αζ2ζ1 =αζ2αζ1
for anyζ1, ζ2 ∈Bn. Since we know a bit aboutSn, the mapping tells us some-
thing about what the groupBn looks like. Such a mapping from one group to
another is called ahomomorphism.
Definition 5.6. LetG and H be groups. A mapping f : G → H is called a (group) homomorphism if for anyα, β ∈G,
f(αβ) = f(α)f(β).
Examples 5.7. We have already seen several other mappings between groups which are homomorphisms. At the end of chapter 2 we defined the permutation matrixIα ∈ GL(n,R)belonging to a permutationα ∈ Sn. Recall that thejth
row ofIα is just theα(j)th row of the identity matrix. We saw that
Iαβ = IαIβ
for anyα, β ∈Sn. Thus the mappingp:Sn→GL(n,R)given by
p:α7→Iα
is a homomorphism. We also know that for any fieldF, the determinant det:GL(n, F)→F×
satisfies
det(αβ) = det(α)det(β)
for any α, β ∈ GL(n, F). So det is also a homomorphism. A final example: addition inZ/nZwas defined so that the canonical mapping Z→Z/nZ given by a7→a¯ is a homomorphism.
Homomorphisms have the formal properties that you would expect. Theorem 5.8. Letf :G→Hbe a group homomorphism. Then
(i) f(1G) = 1H ;
(ii) f(α−1) = f(α)−1 for anyα∈G;
(iii) iff is bijective, thenf−1 :H →Gis also a homomorphism. Proof. We have
f(1G) =f(1G1G) = f(1G)2.
Multiply on the left byf(1G)−1:
1H =f(1G)−1f(1G) =f(1G)−1f(1G)2 =f(1G).
It follows that for anyα∈G,
1H =f(1G) = f(αα−1) = f(α)f(α−1).
Therefore f(α−1) = f(α)−1. Lastly suppose that f is bijective. For any
σ, τ ∈H, we have
f(f−1(στ)) = στ
On the other hand,
f(f−1(σ)f−1(τ)) = f(f−1(σ))f(f−1(τ)) = στ
as well. Sincef is injective, it follows that
f−1(στ) = f−1(σ)f−1(τ).
A homomorphism which is bijective is called anisomorphism. Two groupsG
andHare said to be isomorphic, writtenG∼=H, if there exists an isomorphism
5.2. HOMOMORPHISMS 75
Example 5.9. LetR+ denote the positive real numbers. Under multiplication R+is an abelian group. Let
f :R→R+
be the exponential mapping
f :x7→ex .
Sinceex+y =exey,f is a homomorphism. It has an inverse, the logarithm:
log:R+ →R
So the exponential is an isomorphism of groups. The formula log(zw) =log(z)+
log(w)just says that log is a homomorphism.
△
Given two groupsGandHwe define theirdirect productto be the setG×H
with the operation: (
(α1, β1),(α2, β2)
)
7→(α1α2, β1β2)
whereα1, α2 ∈Gandβ1, β2 ∈H. It is not hard to see that this operation makes
G×H into a group (see exercise16). For example,R2 with vector addition is the direct product ofRwith itself. The two projections:
p1 :G×H →G
p1 : (α, β)7→α
and
p2 :G×H →H
p2 : (α, β)7→β ,
Example 5.10. (i) Let's show thatGL(3,R)is isomorphic to SL(3,R)×
R×. We define a mapping
f :SL(3,R)×R×→GL(3,R)
by
f : (α, a)7→aα
forα ∈SL(3,R)anda∈R×. We see immediately thatf is a homomor- phism. To define an inverse tof, first notice that the mapping
h:R×→R×
given by
h(a) =a3
is an isomorphism. And ifα ∈ GL(3,R), then (detα)−1/3α has deter- minant1. Therefore, since det is a homomorphism, the mapping
g :α7→((detα)−1/3α,(detα)1/3)
is a homomorphism fromGL(3,R)toSL(3,R)×R×. It is clearly inverse tof.
(ii) Suppose thatm, n∈ Nand(m, n) = 1. For anya ∈ Z, let¯adenote its residue class modmn, ¯a1 denote its residue class modm anda¯2 denote
its residue class modn. Define a mapping
g :Z/mnZ→(Z/mZ)×(Z/nZ)
by
g(¯a) := (¯a1,a¯2).
This mapping is well-defined because for anyk ∈Z,
5.3. EXERCISES 77
We have
g(¯a+ ¯b) = (¯a1+ ¯b1,¯a2+ ¯b2) = (¯a1,¯a2) + (¯b1,¯b2) =g(¯a) +g(¯b).
Sogis a homomorphism. Now the Chinese remainder theorem (theorem 1.12) says precisely thatgis an isomorphism. Why is this? Well, what does it mean forg to be surjective? Given integersaandb, we must show that there exists an integercsuch that
c≡a (mod m) and c≡b (mod n).
And to say thatgis injective is to say that such an integercis unique mod
mn.
△
5.3
Exercises
1. Which of the following groups are abelian: the permutation group V, the linear groupsT,F(p)orG(p)?
2. LetGbe a group. Forα ∈ Gletαn be then-fold product of αwith itself,
forn > 0, and then-fold product ofα−1 with itself, forn <0. Show that
αmαn=αm+n
(αm)n=αmn
for anym, n∈Z.
3. What is the order of(Z/16Z)×? (Z/24Z)×? For each group make a table which gives the order of the elements in the group.
4. What is the order of any element in the additive group ofFp? What are the
5. Letαbe an element of order n in a groupG. Suppose that αr = 1. Show
thatn |r.
6. • Suppose that αis an element of order n in a group G. For any m ∈ N,
prove that
|αm|=n/(m, n).
7. • Let α and β be elements of prime order in a group G, with |α| ̸= |β|. Suppose thatαβ =βα. Prove that
|αβ|=|α||β|.
8. •Letαandβ be elements of finite order in a groupG, with(|α|,|β|) = 1. Suppose thatαβ =βα. Prove that
|αβ|=|α||β|.
9. Determine the structure of the braid groupB2.
10. Show that an isometry is injective.
11. •Let exp :R→S be the exponential mapping exp(x) = e2πix.
Verify that exp is a homomorphism. Is exp injective? Surjective? 12. Show that the mapping
h:R× →R×
given by
h(a) = a3
5.3. EXERCISES 79
13. Check that the mapping in example 4.2(i)is a homomorphism.
14. Given a natural numbern, prove that ifGis an abelian group, then the map- pingf :G→G, where
f :α 7→αn,
is a homomorphism.
15. Prove that the permutation groupsV andV′ are isomorphic.
16. Verify that the direct product of two groups with the operation given is a group.
17. Give an example of two finite groups of the same order which are not iso- morphic.
18. a) Show that
(Z/16Z)× ∼= (Z/2Z)×(Z/4Z).
b) Show that
(Z/24Z)× ∼= (Z/2Z)3.
19. Let f : G → H be a homomorphism of groups and letg ⊂ Gbe a set of generators ofG. Suppose that
h:={f(α)|α∈g}
generatesH. Prove thatf is surjective.
20. •Suppose thatGandHare finite groups with the same order. Letf :G→ Hbe a homomorphism. Show that
a) iff is surjective, then it is injective; b) iff is injective, then it is surjective.
21. •SupposeGis a group. An isomorphismf :G→Gis called anautomorphism . Let Aut(G)denote the set of automorphisms ofG. Prove thatGis a group under composition of mappings.
22. Prove that Aut(Z/nZ)∼= (Z/nZ)×.
23. Suppose thatm, n∈Nand(m, n) = 1. Define a mapping
h: (Z/mZ)×(Z/nZ)→Z/mnZ
by
h: (¯a,¯b)7→na+mb .
Show thathis well-defined and is a homomorphism. Prove thathis an iso- morphism.
24. •Recall that the order of(Z/mZ)×isφ(m)(see page14). a) Suppose that(m, n) = 1. Prove that
g : (Z/mnZ)× −−−→∼= (Z/mZ)××(Z/nZ)×,
wheregis the mapping defined in example5.10(ii). b) Show thatφ(mn) =φ(m)φ(n). c) Suppose that n=pj1 1 · · ·p jr r ,
wherep1, . . . , prare distinct primes andj1, . . . , jr >0. Show that
φ(n) = (pj1
1 −p
j1−1
1 )· · ·(pjrr −pjrr−1) =n(1−1/p1)· · ·(1−1/pr)
6
Subgroups
6.1
Definition
Subgroups are subsets of groups which are groups themselves under the oper- ation inherited from the group. Of course for this to be possible, the product of two elements of the subgroup must lie in it, and so must the inverse of every element. It turns out that this is in fact enough.
Definition 6.1. Let Gbe a group. A non-empty subsetH ⊆ Gis a subgroup ofG, writtenH < G, if for allα, β ∈H
αβ ∈H (i)
α−1 ∈H . (ii)
Condition (i) ensures that the group operation onGgives us an operation on
H. It is associative because the operation onGis. SinceHis non-empty, there exists an elementα ∈H. By (ii),α−1 ∈ Has well. Therefore1 = αα−1 ∈ H
by (i). And by (ii) again, the inverse of every element inHlies inH. SoHwith the operation inherited fromGis a group.
Looking back at chapters 3 and 4 we see that permutation groups were de- fined as subgroups ofSnand linear groups as subgroups ofGL(n, F). Looking
at the list of examples in chapter 5, we see that in example 3 Z<Q<R<C.