Group theory notes

Group Theory Notes

1. M2PM2 Notes on Group Theory
Here are some notes on the M2PM2 lectures on group theory.
1 Revision from M1P2
Would be a good idea to refresh your memory on the following topics from
group theory.
(a)Group axioms: closure, associativity, identity, inverses
(b)Examples of groups:
(Z,+), (Q,+), (Q
Λ
,×), (C
Λ
,×), etc
GL(n,R), the group of all invertiblen×nmatrices overR, under matrix
multiplication
S
n
, the symmetric group, the set of all permutations of{1,2, . . . , n},
under composition. Recall the cycle notation for permutations – every per-
mutation can be expressed as a product of disjoint cycles.
ForpprimeZ
Λ
p
={[1],[2], . . . ,[p−1]}is a group under multiplication
modulop.
C
n
={x2C:x
n
= 1}={1, ω, ω
2
, . . . , ω
n−1
}is a cyclic group of sizen,
whereω=e
2πi/n
.
(c)Some theory:
Criterion for subgroups:His a subgroup ofGiff (1)e2H; (2)x, y2
H)xy2H, and (3)x2H)x
−1
2H.
Fora2G, we define the cyclic subgrouphai={a
n
:n2Z}. The size
ofhaiis equal too(a), theorderofa, which is defined to be the smallest
positive integerksuch thata
k
=e.
Lagrange: ifHis a subgroup of a finite groupGthen|H|divides|G|.
Consequences: (1) For any elementa2G,o(a) divides|G|.
(2) If|G|=nthenx
n
=efor allx2G
(3) If|G|is prime thenGis a cyclic group.
1

2. 2 More examples: symmetry groups
For any object in the planeR
2
(laterR
3
) we’ll show how to define a group
called the symmetry group of the object. This group will consist of functions
calledisometries, which we now define. Recall forx= (x
1
, x
2
),y= (y
1
, y
2
)2
R
2
, the distance
d(x, y) =
p
(x
1
−y
1
)
2
+ (x
2
−y
2
)
2
.
We define anisometryofR
2
to be a bijectionf:R
2
→R
2
which preserves
distance, i.e. for allx, y2R
2
,
d(f(x), f(y)) =d(x, y).
There are many familiar examples of isometries:
(1) Rotations: letρ
P,θ
be the functionR
2
→R
2
which rotates every
point aboutPthrough angleθ. This is an isometry.
(2) Reflections: iflis a line, letσ
l
be the function which sends every
point to its reflection inl. This is an isometry.
(3) Translations: fora2R
2
, letτ
a
be the translation sendingx→x+a
for allx2R
2
. This is an isometry.
Not every isometry is one of these three types – for example a glide-reflection
(i.e. a function of the formσ
l
◦τ
a
) is not a rotation, reflection or translation.
DefineI(R
2
) to be the set of all isometries ofR
2
. For isometriesf, g, we
have the usual composition functionf◦gdefined byf◦g(x) =f(g(x)).
Proposition 2.1I(R
2
)is a group under composition.
ProofClosure: Letf, g2I(R
2
). We must showf◦gis an isometry. It is
a bijection asf, gare bijections (recall M1F). And it preserves distance as
d(f◦g(x), f◦g(y)) =d(f(g(x)), f(g(y)))
=d(g(x), g(y) (asfis isometry)
=d(x, y) (asgis isometry).
Assoc: this is always true for composition of functions (sincef◦(g◦h)(x) =
(f◦g)◦h(x) =f(g(h(x)))).
Identityis the identity functionedefined bye(x) =xfor allx2R
2
, which
is obviously an isometry.
2

3. Inverses: letf2I(R
2
). Thenf
−1
exists asfis a bijection, andf
−1
preserves distance since
d(f
−1
(x), f
−1
(y)) =d(f(f
−1
(x)), f(f
−1
(y))) =d(x, y).
So we’ve checked all the axioms andI(R
2
) is a group.Λ
Now let Π be a subset ofR
2
. For a functiong:R
2
→R
2
,
g(Π) ={g(x)|x2Π}
Example:Π =square with centre in the origin and aligned with axes,
g=ρ
π/4
. Theng(Π) is the original square rotated byπ/4.
DefinitionThesymmetry groupof Π isG(Π) – the set of isometriesgsuch
thatg(Π) = Π, i.e.
G(Π) =
Φ
g2I(R
2
)|g(Π) = Π

.
Example:For the square from the previous example,G(Π) containsρ
π/2
,
σ
x
. . .
Proposition 2.2G(Π)is a subgroup ofI(R
2
).
ProofWe check the subgroup criteria:
(1)e2G(Π) ase(Π) = Π.
(2) Letf, g2G(Π), sof(Π) =g(Π) = Π. So
f◦g(Π) =f(g(Π)) (1)
=f(Π) (2)
= Π. (3)
Sof◦g2G(Π).
(3) Letf2G(Π), so
f(Π) = Π.
Applyf
−1
to get
f
−1
(f(Π)) =f
−1
(Π) (4)
Π =f
−1
(Π) (5)
andf
−1
2G(Π).Λ
3

4. So we have a vast collection of new examples of groupsG(Π).
Examples
1.Equilateral triangle(= Π)
HereG(Π) contains
3rotations:e=ρ
0
,ρ=ρ
2π/3
,ρ
2
=ρ
4π/3
,
3reflections:σ
1
=σ
l
1
,σ
2
=σ
l
2
,σ
3
=σ
l
3
.
Each of these corresponds to a permutation of the corners 1, 2, 3:
eξe, (6)
ρξ(1 2 3), (7)
ρ
2
ξ(1 3 2), (8)
σ
1
ξ(2 3), (9)
σ
2
ξ(1 3), (10)
σ
3
ξ(1 2). (11)
Any isometry inG(Π) permutes the corners. Since all the permu-
tations of the corners are already present, there can’t be any more
isometries inG(Π). So the Symmetry group of equilateral triangle is
Φ
e, ρ, ρ
2
, σ
1
, σ
2
, σ
3

,
called thedihedral groupD
6
.
Note that it is easy to work out products inD
6
: e.g.
ρσ
3
ξ(1 2 3)(1 2) = (1 3) (12)
ξσ
2
. (13)
2.The square
HereG=G(Π) contains
4rotations:e, ρ, ρ
2
, ρ
3
whereρ=ρ
π/2
,
4reflections:σ
1
, σ
2
, σ
3
, σ
4
whereσ
i
=σ
l
i
.
So|G| ≥8. We claim that|G|= 8: Anyg2Gpermutes the corners
1,2,3,4 (asgpreserves distance). Sogsends
1→i, (4 choices ofi)
4

5. 2→j, neighbour ofi, (2 choices forj)
3→oppositeofi,
4→oppositeofj.
So|G| ≤(num. of choices fori)×(forj) = 4×2 = 8. So|G|= 8.
Symmetry group of the square is
Φ
e, ρ, ρ
2
, ρ
3
, σ
1
, σ
2
, σ
3
, σ
4

,
called thedihedral groupD
8
.
Can work out products using the corresponding permutations of the
corners.
e¸e, (14)
ρ¸(1 2 3 4), (15)
ρ
2
¸(1 3)(2 4), (16)
ρ
3
¸(1 4 3 2), (17)
σ
1
¸(1 4)(2 3), (18)
σ
2
¸(1 3), (19)
σ
3
¸(1 2)(3 4), (20)
σ
4
¸(2 4). (21)
For example
ρ
3
σ
1
→(1 4 3 2)(1 4)(2 3) = (1 3) (22)
→σ
2
. (23)
Note thatnotall permutations of the corners are present inD
8
, e.g.
(1 2).
More onD
8
:DefineHto be the cyclic subgroup ofD
8
generated by
ρ, so
H=hρi=
Φ
e, ρ, ρ
2
, ρ
3

.
Writeσ=σ
1
. The right coset
Hσ=
Φ
σ, ρσ, ρ
2
σ, ρ
3
σ

is different fromH
.
H Hσ
5

6. So the two distinct right cosets ofHinD
8
areHandHσ, and
D
8
=H∪Hσ.
Hence
Hσ=
Φ
ρ, ρσ, ρ
2
σ, ρ
3
σ

(24)
={σ
1
, σ
2
, σ
3
, σ
4
}. (25)
So the elements ofD
8
are
e, ρ, ρ
2
, ρ
3
, σ, ρσ, ρ
2
σ, ρ
3
σ.
To work out products, use the “magic equation” (see Sheet 1, Question
2)
σρ=ρ
−1
σ.
3.Regularn-gon
Let Π be the regular polygon withnsides. Symmetry groupG=G(Π)
contains
nrotations:e, ρ, ρ
2
, . . . , ρ
n−1
whereρ=ρ
2π/n
,
nreflectionsσ
1
, σ
2
, . . . , σ
n
whereσ
i
=σ
l
i
.
So|G| ≥2n. We claim that|G|= 2n.
Anyg2Gsends corners to corners, say
1→i, (nchoices fori)
2→jneighbour ofi. (2 choices forj)
Thengsendsnto the other neighbour ofiandn−1 to the remaining
neighbour ofg(n) and so on. So oncei, jare known, there is only one
possibility forg. Hence
|G| ≤number of choices fori, j= 2n.
Therefore|G|= 2n.
Symmetry group of regularn-gon is
D
2n
=
Φ
e, ρ, ρ
2
, . . . , ρ
n
, σ
1
, . . . , σ
n

,
thedihedral groupof size 2n.
Again can work inD
2n
using permutations
ρ→(1 2 3∙ ∙ ∙n) (26)
σ
1
→(2n)(3n−1)∙ ∙ ∙ (27)
6

7. 4.Benzene molecule
C
6
H
6
. Symmetry group isD
12
.
5.Infinite strip ofF’s
. . .F F F . . .
−1 0 1
What is symmetry groupG(Π)?
G(Π) contains translation
τ
(1,0)
:v7→v+ (1,0).
Writeτ=τ
(1,0)
. ThenG(Π) contains all translationsτ
n
=τ
(n,0)
. Note
G(Π) is infinite. We claim that
G(Π) ={τ
n
|n2Z} (28)
=hτi, (29)
infinite cyclic group.
Letg2G(Π). Must show thatg=τ
n
for somen. Saygsends F at 0
to F atn. Note thatτ
−n
sendsFatntoFat 0. Soτ
−n
gsendsFat 0
toFat 0. Soτ
−n
gis a symmetry of theFat 0. It is easy to observe
thatFhas only symmetrye. Hence
τ
−n
g=e (30)
τ
n
τ
−n
g=τ
n
(31)
g=τ
n
. (32)
NoteVarious other figures have more interesting symmetry groups, e.g.
infinite strip ofE’s, square tiling of a plane, octagons and squares tiling of
the plane, 3 dimensions – platonic solids. . . later.
3 Isomorphism
LetG=C
2
={1,−1},H=S
2
={e, a}(wherea= (1 2)). Multiplication
tables:
OfG
: 1 −1
1 1 −1
−1−11
7

8. OfH: ea
e e a
a ae
These are the same, except that the elements have different labels (1¸e,
−1¸a).
Similarly forG=C
3
={1, ω, ω
2
},H=hai={e, a, a
2
}(wherea=
(1 2 3)2S
3
):
OfG
: 1 ω ω
2
1 1 ω ω
2
ω ω ω
2
1
ω
2
ω
2
1
ω
OfH: e a a
2
e e a a
2
a a a
2
e
a
2
a
2
e
a
Again, these are same groups with relabelling
1¸e,
ω¸a,
ω
2
¸a
2
.
In these examples, there is a flelabelling” functionφ:G→Hsuch that if
g
1
7→h
1
,
g
2
7→h
2
,
then
g
1
g
2
7→h
1
h
2
.
DefinitionG, Hgroups. A functionφ:G→His anisomorphismif
(1)φis a bijection,
(2)φ(g
1
)φ(g
2
) =φ(g
1
g
2
) for allg
1
, g
2
2G.
If there exists an isomorphismφ:G→H, we sayGisisomorphictoH
and writeG
¸
=H.
Notes1. IfG
¸
=Hthen|G|=|H|(asφis a bijection).
2. The relation
¸
=is an equivalence relation, i.e.
•G
¸
=G,
8

9. •G
¸
=H)H
¸
=G,
•G
¸
=H, H
¸
=K)G
¸
=K.
ExampleWhich pairs of the following groups are isomorphic?
G
1
=C
4
=hii={1,−1, i,−i},
G
2
= symmetry group of a rectangle ={e, ρ
π
, σ
1
, σ
2
},
G
3
= cyclic subgroup ofD
8
hρi=
Φ
e, ρ, ρ
2
, ρ
3

.
1.G
1
¸
=G
3
? To prove this, defineφ:G
1
→G
2
i7→ρ,
−17→ρ
2
,
−i7→ρ
3
,
17→e,
i.e.φ:i
n
7→ρ
n
. To check thatφis an isomorphism
(1)φis a bijection,
(2) form, n2Z
φ(i
m
i
n
) =φ(i
m+n
)
=ρ
m+n
=ρ
m
ρ
n
=φ(i
m
)φ(i
n
).
Soφis an isomorphism andG
1
¸
=G
3
.
Note that there exist many bijectionsG
1
→G
3
which are not isomor-
phisms.
2.G
2
¸
=G
3
orG
2
¸
=G
1
? Answer:G
2
6
¸
=G
1
. By contradiction. Assume
there exists an isomorphismφ:G
1
→G
2
. Sayφ(i) =x2G
2
,φ(1) =y2
G
2
. Then
φ(−1) =φ(i
2
) =φ(i∙i) =φ(i)φ(i) =x
2
=e
asg
2
=efor allg2G
2
. Similarlyφ(1) =φ(1∙1) =φ(1)φ(1) =y
2
=e. So
φ(−1) =φ(1), a contradiction asφis a bijection.
In general, to decide whether two groupsG, Hare isomorphic:
•If you thinkG
¸
=H, try to define an isomorphismφ:G→H.
•If you thinkG6
¸
=H, try to use the following proposition.
9

10. Proposition 3.1LetG, Hbe groups.
(1) If|G| 6=|H|thenG6
¸
=H.
(2) IfGis abelian andHis not abelian, thenG6
¸
=H.
(3) If there is an integerksuch thatGandHhave different number of
elements of orderk, thenG6
¸
=H.
Proof(1) Obvious.
(2) We show that ifGis abelian andG
¸
=H, thenHis abelian (this
gives (2)). SupposeGis abelian andφ:G→His an isomorphism. Let
h
1
, h
2
2H. Asφis a bijection, there existg
1
, g
2
2Gsuch thath
1
=φ(g
1
)
andh
2
=φ(g
2
). So
h
2
h
1
=φ(g
2
)φ(g
1
)
=φ(g
2
g
1
)
=φ(g
1
)φ(g
2
)
=h
1
h
2
.
(3) Let
G
k
={g2G|o(g) =k},
H
k
={h2H|o(h) =k}.
We show thatG
¸
=Himplies|G
k
|=|H
k
|for allk(this gives (3)).
SupposeG
¸
=Hand letφ:G→Hbe an isomorphism. We show that
φsendsG
k
toH
k
: Letg2G
k
, soo(g) =k, i.e.
g
k
=e
G
,andg
i
6=e
G
for 1≤i≤k−1.
Nowφ(e
G
) =e
H
, since
φ(e
G
) = φ(e
G
e
G
)
=φ(e
G
)φ(e
G
)
φ(e
G
)
−1
φ(e
G
) =φ(e
G
)
e
H
=φ(e
G
).
Also
φ(g
i
) =φ(gg∙ ∙ ∙g) (itimes)
=φ(g)φ(g)∙ ∙ ∙φ(g)
=φ(g)
i
.
Hence
φ(g)
k
=φ(e
G
) =e
H
,
φ(g)
i
6=e
H
for 1≤i≤k−1.
10

11. In other words,φ(g) has orderk, soφ(g)2H
k
. SoφsendsG
k
toH
k
. Asφ
is 1-1, this implies|G
k
| ≤ |H
k
|.
Alsoφ
−1
:H→Gis an isomorphism and similarly sendsH
k
toG
k
,
hence|H
k
| ≤ |G
k
|. Therefore|G
k
|=|H
k
|.Λ
Examples1. LetG=S
4
,H=D
8
. Then|G|= 24,|H|= 8, soG6
¸
=H.
2. LetG=S
3
,H=C
6
. ThenGis non-abelian,His abelian, soG6
¸
=H.
3. LetG=C
4
,H= symmetry group of the rectangle ={e, ρ
π
, σ
1
, σ
2
}.
ThenGhas 1 element of order 2,Hhas 3 elements of order 2, soG6
¸
=H.
4. Question: (R,+)
¸
=(R
Λ
,×)? Answer: No, since (R,+) has 0 elements
of order 2, (R
Λ
,×) has 1 element of order 2.
Cyclic groups
Proposition 3.2(1) IfGis a cyclic group of sizen, thenG
¸
=C
n
.
(2) IfGis an infinite cyclic group, thenG
¸
=(Z,+).
Proof(1) LetG=hxi,|G|=n, soo(x) =nand therefore
G=
Φ
e, x, x
2
, . . . , x
n−1

.
Recall
C
n
=
Φ
1, ω, ω
2
, . . . , ω
n−1

,
whereω=e
2πi/n
. Defineφ:G→Gbyφ(x
r
) =ω
r
for allr. Thenφis a
bijection, and
φ(x
r
x
s
) =φ(x
r+s
)
=ω
r+s
=ω
r
ω
s
=φ(x
r
)φ(x
s
).
Soφis an isomorphism, andG
¸
=C
n
.
(2) LetG=hxibe infinite cyclic, soo(x) =∞and
G=
Φ
. . . , x
−2
, x
−1
, e, x, x
2
, x
3
, . . .

,
all distinct. Defineφ:G→(Z,+) byφ(x
r
) =rfor allr. Thenφis an
isomorphism, soG
¸
=(Z,+).Λ
This proposition says that if we think of isomorphic groups as being
Ψhe same”, then there is onlyonecyclic group of each size. We say: “up
to isomorphism”, the only cyclic groups areC
n
and (Z,+).
11

12. ExampleCyclic subgrouph3iof (Z,+) is{3n|n2Z}, infinite, so by the
propositionh3i
¸
=(Z,+).
4 Even and odd permutations
We’ll classify each permutation inS
n
as either “even” or “odd” (reason given
later).
ExampleForn= 3. Consider the expression
Δ = (x
1
−x
2
)(x
1
−x
3
)(x
2
−x
3
),
a polynomial in 3 variablesx
1
,x
2
,x
3
. Take each permutation inS
3
to
permutex
1
, x
2
, x
3
in the same way it permutes 1,2,3. Then eachg2S
3
sends Δ to±Δ. For example
fore,(1 2 3),(1 3 2) : Δ7→+Δ,
for (1 2),(1 3),(2 3) : Δ7→ −Δ.
Generalizing this: for arbitraryn≥2, define
Δ =
Y
i<j
(x
i
−x
j
),
a polynomial innvariablesx
1
, . . . , x
n
.
If we let each permutationg2S
n
permute the variablesx
1
, . . . , x
n
just
as it permutes 1, . . . , nthengsends Δ to±Δ.
DefinitionForg2S
n
, define thesignaturesgn(g) to be +1 ifg(Δ) = Δ
and−1 ifg(Δ) =−Δ. So
g(Δ) = sgn(g)Δ.
The function sgn :S
n
→ {+1,−1}is thesignature functiononS
n
. Callg
anevenpermutation if sgn(g) = 1, andoddpermutation if sgn(g) =−1.
ExampleInS
3
e,(1 2 3),(1 3 2) are even and (1 2),(1 3),(2 3) are odd.
Given (1 2 3 5)(6 7 9)(8 4 10)2S
10
, what’s its signature ? Our next
aim is to be able answer such questions instantaneously. This is the key:
Proposition 4.1(a)sgn(xy) = sgn(x)sgn(y)for allx, y2S
n
12

13. (b)sgn(e) = 1,sgn(x
−1
) = sgn(x).
(c) Ift= (i j)is a 2-cycle thensgn(t) =−1.
Proof(a) By definition
x(Δ) = sgn(x)Δ,
y(Δ) = sgn(y)Δ.
So
xy(Δ) =x(y(Δ))
=x(sgn(y)Δ)
= sgn(y)x(Δ) = sgn(y)sgn(x)Δ.
Hence
sgn(xy) = sgn(x)sgn(y).
(b) We havee(Δ) = Δ, so sgn(e) = 1. So
1 = sgn(e) = sgn(xx
−1
)
= sgn(x)sgn(x
−1
) (by (a))
and hence sgn(x) = sgn(x
−1
).
(c) Lett= (i j),i < j. We count the number of brackets in Δ that are
sent to brackets (x
r
−x
s
),r > s. These are
(x
i
−x
j
),
(x
i
−x
i+1
), . . . ,(x
i
−x
j−1
),
(x
i+1
−x
j
), . . . ,(x
j−1
−x
j
).
Total number of these is 2(j−i−1) + 1, an odd number. Hencet(Δ) =−Δ
and sgn(t) =−1.Λ
To work out sgn(x),x2S
n
here’s what we shall do:
•expressxas a product of 2-cycles
•use proposition 4.1
Proposition 4.2Letc= (a
1
a
2
. . . a
r
), anr-cycle. Thenccan be expressed
as a product of(r−1)2-cycles.
13

14. ProofConsider the product
(a
1
a
r
)(a
1
a
r−1
)∙ ∙ ∙(a
1
a
3
)(a
1
a
2
).
This product sends
a
1
7→a
2
7→a
3
7→ ∙ ∙ ∙ 7→a
r−1
7→a
1
.
Hence the product is equal toc.Λ
Corollary 4.3The signature of anr-cycle is(−1)
r−1
.
ProofFollows from previous two props.Λ
Corollary 4.4Everyx2S
n
can be expressed as a product of2-cycles.
ProofFrom first year, we know that
x=c
1
∙ ∙ ∙c
m
,
a product of disjoint cyclesc
i
. Eachc
i
is a product of 2-cycles by 4.2. Hence
so isx.Λ
Proposition 4.5Letx=c
1
∙ ∙ ∙c
m
a product of disjoint cyclesc
1
, . . . , c
m
of
lengthsr
1
, . . . , r
m
. Then
sgn(x) = (−1)
r
1
−1
∙ ∙ ∙(−1)
r
m
−1
.
ProofWe have
sgn(x) = sgn(c
1
)∙ ∙ ∙sgn(c
m
) by 4.1(a)
= (−1)
r
1
−1
∙ ∙ ∙(−1)
r
m
−1
by 4.3.
Example(1 2 5 7)(3 4 6)(8 9)(10 12 83)(79 11 26 15) has sgn =−1.
Importance of signature
1. We’ll use it to define a new family of groups below.
2. Fundamental in the theory of determinants (later).
14

15. DefinitionDefine
A
n
={x2S
n
|sgn(x) = 1},
the set of even permutations inS
n
. CallA
n
thealternating group(after
showing that it is a group).
Theorem 4.6A
n
is a subgroup ofS
n
, of size
1
2
n!.
Proof(a)A
n
is a subgroup:
(1)e2A
n
as sgn(e) = 1.
(2) forx, y2A
n
,
sgn(x) = sgn(y) = 1,
sgn(xy) = sgn(x)sgn(y) = 1,
soxy2A
n
,
(3) forx2A
n
, we have sgn(x) = 1, so by 4.1(b), sgn(x
−1
) = 1, i.e.
x
−1
2A
n
.
(b)|A
n
|=
1
2
n!: Recall that there are right cosets ofA
n
,
A
n
=A
n
e, A
n
(1 2) ={x(1 2)|x2A
n
}.
These cosets are distinct (as (1 2)2A
n
(1 2) but (1 2)/2A
n
), and have equal
size (i.e.|A
n
|=|A
n
(1 2)|). We show thatS
n
=A
n
∪A
n
(1 2): Letg2S
n
.
Ifgis even, theng2A
n
. Ifgis odd, theng(1 2) is even (as sgn(g(1 2)) =
sgn(g)sgn(1 2) = 1), sog(1 2) =x2A
n
. Theng=x(1 2)2A
n
(1 2).
So|A
n
|=
1
2
|S
n
|=
1
2
n!.Λ
Examples
1.A
3
={e,(1 2 3),(1 3 2)}, size 3 =
1
2
3!.
2.A
4
:
cycle shap
ee(2)(3)(4) (2,2)
inA
4
? yesno yesno yes
no.1 8 3
Total|A
4
|= 12 =
1
2
4!.
15

16. 3.A
5
:
cycle shap
ee(2)(3)(4)(5) (2,2) (3,2)
inA
5
? yesno yesno yes yesno
no.1 20 2415
Total|A
5
|= 60 =
1
2
5!.
5 Direct Products
So far, we’ve seen the following examples of finite groups:C
n
, D
2n
, S
n
, A
n
.
We’ll get many more using the following construction.
Recall: ifT
1
, T
2
, . . . , T
n
are sets, theCartesian productT
1
×T
2
×∙ ∙ ∙×T
n
is the set consisting of alln-tuples(t
1
, t
2
, . . . , t
n
) witht
i
2T
i
.
Now letG
1
, G
2
, . . . , G
n
be groups. Form the Cartesian productG
1
×
G
2
× ∙ ∙ ∙ ×G
n
and define multiplication on this set by
(x
1
, . . . , x
n
)(y
1
, . . . , y
n
) = (x
1
y
1
, . . . , x
n
y
n
)
forx
i
, y
i
2G
i
.
DefinitionCallG
1
× ∙ ∙ ∙ ×G
n
thedirect productof the groupsG
1
, . . . , G
n
.
Proposition 5.1Under above defined multiplication,G
1
× ∙ ∙ ∙ ×G
n
is a
group.
Proof
•ClosureTrue by closure in eachG
i
.
•AssociativityUsing associativity in eachG
i
,
[(x
1
, . . . , x
n
)(y
1
, . . . , y
n
)] (z
1
, . . . , z
n
) = (x
1
y
1
, . . . , x
n
y
n
)(z
1
, . . . , z
n
)
= ((x
1
y
1
)z
1
, . . . ,(x
n
y
n
)z
n
)
= (x
1
(y
1
z
1
), . . . , x
n
(y
n
z
n
))
= (x
1
, . . . , x
n
)(y
1
z
1
, . . . , y
n
z
n
)
= (x
1
, . . . , x
n
) [(y
1
, . . . , y
n
)(z
1
, . . . , z
n
)].
•Identityis (e
1
, . . . , e
n
), wheree
i
is the identity ofG
i
.
•Inverseof (x
1
, . . . , x
n
) is (x
−1
1
, . . . , x
−1
n
).
16

17. Examples
1. Some new groups:C
2
×C
2
, C
2
×C
2
×C
2
, S
4
×D
36
, A
5
×A
6
×S
297
, . . . ,Z×
Q×S
13
, . . ..
2. ConsiderC
2
×C
2
. Elements are{(1,1),(1,−1),(−1,1),(−1,−1)}.
Calling thesee, a, b, ab, mult table
is
eabab
eeabab
aaeabb
bbabea
ababbae
G=C
2
×C
2
is abelian andx
2
=efor allx2G.
3. SimilarlyC
2
×C
2
×C
2
has elements (±1,±1,±1), size 8, abelian,
x
2
=efor allx.
Proposition 5.2(a) Size ofG
1
× ∙ ∙ ∙ ×G
n
is|G
1
||G
2
| ∙ ∙ ∙ |G
n
|.
(b) If allG
i
are abelian so isG
1
× ∙ ∙ ∙ ×G
n
.
(c) Ifx= (x
1
, . . . , x
n
)2G
1
×∙ ∙ ∙×G
n
, then order ofxis the least common
multiple ofo(x
1
), . . . , o(x
n
).
Proof(a) Clear.
(b) Suppose allG
i
are abelian. Then
(x
1
, . . . , x
n
)(y
1
, . . . , y
n
) = (x
1
y
1
, . . . , x
n
y
n
)
= (y
1
x
1
, . . . , y
n
x
n
)
= (y
1
, . . . , y
n
)(x
1
, . . . , x
n
).
(c) Letr
i
=o(x
i
). Recall from M1P2 thatx
k
i
=eiffr
i
|k. Letr=
lcm(r
1
, . . . , r
n
). Then
x
r
= (x
r
1
, . . . , x
r
n
)
= (e
1
, . . . , e
n
) =e.
For 1≤s < r,r
i
6 |sfor somei. Sox
s
i
6=e. So
x
s
= (. . . , x
s
i
, . . .)6= (e
1
, . . . , e
n
).
Hencer=o(x).Λ
Examples
17

18. 1. Since cyclic groupsC
r
are abelian, so are all direct products
C
r
1
×C
r
2
× ∙ ∙ ∙ ×C
r
k
.
2.C
4
×C
2
andC
2
×C
2
×C
2
are abelian of size 8. Are they isomorphic?
Claim: NO.
ProofCount the number of elements of order 2 :
InC
4
×C
2
these are (±1,±1) except for (1,1), so there are 3.
InC
2
×C
2
×C
2
, all the elements exceptehave order 2, so there
are 7.
SoC
4
×C
2
6
ξ
=C
2
×C
2
×C
2
.
Proposition 5.3Ifhcf(m, n) = 1, thenC
m
×C
n
ξ
=C
mn
.
ProofLetC
m
=hαi,C
n
=hβi. Soo(α) =m,o(β) =n. Consider
x= (α, β)2C
m
×C
n
.
By 5.2(c),o(x) = lcm(m, n) =mn. Hence cyclic subgrouphxiofC
m
×C
n
has sizemn, so is whole ofC
m
×C
n
. SoC
m
×C
n
=hxiis cyclic and hence
C
m
×C
n
ξ
=C
mn
by 2.2.Λ
Direct products are fundamental to the theory of abelian groups:
Theorem 5.4Every finite abelian group is isomorphic to a direct product
of cyclic groups.
Won’t give a proof here. Reference: [Allenby, p. 254].
Examples
1. Abelian groups of size 6: by theorem 5.4, possibilities areC
6
, C
3
×C
2
.
By 5.3, these are isomorphic, so there is only one abelian group of size
6 (up to isomorphism).
2. By 5.4, the abelian groups of size 8 are:C
8
,C
4
×C
2
, C
2
×C
2
×C
2
.
Claim: No two of these are isomorphic.
Proof
Group C
2
×C
2
×C
2
C
4
×C
2
C
8
| {x|o(x) = 2}
| 7 3 1
So up to isomorphism, there are 3 abelian groups of size 8.
18

19. 6 Groups of small size
We’ll findallgroups of size≤7 (up to isomorphism). Useful results:
Proposition 6.1If|G|=p, a prime, thenG
¸
=C
p
.
ProofBy corollary of Lagrange,Gis cyclic. HenceG
¸
=C
p
by 2.2.
Proposition 6.2If|G|is even, thenGcontains an element of order 2.
ProofSuppose|G|is even andGhas no element of order 2. List the
elements ofGas follows:
e, x
1
, x
−1
1
, x
2
, x
−1
2
, . . . , x
k
, x
−1
k
.
Note thatx
i
6=x
−1
i
sinceo(x
i
)6= 2. Hence|G|= 2k+ 1, a contradiction.Λ
Groups of size1,2,3,5,7
By 6.1, only such groups areC
1
, C
2
, C
3
, C
5
, C
7
.
Groups of size4
Proposition 6.3The only groups of size 4 areC
4
andC
2
×C
2
.
ProofLet|G|= 4. By Lagrange, every element ofGhas order 1,2 or 4.
If there existsx2Gof order 4, thenhxiis cyclic, soG
¸
=C
4
. Now suppose
o(x) = 2 for allx6=e,x2G. Sox
2
=efor allx2G.
Lete, x, ybe 3 distinct elements ofG. Ifxy=etheny=x
−1
=x, a
contradiction; ifxy=xtheny=e, a contradiction; similarlyxy6=y. It
follows that
G={e, x, y, xy}.
As above,yx6=e, x, yhenceyx=xy. So multiplication table ofG
is
exyxy
eexyxy
xxexyy
yyxyex
xyxyyxe
19

20. This is the same as the table forC
2
×C
2
, soG
¸
=C
2
×C
2
.Λ
Groups of size6
We know the following groups of size 6:C
6
, D
6
, S
3
. RecallD
6
is the
symmetry group of the equilateral triangle and has elements
e, ρ, ρ
2
, σ, ρσ, ρ
2
σ.
satisfying the following equations:
ρ
3
=e,
σ
2
=e
σρ=ρ
2
σ.
The whole multiplication table ofD
6
can be worked out using these equa-
tions. e.g.
σ∙(ρσ) =ρ
2
σσ=ρ
2
.
Proposition 6.4Up to isomorphism, the only groups of size 6 areC
6
and
D
6
.
ProofLetGbe a group with|G|= 6. By Lagrange, every element ofG
has order 1, 2, 3 or 6. If there existsx2Gof order 6, thenG=hxiis cyclic
and thereforeG
¸
=C
6
by 2.2. So assumeGhas no elements of order 6. Then
everyx2G, (x6=e) has order 2 or 3. If all have order 2 thenx
2
=efor all
x2G. So by Sheet 2 Q5,|G|is divisible by 4, a contradiction. We conclude
that there existsx2Gwitho(x) = 3. Also by 6.2, there is an elementyof
order 2.
LetH=hxi=
Φ
e, x, x
2

. Theny /2HsoHy6=Hand
G=H∪Hy=
Φ
e, x, x
2
, y, xy, x
2
y

.
What isyx? Well,
yx=e)y=x
−1
yx=x)y=e
yx=x
2
)y=x
yx=y)x=e
9
>
>
=
>
>
;
a contradiction.
Ifyx=xy, let’s consider the order ofxy:
(xy)
2
=xyxy=xxyy(asyx=xy) =x
2
y
2
=x
2
.
20

21. Similarly
(xy)
3
=x
3
y
3
=y6=e.
Soxydoes not have order 2 or 3, a contradiction. Henceyx6=xy. We
conclude thatyx=x
2
y.
At this point we know the following:
•G=
Φ
e, x, x
2
, y, xy, x
2
y

,
•x
3
=e,x
2
=e,yx=x
2
y.
In exactly the same way as forD
6
, can work out the whole multiplication
table forGusing these equations. It will be the same as the table forD
6
(withx, yinstead ofρ, σ). SoG
¸
=D
6
.Λ
RemarkNote that|S
3
|= 6, andS
3
¸
=D
6
.
Summary
Proposition 6.5Up to isomorphism, the groups of size≤7 are
Size Gr
oups
1C
1
2C
2
3C
3
4C
4
,C
2
×C
2
5C
5
6C
6
, D
6
7C
7
Remarks on larger sizes
Size 8: here are the groups we know:
AbelianC
8
, C
4
×C
2
, C
2
×C
2
×C
2
,
Non-abelianD
8
.
Any others? Yes, thequaterniongroupQ
8
:
Define matrices
A=
`
i0
0−i
´
, B=
`
0 1
−1 0
´
.
21

22. Check equations:
A
4
=I, B
4
=I, A
2
=B
2
, BA=A
4
B.
Define
Q
8
={A
r
B
s
|r, s2Z}
={A
m
B
n
|0≤m≤3,0≤n≤1}.
Sheet 3 Q5:|Q
8
|= 8.Q
8
is a subgroup ofGL(2,C) and is not abelian and
Q
8
6
¸
=D
8
. CallQ
8
thequaternion group. Sheet 3 Q7: The only non-abelian
groups of size 8 areD
8
andQ
8
. Yet more info:
Size
Groups
9 only abelian (Sh3 Q4)
10C
10
, D
10
11C
11
12 abelian,D
12
,A
4
+ one more
13C
13
14C
14
, D
14
15C
15
16 14 groups
7 Homomorphisms, normal subgroups and factor
groups
Homomorphisms are functions between groups which “preserve multiplica-
tion”.
DefinitionLetG, Hbe groups. A functionφ:G→His ahomomorphism
ifφ(xy) =φ(x)φ(y) for allx, y2G.
Note that an isomorphism is a homomorphism which is abijection.
Examples
1.G, Hany groups. Defineφ:G→Hby
φ(x) =e
H
8x2G
Thenφis a homomorphism sinceφ(xy) =e
H
=e
H
e
H
=φ(x)φ(y).
22

23. 2. Recall the signature function sgn :S
n
→C
2
. By 4.1(a), sgn(xy) =
sgn(x)sgn(y), so sgn is a homomorphism.
3. Defineφ: (R,+)→(C
Λ
,×) by
φ(x) =e
2πix
8x2R.
Thenφ(x+y) =e
2πi(x+y)
=e
2πix
e
2πiy
=φ(x)φ(y), soφis a homo-
morphism.
4. Defineφ:D
2n
→C
2
(writingD
2n
=
Φ
e, ρ, . . . , ρ
n−1
, σ, ρσ, . . . , ρ
n−1
σ

)
by
φ(ρ
r
σ
s
) = (−1)
s
.
(soφsends rotations to +1 and reflections to−1). Thenφis a homo-
morphism since:
φ
Γ
(ρ
r
σ
s
)(ρ
t
σ
u
)
Δ
=φ(ρ
r±t
σ
s+u
)
= (−1)
s+u
=φ(ρ
r
σ
s
)φ(ρ
r
σ
u
).
Proposition 7.1Letφ:G→Hbe a homomorphism
(a)φ(e
G
) =e
H
(b)φ(x
−1
) =φ(x)
−1
for allx2G.
(c)o(φ(x))divideso(x)for allx2G.
Proof(a) Note thatφ(e
G
) =φ(e
G
e
G
) =φ(e
G
)φ(e
G
). Multiply byφ(e
G
)
−1
to gete
H
=φ(e
G
).
(b) By (a),e
H
=φ(e
G
) =φ(xx
−1
) =φ(x)φ(x
−1
). Soφ(x
−1
) =φ(x)
−1
.
(c) Letr=o(x). Then
φ(x)
r
=φ(x)∙ ∙ ∙φ(x) =φ(x∙ ∙ ∙x) =φ(x
r
) =φ(e
G
) =e
H
.
Henceo(φ(x)) dividesr.Λ
DefinitionLetφ:G→Hbe homomorphism. Theimageofφis
Imφ=φ(G) ={φ(x)|x2Gg `H.
Proposition 7.2Ifφ:G→His a homomorphism, thenImφis a subgroup
ofH.
23

24. Proof
(1)e
H
2Imφsincee
H
=φ(e
G
).
(2) Letg, h2Imφ. Theng=φ(x) andh=φ(y) for somex, y2G, so
gh=φ(x)φ(y) =φ(xy)2Imφ.
(3) Letg2Imφ. Theng=φ(x) for somex2G. Sog
−1
=φ(x)
−1
=
φ(x
−1
)2Imφ.
Hence Imφis a subgroup ofH.Λ
Examples
1. Is there a homomorphismφ:S
3
→C
3
? Yes,φ(x) = 1 for allx2S
3
.
For this homomorphism, Imφ={1}.
2. Is there a homomorphismφ:S
3
→C
3
such that Imφ= C
3
?
To answer this, supposeφ:S
3
→C
3
is a homomorphism. Consider
φ(1 2). By 7.1(c),φ(1 2) has order dividingo(1 2) = 2. Asφ(1 2)2C
3
,
this implies thatφ(1 2) = 1. Similarlyφ(1 3) =φ(2 3) = 1. Hence
φ(1 2 3) =φ((1 3)(1 2)) =φ(1 3)φ(1 2) = 1
and similarlyφ(1 3 2) = 1. We’ve shown that
φ(x) = 18x2S
3
.
So there is no surjective homomorphismφ:S
3
→C
3
.
Kernels
DefinitionLetφ:G→Hbe a homomorphism. Thenkernelofφis
Kerφ={x2G|φ(x) = e
H
}.
Examples
1. Ifφ:G→Hisφ(x) =e
H
for allx2G, then Kerφ= G.
2. For sgn :S
n
→C
2
,
Ker(sgn) ={x2S
n
|sgn(x) = 1}= A
n
,the alternating group.
24

25. 3. Ifφ: (R,+)→(C
Λ
,×) isφ(x) =e
2πix
for allx2R, then
Kerφ=
Φ
x2R|e
2πix
= 1

=Z.
4. Letφ:D
2n
→C
2
be given byφ(ρ
r
σ
s
) = (−1)
s
. Then Kerφ=hρi.
Proposition 7.3Ifφ:G→His a homomorphism, thenKerφis a sub-
group ofG.
Proof
(1)e
G
2Kerφasφ(e
G
) =e
H
by 7.1.
(2)x, y2Kerφthenφ(x) =φ(y) =e
H
, soφ(xy) =φ(x)φ(y) =e
H
; i.e.
xy2Kerφ.
(3)x2Kerφthenφ(x) =e
H
, soφ(x)
−1
=φ(x
−1
) =e
H
, sox
−1
2Kerφ.
Λ
In fact, Kerφis a very special type of subgroup ofGknown as anormal
subgroup.
Normal subgroups
DefinitionLetGbe a group, andN`G. We sayNis anormal subgroup
ofGif
(1)Nis a subgroup ofG,
(2)g
−1
Ng=Nfor allg2G, whereg
−1
Ng=
Φ
g
−1
ng|n2N

.
IfNis a normal subgroup ofG, writeNCG.
Examples
1.Gany group. Subgrouphei={e}CGasg
−1
eg=efor allg2G. Also
subgroupGitself is normal, i.e.GCG, asg
−1
Gg=Gfor allg2G.
Next lemma makes condition (2) a bit easier to check.
Lemma 7.4LetNbe a subgroup ofG. ThenNCGif and only ifg
−1
Ng`
Nfor allg2G.
25

26. Proof
)Clear.
(Supposeg
−1
Ng`Nfor allg2G. Letg2G. Then
g
−1
Ng`N.
Usingg
−1
instead, we get (g
−1
)
−1
Ng
−1
`N, hence
gNg
−1
`N.
HenceN`g
−1
Ng. Thereforeg
−1
Ng=N.Λ
Examples(1) We show thatA
n
CS
n
. Need to show that
g
−1
A
n
g`A
n
8g2S
n
(this will showA
n
CS
n
by 7.4).
Forx2A
n
, using 4.1 we have
sgn(g
−1
xg) = sgn(g
−1
)sgn(x)sgn(g) = sgn(g
−1
)∙1∙sgn(g) = 1.
Sog
−1
xg2A
n
for allx2A
n
. Hence
g
−1
A
n
g`A
n
.
SoA
n
CS
n
.
(2) LetG=S
3
,N=h(1 2)i={e,(1 2)}. IsNCG? Well,
(1 3)
−1
(1 2)(1 3) = (1 3)(1 2)(1 3) = (2 3)/2N.
So (1 3)
−1
N(1 3)6=NandN6CS
3
.
(3) IfGis abelian, thenallsubgroupsNofGare normal since forg2G,
n2N,
g
−1
ng=g
−1
gn=n,
and henceg
−1
Ng=N.
(4) LetD
2n
=
Φ
e, ρ, . . . , ρ
n−1
, σ, ρσ, . . . , ρ
n−1
σ

. Fix an integerr. Then
hρ
r
iCD
2n
.
Proof – sheet 4. (key: magic equationσρ=ρ
−1
σ, . . . , σρ
n
=ρ
−n
σ).
26

27. Proposition 7.5Ifφ:G→His a homomorphism, thenKerφCG.
ProofLetK= Kerφ. By 7.3Kis a subgroup ofG. Letg2G,x2K.
Then
φ(g
−1
xg) =φ(g
−1
)φ(x)φ(g) =φ(g)
−1
e
H
φ(g) =e
H
.
Sog
−1
xg2Kerφ= K. This showsg
−1
Kg`K. SoKCG.Λ
Examples
1. We know that sgn :S
n
→C
2
is a homomorphism, with kernelA
n
. So
A
n
CS
n
by 7.5.
2. Knowφ:D
2n
→C
2
defined byφ(ρ
r
σ
s
) = (−1)
s
is a homomorphism
with kernelhρi. SohρiCD
2n
.
3. Here’s a different homomorphismα:D
8
→C
2
where
α(ρ
r
σ
s
) = (−1)
r
.
This is a homomorphism, as
α((ρ
r
σ
s
)(ρ
t
σ
u
)) =α(ρ
r±t
σ
s+u
)
= (−1)
r±t
= (−1)
r
∙(−1)
t
=α(ρ
r
σ
s
)α(ρ
t
σ
u
).
The kernel ofαis
Kerα={ρ
r
σ
s
|r even}=
Φ
e, ρ
2
, σ, ρ
2
σ

.
Hence
Φ
e, ρ
2
, σ, ρ
2
σ

CD
8
.
Factor groups
LetGbe a group,Na subgroup ofG. Recall that there are exactly
|G
|
|N|
different right cosetsNx(x2G). Say
Nx
1
, Nx
2
, . . . , Nx
r
wherer=
|G
|
|N|
. Aim is to make this set of right cosets into a group in a
natural way. Here is a Ωatural” definition of multiplication of these cosets:
(Nx)(Ny) =N(xy). (33)
27

28. Does this definition make sense? To make sense, we need:
Nx=Nx
0
Ny=Ny
0
œ
)Nxy=Nx
0
y
0
for allx, y, x
0
, y
0
2G. This property may or may not hold.
ExampleG=S
3
,N=h(1 2)i={e,(1 2)}. The 3 right cosets ofNinG
are
N=Ne, N(1 2 3), N(1 3 2).
Also
N =N(1 2)
N(1 2 3) =N(1 2)(1 2 3) =N(2 3)
N(1 3 2) =N(1 2)(1 3 2) =N(1 3).
According to (33),
(N(1 2 3)) (N(1 2 3)) =N(1 2 3)(1 2 3) =N(1 3 2).
But (33) also says that
(N(2 3)) (N(2 3)) =N(2 3)(2 3) =Ne.
So (33) makes no sense in this example.
How do we make (33) make sense? The condition is thatNCG. Key is
to prove the following:
Proposition 7.6LetNCG. Then forx
1
, x
2
, y
1
, y
2
2G
Nx
1
=Nx
2
Ny
1
=Ny
2
œ
)Nx
1
y
1
=Nx
2
y
2
.
(Hence definition of multiplication of cosets in (33) makes sense whenNC
G.)
To prove this we need a definition and a lemma: forHa subgroup ofG
andx2Gdefine theleft coset
xH={xh:h2H}.
Lemma 7.7SupposeNCG. ThenxH=Hxfor allx2G.
28

29. ProofLeth2H. AsHCG,xHx
−1
=H, and soxhx
−1
=h
0
2H.
Thenxh=h
0
x2Hx. This shows thatxH`Hx. Similarly we see that
Hx`xH, hencexH=Hx.Λ
Proof of Prop 7.6
LetNCG. SupposeNx
1
=Nx
2
andNy
1
=Ny
2
. Then
Nx
1
y
1
=Nx
2
y
1
asNx
1
=Nx
2
=x
2
Ny
1
by Prop 7.7
=x
2
Ny
2
asNy
1
=Ny
2
=Nx
2
y
2
by Prop 7.7.Λ
So we have established that whenNCG, the definition of multiplication
of cosets
(Nx)(Ny) =Nxy
forx, y2Gmakes sense.
Theorem 7.8LetNCG. DefineG/Nto be the set of all right cosetsNx
(x2G). Define multiplication onG/Nby
(Nx)(Ny) =Nxy.
ThenG/Nis a group under this multiplication.
Proof
Closureobvious.
AssociativityUsing associativity inG
(NxNy)Nz= (Nxy)Nz
=N(xy)z
=Nx(yz)
= (Nx)(Nyz)
=Nx(NyNz).
IdentityisNe=N, sinceNxNe=Nxe=NxandNeNx=Nex=
Nx.
InverseofNxisNx
−1
, asNxNx
−1
=Nxx
−1
=Ne, the identity.
29

30. DefinitionThe groupG/Nis called thefactor groupofGbyN.
Note that
|G/N|=
|G|
|N|
.
Examples
1.A
n
CS
n
. Since
|S
n
|
|A
n
|
= 2, the factor groupS
n
/A
n
has 2 elements
A
n
, A
n
(1 2).
SoS
n
/A
n
ξ
=C
2
. Note: in the groupS
n
/A
n
the identity is the coset
A
n
and the non identity elementA
n
(1 2) has order 2 as
(A
n
(1 2))
2
=A
n
(1 2)A
n
(1 2) =A
n
(1 2)(1 2) =A
n
.
2.Gany group. We know thatGCG. What is the factor groupG/G?
Ans:G/Ghas 1 element, the identity cosetG. SoG/G
ξ
=C
1
.
Alsohei={e}CG. What isG/hei? Cosetheig={g}, and multipli-
cation
(heig) (heih) =heigh.
SoG/hei
ξ
=G(isomorphismg7→ heig).
3.G=D
12
=
Φ
e, ρ, . . . , ρ
5
, σ, σρ, . . . , σρ
5

whereρ
6
=σ
2
=e,σρ=
ρ
−1
σ.
(a) Know thathρiCD
12
. Factor groupD
12
/hρihas 2 elements
hρi,hρiσsoD
12
/hρi
ξ
=C
2
.
(b) Know also that

ρ
2

=
Φ
e, ρ
2
, ρ
4

CD
12
. SoD
12
/

ρ
2

has 4
elements, so
D
12
/

ρ
2

ξ
=C
4
orC
2
×C
2
.
Which? Well, letN=

ρ
2

. The 4 elements ofD
12
/Nare
N, Nρ, Nσ, Nρσ.
We work out the order of each of these elements ofD
12
/N:
(Nρ)
2
=NρNρ=Nρ
2
=N,
(Nσ)
2
=NσNσ=Nσ
2
=N,
(Nρσ)
2
=N(ρσ)
2
=N.
30

31. So all non-identity elements ofD
12
/Nhave order 2, henceD
12
/hρi
¸
=
C
2
×C
2
.
(c) Also

ρ
3

=
Φ
e, ρ
3

CD
12
. Factor groupD
12

ρ
3

has 6 elements
so is
¸
=C
6
orD
6
. Which? LetM=

ρ
3

. The 6 elements of
D
12
/Mare
M, Mρ, Mρ
2
, Mσ, Mρσ, Mρ
2
σ.
Letx=Mρandy=Mσ. Then
x
3
= (Mρ)
3
=MρMρMρ =Mρ
3
=M,
y
2
= (Mσ)
2
=Mσ
2
=M,
yx=MσMρ=Mσρ=Mρ
−1
σ=Mρ
−1
Mσ
=x
−1
y.
SoD
12
/M=
Φ
identity, x, x
2
, y, xy, x
2
y

andx
3
=y
2
= identity,yx=
x
−1
y. SoD
12
/

ρ
3

¸
=D
6
.
Here’s a result tying all these topics together:
Theorem 7.9 (First Isomorphism Theorem) Letφ:G→Hbe a ho-
momorphism. Then
G/Kerφ
¸
=Imφ.
ProofLetK= Kerφ. SoG/Kis the group consisting of the cosets
Kx(x2G) with multiplication (Kx)(Ky) =Kxy. We want to define a
Ωatural” functionG/K→Imφ. Obvious choice is the functionKx7→φ(x)
forx2G. To show this is a function, need to prove:
Claim 1.IfKx=Ky, thenφ(x) =φ(y).
To prove this, supposeKx=Ky. Thenxy
−1
2K(asx2Kx)x=ky
for somek2K)xy
−1
=k2K). Hencexy
−1
2K= Kerφ, so
φ(xy
−1
) =e
)φ(x)φ(y
−1
) =e
)φ(x)φ(y)
−1
=e
)φ(x) =φ(y).
By Claim 1, we can define a functionα:G/K→Imφby
α(Kx) =φ(x)
31

32. for allx2G.
Claim 2.αis an isomorphism.
Here is a proof of this claim.
(1)αis surjective: for ifφ(x)2Imφthenφ(x) =α(Kx).
(2)αis injective:
α(Kx) =α(Ky)
)φ(x) =φ(y)
)φ(x)φ(y)
−1
=e
)φ(xy
−1
) =e,
soxy
−1
2Kerφ= K and soKx=Ky.
(3) Finally
α((Kx)(Ky)) =α(Kxy)
=φ(xy)
=φ(x)φ(y)
=α(Kx)α(Ky).
Henceαis an isomorphism.
This completes the proof thatG/K
¸
=Imφ.Λ
Corollary 7.10Ifφ:G→His a homomorphism, then
|G|=|Kerφ| ∙ |Imφ|.
One can think of this as the group theoretic version of the rank-nullity
theorem.
Examples
1. Homomorphism sgn :S
n
→C
2
. By 7.9
S
n
/Ker(sgn)
¸
=Im(sgn),
so
S
n
/A
n
¸
=C
2
.
2. Homomorphismφ: (R,+)→(C
Λ
,×)
φ(x) =e
2πix
.
32

33. Here
Kerφ=
Φ
x2R|e
2πix
= 1

=Z,
Imφ=
Φ
e
2πix
|x2R

=Tthe unit circle.
SoR/Z
¸
=T.
3. Is there a surjective homomorphismφfromS
3
ontoC
3
? Shown pre-
viously – No.
Here’s a better way to see this: suppose there exist suchφ. Then
Imφ= C
3
, so by 7.9,S
3
/Kerφ
¸
=C
3
. So Kerφis a normal subgroup
ofS
3
of size 2. ButS
3
has no normal subgroups of size 2 (they are
h(1 2)i,h(1 3)i,h(2 3)i).
Given a homomorphismφ:G→H, we know KerφCG. Converse
question: Given a normal subgroupNCG, does there exist a homomorphism
with kernelN? Answer is YES:
Proposition 7.11LetGbe a group andNCG. DefineH=G/N. Let
φ:G→Hbe defined by
φ(x) =Nx
for allx2G. Thenφis a homomorphism andKerφ= N.
ProofFirst,φ(xy) =Nxy= (Nx)(Ny) =φ(x)φ(y), soφis a homomor-
phism. Also
x2Kerφ,φ(x) = e
H
,Nx = N,x2N.
Hence Kerφ= N.Λ
ExampleFrom a previous example, we know

ρ
2

=
Φ
e, ρ
2
, ρ
4

CD
12
. We
showed thatD
12

ρ
2

¸
=C
2
×C
2
. So by 7.11, the functionφ(x) =

ρ
2

x
(x2D
12
) is a homomorphismD
12
→C
2
×C
2
which is surjective, with
kernel

ρ
2

.
Summary
There is a correspondence
{normal subgroups ofG} ↔ {homomorphisms ofG}.
33

34. ForNCGthere is a homomorphismφ:G→G/Nwith Kerφ= N. For a
homomorphismφ, Kerφis a normal subgroup ofG.
GivenG, to find allHsuch that there exist a surjective homomorphism
G→H:
(1) Find all normal subgroups ofG.
(2) The possibleHare the factor groupsG/NforNCG.
Example:G=S
3
.
(1) Normal subgroups ofGare
hei, G, A
3
=h(1 2 3)i
(cyclic subgroups of size 2h(i j)iare not normal).
(2) Factor groups:
S
3
/hei
¸
=S
3
, S
3
/S
3
¸
=C
1
, S
3
/A
3
¸
=C
2
8 Symmetry groups in 3 dimensions
These are defined similarly to symmetry groups in 2 dimensions, see chapter
2. AnisometryofR
3
is a bijectionf:R
3
→R
3
such thatd(x, y) =
d(f(x), f(y)) for allx, y2R
3
.
Examples of isometries are: rotation about an axis, reflection in a plane,
translation.
As in 2.1, the set of all isometries ofR
3
, under composition, forms a group
I(R
3
). For Π`R
3
, thesymmetry groupof Π isG(Π) =
Φ
g2I(R
3
)|g(Π) = Π

.
There exist many interesting symmetry groups inR
3
. Some of the most in-
teresting are the symmetry groups of the Platonic solids: tetrahedron, cube,
octahedron, icosahedron, dodecahedron.
Example:The regular tetrahedron
Let Π be regular tetrahedron inR
3
, and letG=G(Π).
•Rotations inG: LetRbe the set of rotations inG. Some elements of
R:
34

35. (1)e,
(2) rotations of order 3 fixing one corner: these are
ρ
1
, ρ
2
1
, ρ
2
, ρ
2
2
, ρ
3
, ρ
2
3
, ρ
4
, ρ
2
4
(whereρ
i
fixes corneri),
(3) rotations of order 2 about an axis joining the mid-points of op-
posite sides
ρ
12,34
, ρ
13,24
, ρ
14,23
.
So|R| ≥12. Also|R| ≤12: can rotate to get any faceion bottom (4
choices). Ifiis on the bottom, only 3 possible configurations. Hence
|R| ≤4∙3 = 12. Hence|R|= 12.
Claim 1:R
¸
=A
4
.
To see this, observe that each rotationr2Rgives a permutation of
the corners 1,2,3,4, call itπ
r
:
e →π
e
= identity permutation
ρ
i
, ρ
2
i
→all 8 3-cycles inS
4
(1 2 3),(1 3 2), . . .
ρ
12,34
→(1 2)(3 4)
ρ
13,24
→(1 3)(2 4)
ρ
14,23
→(1 4)(2 3).
Notice that{π
r
|r2R}consists of all the 12evenpermutations in
S
4
, i.e.A
4
. The mapr7→π
r
is an isomorphismR→A
4
. SoR
¸
=A
4
.
Claim 2:The symmetry groupGisS
4
.
ObviouslyGcontains a reflectionσwith corresponding permutation
π
σ
= (1 2). SoGcontains
R∪Rσ.
So|G| ≥ |R|+|Rσ|= 24. On the other hand, eachg2Ggives a
unique permutationπ
g
2S
4
, so|G| ≤ |S
4
|= 24. So|G|= 24 and the
mapg7→π
g
is an isomorphismG→S
4
.
9 Counting using groups
Consider the following problem. Colour edges of an equilateral triangle with
2 coloursR, B. How many distinguishable colourings are there?
Answer: There are 8 colourings altogether:
35

36. (1) all the edges red – RRR,
(2) all the edges blue – BBB,
(3) two reds and a blue – RRB,RBR,BRR,
(4) two blues and a red – BBR,BRB,RBB.
Clearly there are 4 distinguishable colourings. Point: Two colourings are
not distinguishable iff there exists a symmetry of the triangle sending one
to the other.
To bring groups into the picture: callCthe set of all 8 colorings. So
C={RRR, . . . , RBB}.
LetGbe the symmetry group of the equilateral triangle,D
6
=
Φ
e, ρ, ρ
2
, σ, ρσ, ρ
2
σ

.
Each element ofD
6
gives a permutation ofC, e.g.ρgives the permutation
(RRR) (BBB) (RRB RBR BRR ) (BBR BRB RBB ).
Divide the setCinto subsets calledorbitsofG: two colouringsc, dare
in the same orbit if there existsg2D
6
sendingctod. The orbits are the
sets (1) - (4) above. The number of distinguishable colourings is equal to
the number of orbits ofG.
General situation
Suppose we have a setSand a groupGconsisting of some permutations
ofS(e.g.S=C,G=D
6
above). PartitionSintoorbitsofG, by saying
that two elementss, t2Sare in the same orbit iff there exists ag2Gsuch
thatg(s) =t. How many orbits are there?
Lemma 9.1 (Burnside’s Counting Lemma) Forg2G, define
fix(g) =number of elements ofSfixed byg
=|{s2S|g(s) =s}|.
Then
number of orbits ofG=
1
|G|
X
g2G
fix(g).
I won’t give a proof. Look it up in the recommended book by Fraleigh
if you are interested.
Examples
36

37. (1)C= set of 8 colourings of the equilateral triangle.G=D
6
. Here are
the values of fix(g):
geρ ρ
2
σρσ ρ
2
σ
fix(g)822444
By 9.1, number of orbits is
1
6
(8 + 2 + 2 + 4 + 4 + 4) = 4.
(2) 6 beads coloured R, R, W, W, Y, Y are strung on a necklace. How
many distinguishable necklaces are there?
Each necklace is a colouring of a regular hexagon. Two colourings are
indistinguishable if there is a rotation or reflection sending one to the
other (a reflection is achieved by turning the hexagon upside down).
LetDbe the set of colourings of the hexagon andG=D
12
.
g e ρ ρ
2
ρ
3
ρ
4
ρ
5
fix(g
)
Γ
6
2
Δ
×
Γ
4
2
Δ
00600
gσρσ ρ
2
σ ρ
3
σ ρ
4
σ ρ
5
σ
fix(g)666 6 6 6
So by 9.1
number of orbits =
1
12
(90 + 42) = 11.
So the number of distinguishable necklaces is 11.
(3) Make a tetrahedral die by putting 1, 2, 3, 4 on the faces. How many
distinguishable dice are there?
Each die is a colouring (colours 1, 2, 3, 4) of a regular tetrahedron. Two
such colourings are indistinguishable if there exists arotationof the
tetrahedron sending one to the other. LetEbe the set of colourings,
andG= rotation group of tetrahedron (so|G|= 12,G
¸
=A
4
by
Chapter 8). Here forg2G
fix(g) =
æ
24 ifg=e,
0 ifg6=e.
So by 9.1, number of orbits is
1
12
(24) = 2. So there are 2 distinguishable
tetrahedral dice.
37