Download ALGEBRA 1, D. CHAN 1. Introduction 1Introduction to groups via

ALGEBRA 1, D. CHAN
1. Introduction
1
Introduction to groups via symmetry. A symmetry of F is a surjective isometry which preserves F .
Definition 1.1. A set G is a group when given an operation G × G −→ G, satisfying
1. associativity, (ab)c = a(bc),
2. identity, there exists a 1 such that a1 = a
3. inverse, there exists b for all a such that ba = 1, we write b = a−1 .
Example 1. G = R − {0} the real numbers, with the usual multiplication forms a group.
but G = Z the integers do not form a group under multiplication, Z is a group under addition.
2. Symmetry groups
Symmetric group, alternating groups, cycles.
Disjoint cycles commute, and generate the symmetric group.
Example 2.
1 2 3
2 4 6
This can be written as σ = (124)(36)(5) = (124)(36)
σ=
4
1
5
5
6
3
Lemma 2.1. Any element of Σn can be written
S as a product of disjoint cycles. I.e. Let σ ∈ Σn , then for
some set S, where |S| = n, we can write S = Si such that σ permutes elements of Si cyclically.
i
Proof. see handout.
Lemma 2.2. Any element of Σn can be written as a product of transpositions.
2
2
Proof. (a1 a2 . . . an ) = (a1 an )(a1 an−1 ) . . . (a1 a3 )(a1 a2 )
3. Dihedral groups, subgroups generated by subsets
1 2
... n
Consider σ, τ ∈ Σn , where σ = (12 . . . n), and τ =
, τ 2 = σ n = 1Σn . What is the
n n−1
1
smallest subgroup H containing σ and τ ? This group is called the dihedral group, Dn .
Definition 3.1.
hσ, τ i = Dn
We will use τ σ = σ −1 τ (show this) to simplify products of σ and τ by swapping τ and σ around in the
product.
Proposition 3.2. H is the set of 2n elements, σ i τ , σ i , for i ∈ {0 . . . n − 1}.
Proof. Show closure under multiplication, inversion, identity.
T
Proposition 3.3. Let {Hi } be a set of subgroups of a group G, then H = Hi is also a subgroup.
2
i
2
Proof. Trivial
Proposition 3.4. Let S ⊆ G, there exists a unique smallest subgroup H containing S, this is called the
subgroup generated by S.
Proof. Use above and take intersections of all subgroups containing S.
Last lecture. For the dihedral group, Dn , we usually require n > 3, and |Dn | = 2n.
Alternating group, abelian groups.
Symmetric function
1
The first few lectures are a bit sketchy, my apologies.
1
2
2
ALGEBRA 1, D. CHAN
3.1. Van der Monde determinant. ...symmetries of. (12)∆ = −∆
Lemma 3.5. For σ, τ ∈ Σn , f (x1 . . . xn ).
(στ )f = σ(τ f )
2
Proof. Work from outside in..
Definition 3.6. Let ∆ be the difference product, σ ∈ Σn . By proposition in lecture 3, σ can be written
as a product of transpositions, τi . . . τm . Then
−∆
if m is odd
σ∆ =
∆
if m is even
Proof. It suffices to prove that for σ = (ij), i.e. m = 1, since
σ∆ = τ1 . . . τm ∆ = τ1 . . . τm−1 (τm ∆)
Now we need to account for all the factors,
•
•
•
•
•
σ(xi − xj ) = −(xi − xj )
σ(xr − xs ) = (xr − xs )
For r < i < j, σ(xr − xi )(xr − xj ) = (xr − xj )(xr − xi )
For i < r < j, σ(xi − xr )(xr − xj ) = (xi − xr )(xr − xj )
For j < i < r, σ(xi − xr )(xj − xr ) = (xi − xr )(xj − xr )
This implies σ∆ = −∆, and the lemma follows.
2
Corollary 3.7. Let An = {σ ∈ Sn : σ∆ = ∆}, be all the even permutations. It is a proper subgroup of
Σn called the alternating group. It is the subgroup generated by the subset,
S = {τ1 τ2 : τ12 = τ22 = 1}
2
Proof. S clearly contains all the even permutations...
Abelian groups, cosets and Lagrange’s Theorem
Examples. A transposition coset of the alternating group,
1
0
0
d
coset of SL2
Example 3. (Disjoint union of cosets implies...)
S
• Z = m−1
r=0 (r +mZ)
S
1 0
• GL2 = d∈R∗
SL2
0 d
• Σn = An ∪ τ An , where τ is any transposition.
Lemma 3.8. Any left/right coset, gH/Hg of H, has the same cardinality as H.
Definition 3.9. (index)
Theorem 3.10. (Lagrange)
Proposition 3.11. There is a 1 − 1 correspondence between the left and right cosets
Proof. by inversion
2
ALGEBRA 1, D. CHAN
3
4. Normal groups and Quotient groups
Example 4. Let G = Dn < Sn , recall that Dn = hσ, τ i, σ = (12 . . . n), τ =
1
n
2
n−1
...
...
n
1
. Also
σ n = 1 = τ 2 , τ σ = σ −1 τ . Let
H
N
Dn
hτ i
hσi
hτ, σi
=
=
=
=
=
=
{1, τ }
{σ i }n−1
i=0
{σ i , σ i τ }n−1
i=0
Definition 4.1. Let N P G, then the subset product makes G/N into a group, the quotient group of G
mod N .
2
Proof. trivial
Example 5. Modulo arithmetic, GLn (C)/ SLn (C)
5. Isomorphisms and homomorphisms
Definition 5.1. Let ϕ : H −→ G be an homomorphism of groups, the following statements are equivalent,
1. There exists a homomorphism ϕ0 : G −→ H which is an inverse to ϕ.
2. ϕ is bijective.
∼
In this case ϕ is an isomorphism (write ϕ : G −→ H, and G ∼
= H are isomorphic)
Proof. (2 =⇒ 1) ϕ0 is the inverse bijection to ϕ.
(1 =⇒ 2) Let ϕ0 = ϕ−1 for h, h0 ∈ H we know that ϕ(h0 h) = ϕ(h0 )ϕ(h). For g, g 0 ∈ G let h = ϕ−1 (g), h0 =
ϕ−1 (g 0 ). ϕ(ϕ−1 (g)ϕ−1 (g 0 )) = gg 0 . Take inverse to show that ϕ−1 is a homomorphism.
2
1
0
Example 6. ϕ : R −→ H, where H =
a
1
1
∈ GL2 (R) : a ∈ R . ϕ : a 7−→
0
a
1
. ϕ is a
group isomorphism.
Show homormophicity and bijectivity.
Proposition 5.2. Let ϕ : H −→ G be an homomorphism of groups.
1. ϕ(1H ) = 1G
2. ∀h ∈ H, ϕ(h−1 ) = ϕ(h)−1
3. ϕ(H) is a subgroup of G.
Proof. all trivial
Proposition 5.3. Let ϕ : H −→ G, ker(ϕ) = 1H =⇒ ϕ is injective.
2
Proposition 5.4. Let G be a group, and g ∈ G, then
Cg
:
G −→ G
:
h 7−→ g −1 hg
this is known is the conjugation by g, and is an isomorphism.
Proof. We have Cg −1 as the inverse map (check). Check Cg is an homomorphism.
Example 7. Let f : S −→ T be a set bijection, then Perm(S) ∼
= Perm(T ).
2
Proof. In fact,
ϕ
:
Perm(S) −→ Perm(T )
:
σ 7−→ f σf −1
is a group isomorphism, with inverse being ψ : τ 7−→ f −1 τ f . Exercise, show that the previous calculation
implies ϕ, ψ are inverse isomorphisms.
2
Example 8. Let F be an equilateral triangle in R2 , G be the group of symmetries of F , and V be the set
of vertices of F . We wish to define an isomorphism
ϕ : G −→ Perm(V ) ∼
= Σ3
σ 7−→ σ|V
ϕ : G −→ Perm(V ) ∼
= Σ3 .
4
ALGEBRA 1, D. CHAN
6. Homomorphims, Quotient Homomorphisms
As for linear maps, we can compose homomorphisms
Proposition 6.1. Let ϕ : G −→ G0 ϕ0 : G0 −→ G00 then ϕ0 ϕ : G −→ G00 is an homomorphism.
Proof. For h, g ∈ G (ϕ0 ϕ)(hg) = ϕ0 (ϕ(h)ϕ(g)) = ϕ0 ϕ(h)ϕ0 ϕ(g)
2
Example 9. Let ϕ : Σn −→ GLn (R),
ϕ(σ)
Rn −→ Rn
:
ei 7−→ eσ(i)
0
ϕ is an homomorphism. Let ϕ = det(GLn (R)) −→ R∗ what is ϕ0 ϕ : Σn −→ R∗ ? ...
ϕ0 ϕ(σ)
=
(ϕ0 ϕ)(τ1 . . . τm )
=
(ϕ0 ϕ)(τ1 ) . . . (ϕ0 ϕ)(τm )
=
(−1)m
1
−1
=
σ ∈ An
otherwise
Definition 6.2. A group homomorphism ϕ : G −→ G0 is said to be an
i. epimorphism if ϕ is surjective, e.g. det
ii. monomorphism if ϕ is injective
iii. automorphism if ϕ is an isomorphism from G −→ G
Definition 6.3. (Kernel) This gives a condition when injectivity fails. Let ϕ : G −→ G0 be a group
homomorphism, then we define the kernel of ϕ to be
ker(ϕ)
=
ϕ−1 (1)
=
{g ∈ G : ϕ(g) = 1}
Proposition 6.4. Let ϕ be as above.
(1) ker(ϕ) P G
(2) The non empty fibres of ϕ is sets of the form, for g 0 ∈ G0
ϕ−1 (g 0 ) = {g ∈ G : ϕ(g) = g 0 }
are the cosets of ker(ϕ).
(3) ϕ is injective iff ker(ϕ) = 1G .
Proof.
(1) ϕ(1) = 1 ∈ ker(ϕ), for h, g ∈ ker(ϕ), then ϕ(hg) = ϕ(h)ϕ(g) = (1)(1) = 1 =⇒ hg ∈ ker(ϕ).
ϕ(g −1 ) = ϕ(g)−1 = 1−1 = 1. So ker(ϕ) 6 G.We now check normality, i.e. for any g ∈ G,
g −1 ker(ϕ)g ⊆ ker(ϕ). Let k ∈ ker(ϕ)
ϕ(g −1 kg)
=
ϕ(g)−1 (1)ϕ(g)
=
1
−1
implying g ker(ϕ)g ⊆ ker(ϕ).
(2) Suppose g ∈ ϕ−1 (g 0 ) we must show, g ker(ϕ) = ϕ−1 (g 0 ), K = ker(ϕ)
ϕ(gK)
=
ϕ(g)ϕ(K)
=
ϕ(g)
=
g0
=
ϕ(g)−1 ϕ(h)
=
g 0−1 g 0
=
1
Suppose h ∈ ϕ−1 (g 0 ), then
ϕ(g −1 h)
(3) Follows from 2.
2
ALGEBRA 1, D. CHAN
5
Example 10. ϕ = det : GL2 (R) −→ R∗ . We saw before that ker(ϕ) = SL2 (R) is a normal subgroup. We
also computed the cosets to be
1 0
SL2 = set of matrices of determinant d = ϕ−1 (d)
0 d
clearly ϕ is not injective.
Example 11. Consider ϕ0 ϕ as in example 1, ϕ0 ϕ : Σn −→ GLn (R) −→ R∗ . ker(ϕ0 ϕ) = An , which is a
normal subgroup. Fibres of ϕ0 ϕ were An , τ An , where τ is a transposition.
Proposition 6.5.
exercise.
Let H 6 G, the inclusion function H ,→ G, h 7−→ h is a monomorphism. Proof as
Proposition 6.6. (Quotient homomorphism) Let N P G, there is an epimorphism
π
:
G −→ G/N
:
g 7−→ gN
Also ker(π) = N . π is called a quotient homomorphism.
Proof. Note that π is surjective. We check that π is a homomorphism, let h, g ∈ G
π(hg)
=
π(h)π(g)
=
hN gN
=
(hg)N
We check the kernel of π
ker(π)
=
π −1 (1G/N )
=
π −1 (N )
=
{g : π(g) = gN = N }
=
N
2
Example 12. Let V and W be vector spaces, V 6 W over some field F . Recall vector spaces can be
considered as abelian groups, therefore we can form the quotient group V /W . In fact V /W can be made
into a vector space. We describe V /W geometrically, let V = R3 , and W be some plane. The cosets of W
will be v + W for some v ∈ V , giving a plane parallel to W . So V /W is the set of all these parallel planes.
7. First isomorphism theorem
Remark 1. how much does an homomorphism deviate from being an isomorphism?
Theorem 7.1. (Universal property for quotient homomorphisms) Let ϕ : G −→ G0 be a group homomorphism, and N P G. Let π : G −→ G/N be the quotient homomorphism.
ϕ
G
−→
π↓
G/N
%ψ
G0
If N 6 ker(ϕ) then there is a group homomorphism ψ : G/N −→ G0 such that ϕ = ψπ. Call ψ the induced
homomorphism and say ϕ factors through ψ or π. In this case, ψ is uniquely defined by (*)
ψ(gN ) = ϕ(g)
Proof. Frist note that if the diagram above commutes, i.e. ϕ = ψπ then (*) holds and thus determines ψ
uniquely.
• ψ is an homomorphism.
For g, h ∈ G
so ψ is an homomorphism.
ψ(gN hN )
=
ψ(gN )ψ(hN )
ψ(gN hN )
=
ψ((gh)N )
=
ψ(g)ψ(h)
6
ALGEBRA 1, D. CHAN
• Check (*) is well defined.
(*) seems to depend on the choice of the coset representative, g, we must show it is independent
of this choice, i.e. if gN = hN (for h ∈ G) we require ϕ(g) = ϕ(h). Suppose h = gn for some
n ∈ N , ϕ(h) = ϕ(g)ϕ(n) = ϕ(g)(1) = ϕ(g) since n ∈ N 6 ker(ϕ). Hence ϕ is well defined.
2
Example 13. Let ϕ = det : GL2 (R) −→ R∗ , ker(ϕ) = SL2 =: N . The theorem above implies ∃ψ :
GL2 (R)/ SL2 (R) −→ R∗ , where ψ maps
1 0
1 0
ψ:
SL2 7−→ det
=d
0 d
0 d
Example 14. (Classification of cyclic groups) Let G = hgi be a cyclic group. Consider the function
ϕ
:
Z −→ G
:
n 7−→ g n
Check that ϕ is a homomorphism. Let N in the theorem be ker(ϕ), the thereom implies ∃ψ : Z/N −→ G.
The claim is that ψ is bijective and therefore an isomorphism.
Proof.
• ψ is surjective, since ψ(n + N ) = ψ(n) = g n .
• ψ is injective. Let n + N ∈ ker(ψ) i.e. ψ(n + N ) = ψ(n) = 1 =⇒ ker(ϕ) = N . n + N = N = 0Z/N .
So ker(ψ) = {0N/Z }.
2
Proposition 7.2. The subgroups of Z all have form N = mZ, where m ∈ N. (Note that m = 0 =⇒ N =
{0})
Proof. Suppose N = 0, (otherwise m = 0). Closure under negation implies ∃ minimum positive integer
m ∈ N . Note that N is closed under addition and subtraction, implying N > mZ. Suppose n ∈ N , divide
to get n = qm + r, with q, r ∈ N, and 0 6 r < m. This implies N 3 n − qm = r which contradicts the
minimality of m unless r = 0, therefore n = qm ∈ mZ, and N = mZ. So any cyclic group is isomorphic
to Z/mZ and conversely every such group is cyclic, Z/mZ = h1 + mZi. This is part of a more general
phenomenon.
2
Theorem 7.3.
(First isomorphism theorem) Let ϕ : G −→ G0 , be a group homomorphism, ψ :
0
G/ ker(ϕ) −→ G be the induced homomorphism as in theorem on universal properties of quotient homomorphisms. Then ψ is a monomorphism and G/ ker(ϕ) ∼
= im(ψ).
Proof. It suffices to check that ker(ψ) = 1G/ ker(ϕ) = 1(ker(ϕ)). Let K = ker(ϕ), gK ∈ ker(ϕ), i.e.
ϕ(gK)
=
1
ϕ(gK)
=
ϕ(g)(1)
so, g ∈ K =⇒ gK = K =⇒ ker(ψ) = K, so ψ is injective.
2
Corollary 7.4. (Rank-Nullity Theorem) Let T : V −→ W be a linear map of vector spaces and K =
ker(T ). (Think T as a projection onto a line in R3 ) Given a subspace U < V , V /U is also a vector space
of dimension dim(V ) − dim(U ). We have the induced homomorphism,
ψ
:
∼
=
V /K
V /K −→ W
im(T )
We take dimensions of both sides and we have, dim(V ) − null(T ) = rank(T ).
Corollary 7.5. (from last week) Any group homomorphism ϕ : G −→ G0 can be factored as
π
ϕ : G −→ G/ ker(ϕ)−→G
˜ 0
ALGEBRA 1, D. CHAN
7
8. Second and third isomorophism theorems. Subgroups of quotient groups
Proposition 8.1. Let ϕ : G −→ G0 be a group homomorphism and H 0 6 G0 , then
1. ϕ−1 (H 0 ) 6 G
2. If H 0 P G0 then ϕ−1 (H 0 ) P G
(1) 1 ∈ ϕ−1 (1) ⊆ ϕ−1 (H 0 ) (identity). Suppose g, g 0 ∈ ϕ−1 (H 0 ), ϕ(gg 0 ) = ϕ(g)ϕ(g 0 ) ∈ H 0 =⇒
gg ∈ ϕ−1 (H 0 ) (closure). g ∈ ϕ−1 (H 0 ), ϕ−1 (g −1 ) = ϕ(g)−1 ∈ H 0 =⇒ g −1 ∈ ϕ−1 (H 0 ) (inverse).
Hence ϕ−1 (H 0 ) 6 G.
(2) Let g ∈ G, h ∈ ϕ−1 (H 0 ). We require g −1 hg ∈ ϕ−1 (H 0 ) for normality. ϕ(g −1 hg) = ϕ(g)−1 ϕ(h)ϕ(g) ∈
H 0 . therefore H 0 P G0 . So g −1 hg ∈ ϕ−1 (H 0 ) and ϕ−1 (H 0 ) P G.
2
Proof.
0
This is analogous to morphisms, continuous maps, in the category of topological spaces. Let (X, τ ), (X 0 , τ 0 )
be topological spaces, suppose
f : (X, τ ) −→ (X 0 , τ 0 )
f is continuous iff U 0 ∈ τ 0 =⇒ f −1 (U 0 ) ∈ τ . The same applies in morphisms, group homomorphisms, in
the category of groups. Let G and G0 be groups, suppose
φ : G −→ G0
φ is a homomorphism iff H 0 6 G0 =⇒ φ−1 (H 0 ) 6 G.
Corollary 8.2. Let N P G, π : G −→ G/N be the quotient homomorphism,
1. The subgroups H̃ of G/N are the groups of the form H/N where H is such that
N 6H6G
−1
also, H = π (H̃)
2. In 1., H is normal in G iff H̃ is normal in G/N .
Proof. (⇐=)We show H/N 6 G/N .
ϕ
:
:
i
H
h
,−→
7−→
G
h
π
−→
−→
G/N
hN
Then im(ϕ) is a group which is {hN : h ∈ H} = H/N . Check 2. normality if H P G. Let g ∈ G, h ∈ H.
we require,
(gH)−1 (hN )(gH) ∈ H/N
but (g −1 hg)N ∈ H/N , (since H P G)
(=⇒) H/N P G/N if H P G. Now =⇒ for 1. and 2. Consider H 6 G/N . Then let H = π −1 (H̃) is a
subgroup of G by proposition. We require H̃ = H/N .
Now π is surjective implies, gN ∈ H̃ ⇐⇒ gN = hN for some h ∈ π −1 (H̃) = H ⇐⇒ gN ∈ H/N . So
H̃ = H/N and H is a subgroup. The previous proposition states that H P G if H/N = H̃ P G/N .
2
Proposition 8.3. Let m be a positive integer. The subgroups of Z/mZ are those of the form nZ/mZ
where n|m, moreover nZ/mZ is cyclic of order n/m.
Proof. By the previous corollary, subgroups have the form H/mZ where mZ 6 H 6 Z. But all subgroups
of Z have the form, nZ for some n ∈ Z + . So subgroups of Z/mZ has the form nZ/mZ where nZ 6
mZ ⇐⇒ n|m. Note that nZ/mZ is generated by n + mZ and therefore has order m/n.
2
Note 1. The second and third isomorphism are consequences of the first isomorphism theorem when we
have a subgroup H 6 G. If N P H 6 G, we can construct 2 subsequent projections onto the quotient
G/N
groups G/N and H/N
for the second theorem (note that H/N P G/N ). For the third H does not have
to be a subgroup of N , we form an inclusion map form H to G then a projection onto the quotient group
G/N .
More precisely, consider N P G as before and let H 6 G.
i
i. ϕ : H ,−→G −→ G/N
ii. If N P H 6 G then H/N P G/N
ϕ
:
G
π
N
−→
G/N
we apply the first isomorphism theorem for the following.
πH/N
−→
G/N
H/N
8
ALGEBRA 1, D. CHAN
Theorem 8.4. (Second isomorphism theorem.) Assume the definitions of ii. Then
G/N
' G/H
H/N
Proof. We use the first isomorphism theorem. note that
ϕ
:
:
π
N
−→
7−→
G
g
πH/N
G/N
H/N
−→
G/N
gN
is surjective and πN and πH/N are surjective also. We find ker(ϕ). Let g ∈ ker(ϕ), gN ∈ ker(πH/N ) =
G/N
H/N ⇐⇒ g ∈ H, so ker(ϕ) = H. By the first isomorphism theorem, H/N
= im(ϕ) ' G/ ker(ϕ) = G/H.
2
Theorem 8.5. (Third isomorphism theorem) Assume the definitions of i. recall that N C G, N 6 G so
HN 6 G and H ∩ N P H, we have the following
H/H ∩ N ' HN/N
Note: H ⊇ N =⇒ HN = H. A device for remembering this is the following Hasse diagram
HN
.
&
&
.
H
N
H ∩N
Proof. We apply the first isomorphism theorem to
ϕ
:
:
H
h
i
,−→
7−→
π
−→
7−→
G
h
G/N
hN
Note that ker(ϕ) = ϕ−1 (N ) = i−1 (N ) = H ∩ N P H, since H ∩ N is a kernel. We require im(ϕ) = {hN :
h ∈ H} =: H̃ 6 G/N . From the corollary on subgroups of quotient groups
H̃ = π −1 (H̃)/N
S
π −1 (H̃) = {g : gN = hN, for some h ∈ H} = h∈H hN = HN . (This is necessary since H is not necessarily
a subgroup of N ) Therefore H̃ 6 G/N =⇒ HN = π −1 (H̃) 6 G and the first isomorphism theorem implies
HN/N = π(H̃)/N = im(ϕ) ' H/ ker(ϕ) = H/H ∩ N .
2
9. Products
For groups G1 and G2 we wish to construct a bigger group G with a normal subgroup isomorphic to G1
and G/G1Q' G2 . Let I be an index set and Gi be a group for all i ∈ I.
Let G = Gi = {(gi )i∈I : gi ∈ Gi }. The key example would be I = {1 . . . n}, G = G1 × G2 . . . × Gn =
{(g1 . . . gn ) : gi ∈ Gi }.
Proposition 9.1. Endow G with a multiplication
µ
:
G×G
((gi ), (gi0 ))
−→
7−→
G
(gi gi0 )i∈I
Then G is a group called the product of the G0i s.
Proof. Check associativitiy. The identity is 1G = (1Gi )i∈I , indeed G 3 (gi )(1Gi ) = (gi 1Gi ) = 1G (gi ). For
inverse, we let (gi )−1 = (gi−1 ), it is easy to see this works.
2
Example 15. Let

G = GLn (C)
d1
0
..

D = { 0


∗
 : di ∈ C }
.
dn

d1

Claim ϕ : D−→C
˜
× . . . × C n-times, ϕ :  0
∗
∗
0
..


 7−→ (d1 . . . dn ) is an isomorphism.
.
dn
ALGEBRA 1, D. CHAN
9
Proof. Clearly ϕ is bijective. We require check it is am homomorphism.




d1 e1
d1
0
e1
0




.
.
..
..
ϕ  0
 0
 = ϕ  0
dn
en
0
..



.
dn en
=
(d1 e1 . . . dn en )
=
(d1 . . . dn )(e1 . . . en )
2
Example
16. Canonical injections and projections. Let I be an index set, Gi be a group ∀i ∈ I, G =
Q
i∈I Gi . consider the canonical maps,
1. Projection
πi
:
−→
7−→
G
(gk )k∈I
Gi
gi
2. Injection
ιi
:
Gi
g
−→
7−→
G
(1, 1, . . . gi , . . . 1)
Check that these are group homomorphisms. Consider the case where G = G1 × G2 , then we have
π2
:
−→
7−→
G
(g1 , g2 )
G2
g2
is an epimorphism, the first isomorphism theorem states that G2 ' G/ ker(π2 ). The kernel is normal
and ker(π2 ) = {(g1 , 1) : gi ∈ G} = im(ι1 : G1 −→ G) ' G1 . In other words, G has a normal subgroup
isomorphic to G1 whose quotient is G2 .
Proposition 9.2. (Recognising products) Let G be generated by subgroups G1 . . . Gn . Suppose also that
1. For i 6= j, gi ∈ G1 , gj ∈ Gj , gi gj = gj gi .
2. Suppose for any i,
Gi ∩ hGi : j 6= ii = 1
Q
Then ϕ : G1 × G2 . . . × Gn −→G,
˜
ϕ :7−→ (gi )i∈I 7−→ i∈I gi .
Proof. Check that ϕ is an homomorphism
ϕ((gi )i∈I (gi0 )i∈I )
=
Y
=
Y
gi gi0
i∈I
i∈I
gi
Y
gi0
i∈I
since elements of different Gi ’s commute. Note that ϕQis surjective, since the Gi ’s Q
generate G. We require
injectivity, i.e. ker(ϕ) = 1. Let (gi )i∈I ∈ ker(ϕ), then i∈I gi = 1, therefore gk−1 = i∈I,i6=k gk ∈ Gk ∩hGj :
j 6= ki = 1 =⇒ gi = 1, ∀i ∈ I. So ker(ϕ) = 1 implies ϕ is bijective and hence an isomorphism.
2
Definition 9.3. Let G be a group then the exponent of G is the smallest positive integer, n, such that
g n = 1, ∀g ∈ G.
Proposition 9.4. Let G be any finite group. The order of any g ∈ G divides |G| hence the exponent n,
divides |G| as well.
Proof. Consider hgi 6 G. Lagrange’s theorem implies |hgi||o(G) but |hgi| = o(g). The exponent is the
l.c.m. of all |hgi| so divides |G| too.
2
Example 17. Any finite group G of exponent 2 is isomorphic to Z/2Z × Z/2Z . . . × Z/2Z.
Proof. G is finite implies G is finitely generated. We can pick a minimal set of generators and write
G = hgi in
i=1 . We require G ' G1 × . . . × Gn where Gi = hgi i by applying the previous proposition. Note
that gi has order 2, since G has exponent 2, so Gi ' Z/2Z.
2
The Gi ’s generate G, since G = hgi in
i=1 . G is commutative since its exponent is 2, for a, b ∈ G, (ab) = abab.
If 2. in proposition (recognising products) then gi ∈ hGj : j 6= ii = hGj i = G, contradicting the minimality
of n. Therefore 2 holds and G ' Z/2Z × Z/2Z . . . × Z/2Z.
2
10
ALGEBRA 1, D. CHAN
Q
Theorem 9.5. (Universal property of products.) Let I be an index set, Gi be a group ∀i ∈ I, G = i∈I Gi .
Consider a group H and a homomorphism ϕi : H −→ Gi for each i ∈ I. Then there is a unique group
homomorphism ϕ : H −→ G such that ϕi is the composite
ϕ
:
H
h
ϕ
−→
7−→
π
i
−→
7−→
G
ϕ(h)
Gi
(ϕ(h))i = ϕi (h)
Proof. If the function ϕ satisfies the above relation, then ϕ must be
Q
ϕ : H −→ G
=
i∈I Gi
h 7−→ (ϕi (h)) i ∈ I
2
We require the above is a homomorphism, this is easy to check.
Remark 2. It is not difficult to show that any group with this universal property is a unique up to
isomorphism. This leads to an alternative definition of products.
10. Classification of Isometries
Recall that an isometry is a function T : Rn −→ Rn such that
kT a − T bk = ka − bk
n
n
n
Example 18. For v ∈ R , Tv : R −→ R ,, Tv : x 7−→ x + v (translation by v) is an isometry.
Proposition 10.1. Let T : Rn −→ Rn be an isometry and v = T (0). Then T = Tv ◦ T 0 where T 0 is an
isometry with T 0 (0) = 0.
Proof. Recall from lecture 1 that
T 0 = Tv−1 ◦ T
is also an isometry (being composite of such) check
T 0 (0)
=
Tv−1 (T (0))
=
T (0) − v
=
0
2
Theorem 10.2. Let T : Rn −→ Rn be an isometry with T (0) = 0, then T is linear so T ∈ On .
Heuristic argument: recall from lecture that T is injective, also T is continuous. We check additivity and
scalar multiplication.
n
Definition 10.3. Consider a finte set V = {v i }m
i=1 ⊆ R , the centre of mass of V is
m
1 X i
v
c(V ) =
m i=1
Proposition 10.4. Consider the function
εV (x) =
m
X
kv i − xk2
i=1
εV (x) attains a minimum at precisely 1 point, x = c(V ).
Proof. Write x = (x1 , x2 . . . xn )
εV (x)
=
X i
(vj − xj )2
=
X
((xj )2 − 2vji xj + (vji )2 )
=
X
(m(xj )2 − xj
i,j
i,j
j
2
m
X
i=1
!
vji
+
m
X
(vji )2 )
i=1
Each (j-th) component is a quadratic in xj , so the unique minimum occurs when,
m
1 X i
xj =
vj
m i=1
i.e. x = c(V )
2
ALGEBRA 1, D. CHAN
11
n
Corollary 10.5. Let T : Rn −→ Rn be an isometry. Suppose V = {v i }m
i=1 ⊆ R is a finite set such that
T (V ) = V . Then T (c(V )) = c(V )
Proof.
εV (T (c(V )))
X
=
kv i − T (c(V ))k2
v i ∈V
X
=
kT (v i ) − T (c(V ))k2
v i ∈V
X
=
kv i − c(V )k2
v i ∈V
=
εV (c(V ))
line 2 permutes the summands, so by uniqueness of previous proposition. T (c(V )) = c(V )
2
2
Symmetries of regular n-gons. Let F be a regular n-gon in R , for n > 3, and G be the group of symmetries
of F , i.e. isometries T of R2 such that T (F ) = F , such that T preserve V = the set of vertices.
For consistency, we pick coordinates so c(V ) = 0 and so by corollary T (0) = 0 for any T ∈ G. Set
the positve x-axis through the midpoint of some side. Label the vertices with {1 . . . n}, counterclockwise
beginning from the positive x-axis.
Recall (from lecture 8) that this labelling induces a natural isomorphism
ψ : Σ(V )−→Σ
˜ n
Each isometry g ∈ G permutes the element set V giving a function
ϕ
:
−→
7−→
G
g
Σ(V )
(g|V : v 7−→ gv)
Proposition 10.6. ϕ is a monomorphism (injective) and the composition
ϕ
ψ
G −→ Σ(V ) −→ Σn
gives an isomorphism of G with im(ψ ◦ ϕ) = Dn 6 Σn
Proof. We check that ϕ is an homomorphism, for g, h ∈ G and v ∈ V
ϕ(gh)v
=
ϕ(gh)|V v
=
(gh)v
=
ϕ(g)vϕ(h)v
Let g ∈ ker(ϕ) so g fixes all the vertices, but g ∈ O2 so g = 1O2 =⇒ ϕ is injective. Note that ψ ◦ ϕ is also
injecive, so it induces an isomorphism of G with its image.
We now compute the im(ψ ◦ ϕ). Let s ∈ G be a rotation such that the vertices permute cyclically, and t
be the reflection about the x-axis. We show that this generates G.
Let g ∈ G
det(g) = 1
=⇒ g is a rotation
=⇒ g = si
det(g) = −1 =⇒ g is an improper rotation =⇒ g = si t
Recall that Dn = hσ, τ i for σ n = 1, τ 2 = 1. Now note that
(ψ ◦ ϕ)(s)
=
σ
(ψ ◦ ϕ)(t)
=
τ
(ψ ◦ ϕ)(si tj )
=
σi τ j
im(ψ ◦ ϕ)
=
Dn
2
11. Abstract symmetry and group actions
Example 19. (Motivation) Consider the group of symmetries of a smiley face and the roman letter Z,
(i.e. let S:) and SZ be sets of points in R2 ). Both of these groups are isomorphic to Z/2Z, but clearly there
is a fundamental difference between these symmetries. The group of symmetries only gives half the picture
in a symmetry, we shall see a way to deal with this difference.
12
ALGEBRA 1, D. CHAN
Definition 11.1. Let G be a group. A G-set is a set S and a map
α
:
G×S
(g, s)
−→
7−→
S
α(g, s)
:= g · s
which satisfies
1. For s ∈ S, 1 · s = s.
2. For g, h ∈ G, s ∈ S
(gh) · s = g · (h · s)
This map is called the group action, or the group operation. We also say that G acts or operates on S.
Example 20. Let G = Z/2Z = {1, τ }, (x, y) ∈ S:) = R2 . We check that G satisfies the conditions above,
(draw the smiley face, :), set the y-axis through the eyes, and the origin halfway between the eyes, τ acts
via reflection about the x-axis)
1(x, y)
=
(x, y)
τ∗i+j (x, y)
=
(x, (−1)i+j y)
=
τ∗i (x, (−1)j y)
=
τ∗i (τ∗j (x, y))
So S:) is a G-set. Define α : G × S −→ S by τ i s = (−1)i s since calculation shows that this is well defined
and thus defines a group action.
Example 21. (Conjugation) Let G = GLn (C) let S = Mn (C), define α : G × S −→ S by A∗ M = AM A−1 .
We check the axioms
I ·M
=
M
(AB) · (M )
=
A(BM B −1 )A−1
=
A · (B · M )
so S is a G-set.
Definition 11.2. (Orbit) Let G be a group and S a G-set. Define a relation on S by x ∼ y iff x = g∗ y
for some g ∈ G
Proposition 11.3. The above relation is an equivalence relation.
Proof. 1. Reflexivity, x = 1∗ x ⇐⇒ x ∼ x
2. Symmetry, x ∼ y ⇐⇒ x = g∗ y for some g ∈ G ⇐⇒ g∗−1 x = y ⇐⇒ y ∼ x.
3. Transitivity, x ∼ y, y ∼ z ⇐⇒ x = g ·y, y = h·z for some g, h ∈ G ⇐⇒ x = g ·(h·z) = (gh)·z ⇐⇒ x ∼ z.
2
Corollary 11.4. We call these equivalence classes of S G-orbits. A G-set S is thus a disjoint union of
G-orbits. We write the G-orbit containing s ∈ S by Gs .
Note 2. When G is the conjugation map, we call the G orbits conjugacy classes.
Example 22. G = SO3 and S = S 2 ,
a
:
G×S
(A, v)
−→
7−→
S
v
Consider v, w ∈ S We can rotate v onto w, i.e. ∃A ∈ SO3 = G, such that Av = w. There is only 1 G-orbit,
the whole of S. (If S = R3 , then the G-orbits of containing s are spheres of radius equal to |s|.)
Definition 11.5. (Stabiliser) Let G be a group and S be a G-set, let x ∈ S, then the stabiliser of x is
stab(x) = {g ∈ G : gx = x} ⊆ G
Proposition 11.6. stab(x) 6 G.
Proof. We check axioms. 1 · x = x =⇒ 1 ∈ stab(x). Given g, h ∈ stab(x), then (gh) · x = g · (h · x) = x, so
gh ∈ stab(x). Given g ∈ stab(x), g · x = x ⇐⇒ x = g −1 · g · x = g −1 · x so g −1 ∈ stab(x). So stab(x) 6 G.
2
Example 23. G = SO3 and S 2 , for x ∈ S 2 , stab(x) =rotations about the axis through x, −x.
ALGEBRA 1, D. CHAN
13
Definition 11.7. Homomorphisms of G-sets. Let G be a group, and ϕ : S −→ S 0 be a function between
2 G-sets. ϕ is an homomorphism of G-sets if ∀g ∈ G, s ∈ S
ϕ(g · s) = g · ϕ(s)
Proposition 11.8. Let ϕ : S −→ S 0 be a bijective homomorphism of G-sets, then ϕ−1 is also an
homomorphism of G-sets. We call ϕ an isomorphism.
2
Proof. Same as for isomorphism of groups.
Example 24. id : S −→ S is an homomorphism.
Example 25. We now return to the motivating example. S:) = SZ = R2 , these are G-sets, for G = Z/2Z,
but they are not isomorphic, hence they represent different symmetries. To show this, suppose
ϕ : S:) −→ SZ
is an isomorphism of G-sets, then ϕ must preserve fixed points. i) Suppose x0 ∈ S:) such that g∗ x0 =
x0 , ∀g ∈ G. So ϕ(g · x0 ) = g · ϕ(x0 ) =⇒ ϕ(g · x0 ) = ϕ(x0 ) = g · ϕ(x0 ), ϕ(x0 ) must be a fixed point of SZ
w.r.t. G.
S:) has fixed points along the x-axis and SZ has a fixed point through the centre of rotation, so there exists
no G-set isomorphism between S:) and SZ .
12. Classification of G-orbits
Definition 12.1. Suppose ϕ : S −→ S 0 is a homomorphism of G-sets, ϕ is also called a G-equivariant
map or one compatible with the group action.
Note 3. For the rest of this section G will be a group, and X will be a G-set, where there is no ambiguity
or indicated otherwise.
Proposition 12.2. Y ⊆ X is said to be G-stable or a G-subset (G-invariant) if for any g ∈ G, y ∈ Y
g·y ∈Y
In this case Y inherits from X the structure of a G-set.
2
Proof. check axioms of subgroups
Example 26. Any G-orbit
Gx = {g · x|g ∈ G} ∈ X
is G-stable. Reason: For g, g 0 ∈ G, g · (g 0 · x) = (gg 0 ) · x ∈ Gx, therefore Gx is a G-orbit.
Definition 12.3. We say the group action of G on X is transitive if X is a single orbit.
Example 27. From last lecture G = SO(3), and X = S 2 . The action
G×X
(A, x)
−→
7−→
X
Ax
is a group action, the action is transitive as there is only 1 orbit.
The G-set G/H
Let H 6 G We can make G/H into a G-set, by defining a group action by g 0 · (gH) := g 0 gH. We check
axioms quickly.
i. 1 · (gH) = gH
ii. For g1 , g2 ∈ G, g1 · (g2 · gH) = g1 · (g2 gH) = g1 g2 gH = (g1 g2 ) · gH.
iii. Note that ·is well defined as it is the set product.
Theorem 12.4. (Classification of G-orbits) Suppose the action of G on X is transitive. Let x ∈ X = Gx.
There is a well defined isomorphism of G-sets.
ϕ
:
G/H
gH
−→
7−→
X
g·x
where H = stab(x). Conversely, every G/H is a G-orbit. (every G-orbit of the form Gx is isomorphic to
G/ stab(x))
14
ALGEBRA 1, D. CHAN
Proof. We firstly check this is well defined. Let h ∈ H.
ϕ(g)
=
g·x
=
g · (h · x)
=
(gh) · x
0
We check G-equivariance, let g, g ∈ G
g 0 · φ(gH)
=
g 0 · (g · x)
=
(g 0 g) · x
=
ϕ(g 0 gH)
=
ϕ(g 0 · gH)
Surjectivity is clear. We check injectivity. Suppose ϕ(gH) = g · x = ϕ(g 0 H) = g 0 · x, g −1 · (g 0 · x) =
g −1 · g · x = 1. Therefore g −1 g 0 ∈ stab(x) = H, so g 0 ∈ g 0 H. g 0 H = gH so ϕ is injective. The converse is
clear.
2
Proposition 12.5. Suppose x ∈ X, g ∈ G, then stab(g · x) = g(stab(x))g −1 .
Proof. It suffices to prove ⊇, since applying this result to g · x, g −1 , giving
stab(g −1 · (g · x))
g stab(x)g −1
⊇
⊇
g −1 stab(g · x)g
stab(g · x)
which is the reverse inclusion. Let’s prove ⊇. Let h ∈ stab(x)
(ghg −1 )(g · x)
=
(ghg −1 g) · x
=
g · (h · x)
=
g·x
2
13. Application to platonic solids
A platonic solid, X, is a solid such that all faces are congruent to some fixed regular polygon and the same
number of faces meet at each vertex. There are 5 such solids.
T
C
O
D
I
:=
:=
:=
:=
:=
tetrahedron
cube
octahedron
dodecahedron
isocahedron
4 triangular faces
6 square
8 triangular
12 pentagonal
20 triangular
Definition 13.1. Let X be a platonic solid and assume in R3 with centre of mass at 0. The rotational
symmetric group consists of those symmetries which are (exclusively) in SO(3), i.e. the proper rotations.
Corollary 13.2. (..to classification theorem) Let G be the rotational symmetry group of X (a platonic
solid). Then |G| = (f )(s), where f is the number of faces of X, and s is the number of sides of X, i.e.
|G|
T
12
C
24
O
24
D
60
I
60
Proof. Let F = {Fi }fi=1 be the faces of X, G permutes these faces giving a G-set, namely, for T ∈ G, T ·Fi :=
T (Fi ). Exercise: check that this is a G-set. Note that this action is transitive (one can rotate any face
onto another). Therefore F = G · F . Let l be the axis through the centre of F1 . stab(F1 ) =rotations in
G about l ' Z/sZ (spin each of the edges around), so | stab(F1 )| = s. The classification theorem implies
F ' G/ stab(F ),
|F |
sf
as required.
=
|G/ stab(F1 )|
=
|G|/| stab(F1 )
=
|G|/s
=
|G|
2
ALGEBRA 1, D. CHAN
15
Remark 3. Turns out, rotational symmetry groups are
T
4!/2
A4
G
|G|
C
4!
S4
O
4!
S4
D
5!/2
A5
I
5!/2
A5
Counting number of orbits and permutation representations
Let G be a group and X be a G-set.
Definition 13.3. Let J ⊆ G the fixed point set of J is defined as
X J = {x ∈ X|g · x = x, ∀g ∈ J} ⊆ X
Proposition 13.4. If H = hJi, then X J = X H .
Theorem 13.5. (Number of orbits) Let G be a finite group, and X be a finite G-set. Then the number
of orbits in X is
1 X g
|X |
n=
|G| g∈G
Proof. Suppose X =
S
Xi , is the disjoint union of G-stable subsets of Xi .
l.h.s =
X
number of orbits of Xi
i
since X g =
S
Xig .
r.h.s =
X 1 X g
|X |
|G| g∈G i
i
It suffices to prove equality for Xi ’s. Now X is a disjoint union of G-orbits and these are isomorphic to
G-sets of the form G/H, so w.l.o.g. we let Xi = G/H, for some H 6 G.
1 X g
|X |
|G| g∈G i
=
=
1
|{(g, x) ∈ G × Xi |g · x = x}|
|G|
1 X
| stab(x)|
|G|
x∈G/H
this is analogous to switching the sum over G to over Xi ' G/H. Note that the action is transitive, so a
previous proposition implies stab(x) are all conjugate to each other, hence, stab(1 · H) = H.
=
1 X
|H|
|G|
x∈G/H
=
1
|G/H| · |H|
|G|
1
=
number of orbits of G/H
=
2
this gives the theorem.
Example 28. (Cake) In each of 8 equal sectors, we place a white or orange candle at centre. How many
different ways are there of doing this? (two arrangements are the same if we can rotate the cake to get the
other). Formulation: Let G = hσi = Z/8Z, X = (Z/2Z)8 . Define a group action by
σ · (x1 . . . x8 )
=
(x2 . . . x8 , x1 )
16
ALGEBRA 1, D. CHAN
Check that this makes X a G-set. The question becomes, how many orbits are there? Using the above
P
σi
1
theorem. n =number of orbits= |G|
σ i ∈G |X |
X1
=
X
1
=
28
σ
=
|{(x1 . . . x8 )|x1 = 0, 1, all xi = xj }|
|X |
|X |
=
2
=
|{(x1 . . . x8 )|x1 , x2 = 0, 1, x2i = x2 , x2i+1 = x1 }|
=
4
=
2
|
=
24
|X σ |
=
2
|
=
4
|
=
2
n
=
i
1 X
|X σ |
|G| i
2
|X σ |
3
|X σ |
|X
σ4
5
|X
|X
σ6
σ7
σ ∈G
=
=
1 8
(2 + 2 + 4 + 2 + 24 + 2 + 4 + 2)
23
36
14. Permutation Representations
Definition 14.1. Let G be a group, X be a set. A premutation representation is a group homomorphism
ϕ : G −→ Σ(X)
There is a 1:1 correspondence between G-sets and permutation representations, which is as follows.
Permutation representation -set.
Let ϕ : G −→ Σ(X) be an homomorphism. Define a corresponding G-set as a set X with its group action
defined by
G×X
(g, x)
−→
7−→
X
ϕ(g)x
We check the axioms
1. 1 · x = ϕ(1)x = id x = x.
2. Associativity. For g, g 0 ∈ G, x ∈ X, g 0 (gx) = ϕ(g 0 )(ϕ(g)x) = ϕ(g 0 )ϕ(g)x = ϕ(g 0 g)x.
So X is a G-set.
G-set−→Permutation representation
Suppose X is a G-set, we require a group homomorphism,
ϕ
:
−→
7−→
G
g
Σ(X)
ϕ(g)
Let ϕ(g) be defined by
ϕ(g)
:
X
x
−→
7−→
X
g·x
We require ϕ(g) ∈ Σ(X). ϕ(g) has inverse ϕ(g −1 ), since ϕ(g −1 )ϕ(g) · x = g −1 · (g · x) = x. So ϕ(g)ϕ(g −1 ) =
id. We now check ϕ is a homomorphism. Suppose g, g 0 ∈ G, x ∈ X
ϕ(g 0 g)x
=
(g 0 g) · x
=
g 0 · (g · x)
=
ϕ(g)(ϕ(g 0 )x)
Therefore ϕ is a permutation representation. Exercise: check these constructions are inverses of each other.
The key to the correspondence is g · x = ϕ(g)x.
Example 29. Refer to alglec7.tm, smiley face example. Consider the G-set, S:) , G = {1, σ} ' Z/2Z.
ϕ(x, y)T = (x, −y)T , 1(x, y)T = (x, y)T . We seek the corresponding permutation representation. The
ALGEBRA 1, D. CHAN
17
composition
G
−→
1
7−→
ϕ
7−→
−→
GL
2 (R) 1 0
0 1 1
0
0 −1
Σ(R)
We check the action of ϕ
ϕ(x, y)
=
1
0
0
−1
=
(x, −y)
x
y
Definition 14.2. A permutation representation ϕ : G −→ Σ(X) is faithful if ϕ is injective.
Theorem 14.3. (Cayley) Let G be a finite group, then G is isomorphic to a subgroup of Σ(G), (i.e. if
n = |G| < ∞, then G is isomorphic to Σn ??).
Proof. Consider G-set, G(= G/H with H = 1) gives premutation representation
ϕ : G −→ Σ(G)
It suffices to check ϕ is faithful by the first isomorphism theorem. Let g ∈ ker(ϕ). g = g · 1 = ϕ(g) · 1 =
id 1 = 1. Thereofore ker(ϕ) = 1 so ϕ is injective. The first isomorphism theorem gives correspondence with
a subgroup of Σ(G).
2
15. Finite groups of isometries
Let (A for affine) A GLn =set of isometries of T : Rn −→ Rn of Rn = V . A GLn ⊆ Σ(V ). Let G 6 A GLn
be a finite subgroup. The inclusion homomorphism G ,→ Σ(V ) defines a permutation representation. The
corresponding G-set V , has action, T ∈ G, x ∈ V = Rn , T · x = T (x).
Proposition 15.1. Let finite G < A GLn act on V as above. Then V G is nonempty and changing
coordinates so that a fixed point is 0, we see that G is isomorphic to a subgroup of On .
Proof. Consider any G-orbit F = Gv. F is finite so G fixes the centre of mass c of F , therefore c ∈ V G .
From previous lecture, any isometry T with T (0) = 0 is linear, so G is isomorphic to a subgroup of On . 2
Theorem 15.2. Let G < O2 be finite, then G is either cyclic or dihedral (recall D2 = Z/2Z × Z/2Z)
Proof. Let H = SO2 ∩G. We claim that H is cyclic. H is a finite subgroup of SO2 =set of proper rotations
in the plane. Let h ∈ H be a rotation by θ such that θ is minimally positive. Suppose h0 ∈ H rotates by
θ0 where nθ 6 θ0 < (n + 1)θ for some n ∈ Z. Note that h0 h−n rotates by 0 6 θ0 − nθ < θ Minimality of θ
implies that θ0 − nθ = 0, so h0 = hn ⇐⇒ H = hhi is cyclic.
Suppose G * SO2 , let τ ∈ G − SO2 , note that SO2 C O2 since [O2 : SO2 ] = 2. The third isomorphism
theorem implies
G
H
=
G
G∩SO2
'
(G)(SO2 )
SO2
6
O2
SO2
=
{±1}
'
Z/2Z
Since H 6= G, G/H ' Z/2Z, therefore G = H ∪ τ H. Now τ is a reflection in O2 − SO2 . Changing
coordinates such that this is a reflection about the x-axis. Consider a regular n-gon, H is a cyclic group of
rotations of the n-gon. So G is the set of 2n symmetries of the regular n-gon. Therefore G ' D2n , at least
for n > 3. Exercise: check the degenerate case, n = 2.
2
Definition 15.3. Poles. Let G < SO3 be finite. Consider a G-set R2 = V . Note that W = S 2 be the
unit sphere in R3 is a G-stable subset. A point x ∈ W is a pole if stab(x) 6= 1. The set X ⊂ R3 = V of
poles is G-stable since for x ∈ X, g ∈ G, stab(g · x) = g stab(x)g −1 6= 1.
Theorem 15.4. The finite subgroups of SO3 are the cyclic groups, dihedral groups, or the rotational
groups of symmetries of a Platonic solids.
Proof. We prove the case for platonic solids. Let G be the group of rotational symmetries of the Platonic
solid S with centre of mass 0. It seems we have poles through the centres of opposite faces, opposite
vertices, and midpoints of opposite edges. Note that for x ∈ X, stab(x) is a cyclic group of rotations about
the axis {±x}. So it is reasonable to surmise that the 3 orbits correpsond to {face poles}, {edge poles}, and
{vertex poles}. We also require the concept of duality (draw a picture :¿), which maps faces to vertices of
dual solids. Since an isometry of a Platonic solid preserves the centres of faces, the rotational symmetry
groups of dual solids are isomorphic.
18
ALGEBRA 1, D. CHAN
Let G < SO3 be finite, X = G-set of poles. From last lecture we have a formula for the number of orbits,
r. Suppose X = Gx1 ∪ Gx2 ∪ . . . ∪ Gxr , then
1 X g
r =
|X |
|G| g∈G
X
if g = 1
Xg =
pair of poles of g if g 6= 1
1
r =
((|G| − 1)2 + |X|)
|G|
!
r
X
1
=
(|G| − 1)2 +
|Gxi |
|G|
i=1
!
r
X
1
1
=
1−
2+
|G|
| stab(x)|
i=1
2−
2
|G|
r
X
r−
=
i=1
r
X
=
1−
i=1
1
1−
| stab(x)|
so r
>
1−
6
3
1
| stab(x)|
1
| stab(x)|
1
1
=
2
2
r 6= 1, since the formula above fails. If r = 2, then G is cyclic. Since
r
=
2
=
|X|
=
1
((|G| − 1)2 + |X|)
|G|
|X|
2
2−
+
|G|
|G|
2
therefore there are only 2 poles, say ±x, G mmust be some cyclic group of rotations about the axis x, −x.
For r = 3, let ni = | stab(x)|, the equation bounds the ni ’s as follows.
If n1 6 n2 6 n3 , (n1 , n2 , n3 ) is a Platonic triple, i.e. (we require |G| ∈ Z+ )
a.
b.
c.
d.
(2, 2, n)
(2, 3, 3)
(2, 3, 4)
(2, 3, 5)
n>2
We examine case d, (n1 , n2 , n3 ) = (2, 3, 5) ⇐⇒ |G| = 60. Let us consider Gx3 ' G/ stab(x3 ). |Gx3 | =
60
= 12 =number of vertices of isocahedron I. The claim is Gx3 are the vertices of the same I.
n3
stab(x3 )
'
Z/5Z
=
hgi
since g ∈ stab(x3 ). If y ∈ Gx3 − {±x3 }, then we get distinct {y, g(y), . . . , g 4 (y)}. But |Gx3 | = 12. the only
possible configuration for Gx3 is as in picture ... so
Gx3 = {±x3 , y, g(y), . . . , g 4 (y), z, g(z), . . . g 4 (z)}
the corners of an isocahedron. Note that kx3 −yk < kx3 −zk, and kx3 −yk = kx3 −g(y)k = . . . = kx3 −g 4 (y)k.
This correpsonds to saying ‘adjacent’ vertices are the same distance apart. Since x3 can be an arbitrary
element of Gx3 , this will show adjacent vertices are the same distance apart. From the picture we see 5
equilateral triangles meet at each vertex. Therefore Gx3 is the set of vertices of an isocahedron I, and G
is a subgroup of rotational symmetries of I since Gx3 is G-stable, but |G| = 60 and the group of rotational
symmetries of I is 60, the equality holds.
For case c, (n1 , n2 , n3 ) = (2, 3, 4), the formula implies |G| = 24.
|Gx3 | = |G/ stab(x3 )| = 6
Suppose hgi = stab(x3 ) = Z/4Z. Symmetry implies for y on the equator, we can see as before, that Gx3 is
the set of vertices of an octahedron O, and G is a group of rotational symmetries of G.
For case b, (n1 , n2 , n3 ) = (2, 3, 3), the formula implies |G| = 12.
|Gx3 | = |G/ stab(x3 )| = 4
Suppose hgi = stab(x3 ) = Z/3Z. Check that we have a tetrahedron, as before.
ALGEBRA 1, D. CHAN
19
For case a, (n1 , n2 , n3 ) = (2, 2, n). Assume n > 2 to avoid the non degenerate case, the formula implies
|G| = 2n. Consider Gx3 ' G/ stab(x3 )
2n
=2
n
stab(x3 ) = hσi ' Z/nZ, let y ∈ Gx3 − {±x3 }, then Gx3 ⊃ {±x3 , y, g(y) . . .}, but |Gx3 | = 2, so we are left
with Gx3 = {±x3 }. Consider Gx2 ' G/ stab(x2 )
|Gx3 | = |G/ stab(x3 )| =
2n
=n
2
stab(x2 ) = hτ i = {1, τ }. Now τ (x3 ) ∈ Gx3 = {±x3 }, implying x2 is on the equator w.r.t. the poles x3 , −x3
since G(x3 ) = {±x3 }. Similarly x1 is on the equator as well. stab(x3 ) is a group of order index 2
|Gx2 | = |G/ stab(x2 )| =
G
=
hσi ∪ τ hσi
=
symmetries of the n − gon, {x2 , g(x2 ), . . . , g n−1 (x2 )}
exercise: check the degenerate case, n = 2.
This finishes the proof of the theorem.
Lemma 15.5. The formula r =
1
|G|
2
((|G| − 1)2 + |X|) and r = 3 implies (n1 , n2 , n3 ) ∈ {(2, 2, n), (2, 3, 3), (2, 3, 4), (2, 3, 5)}.
Proof. The formula implies
1
1
1
+
+
n1
n2
n3
>
1
if n3 > 6
1
1
+
n1
n2
>
1−
1
6
n1 = n2
=
5
6
1
1
+
2
3
2
1
1
+
n1
n2
>
1−
=
=
if n3 = 3, 4, 5.
1
6
1
1−
3
2
3
=
=
2
finish as exercise..
16. Class equation and conjugacy
Aim: study of groups via symmetry. Let Aut(G) be the set of automorphisms of G, Aut(G) ⊆ Perm(G).
Proposition 16.1. Aut(G) 6 Perm(G). Proof is trivial.
Definition 16.2. Let g, h ∈ G. Define Cg (h) = ghg −1 , called conjugation by g. Recall that Cg : G −→ G
is an automorphism of G. If H 6 G, then Cg (H) = gHg −1 is a subgroup of G, called a conjugate of H.
Proposition 16.3. The canonical map C : G −→ Aut(G) is a group homomorphism.
Proof. For g1 , g2 , h ∈ G, Cg1 g2 (h) = g1 g2 hg2−1 g1−1 = Cg1 (g2 hg2−1 ) = Cg1 Cg2 (h) for all h.
Consider the composite group homomorphism
G
C
−→
Aut(G)
ι
,−→
2
Perm(G)
this defines a permutation representation. The corresponding G-set is G with action defined by (for
g, g 0 ∈ G)
g · g 0 = Cg (g 0 ) = gg 0 g −1
we say the group G acts on itself by conjugation.
Definition 16.4. Let G act on G by conjugation. The G-orbits are called conjugacy classes, which we
denote by G · x to distinguish it from Gx. The set of fixed points is called the centre, denoted, Z = Z(G).
20
ALGEBRA 1, D. CHAN
Note 4. We can write the centre as
Z(G)
=
{z ∈ G | gzg −1 = z, for any g ∈ G}
=
{z ∈ G | |G · z| = 1}
=
{z ∈ G | gz = zg, for any g ∈ G}
=
{z ∈ G | g = zgz −1 = Cz (g), for any g ∈ G}
=
{z ∈ G | Cz = id}
=
ker(C : G −→ Aut(G))
Corollary 16.5. Z(G) P G and Z(G) = G iff G is abelian.
Example 30. What is the centre of G = GL2 (R)?
b
∈ Z. We know that
d
1 0
0 0
AB = BA for any B ∈ G and by continuity it is true for any 2×2 real B. Sub in B =
,
.
0 0
1 0
a b
1 0
1 0
a b
=
c d
0 0
0 0
c d
a 0
a b
=
c 0
0 0
a b
0 0
0 0
a b
=
c d
1 0
1 0
c d
b 0
0 0
=
d 0
d b
Proof. The claim is Z(G) = {aI|a ∈ R∗ }. Note that ⊇ holds. Suppose A =
a
c
2
so a = d, and A is scalar so the claim is true.
Definition 16.6. (Class equation) Let (finite) G act on itself by conjugation. Decomposition into orbits
gives
˙ · x1 ∪˙ . . . ∪G
˙ · xr
G = Z ∪G
where G · xi are orbits with |G · xi | > 1. Note that these are disjoint unions. Using the classification of
orbits, and the disjoint union, we have
r
X
|G|
|G| = |Z| +
|
stab(x
i )|
i=1
Definition 16.7. (p-group and Sylow p-group) A group H is said to be a p-group, where p is a prime, if
the o(H) is a power of p. Suppose G is a finite group with |G| = pl q and gcd(p, q) = 1. Then H 6 G is
said to be a Sylow p-subgroup if |H| = pl . I.e. a Sylow p-subgroup G is a maximal p group of G.
Example 31. The conjugate of a Sylow p-subgroup is a Sylow p-subgroup.
Proposition 16.8. Let H be a nontrivial (H 6= 1) p-group, then Z(H) 6= 1.
Proof. (by class equation)
|H|
=
|Z| +
r
X
i=1
|H|
| stab(xi )|
p divides |H|, | stab(xi )| by Lagrange’s theorem, and in fact
Corollary 16.9. Suppose p is prime, then if |G| is
a) p, then G ' Z/pZ.
|H|
| stab(xi )|
6= 1. Therefore p||Z| and Z(G) 6= 1.
2
Proof. Let g ∈ G − {1}, Lagrange’s theorem implies |hgi| | p therefore p = |hgi|. So G = hgi is cyclic,
hence G ' Z/pZ.
2
b) p2 , then G ' Z/p2 Z or Z/pZ × Z/pZ.
Proof. Suppose G Z/p2 Z, let z ∈ Z(G) − {1} (exists by proposition). We assumed hzi 6= G so
Lagrange’s theorem implies |hzi| = p, so hzi ' Z/pZ. Let x ∈ G − hzi, again by the same argument,
hxi ' Z/pZ.
Claim G ' hzi × hxi. It suffices to check criterion for products from lecture 12
ALGEBRA 1, D. CHAN
21
• Since hzi < Z(G), elements of hzi and hxi commute.
• |hz, xi| > |hzi| + 1 = p + 1. Lagrange’s theorem implies |hz, xi| = p2 , then z, x generates G, i.e.
hz, xi = G
• Lagrange’s theorem implies |hzi ∩ hxi| = 1 so hzi ∩ hxi = 1. Proposition from lecture 12 implies
G ' hzi × hxi we should check that hzi × hxi ' Z/pZ × Z/pZ.
2
Theorem 16.10. (Sylow) Let G be a group of order pl q with p prime and gcd(p, q) = 1, let Cl(K) denote
the conjugacy class of K ⊆ G, and Sylp (G) be the collection of all Sylow p-subgroups of G.
1. Sylp (G) is nonempty, suppose P ∈ Sylp (G).
2. Let H 6 G be a p-subgroup, then it is contained in a conjugate of P . In particular, Cl(P ) = Sylp (G),
and hence the members of Sylp (G) isomorphic.
3. Let m = | Sylp (G)|, then m|o(G) and m = 1 + kp for some k ∈ N.
Example 32. Let p be prime and G = Dp = hσ, τ i with σ p = τ 2 = 1, τ σ = σ −1 τ . Then the only Sylow
p-subgroup is hσi, the Sylow 2-subgroups are hτ i, hτ σ i i.
17. Proof of Sylow’s theorem
l
Proof. Recall that |G| = p q, and gcd(p, q) = 1.
1. Let X = {A ⊆ G | |A| = pl }, then |X| =
makes X a G-set.
pl q
pl
. Let G act on X by left translation, check that this
Lemma 17.1. p - |X|.
Proof. For i ∈ {1 . . . pl } the same powers of p divides i as divides pl q − i,
!
(pl q)!
pl q
= l
l
(p (q − 1))!pl !
p
2
S
Write X as its orbit decomposition X = ˙ G · Xi , and by above, there exists Xi such that p - |G · Xi |.
The claim is that P = stab(Xi ) is a Sylow p-subgroup.
• p - |G · Xi |, and |G · Xi | = |G|/|P | = pl q/|P | implies pl | |P |, since |P | has to kill the pl factor.
• Let x ∈ Xi then P x ⊆ P Xi = Xi . Therefore |P | 6 |Xi x−1 | = pl , containment gives equality then
P is a Sylow p-subgroup. (or the fact that |G|/|P | is an integer.)
2. We have to show that all Sylow p-subgroups are conjugate, with |P | = pl , and maximal.
Lemma 17.2. Let H be a p-group acting on a set S. Then p | |S| − |S H |
Proof. First note that |S| − |S H | = |S − S H |. S H is the fixed point group of H in S. S − S H is the
set of all points not fixed by H, so is an orbit, which we denote H si . By classification of orbits, this is
isomorphic to H/ stab(si ). Lagrange’s theorem implies p | |H/ stab(si )| therefore p | |S − S H |.
2
Let H 6 G be a p-group, and P a Sylow p-subgroup. Consider H-set G/P where the action is given
by left translation,
h · gP = hgP
for h ∈ H, g ∈ G. Check that this is an H-set. |G/P | = pl q/pl = q so p - |G/P |, the lemma above
implies 1 6 |(G/P )H | < p, so (G/P )H is non-empty.
Let g P ∈ (G/P )H , so for any h ∈ H, hgP = gP , therefore g −1 hg ∈ P , and h ∈ gP g −1 . h
is arbitrary implies H ⊆ gP g −1 , with equality iff |H| = |P |, i.e. equality occurs iff H is a Sylow
p-subgroup. Therefore Cl(P ) = Sylp (G).
3. Let by above. Both G and P act on S by conjugation, namely for h ∈ G or P and g ∈ G.
h · (gP g −1 )
:=
Ch (gP g −1 )
=
hgP g −1 h−1
Check that this is an action. The action of G on S is transitive, so S ' G/ stab(P ) therefore |S| |
|G|/| stab(P )| = q =⇒ |S| - p. It remains to show |S| = 1 + kp for some k ∈ N.
Consider now P -action on S. By above lemma, p | |S − S P | so it suffices to show that |S P | = 1 or
S P = {P }. Note that P is fixed by P -action, so P ∈ S P . Suppose P 0 ∈ Sylp (G) with P 0 6= P , we
require J := stab(P 0 ) 6 P . Given j ∈ J, jP 0 j −1 = P 0 implies jP 0 = P 0 j so JP 0 = P 0 J. We now need
JP 0 to be a p-group.
Proof. We check that this is a subgroup
• 1 ∈ JP 0
22
ALGEBRA 1, D. CHAN
• JP 0 JP 0 = JJP 0 P 0 = JP 0 so JP 0 is closed under product.
• Let j ∈ J, h ∈ P then (jh)−1 = h−1 j −1 = P 0 J = JP 0 so we have closure under inverses.
We need |JP 0 | to be a power of p.
|JP 0 |
=
3 rd isomorphism theorem
=
|
|P 0 ||JP 0 /P 0 |
J l
p J ∩ P0 p2l
So JP 0 is a p-subgroup but JP 0 > P 0 , together with |P 0 | = pl implying that JP 0 = P 0 . So J 6 P 0 ∩ P
therefore J 6= P and P 0 6∈ S P . This completes the proof of 3.
2
2
Corollary 17.3. Let p be an odd prime. Suppose G is a group of order 2p then
Z/2pZ
G'
Z/pZ × Z/2Z or Dp
Proof. Assume G is not cyclic, then by Sylow’s theorem ∃P ∈ Sylp (G) with o(P ) = p =⇒ P ' Z/pZ. So
˙ P . Also P C G as [G : P ] = 2.
P = hσi for some σ ∈ P . Let τ ∈ G − P , note that |G/P | = 2 so G = P ∪τ
Let e be the order of τ . τ P has order 2 so 2|e. Also e 6= 2p as G is not cyclic. Lagrange’s theorem
implies e|2p so e = 2. Similarly, the order of τ σ is 2, i.e. τ στ σ = 1 =⇒ τ σ = σ −1 τ . This gives the same
multiplication table for Dp , so G ' Dp .
2
Note 5. Compositions of G-set homomorphisms are G-set homomorphisms.
18. Abelian groups
Let {Ai }i∈I be a family of abelian groups indexed by I.
Definition 18.1. The direct sum of the Ai ’s is the subgroup of Πi∈I Ai ,
M
Ai = {(ai ∈ Πi∈I Ai | only finitely many Ai 6= 0}
i∈I
Check that this is a subgroup.
Note 6. If I is finite then
L
i∈I
Ai = Πi∈I Ai .
Recall from lecture 12, the canonical injections
ιj
:
Aj
a
ai
=
,−→
7−→
Πi∈I Ai
(ai )
0
a
i 6= j
i=j
L
L
The image lies in
we have a canonical
i∈I Ai , so L
L injection Aj ,−→ i∈I Ai . We sometimes use this to
identify Aj with a subgroup of i∈I Ai . Hence i∈I Ai is also the set of formal finite sums
aj1 + aj2 + . . . + aj r
where ji ’s are all distinct and aj i ∈ Aj i .
Theorem
18.2. (Universal property for direct sums) Let {Ai }i∈I , B be abelian groups. Further let
L
λ : i∈I Ai −→ B be a group homomorphism. Then the restrictions to Aj
λ|Aj
:
Aj
−→
B
are group homomorphisms. Conversely, given a family of group homomorphisms, λi : Ai −→ B, i ∈ I, the
map
L
λ :
−→ B
i∈I Ai
P
(ai )
7−→
i∈I λi (ai )
this makes sense as all but finitely many
Lai ’s are zero, so λ is a homomorphism. There is a 1-1 correspondence between group homomorphisms
i∈I Ai −→ B and families of group homomorphisms λi : Ai −→
B, i ∈ I
ALGEBRA 1, D. CHAN
23
2
Proof. Routine.
Definition 18.3. Let A, B be abelian groups. Define the Hom group Hom(A, B) to be the set of group
homomorphisms λ : A −→ B.
Proposition 18.4.
The Hom group is an abelian group when endowed with addition, for g, h ∈
Hom(A, B), (f + g)(a) = f (a) + g(a).
Proof. As for vector spaces.
L
2
Definition
˜ .
L 18.5. An abelian group, F , is free abelian if there exists a group isomorphism λ : i∈I Z−→F
Let 1i ∈
Z be the element with 1 in the i-th slot and 0’s elsewhere. The set {λ(1i )}i∈I is called a basis
of F .
Proposition 18.6. Let A be an abelian group, then
Φ
Ψ
Φ
(1) Hom(Z, A) A, f 7−→ f (1), a 7−→ (f : n −→ n a) are inverse homomorphisms.
Ψ
Proof. Exercise, check group homomorphism
(2) There is an isomorphism of groups
2
Hom(Zs , A)−→A
˜ s
with correspondence given by
left mul x ∈ Zs by(a1 . . . as ) ←→ (a1 . . . as )
Proof. Exercise, universal property.
2
Example 33. Hom(Zs , Zt ) is just t × s-matrices over Z.
Corollary 18.7.
1. Aut(Zn ) = GLn (Z) = {A ∈ GLn (Q) | A−1 have integer entries}.
2
Proof. Follows from proposition.
m
2. If Z
n
' Z then m = n.
Proof. Suppose we have inverse isomorphisms Zm −→ Zn and Zn −→ Zm given by m × n matrix M
and n × m matrix N . Then M N = I and N M = I so N and M are square and m = n.
2
Definition 18.8. Let A be an abelian group. The torsion subgroup is
Ators = {a ∈ A | a has finite order, n a = 0, for some n > 1}
Also define
An = {a ∈ A | n a = 0, n ∈ N}
For prime p,
Ap∞ = {a ∈ A | pr a = 0, for some r}
I.e. order of a is some power of p.
Proposition 18.9. Ators , An and Ap∞ are subgroups.
Proof. (For Ap∞ .) Suppose a, b ∈ Ap∞ , say pm a = 0 = pn b. W.l.o.g. m > n then pm (a + b) = pm a + pm b =
0, so a + b ∈ Ap∞ . The proofs of the others follow similarly, note that this depends on abelian property.
2
Example 34. (C∗ )tors = µ, group of roots of unity.
Definition 18.10. An abelian group A is said to be
torsion
A
if Ators =
torsion free
0
Example 35. Z, Q are torsion-free, Z/nZ is torsion.
Proposition 18.11. Let A be an abelian group, then B = A/Ators is torsion free.
24
ALGEBRA 1, D. CHAN
Proof. Let a + Ators ∈ (A/Ators )tors , i.e. ∃n ∈ Z+ such that n a ∈ Ators . Hence n a has finite order and
∃m ∈ Z+ with m n a = 0, this implies a ∈ Ators and a + Ators = Ators = 0B . Therefore Btors = 0 so B is
torsion free.
2
Proposition 18.12. Let f : A −→ B be an homomorphism of abelian groups. Then f (Ators ) ⊆ Btors .
The universal property of quotients, applied to A −→ B −→ B/Btors , shows that there is an induced map
f˜ : A/Ators −→ B/Btors
Hence if f is an isomorphism, f |Ators : Ators −→ Btors and f˜ : A/Ators −→ B/Btors are isomorphisms.
Proof. Let a ∈ Ators say n a = 0 for n ∈ Z+ . Then n f (a) = f (n a) = 0 so f (a) ∈ Btors . Therefore
f (Ators ) ⊆ Btors . If f is an isomorphism, f −1 induces an inverse isomorphism
f −1 |Btors : Btors −→ Ators
to f |Ators , and
−1 : B/B
fg
tors −→ A/Ators
to f˜.
2
19. Structure of finitely generated abelian groups
Theorem 19.1. Let A be a finitely generated abelian group, then
A ' Z/h1 Z × Z/h2 Z × . . . × Z/hr Z × Zs
where hi |hi+1 , i ∈ {1 . . . r − 1}.
Proof. (by linear algebra over Z) Let A = ha1 . . . an i. By proposition above (8 by texmacs numbering, 2 if
you are reading pdf), there exists a surjective group homomorphism
f : Zn
(a1 ...an )
−→
A
Write f = (a1 . . . an ) and let K = ker(f ), such that A ' Zn /K. Thus proof of the above amounts to
proving the following lemma.
Lemma 19.2. Let K 6 Zn . By changing the basis of Zn , K



 
h1
0

 0   h2 

 



 .   0 
...
 ..  , 

  . 



  .. 

0
0
is generated by r 6 n elements of the form
.. 
. 
0 

hr 

0 

..
.
Proof. Assume firstly K 6 Z n is finitely generated† . Let K = ha1 . . . ar i. Write the generators of K in
vector form and form a matrix, A, with these vectors. Clearly A : Zr −→ Zn and K = im(A). A change
of basis in Zn corresponds to left multiplication by some M ∈ GLn (Z). Note also that for N ∈ GLr (Z),
im(A N ) = im(A) = K. Rephrasing this,
2
Lemma 19.3. There exist M ∈ GLn (Z), N ∈ GLr (Z) such that


h1


..
M AN = 

.
hr
where hi |hi+1 , note that we allow hj = 0.
Note 7. After the n-th column, all entries are zero so im(A N ) is generated by 6 n elements.
Recall that GLm (Z) contains elementary matrices, Ei j (α), Pi j , Si for multiplying rows by a constant, α,
swapping two rows, and negating a row respectively. Left and right multiplication by these correspond to
elementary row operations (ERO’s) and elementary column operations (ECO’s).
Definition 19.4. For a matrix B with entries as integers define
gcd(B) = gcd({bij })
ALGEBRA 1, D. CHAN
25
Lemma 19.5.
Let M ∈ GLm (Z), N ∈ GLs (Z), B is an m × s matrix with integer entries, then
gcd(M B N ) = gcd(B).
Proof. Entries of M B N are integer linear combinations of the bi j ’s and is a multiple of gcd(B), this is
true for every entry so gcd(B) | gcd(M B N ) and similarly gcd(M B N ) | gcd(B).
2
Lemma 19.6. Let x ∈ Zm , then ∃M ∈ GLm (Z) such that M x = (gcd(x), 0 . . . 0)T .
P
Proof. (sketch) Apply (integer) ERO’s to x = (x1 . . . xn )T to reduce
|xi | until there is only 1 non zero
entry remaining, and swap it up to top. The following example illustrates this procedure.
Example 36.








6
6
0
3
 −9  −→  −3  −→  3  −→  0 
12
0
0
0
2
Lemma 19.7. A is an n × r matrix as before, there exists N ∈ GLr (Z) so that the first column of A N ,
say a, satisfies
gcd(a) = min{gcd(A x)|x ∈ Zr }
we will later see that this is equal to gcd(A).
Proof. Let y be suchthat gcd(A
y) minimises gcd(A x). Note gcd(y) = 1, otherwise we can replace y with
Ay
y
r
∈ Z and gcd gcd(y) < gcd(A y) would contradict minimality of gcd(A y). By the previous lemma,
gcd(y)
∃N ∈ GLr (Z) with N −1 y = (1, 0 . . . 0)T . Then the first column of A N is
a = A N (1, 0 . . . 0)T = A N N −1 y = A y
2
By this, a = A N e1 is such that h1 := gcd(a) = min{gcd(A x)}. We use M from the second previous
lemma, so
Ma
=
(h1 , 0 . . . 0)T
B
=
M AN

h1 b12
 0
b22

 .
 ..
0 bm2
=
h1r





...
bmr
Using ECO’s we can replace b1j ’s with remainders modulo h1 . But the third previous lemma, gcd(M B N ) =
gcd(B), and h1 = min{gcd(A x)} =⇒ 0 = h12 = h13 . . . = h1r , giving


h1
0
...
0
 0
b22
b2r 


M AN =  .

.
..
 ..

0 bm2
bm r
Applying elementary row and column operations, we can replace any bi j ’s by remainders mod h1 , this
implies h1 |bi j for all i, j. Therefore we can modify M and N such that
B
=
=
M
AN
h1
0T
0
0 h1 B
Now use induction on the size of A to rewrite B 0 in diagonal form, this completes proof of the key lemma,
and proves the theorem. We are left to show that,
Lemma 19.8.
†
K 6 Zn is finitely generated.
Proof. Let {(a1 . . . an ) ∈ Zn | an = 0} = A 6 Zn , so A ' Zn−1 . We argue by induction.
Suppose K ∩ A = ha1 . . . al i 6 A ' Zn−1 . By the 3rd isomorphism theorem, we have
K
KA
'
K ∩A
A
so there exists an epimorphism
K
−→ Zn /A ' Z
K ∩A
26
ALGEBRA 1, D. CHAN
K
K
thus K∩A
6 Z. Classification of subgroups of Z shows that K∩A
is cyclic, say generated by a0 + K ∩ A,
for some a0 ∈ K.
We require a0 . . . al to generate K. Let a ∈ K,
0 a0 + K ∩ A for some n0 ∈ Z. Therefore
Pla + K ∩ A = n
P
l
a − n0 a0 ∈ K ∩ A, this implies a − n0 a0 =
i=1 ni ai , a =
i=0 ni ai ∈ ha0 , . . . , al i. So K is finitely
generated.
2
2
20. Solvable Groups
G will denote a finite group for this section.
Definition 20.1. A normal series (normal chain of subgroups) of G, is defined as
G∗ := 1 = G0 C G1 C . . . C Gr = G
The quotients Gi+1 /Gi are called factors.
Note 8. It is possible that G1 is not normal in Gi+2 or higher indecies.
Example 37. Let G = H × J × K where H, J, K are groups. Note that H × J × 1 = ker(π3 ) C G, where
π3 : H × J × K −→ K is the canonical projection. Similarly H × 1 × 1 C H × J × 1, we have a normal
series,
1CH ×1×1CH ×J ×1CH ×J ×K CG
The factors are H, J and K respectively.
Example 38. Let G = Z/nZ, where n ∈ Z+ , consider the prime factorisation of n,
n = p1 p2 . . . pn
with pi ’s not necessarily distinct. The series
nZ = p1 p2 . . . pn Z 6 p1 p2 . . . pn−1 Z 6 . . .
gives a normal chain of subgroups
0 C p1 p2 . . . pn Z/nZ C p1 p2 . . . pn−1 Z/nZ C . . . C Z/nZ = G
The factors are
p1 p2 . . . pi−1 Z/nZ
p1 p2 . . . pi−1 Z
'
' Z/pi Z
p1 p2 . . . pi Z/nZ
p1 p2 . . . pi Z
Definition 20.2. A finite group G is solvable if there exists a normal series with all factors (cyclic) of
prime order.
Example 39. Z/nZ is solvable.
Lemma 20.3. Let G be a finite group and N C G, then G is solvable iff N and G/N are both solvable.
Proof. (=⇒) Suppose G is solvable, consider a normal series as defined above, with factors of prime order.
It suffices to show that the series
a) 1 = N ∩ G0 6 N ∩ G1 6 . . . 6 N ∩ Gr = N
b) 1 = G0 N/N C G1 N/N C . . . C Gr N/N = GN
are both normal series with factors Z/pi Z or 0.
H̃
' H̃ÑÑ .
Recall the third isomorphism theorem states that H̃ 6 G̃, Ñ C G̃, H̃∩
Ñ
We show a) using the third isomorphism theorem with H̃ = N ∩ Gi , Ñ = Gi−1
(Gi ∩ N )Gi−1
N ∩ Gi
'
6 Gi /Gi−1
N ∩ Gi−1
Gi−1
Lagrange’s theorem implies the factor group is either Z/pZ or 0. b can be proved by a similar approach.
(⇐=) Suppose
H∗
1 = N0 C N1 C . . . C Nr = N
G/N∗ 1 = N/N C H1 /N C . . . C Hs /N = G
have prime factors, then
1 = N0 C N1 C . . . C Nr = N C H1 C H2 C . . . C Hs = G
is a normal series with prime power quotients.
2
Note 9. There is a kind of ‘induction’ that is often used in arguments in group theory. If a property P
holds for all groups of order less than o(G), and if whenever there exists a normal subgroup N P G such
that N, G/N satisfying P implies G satisfies P , then if G has a nontrivial proper normal subgroup, then
G satisfies P .
ALGEBRA 1, D. CHAN
27
Proposition 20.4. A finite group G is solvable iff there exists a normal series with abelian factors.
Proof. (=⇒) By definition
(⇐=) By induction on |G|. If r > 1 and Gi ’s distinct. If Gr−1 and Gr /Gr−1 are solvable then Gr is
solvable. Suppose r = 1 so G is abelian. Pick g ∈ G − {1}, let H = hgi then G/H is solvable by induction
and H is solvable since it is cyclic. G is thus solvable by the previous proposition.
2
Proposition 20.5. A p-group is solvable.
Proof. It was proved previously that the centre of a p group is non trivial. If Z(G) is abelian then G is
solvable. By induction G/Z(G) is also a p-group, so it is solvable. Hence G is also solvable.
2
Definition 20.6. Let g, h ∈ G, we define the commutator of G to be
[g, h] = g −1 h−1 gh
The commutator, or derived group, G0 = [G, G] is the subgroup of G generated by all such commutators.
Note 10. G0 = 1 iff G is abelian.
Proposition 20.7. G0 C G
Proof. For k ∈ G
k[g, h]k−1
=
[kgk−1 , kh k−1 ]
∈
G0
implying kG0 k−1 = G0 i.e. G0 C G
2
Note 11. G/G0 is abelian and is called the abelianisation of G.
Proposition 20.8. Let G be a group, A be an abelian group and f : G −→ A be an homomorphism.
Then ker(f ) > G0 . So f factors through some homomorphism, f˜
G
π
f
−→
&
%
A
f˜
G/G0
Proof.
f ([g, h])
[g, h]
=
f (g −1 )f (h−1 )f (g)f (h)
=
1
∈
ker(f )
as required.
Corollary 20.9. G/N is abelian iff N > G0 .
2
Proposition 20.10. For n > 5, G = An is not solvable.
Proof. Suppose the contrary, and
G∗ := 1 = G0 C G1 C . . . C Gr = G
has abelian factors. We arrive at a contradiction if we can show by downward induction on r, that all Gi
contain every 3-cycle. Gr contains all 3 cycles since 3-cycles are even.
Suppose Gi contains all 3-cycles. Gi /Gi−1 is abelian implies Gi−1 > G0i by previous proposition applied to
Gi −→ Gi /Gi−1 . Let a, b, c, d, e ∈ {1 . . . n} be distinct,
(abc)−1 (cde)−1 (abc)(cde) = (ecb) ∈ Gi−1
therefore Gi−1 contains all 3-cycles and we have a contradiction.
2
28
ALGEBRA 1, D. CHAN
21. Jordan Holder Theorem
c.f. http://www.uoregon.edu/˜brundan/math647fall99/ch1.pdf
Let G be a finite group, and we have a normal series
G∗ := 1 = G0 C G1 C . . . C Gi C Gi+1 C . . . C Gr = G
Suppose H/Gn−1 C Gi+1 /Gi is non trivial, then we can obtain a non trivial refinement of the series,
1 = G0 C G1 C . . . C Gi C H C Gi+1 C . . . C Gr = G
The process can be repeated but must stop since G is finite.
Definition 21.1. A series G∗ is called a composition series if no such non trivial refinement exists, i.e.
all the factors Gi+1 /Gi are simple, in the sense that they have no nontrivial normal subgroups, and are
not 1.
Example 40. G = Z/6Z there are composition series
0 C 3Z/6Z C Z/6Z
The factors are Z/2Z and Z/3Z. Also
0 C 2Z/6Z C Z/6Z
with factors being Z/3Z and Z/2Z. Both these factors are clearly simple.
Definition 21.2. The factors of a composition series are called composition factors. In the above example,
they are the same in both series up to permutation.
We will see that this is true in general.
Theorem 21.3. (Jordan Holder) Let G be a finite group, and consider 2 composition series,
G∗ := 1 = G0 C G1 C . . . C Gr = G
H∗ := 1 = H0 C H1 C . . . C Hs = G
Then r = s (called the length of G) and composition factors of G∗ and H∗ are the same up to isomorphism
and permutation.
Lemma 21.4. (Zassenhaus’ Butterfly lemma) Let G be a group, and
U1
∇
U0
V1
∇
V0
6
G
where the triangles stand for P on its side. Clearly we have the following Hasse diagram.
DIAGRAM 1
Whenever two groups are connected by a segment to a point lying directly above, this point above represents
their product. Whenever .. directly below, this represents their interesection. The double lines indicate
normal subgroup inclusion.
Then,
1. Note that the following are groups as U0 P U1 , V0 P V1
a) (U1 ∩ V0 )U0 P (U1 ∩ V1 )U0
b) (U0 ∩ V1 )V0 P (U1 ∩ V1 )V0
2. The quotient groups formed along the three central vertical double lines are all isomorphic, namely
(U1 ∩ V1 )U0
(U1 ∩ V1 )V0
U1 ∩ V 1
'
'
(U1 ∩ V0 )U0
(U1 ∩ V0 )(U0 ∩ V1 )
(U0 ∩ V1 )V0
Proof. By symmetry it suffices to show 1a, and the following,
†
(U1 ∩ V1 )U0
U1 ∩ V 1
'
(U1 ∩ V0 )U0
(U1 ∩ V0 )(U0 ∩ V1 )
(1a) Let a ∈ U1 ∩ V1 , b ∈ U0 , we require ab(U1 ∩ V0 )U0 b−1 a−1 = (U1 ∩ V0 )U0 for (U1 ∩ V0 )U0 P (U1 ∩ V1 )U0 .
Rememeber that N is normal in G imply that xN x−1 = N, ∀x ∈ G.
ab(U1 ∩ V0 )U0 b−1 a−1
as required (since a ∈ U1 and U0 C U1 ). For
†
=
ab(U1 ∩ V0 )b−1 bU0 b−1 a−1
=
a(U1 ∩ V0 )U0 a−1
=
(a(U1 ∩ V0 )a−1 )(aU0 a−1 )
=
(U1 ∩ V0 )U0
we require the following lemma.
ALGEBRA 1, D. CHAN
29
Lemma 21.5.
(U1 ∩ V0 )U0 ∩ (U1 ∩ V1 ) = (U1 ∩ V0 )(U0 ∩ V1 )
Proof. ⊇ is clear. For ⊆, let a ∈ U1 ∩ V0 , b ∈ U0 suppose ab ∈ U1 ∩ V1 . We require ab ∈ r.h.s. Note
a−1 (U1 ∩ V1 ) ⊆ V1
b
∈
a−1 (U1 ∩ V1 ) ∩ U0
⊆
V 1 ∩ U0
Therefore ab ∈ (U1 ∩ V0 )(V1 ∩ U0 ) = r.h.s.
2
To finish off the butterfly lemma apply the third isomorphism theorem with G = (U1 ∩ V1 )U0 , N =
(U1 ∩ V0 )U0 , H = U1 ∩ V1 . This gives † .
2
Proof. (Of Jordan Holder) Define
Gij
=
(Gi+1 ∩ Hj )Gi
Hij
=
(Hi+1 ∩ Gj )Hi
This gives the following (usually trivial) refinement of G∗ , G∗∗ :
1 = G0
=
G00
G10
P
P
G01
G11
P
G02
P
...
P
P
..
.
G1
G2
=
G0s
Gr
G∗
H∗
we have a similar refinement H∗∗ of H∗ . The composition factors of
are the non trivial factors
G∗∗
of
. It remains only to show that factors of G∗∗ are the same as those for H∗∗ . By the butterfly
H∗∗
lemma
(U1 ∩ V1 )V0
(U1 ∩ V1 )U0
'
(U1 ∩ V0 )U0
(U0 ∩ V1 )V0
We let U1 = Gi+1 , U0 = Gi , V1 = Hj+1 , V0 = Hj
Gi,j+1
Hj,j+1
'
Gi,j
Hj,i
2
This proves the Jordan Holder theorem.
22. Free groups
Let X be a set, formally define a set X −1 = {x−1 | x ∈ X}, where (X −1 )−1 = X.
Definition 22.1. A word in X is an expression of the form
w = x1 x2 . . . xn
with n > 0, xi ∈ X ∪ X
−1
, note that n = 0 gives an empty word denoted 1.
Say w is reduced if xi 6= x−1
i+1 for any i. the set of reduced words is denoted W (X).
Define a mutiplication map
µ
:
W (X) × W (X)
(x1 . . . xn , y1 . . . yn )
−→
7−→
W (X)
x1 . . . xn y1 yn
Note that µ is a associative and 1 is an identity. For inverses, we need an equivalent relation on W (X).
For v, w ∈ W (X), write v ∼ w iff we can obtain v from w by insertion or deletion of adjacent pairs of xx−1
where x ∈ X ∪ X −1 .
We will denote equivalence classes of w by [w] (the square brackets are usually omitted.)
Proposition 22.2. The free group generated by X is F (X) = W (X)/ ∼. µ descends to a well defined
multiplication on F (X).
µ̃ : F (X) × F (X) −→ F (X)
([w1 ], [w2 ])
7−→ [µ(w1 , w2 )]
which makes F (X) a group.
30
ALGEBRA 1, D. CHAN
Proof. For µ̃ to be well defined we require
w1 ∼ w10 , w2 ∼ w20 =⇒ µ(w1 , w2 ) ∼ µ(w10 , w20 )
This is easy to see from any example.
w1 ∼ w1 x−1 x, w2 ∼ y −1 yw2 for x, y ∈ X ∪ X −1
Note
µ(w1 , w2 )
=
w1 w2
∼
w1 x−1 xy −1 yw2
=
µ(w1 x−1 x, y −1 yw2 )
For group axioms, note µ is associative implies µ̃ is associative. [1] is an identity as before. We check
inverse, and it is easy to see
−1
−1
[x1 x2 . . . xn ]−1 = [x−1
n xn−1 . . . x1 ]
This ends the proof.
2
Example 41. Let X = {x}, F (X) = {xn |n ∈ Z} ' Z
Proposition 22.3. (Universal property) Let X be a set and G be a group, and f : X −→ G be a (set)
map. Then
f˜ : F (X)
−→ G
±1
±1
x±1
x
.
.
.
x
7−→ f (x1 )±1 f (x2 )±1 . . . f (xn )±1
n
1
2
with xi ∈ X is a well defined group homomorphism.
Proof. (sketch) If f˜ is well defined, it is multiplicative. Showing f˜ is well defined is easy.
Example 42.
f˜(w1 xx−1 w2 )
=
f˜(w1 )f (x)f (x)−1 f˜(w2 )
f˜(w1 )f˜(w2 )
=
f˜(w1 w2 )
=
2
Defining groups by generators and relations. Recall that we defined a dihedral group by its two generators,
σ, τ , and the relation στ = τ σ −1 . Generally we can define groups in this fashion.
Lemma 22.4. Let G be a group and {Ni }i∈I be normal subgroups. Then N = ∩i∈I Ni C G. Hence given
a subset S ⊆ G the intersection of all normal subgroups of G containing S is the smallest normal subgroup
of G containing S. This is known as the normaliser of S, denoted N (S).
Proof. For g ∈ G
!
g
\
Ni
g −1
=
i∈I
\
gNi g −1
i∈I
=
\
Ni
i∈I
Let X be a set of generators. Consider a set of relations {wi = vi }i∈I where wi , vi ∈ F (X).
2
Definition 22.5. The group generated by X with defining relations {wi = vi }i∈I is
hX | wi = vi , i ∈ Ii = F (X)/N
where N is the smallest normal subgroup of F (X) containing {wi vi−1 }i∈I .
Note 12. F (X)/N is generated by the image of X in F (X)/N and the relations wi = vi hold on them
Proposition 22.6. (Universal property) Use notation as defined above. Let G be a group and f : X −→ G
a map of sets, so we have an induced group homomorphism
f˜ : F (X) −→ G
Then f˜ induces a group homomorphism
fˆ :
iff f˜(wi ) = f˜(vi ), ∀i
F (X)/N
wN
−→
7−→
G
f˜(w)
ALGEBRA 1, D. CHAN
31
Proof. Use universal property of quotient groups. Note that f˜(wi ) = f˜(vi ), ∀i iff wi v −1 ∈ ker(f ), ∀i
2
Example 43.
Dn ' hg, h | g n = h2 = 1, gh = hg −1 i := H
Proof. Here H = F (X)/N where X = {g, h}. Use universal property on
X
g
h
−→
7−→
7−→
G
σ
τ
gives a surjective group homomorphism, because σ, τ are generates Dn .
H
:
−→
F (X)/N
G
To show that this is an isomorphism, it suffices to show that |H| 6 2n.
2
Note 13. Any k ∈ N can be written using relations in the form g i or hg i where i ∈ {0 . . . n − 1}. There
are only 2n such elements so the map is an isomorphism.
Theorem 22.7. (Normal Form) Let X be a set. Every element of F (X) can be written uniquely as a
reduced word. This implies, for example if X = {x, y}.
xyx 6= yxx
reduced words are distinct.
23. Graphs
DIAGRAM 1
A graph Γ = Γ(V, E) is a set of vertices, V , a set of edges, E, and three maps
˜·
s, t
:
:
E
E
=
c
and
−→
−→
E
V
satisfying two axioms
ẽ = α, ∀α ∈ E
1. α
2. s(α) = t(α̃), ∀α ∈ E.
In the above example.
s(ε)
t(ε)
=
d
Definition 23.1. Let Γ = Γ(V, E) be a graph. A path in Γ is a string of edges
p = α1 α2 . . . αn , n > 0
where t(αi ) = s(αi+1 ) for each i. Sometimes we write
s(αi )
p
−→
t(αn )
we say that the path, p, is irreducible if αi 6= α̃i+1 for any i. (read no loops)
Example 44. In above picture, αβ̃β is not irreducible.
p
Definition 23.2. A graph Γ = Γ(V, E) is connected if for any v, w ∈ V there exists a path v −→ w. A
p
connected graph is called a tree if for each v, w ∈ V there is a unique irreducible p with v −→ w.
Example 45. DIAGRAM 2
is not a tree as
a
a
αβγ
−→
empty path
−→
a
a
DIAGRAM 3
is a tree
Definition 23.3. Let Γ = Γ(V, E) be a graph. A subgraph Λ(V 0 , E 0 ) of Γ is a pair of subsets V 0 ⊆
V, E 0 ⊆ E such that Ẽ 0 ⊆ E 0 and s(E 0 ), t(E 0 ) ⊆ V 0 .
Example 46. DIAGRAM 4
is a subgraph of
DIAGRAM 5
32
ALGEBRA 1, D. CHAN
Proposition 23.4. Let Γ = Γ(V, E) be a connected graph, there exists a subgraph Λ(V 0 , E 0 ) which is a
tree and is maximal (i.e. there exists no subgraph Λ00 (V 00 , E 00 ) with V 0 ⊆ V 00 , E 0 ⊆ E 00 one of these not
equal and Λ00 a tree)
Aside on partial order and total order.
Lemma 23.5. (Zorn) Let S be a nonempty set, partially ordered by 6. Suppose every nonempty chain
has an upper bound in S, then S has a maximal element.
Note 14. This is equivalent to the axiom of choice.
Proof. (of proposition) Let T the set of subgraphs of Γ = Γ(V, E) which are trees. Consider a partial order
6 defined by
Λ1 (V1 , E1 ) 6 Λ2 (V2 , E2 ) ⇐⇒ V1 ⊆ V2 , E1 ⊆ E2
Check tha this is a partial order, we require the maximal element of this poset.
Let {Λi (Ei , Vi )})i∈I ⊆ T , which is totally ordered, we require an upper bound, the candidate is
Λ̃ = Λ̃(Ṽ , Ẽ)
S
S
such that Ṽ = Vi , Ũ = Ui . It suffices to check that Λ̃ is a subgraph and a tree. The method is similar
for both so we demonstrate the latter.
Suppose we have irreducible paths.
DIAGRAM 6
note since Λi ’s are totally ordered, p1 , p2 actually lie in some Λi . But Λi is a tree so p1 = p2 as desired.
Now Zorn’s lemma gives the existence of a maximal subtree. Let Λ0 (V 0 , E 0 ) be one such subtree, we require
V 0 = V . We can show maximality easily.
2
24. Fundamental group
p
q
Let Γ = Γ(V, E) and p = α1 . . . αm , q = β1 . . . βn be two paths in Γ. Suppose u −→ v −→ w, then we can
define the product of p and q to be their concatenation, pq = α1 . . . αm β1 . . . βn . We require an equivalent
relation on paths in Γ which have the same source and target.
Definition 24.1. (Homotopy, graph) Let p = α1 . . . αm , q = β1 . . . βn be two paths in Γ with s(α1 ) =
s(β1 ), t(αm ) = t(βn ). We say that p and q are homotopic if we can obtain p from q by inserting or deleting
a sequence of paths of the form αα̃.
Alternatively, we can say that two paths p and q in Γ are edge related if we can write them as
p
=
p1 p2
q
=
p1 αα̃p2
for some α ∈ E. Two paths p and q in Γ are homotopic if there exists a sequence of paths
p = p0 , p1 , . . . , pr = q
where pi is edge related to pi−1 for i ∈ {1 . . . r}.
Proposition 24.2. The homotopy relation is an equivalence relation.
Definition 24.3. (Fundamental group) Let Ω(a) = {p ∈ E | s(p) = t(p) = a}, the set of paths which
begin and end at a, then the fundamental group is defined as π1 (Γ, a) = Ω(a)/ ∼.
Proposition 24.4. The multiplication of paths descends to a multiplication map
µ
:
π1 (Γ, a) × π1 (Γ, a)
−→
π1 (Γ, a)
which makes π1 (Γ, a) a group. Moreover, the isomorphism class of this group is independent of a.
Remark 4. This independence on a means we can write π1 (Γ) ' π1 (Γ, a).
Proof. The path multiplication making π1 (Γ, a) a group is analogous to free groups.
Independence on a. We require π1 (Γ, a) ' π1 (Γ, b), since Γ is connected, there exists p = α1 . . . αn from a
to b. We wish to show that
φ : π1 (Γ, a) −→ π1 (Γ, b)
q 7−→ p̃qp
and ψ : π1 (Γ, b) −→ π1 (Γ, a)
q 7−→ pq p̃
are inverse homomorphisms. Note that both are well defined. First we show that φ is a group homomorphism, let q and q 0 be paths in Γ, then
φ(qq 0 ) = p̃qq 0 p ∼ p̃qpp̃q 0 p = φ(q)φ(q 0 )
ALGEBRA 1, D. CHAN
33
therefore φ is a group homomorphism. To see that the maps defined are inverses, note
ψφ(q) = pp̃qpp̃ ∼ q
so ψφ = id. We can reverse p and p̃ to see that φψ = id. Therefore φ and ψ are inverse group homomorphisms, and the isomorphism class of π1 (Γ, a) is independent of a.
2
p,q
Lemma 24.5. Let Λ = Λ(V, E) be a tree. Any two paths p, q from a to b a −→ b are homotopic.
p̂
Proof. Delete αα̃’s, with α ∈ E, from p until we have an irreducible path, a −→ b. Note that p ∼ p̂.
q̂
Similarly we can an find irreducible path a −→ b such that q ∼ q̂. Uniqueness property for irreducible
paths in trees imply that p̂ = q̂ so p ∼ q.
2
Corollary 24.6. If Λ is a tree, π1 (Λ) = 1.
p
Proof. All paths a −→ a are homotopic to the empty path.
2
Theorem 24.7. Let Γ = Γ(V, E) be a connected graph. Pick a maximal subtree Γ0 = Γ0 (V, E 0 ) as in
proposition in last lecture. For each pair {α, α̃} of edges not in Γ0 , pick one, say α, and let X be the set of
such α’s, then
π1 (Γ) ' F (X)
Proof. Denote α−1 = α̃ for convenience. (We require a set map from the generators to the edges.) Pick,
fv
for each v ∈ V a path a −→ v which is in Γ0 . Define a set map
ϕ
:
†
−→
X
α
v −→ w
π1 (Γ)
fv αff
w
7−→
a→v→w→a
The universal property shows that this extends to a group homomorphism
ϕ̃
:
F (X)
−→
π1 (Γ, a)
Note that ϕ̃(α−1 ) = fw α−1 fev , so † gives ϕ̃ for α ∈ X −1 as well. It suffices now to show that ϕ̃ is an
isomorphism by exhibiting an inverse, ψ. Consider a path p ∈ Ω (t(p) = s(p) = a), we write
p
p
p
p
p2n−1
p
0
1
2
3
4
w0 −→
v0 −→
w1 −→
v1 −→
w2 −→
. . . −→ wn−1 −→ vn
where αi ∈ X ∪ X −1 and pi ’s are paths in Γ0 . Note that by lemma, ∼
fv
α
fg
w fv
α
1
1
1
2
2
a = w0 −→
v1 −→
w1 −→
v2 −→
w2 . . .
with w1 → q → v2 .
=
ϕ̃(α1 )ϕ̃(α2 )
=
ϕ̃(α1 α2 . . . αn−1 )
The candidate for the inverse is
ψ
:
p
7−→
α1 α2 . . . αn−1
This is an inverse provided it is well defined. If we changed p in the homotopy class by inserting or deleting
αα̃ with
1. α ∈ Γ0 then the expression α1 α2 . . . αn−1 does not change
2. α ∈ X ∪ X −1
Then we have inserted αα−1 into the expression which is the same element in the free gorup, then it is well
defined. This completes the theorem
2
34
ALGEBRA 1, D. CHAN
25. Nielsen-Schreier Theorem
Theorem 25.1. (N-S) Any subgroup H of a free group say G = F (X) generated by X is a free group.
Proof. We construct a graph, Γ = BH = BH(V, E), as follows. Let V = H \ G be the set of right cosets of
αHg,x
H in G, and E = {Hg −→ Hgx | x ∈ X, Hg ∈ H \ G} ∪ {α̃Hg,x = Hgx
x
We will abbreviate the notation to Hg −→ Hgx.
αHg,x−1
−→
Hg | x ∈ X, Hg ∈ H \ G}.
Example 47. Let Z ' hxi = F ({x}) > H = hx3 i, we determine BH. We have V = Z/3Z, and we have
the edges H −→ Hx −→ Hx2 −→ H.
To prove the theorem, it suffices by the previous theorem to show that H ' π1 (BH)
Define a group homomorphism
ϕ : π1 (BH) −→ H
First let a = H be the base point, and consider paths from H to H (forming the set Ω).
x
x
x
x
n
1
2
3
Hx1 . . . xn = H
Hx1 −→
Hx1 x2 −→
. . . −→
H −→
where xi ∈ X ∪ X −1 . Note x1 . . . xn ∈ H. We require ϕ(p) = x1 . . . xn , note that this is well defined
as adding and deleting a path of the form xx−1 does not change the r.h.s. Also note that ϕ is a group
homomorphism. It suffices now to construct an inverse ψ : H −→ π1 (BH, H). Let x1 . . . xn ∈ H with
xi ∈ H ∪ H −1 , let ψ(x1 . . . xn ) be the path
x
x
x
x
1
2
3
n
H −→
Hx1 −→
Hx1 x2 −→
. . . −→
Hx1 . . . xn = H
which represents an element of π1 (BH), clearly the inverse is well defined. We check this by inserting
xx−1 , x ∈ X ∪ X −1 into the word x1 . . . xn to obtain a new word w, then ψ(w) changes only in homotopy
class. This proves the Nielsen-Schreier theorem.
2