Sep. 22, 2006 (Fri.) 1. Set Theory
We recall some set theory that will be frequently used in the sequel or that is not covered in the basic college course. However, we will keep this chapter as minimum as possible.
We assume the notion of ”set” and some basic operation of sets without bothering their definition.
1.1. Zorn’s Lemma.
Definition 1.1.1. A set S is said to be partially ordered if there is a relation ≤ such that
(1) (reflexive) x ≤ x
(2) (anti-symmetric) if x ≤ y and y ≤ x, then x = y.
(3) (transitive) if x ≤ y and y ≤ z then x ≤ z.
We usually call a partially ordered set to be a POSET.
Definition 1.1.2. A pair of elements in an POSET is said to be com- parable if either x ≤ y or y ≤ x. A set is said to be totally ordered (or linearly ordered) if every pair is comparable.
We also need the following definition:
Definition 1.1.3. A maximal element of an poset S is an element m ∈ S such that if m ≤ x then m = x.
Foe a given subset T ⊂ S, an upper bound of T is an element b ∈ S such that x ≤ b for all x ∈ T .
One has the following
Theorem 1.1.4 (Zorn’s lemma). Let S be a non-empty poset. If every non-empty totally ordered subset (usually called a ”chain”) has an up- per bound, then there exists a maximal element in S.
Zorn’s Lemma could be taken as an axiom of set theory. It can be proved to be equivalent with the Axiom of Choice. It’s also equivalent to the Well-ordering Principle. We simply give the statement of these.
The reader can find the proof in most of the books on set theory.
An ordered set is said to be well-ordered if it is totally ordered and every non-empty subset B has a least element, i.e. an element a ∈ B such that a ≤ x for all x ∈ B.
Theorem 1.1.5 (Well-ordered Principle). Every non-empty set can be well-ordered.
1
One might wondering that (Q, ≤) equipped with the usual ordering is not well-ordered. So the statement says that there is another ordering which make the set Q well-ordered.
Example 1.1.6. Let R be a non-zero commutative ring. One can prove that there exists a maximal ideal by using Zorn’s lemma. The proof goes as following: Let S = {I C R|I 6= R} equipped with the ⊂ as the partial ordering. S 6= ∅ because 0 ∈ S. For a chain {Ij}j∈J, one has a upper bound I = ∪Ij. Then we have a maximal element in S by Zorn’s lemma. One can easily show that the maximal element corresponds to a maximal ideal.
1.2. cardinality. In order to compare the ”size of sets”, we introduce the cardinality.
Definition 1.2.1. Two sets A, B are said to have the same cardinality if there is a bijection between them, denoted |A| = |B|.
And we say |A| ≤ |B| if there is a injection from A to B.
It’s easy to see that the cardinality has the properties that |A| ≤ |A|
and if |A| ≤ |B|, |B| ≤ |C|, then |A| ≤ |C|. So It’s likely that the
”cardinality are partially ordered” or even totally ordered.
Lemma 1.2.2. Given two set A, B, either |A| ≤ |B| or |B| ≤ |A|.
Sketch. Consider
S = {(C, f )|C ⊂ A, f : C → B is an injection}.
Apply Zorn’s lemma to S, one has an maximal element (D, g), then one claim that either D = A or im(g) = B.
We leave it as an exercise for the readers. ¤ Theorem 1.2.3 (Schroeder-Bernstein). If |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|.
Sketch. Let f : A → B (resp. g : B → A) be the given injections respectively. One needs to construct a bijection by using f and g.
Some parts of A use f and some parts not. So we consider the partition
A1 := {a ∈ A|a has parentless ancestor in A}, A2 := {a ∈ A|a has parentless ancestor in B},
A3 := {a ∈ A|a has infinite ancestor}.
And so does B.
Then we claim that f restricted to A1, A3 are bijections to B1, B3. And g restricted to B2, B3 are bijections to A2, A3. So the desired
bijection can be constructed. ¤
We need some more properties of cardinality. If |A| = |{1, .., n}|, then we write |A| = n. And if |A| = |N| then we write |A| = ℵ0.
Proposition 1.2.4. If A is infinite, then ℵ0 ≤ |A|.
Sketch. By Axiom of Choice. ¤
Definition 1.2.5.
|A| + |B| := |A q B|,
|A| · |B| := |A × B|.
We have the following properties:
Proposition 1.2.6.
(1) If |A| is infinite and |B| is finite, then |A + B| = |A|.
(2) If |B| ≤ |A| and |A| is infinite, then |A + B| = |A|.
(3) If |B| ≤ |A| and |A| is infinite, then |A × B| = |A|.
Proof. For (1), take a countable subset A0 in A by Proposition 1.2.4.
One sees that |A0| = |A0| + |B| by shifting the index by |B|. Then we have
|A| = |A − A0| + |A0| = |A − A0| + |A0| + |B| = |A| + |B|.
For (2), it’s enough to see that |A + A| ≤ |A| since clearly
|A| ≤ |A + B| ≤ |A + A|.
Pick an maximal subset X ⊂ A having the property that |X +X| ≤ |X|
by Zorn’s Lemma. One claim that A − X is finite, and then we are done by (1).
To see the claim, if A − X is infinite, then there is a countable subset A0 ⊂ A−X. One can construct an injective function AqX qAqX → A q X which contradicts to the maximality of X.
For (3), it suffices to show that |A × A| = |A|. We sketch the proof.
Let
S = {(B, f )|B is an infinite subset of A, f : B → B × B an bijection}.
S is non-empty because S contains an infinite countable subset. S can be equipped with natural partial ordering and by Zorn’s Lemma, there exists a maximal element, say (M, g).
Let C be the complement of M in A. If |C| ≤ |M|, then by (2),
|A| = |M|. Hence there is a bijection h : A → M. It follows that there is a bijection A → Mh → M × MG (h−1−→,h−1)A × A.
Finally, if |C| ≥ |M|, then there is a subset M1 ∈ C such that
|M1| = |M|. Let M0 = M ∪ M1. One can construct a bijection from M0 → M0× M0. This contradicts to the maximality of M. Hence we
are done. ¤
Sep. 22, 2006 (Fri.) 2. Group Theory
The concept of groups is a very fundamental one in Mathematics.
It arise as automorphism of certain sets. For example, some geometry can be described as the groups acting on the geometric objects.
In the first section, we are going to recall some definition and basic properties of groups in general. In the second section, we introduce the acting of groups. The groups action can find many applications in geometry, algebra, and the theory of groups itself. In the third section, we are would like to take care of various aspects of reducing or factoring groups into simple ones.
2.1. Basic group theory.
Definition 2.1.1. A group G is a set together with a binary operation
◦ : G × G → G satisfying:
(1) there is an e ∈ G such that e ◦ g = g ◦ e = g for all g ∈ G.
(2) for all g ∈ G, there is an g−1 ∈ G such that g ◦g−1 = g−1◦g = e (3) for all g1, g2, g3 ∈ G, we have (g1 ◦ g2) ◦ g3 = g1◦ (g2◦ g3).
A group is said to be abelian if x ◦ y = y ◦ x for all x, y ∈ G.
For simplicity, we will denote xy for x ◦ y.
A subset H ⊂ G is a subgroup if H is a group by using the binary operation of G, denoted H < G.
A group homomorphism f : G → H is a function between groups that respects the structure of groups. That is, a function satisfying f (xy) = f (x)f (y).
The kernel of f , denote ker(f ), is defined to be {x ∈ G|f (x) = eH}.
A subgroup H < G is said to be normal if gHg−1 = H for all g ∈ G, denoted H C G. Given a subgroup H < G, we note G/H the set of left cosets, i.e. G/H = {gH|g ∈ G}. When H C G is normal, then G/H has induced group structure given by xH ◦ yH := xyH. This is called the quotient group.
We remark that a subgroup is normal if and only if it is the kernel of some homomorphism. The following lemma is useful
Lemma 2.1.2. Let f : G → H be a group homomorphism. Let N be a normal subgroup of G contained in ker(f ), then there is an induced homomorphism ¯f : G/N → H.
Proof. Define ¯f : G/N → H by ¯f (gN ) = f (g). Then it’s routine to verify it’s well-defined and it’s a group homomorphism. ¤
Regarding group homomorphisms, there are some useful facts:
Theorem 2.1.3 (First isomorphism theorem). Let f : G → H be a group homomorphism, then there is an induced isomorphism ¯f : G/ ker(f ) ∼= im(f ).
In particular, if f is surjective, then ¯f : G/ ker(f ) ∼= H.
Proof. Define ¯f : G/ ker(f ) → H by ¯f (g ker(f )) = f (g). Then it’s routine to verify it’s well-defined and it’s a injective group homomor-
phism. ¤
Example 2.1.4. Let G be the set of all maps Ta,b : R → R such that Ta,b(x) = ax + b with a 6= 0. Then G is a group under composition.
There are two natural subgroups:
A := {Ta,0} ∼= R∗, the multiplication group.
N := {T1,b} ∼= R, the translation group.
There is a group homomorphism f : G → R∗ by f (Ta,b) = a. Its kernel is N, which is a normal subgroup of G. So we have G/N ∼= A.
Moreover, G = NA = AN and N ∩ A = {e}. So in fact G is the semidirect product of A and N.
Theorem 2.1.5 (Second isomorphism theorem). Let H, K be sub- groups of G. Then we have group isomorphism
H/(H ∩ K) ∼= HK/K, when
1. H < NG(K) or especially, 2. K C G.
Sketch. Recall that NG(K) := {x ∈ G|xKx−1 = K} denotes the nor- malizer of K in G. It is the maximal subgroup of G in which K is normal. In particular K CNG(K). So K CG if and only if G = NG(K).
Also one can check that if H < NG(K), then HK = KH < NG(K) is a subgroup of G. Moreover, K C HK.
On the other hand, if H < NG(K), then H ∩ K C H. Thus both sides are groups.
Finally, we consider f : H → HK/K by f (h) = hK. It’s easy to check that f is surjective with kernel H ∩ K. By first isomorphism theorem, we proved (1). And (2) is just a special case of (1). ¤ Given a surjective homomorphism f : G → H, by First isomorphism theorem, H ∼= G/N where N = ker(f ) is a normal subgroup. It’s natural to study the group structures between them. It’s easy to see that there is a map
{K < G/N} f→ {L < G|N < L}.−1
In fact, this map is bijective. Moreover, it sends normal subgroups to normal subgroups.
Theorem 2.1.6 (Third isomorphism theorem). Given N CG and K C G containing N. Then K/N C G/N. Moreover, (G/N)/(K/N) ∼= G/K.
Sketch. It’s easy to check that K/N C G/N by definition.
In fact, we consider f : G → G/K. Since N C G and N is contained in ker(F ) = K, by Lemma 2.1.2, we have an induced map ¯f : G/N → G/K which is clearly surjective. One checks that ker(¯g) = K/N and
we are done by Themreom 2.1.3. ¤
2.2. cyclic groups.
Among all groups, perhaps simplest ones are cyclic groups. Let G be a group. We say that G is cyclic if there is an element x ∈ G such that every element g ∈ G can be written as xn for some n ∈ Z.
It’s clear that Z under addition is a cyclic group. By the definition, given a cyclic group G, there is a surjective map f : Z → G, by n 7→ xn. This is indeed a group homomorphism. Therefore, by Theorem 2.1.3, G ∼= Z/ ker(f ).
The reader should find no difficulty showing that subgroups of Z is either {0} or of the form nZ. Since Z is abelian, every subgroup is normal.
Turning back to the discussion of cyclic groups. There are two cases:
1. ker(f ) = 0. Then G ∼= Z. This is called an infinite cyclic group.
2. ker(f ) = nZ. Then G ∼= Z/nZ. This is called a cyclic group of order n, denoted Zn.
There list some properties and leave the proof for the readers.
Proposition 2.2.1. Let G be a cyclic group.
1. Every subgroup is cyclic.
2. Homomorphic image of G is cyclic.
3. If G is a cyclic group of order n, for all d|n there exist a subgroup of order d.
4. If G is a cyclic group of order n with a generator x, then the set of generators consist of {xt|(t, n) = 1}.
Sep. 29, 2006 (Fri.) 2.3. group action.
Group action is one of the most fundamental concept in group theory.
There are many situations that group actions appear naturally. The purpose of this section is to develop basic language of group action and apply this to the study of abstract groups.
We will first define the group action and illustrate some previous known theorem as examples.
Definition 2.3.1. We say a group G acts on a set S, or S is a G- set, if there is function α : G × S → S, usually denoted α(g, x) = gx, compatible with group structure, i.e. satisfying:
(1) let e ∈ G be the idetity, then ex = x for all x ∈ S.
(2) g(hx) = (gh)x for all g, h ∈ G, x ∈ S.
By the definition, it’s clear to see that if y = gx, then x = g−1y.
Because x = ex = (g−1g)x = g−1(gx) = g−1y.
Moreover, one can see that given a group action α : G × S → S is equivalent to have a group homomorphism ˜α : G → A(S), where A(S) denote the group of bijections on S.
Exercise 2.3.2. There is a bijection between {group action of G on S}
with {group homomorphism G → A(S)}.
Example 2.3.3 (Cayley’s Theorem).
Let G be a finite group of |G| = n. Then there is an injective homo- morphism G → Sn.
To see this, we consider G acts on G via the group operation, i.e.
G × G → G. Thus we have a homomorphism ϕ : G → A(G).
It’s clear that A(G) = Sn. It’s easy to see that ϕ is injective. ¤ This give an example of ”permutation representation”. That is, rep- resent a group into permutation groups. We gave another example:
Example 2.3.4.
Let F2be the field of 2 elements. We would like to see that GL(2, F2) ∼= S3.
We consider V the 2 dimensional vector space over F2. There are 3 non-zero vector in V , denoted, W := {v1 := e1, v2 := e2, v3 := e1+ e2}.
It’s clear that GL(2, F2) acts on W . Thus we have a representation GL(2, F2) → A(W ) ∼= S3. One can check that this is indeed an iso-
morphism. ¤
We now introduce two important notions:
Definition 2.3.5. Suppose G acts on S. For x ∈ S, the orbit of x is defined as
Ox:= {gx|g ∈ G}.
And the stabilizer of x is defined as
Gx:= {g ∈ G|gx = x}.
It’s immediate to check the following:
Lemma 2.3.6. Given a group G acting on S. For x, y ∈ S, we have:
1. Gx < G.
2. either Ox = Oy or Ox∩ Oy = ∅.
3. if y = gx, then Gy = gGxg−1. Proposition 2.3.7.
|G| = |Ox| · |Gx|.
Sketch. For given y ∈ Ox, we consider Sy := {g ∈ G|gx = y}. Then G = ∪y∈OxSy,
which is a disjoint union.
Furthermore, for each y ∈ Ox, we can write y = gx. Then one has Sy = gGx. In particular |Sy| = |Gx|. We fix a gy such that y = gyx once and for all. We may define a bijection G → Ox× Gx as sets by g 7→ (gx, g(ggx)−1). Thus
|G| = |Ox| · |Gx|.
¤ Corollary 2.3.8 (Lagrange’s Theorem). Let H < G be a subgroup.
Then |G| = |G/H| · |H|.
Proof. We take S = G/H with the action G × G/H → G/H via α(g, xH) = gxH. For H ∈ S, the stabilizer is H, and the orbit is G/H. Thus we have
|G| = |G/H| · |H|,
which is the Lagrange’s theorem. ¤
Another way of counting is to consider the decomposition of S into disjoint union of orbits. Note that if Ox = Oy if and only if y ∈ Ox. Thus for convenience, we pick a representative in each orbit and let I be a set of representatives of orbits. We have a disjoint union:
S = ∪x∈IOx. In particular,
|S| =X
x∈I
|Ox|.
This simple minded equation actually give various nice application.
We have the following natural applications.
Example 2.3.9 (translation).
Let G be a group. One can consider the action G × G → G by α(g, x) = gx. Such action is called translation. More generally, let H < G be a subgroup. Then one has translation H × G → G by (h, x) 7→ hx. In this setting, Ox = Hx. And the set of orbits is G/H, the right cosets of H in G. Then
|S| =X
x∈I
|Ox| = |G/H| · |H|
gives Lagrange theorem again. ¤
Example 2.3.10 (conjugation).
Let G be a group. One can consider the action G × G → G by α(g, x) = gxg−1. Such action is called conjugation. For a x ∈ G, Gx = C(x), the centralizer of x in G. And Ox = {gxg−1|g ∈ G} the conjugacy classes of x in G. So in general, we have
|G| = X
conj. classes
|C|, which is the class equation.
Now assume that G is finite. The class equation now reads:
|G| =X
x∈I
|G|/|C(x)|,
where I denotes a representative of conjugacy classes.
And Ox = {x} if and only if x ∈ Z(G), the center of G. So, for G finite, the class equation now gives
|G| = |Z(G)| + X
x∈I,x6∈Z(G)
|G|/|C(x)|.
Which is the usual form of class equation. ¤ The class equation is very useful if the group is a finite p-group. We recall some definition
Definition 2.3.11. If p is a prime, then a p-group is a group in which every element has order a power of p.
By a finite p-group, we mean a group G with |G| = pn for some n > 0.
Consider now G is a finite p group acting on S. Let S0 := {x ∈ S|gx = x, ∀g ∈ G}.
Then the class equation can be written as
|S| = |S0| + X
x∈I,x6∈S0
|Ox|.
One has the following
Lemma 2.3.12. Let G be a finite p-group. Keep the notation as above, then
|S| ≡ |S0| (mod p).
Proof. If x 6∈ S0, then 1 6= |Ox| = pk because |G| = |Ox| · |Gx|. ¤ By consider the conjugation G × G → G, one sees that
Corollary 2.3.13. If G is a finite p-group, then G has non-trivial center.
By using the similar technique, one can also prove the important Cauchy’s theorem
Theorem 2.3.14 (Cauchy). Let G be a finite group such that p | |G|.
Then there is an element in G of order p.
sketch. We keep the notation as in Lemma 2.3.12. Let S := {(a1, ..., ap)|ai ∈ G,Y
ai = e}.
And consider a group action Zp×S → S by (1, (a1, .., ap)) 7→ (ap, a1, ..., ap−1).
One claims that S0 = {(a, a, ..., a)|a ∈ G, ap = e}.
By the Lemma, one has |S| ≡ |S0| (mod p). It follows that p | |S0|.
In particular, |S0| > 1, hence there is (a, ..., a) ∈ S0 with a 6= e. One
sees that o(a) = p. ¤
Corollary 2.3.15. A finite group G is a p-group if and only it is a finite p-group.
2.4. Sylow’s theorems. We are now ready to prove Sylow theorems.
The first theorem regards the existence of p-subgroups in a given group.
The second theorem deals with relation between p-subgroups. In par- ticular, all Sylow p-subgroups are conjugate. The third theorem counts the number of Sylow p-subgroups.
Theorem 2.4.1 (First Sylow theorem). Let G be a finite group of order pnm (where (p, m) = 1). Then there are subgroups of order pi for all 0 ≤ i ≤ n.
Furthermore, for each subgroup Hi of order pi, there is a subgroup Hi+1 of order pi+1 such that HiC Hi+1 for 0 ≤ i ≤ n − 1.
In particular, there exists a subgroup of order pn, which is maximal possible, called Sylow p-subgroup. We recall the useful lemma which will be used frequently.
Lemma 2.4.2. Let G be a finite p-group. Then
|S| ≡ |S0| (mod p).
proof of the theorem. We will find subgroup of order pi inductively. By Cauchy’s theorem, there is a subgroup of order p. Suppose that H is a subgroup of order pi. Consider the group action that H acts on
S = G/H by translation, i.e. H × G/H → G/H by h(xH) := hxH.
One shows that xH ∈ S0 if and only if xH = hxH for all h ∈ H if and only if x ∈ NG(H). Thus |S0| = |NG(H)/H|.
If i < n, then
|S0| ∼= |S| = pn−im ≡ 0 (mod p).
By Cauchy’s theorem, the group NG(H)/H contains a subgroup of or- der p. The subgroup is of the form H1/H, hence |H1| = pi+1. Moreover,
H C H1. ¤
Example 2.4.3. If G is a finite p-group of order pn, then one has a series of subgroups {e} = H0 < H1 < ... < Hn = G such that |Hi| = pi and HiC Hi+1, Hi+1/Hi ∼= Zp. In particular, G is solvable.
Definition 2.4.4. A subgroup P of G is a Sylow p-subgroup if P is a maximal p-subgroup of G.
If G is finite of order pnm then a subgroup P is a Sylow p-subgroup if and only if |P | = pn by the proof of the first theorem.
Theorem 2.4.5 (Second Sylow theorem). Let G be a finite group of order pnm. If H is a p-subgroup of G, and P is any Sylow p-subgroup of G, then there exists x ∈ G such that xHx−1 < P .
Proof. Let S = G/P be the set of left cosets and H acts on S by translation. Thus by Lemma 2.3.12, one has |S0| ≡ |S| = m(mod p).
Therefore, S0 6= ∅. One has
xP ∈ S0 ⇔ hxP = xP ∀h ∈ H ⇔ x−1Hx < P.
This completes the proof. ¤
An immedaitely but important consequence is that any two Sylow p-subgroups are conjugate.
Theorem 2.4.6 (Third Sylow theorem). Let G be a finite group of order pnm. The number of Sylow p-subgroups divides |G| and is of the form kp + 1.
Proof. Let S be the conjugate class of a Sylow p-subgroup P (this is the same as the set of all Sylow p-subgroups). We consider the action that G acts on S by conjugation, then the action is transitive, i.e. for any x, y ∈ S, there exists g ∈ G such that y = gx. In particular Ox = S.
Hence |S| | |G| for |G| = |Gx| · |Ox|.
Furthermore, we consider the action P × S → S by conjugation.
Then
Q ∈ S0 ⇔ xQx−1 = Q ∀x ∈ P ⇔ P < NG(Q).
Both P, Q are Sylow p-subgroup of NG(Q) and therefore conjugate in NG(Q). However, Q C NG(Q), Q has no conjugate other than itself.
Thus one concludes that P = Q. In particular, S0 = {P }. By Lemma
2.3.12, one has |S| = 1 + kp. ¤
Example 2.4.7.
Group of order 200 must have normal Sylow subgroups. Hence it’s not simple. To see this, let rp := number of Sylow p-subgroups. Then r5 = 1. So if P is a Sylow 5-subgroup. Since gP g−1 is also a Sylow subgroup, it follows that gP g−1 = P for all g ∈ G. Thus P C G. ¤ Example 2.4.8.
There is no simple group of order 36. To see this, we consider P a Sylow 3-subgroup. Then r3 = 1 or 4. In case that r3 = 4, let S be the set of Sylow 3-subgroups. We have a group action G × S → S by conjugation. Thus we have a group homomorphism ϕ : G → A(S) ∼= S4. Comparing the cardinality of groups, one sees that ϕ must have non-trivial kernel. Hence G is not simple. ¤ 2.5. groups of small order. We can use the technique developed in the previous sections to study group of small order in more detail.
First of all, as a direct consequence of Cauchy’s theorem,
Proposition 2.5.1. Let p be a prime. A group of order p is cyclic.
Example 2.5.2.
Classify groups of order 2p.
If p = 2, then this is well-known. So we may assume that p > 2.
First of all there is a subgroup H < G of order p, generated by x, by Cauchy’s theorem. By Sylow’s third theorem, we have rp = 1, hence H is normal. Similarly, there is an element of order 2, say y. By normality of H, we have yxy−1 = xk for some k. Since
x = y2xy−2 = yxky−1 = xk2, it follows that k2 ≡ 1( mod p). Hence k ≡ 1 or ≡ −1.
Case 1. k ≡ 1, then xy = yx. It follows that G is abelian. By chinese Remainder Theorem, G is cyclic.
Case 2. k ≡ −1, then xy = yx−1. These kind of group is called
dihedral groups, denoted D2p. ¤
Example 2.5.3.
Let p, q be primes. If |G| = pq, then its structure can be determined similarly.
We assume that p > q. Then there are x, y ∈ G of order p, q respec- tively. Moreover, H :=< x > CG. We have yxy−1 = xk for some k.
Since
x = yqxy−q = yxky−1 = xkq,
it follows that kq ≡ 1( mod p). Now the situation depends on the structure of Z∗p. Recall that Z∗p ∼= Zp−1 is cyclic.
Case 1. q - p − 1, then kq ≡ 1( mod p) implies that k ≡ 1. Hence xy = yx. It follows that G is abelian. By chinese Remainder Theorem,
G is cyclic.
Case 2. q | p − 1, then kq ≡ 1( mod p) has q solutions, k ≡ a, a2, ..., aq−1, aq≡ 1. If we pick k ≡ a, then we determined a group G1 which is generated by x1, y1 with y1x1y1−1 = xa1. If we pick k ≡ a2, then we determined a group G2 which is generated by x2, y2 with y2x2y2−1 = xa22. Note that the map ϕ : G2 → G1 by ϕ(y2) = y12, ϕ(x2) = x1 gives an isomorphism. Therefore, for different solution k ≡ a, a2, ..., aq−1,
they determined the same group. ¤
There is a useful construction to produce groups from simple ones called semi-direct product which we now introduce. Given two groups G, H and a homomorphism θ : H → Aut(G). Let G×θH be the set G × H with the binary operation (g, h)(g0, h0) = (g(θ(h)(g0)), hh0).
One can verify that this produce a group.
For example, in the case 2 of above example, we have G = Zp, H = Zq and we consider θ : Zq → Aut(Zp) ∼= Z∗pby θ(1) = a. Then we obtained Zp ×θ Zq. Such group is called a metacyclic groups.
Proposition 2.5.4. Let p be a primes. If |G| = p2, then G is abelian.
We will discuss the structure of finite ableina groups later. In prin- ciple, their structure are pretty easy.
sketch. By class equation, one sees that Z(G) is non-trivial.
Case 1. if |Z(G)| = p2, then G is abelian.
Case 2. if |Z(G)| = p, then G/Z(G) is a group of order p , hence cyclic. We pick x ∈ G such that G/Z(G) is generated by xZ(G). We also pick y ∈ G such that Z(G) is generated by y. It’s easy to check that G is generated by x, y. Note that xy = yx, it follows that G is
abelian. ¤
Using above properties, one can classified groups of order ≤ 15 com- pletely unless for order 8 and 12. In fact groups of order 8 are either abelian or D8 or Q8. Where Q8 is the quaterion group defined by {i, j, k, −i, −j, −k, 1, −1|i2 = j2 = k2 = −1, ijk = −1}.
Easy example of non-abelian groups of order 12 includes A4, D12. In fact there is one more, T =< a, b|a6 = b4 = 1, b2 = a3 = (ab)2 >.
Theorem 2.5.5. every non-abelian group G of order 12 is isomorphic to A4, D12 or T .
sketch. Let P be a Sylow 3-subgroup. We first consider the action G × G/P → G/P by translation. It gives rise to a homomorphism ϕ : G → A(G/P ) ∼= S4. It’s clear that ker(ϕ) < P .
Case 1. ker(ϕ) = {e}, then G ∼= A4.
Case 2. ker(ϕ) = P . Then we need to wok harder. So now, P C G and P is the unique Sylow 3-subgroup. Let P = {x, x2, x3 = e}, then x, x2 are the only element in G of order 3.
Let K be a Sylow 2-subgroup, then K is either V4 or Z4.
Case 2.i. If K ∼= V4, by computing the relation between generators,
one can show that G ∼= D12.
Case 2.ii. If K ∼= Z4, by computing the relation between generators, one can show that G ∼= T .
¤ Groups of order pn, n ≥ 3 could be very complicated. Here just give two more examples.
Example 2.5.6.
Let G < GL(2, C) be the group generated by A =
µ 0 ω ω 0
¶ and B =
µ 0 1
−1 0
¶
, where ω is a primitive 2n−1th root of unity for n ≥ 3.
Then G is a group of order 2n. ¤
Example 2.5.7.
Let G < GL(3, C) be the group generated by A =
1 0 0
0 ω 0
0 0 ω2
and B =
0 1 0 0 0 1 1 0 0
, where ω is a primitive 3th root of unity. Then
G is a group of order 27. ¤
Oct. 13, 2006 (Fri.)
2.6. symmetry of the plane. A map from plane itself is called a rigid motion, or an isometry, if it is distance-preserving. Let S be a subset of the plane. Then the subgroups of rigid motions preserving S is called the symmetry of S. It’s well-known that:
Example 2.6.1.
Let S be the regular n-gon centered at the origin. Then the symme-
try of S id the group D2n. ¤
In order to build this is a more solid foundation, we need to work a little bit more.
A list of rigid motions consists of:
1. Orientation-preserving motions:
a. Translation.
b. Rotation.
2. Orientation-reversing motions:
a. Reflection.
b. Glide reflection, i.e. reflecting about a line l and then translating by a non-zero vector a parallel to l.
Theorem 2.6.2. The above list is complete.
Sketch. We first fix some notations:
ta: translation by a vector a.
ρθ: rotation by an angle θ about the origin.
r: reflection about the x-axis.
Step 1. Orientation preserving motions that fix the origin are {ρθ}.
Step 2. Let m ne an orientation preserving motion. If m(o) = a, then t−am = ρθ for some θ. by Step 1. Thus m = taρθ.
Step 3. If m is not a translation, i.e. θ 6= 0, then m is a rotation about a point p. To see this, first show that m has a fixed point, denoted p, if θ 6= 0. A point on the plane can be written as p + x,
m(p + x) = taρθ(p + x) = ρθ(p + x) + a = ρθ(p) + ρθ(x) + a = p + ρθ(x).
Step 4. Orientation reversing motions that fix the origin are {ρθr}.
For given such m, it’s clear that rm preserves the orientation and fixes the origin. So rm = ρθ for some θ. Thus m = rρθ = ρ−θr. Also note that ρθr is the reflection about l, denoted rl, which is the line obtained by rotating x-axis by 12θ.
Step 5. Let m be an orientation reversing motion. Then m(o) = a for some a. Thus t−am is an orientation reversing motion that fixes origin, hence t−am = rl. Therefore, m = tarl which is a glide reflection. ¤ Indeed, let O(2, R) be the subgroup of motions that fix the origin.
Then O(2, R) is generated by {ρθ, r}. Let M be the groups of plane
rigid motions, then there is a group action M × R2 → R2. The orbit of o is the whole R2 and the stabilizer of o is O(2, R).
For readers who want to know more about symmetry, we refer [Artin], Chapter 5.
2.7. abelian groups. In this section, we are going to study a simple but important category of groups, the abelian groups.
Given an abelian group G, we usually use + to denote the operation.
We say that G can be generated by X ⊂ G, denoted G =< X >, if every element of G can be written as P
nixi for some ni ∈ Z and xi ∈ X. Note that ni 6= 0 for all but finitely many i.
A basis of an abelian group G is a linearly independent generating subset X. That is for distinct x1, ..., xk ∈ X, P
nixi = 0 implies that ni for all i.
An abelian group with a basis is called a free abelian group. And the rank, denoted rk(F ), is |X|.
It’s easy to prove that an abelian group is free if and only if it’s a direct sum of Z.
On the other hand, given a set X, we can always construct a free abelian group on the set X by consider the set
F := {X
nxx|x ∈ X, nx ∈ Z, nx = 0 for all but finitely many x}.
The group operation on F is nothing but P
nxx +P
mxx := P (nx+ mx)x. It’s clear that X is a basis of F in this construction.
Example 2.7.1.
This construction appeared, for example, in algebraic topology. The groups of 1-chains is the free abelian group on the set of simplicial
1-chains. ¤
Example 2.7.2.
Let X be a Riemann surface, then the group of divisors, Div(X), is
the free abelian group on the set X. ¤
It has the following universal property:
Proposition 2.7.3. Let F be a free abelian group with basis X. For any function f : X → G to an abelian group G. There exist a unique homomorphism ϕ : F → G extending f .
Proof. Let ϕ(P
nxx) = P
nxf (x), then verify it. ¤ Corollary 2.7.4. Every abelian group is a quotient of a free abelian group.
Proof. LetG be an abelian group. Let F be the free abelian group on the set G. Consider f : G → G the identity map. Then we are
done. ¤
Example 2.7.5.
Q can be describe as following. Let X = {x1, ..., xn, ...} and F the free abelian group on the set X. Take f : X → Q by f (xi) = 1i. Then
Q is a quotient of F . ¤
We are now ready to state develop to main theorem of this section.
We need the following:
Lemma 2.7.6. If {x1, ..., xn} is a basis of F , then {x1, ..., xj−1, xj + axi, xj+1, ..., xn} is also a basis of F for i 6= j and a ∈ Z.
Theorem 2.7.7. Let F be a free abelian group of rank n and G is a non-zero subgroup of F , then there exists a basis {x1, ...., xn} of F , an integer r (1 ≤ r ≤ n) and positive integer d1, ..., dr such that d1|d2|...|dr and G is free abelian group with basis {d1x1, ..., drxr}.
Sketch. If n = 1, this is easy.
By induction, we assume that the theorem is true for all abelian groups of rank ≤ n − 1. Let
S := {s ∈ Z|sy1+ ...knyn∈ G, for some basis of F , y1, ..., yn}.
Let d1 be the smallest positive integer in S. By changing basis, we may have {x1, y2, ..., yn} basis of F and d1x1 ∈ G.
Let H =< y2, ..., yn >. It’s clear that F = H ⊕ Zx1. We claim that G = (H ∩ G) ⊕ Zd1x1.
Apply induction hypothesis to G ∩ H < H, then we are done. ¤ Corollary 2.7.8 (fundamental theorem of finitely generated abelian groups). Let G be a finitely generated abelian group. Then there exist an integer r and positive integers d1|d2|...|dt such that
G ∼= Zd1 ⊕ ... ⊕ Zdt ⊕ Zr.
Proof. Let X be a finite generating set of G. And let F be the free abelian group on the set X. Then there is a surjective homomorphism
F → G. Apply Theorem 2.7.7 to ker < F . ¤
Now we restrict ourselves to finite abelian groups. Let G be a finite abelian group, by Corollary 2.7.8,
G ∼= Zd1 ⊕ ... ⊕ Zdt.
These d1, ..., dt are called invariant factors. We consider the factor- ization of di into prime factors, then we have for all i,
di = pa1i,1...paki,k.
By Chinese Remainder Theorem, we have for all i, Zdi ∼= Zpai,1
1 ⊕ ... ⊕ Zpai,k k . Therefore,
G ∼= ⊕kj=1(⊕ti=1Zpai,j
j ).
It’s clear that ⊕ti=1Zpai,j
j is the Sylow pj-subgroup. And these paji,j are called elementary divisors.
Example 2.7.9.
Let G = Z100⊕Z40. By Chinese Remainder Theorem, Z100∼= Z4⊕Z25 and Z40∼= Z8⊕ Z5. Thus
G ∼= Z4⊕ Z8⊕ Z5⊕ Z25 ∼= Z20⊕ Z200.
So invariant factors are 20, 200 and elementary divisors are 4, 8, 5, 25.
¤ Example 2.7.10.
Let G = Zm⊕ Zn. Then invariant factors are (m, n), [m, n], the gcd
and lcm of m, n. ¤
Oct. 20, 2006 (Fri.)
Let G be an abelian group, there there is a natural important ho- momorphism m : G → G by m(x) := mx for m ∈ N. The image is denoted mG and kernel is denoted G[m]. Let G(p) = {u ∈ G|o(u) = pn for some n ≥ 0}. One can show that G(p) is the Sylow p-subgroup of G. And G is a direct sum of Sylow subgroups. Thus it remains to study finite abelian p-groups. The only non-trivial part of classical theory is showing that a finite abelian p-group is a direct sum of cyclic p-groups.
We also remark that for a given finitely generated abelian group G, the rank, invariant factors, and elementary divisors are unique. To see this, we proceed as following steps:
1. if Zn∼= Zm, then n = m.
To see this, let G ∼= Zn ∼= Zm. We consider G/2G ∼= Zn2 ∼= Zm2 . Thus n = m.
2. let Gtor := {u ∈ G|mu = 0 for some m}. It’s clear that Gtor < G.
3. If
G1 = Zd1 ⊕ ... ⊕ Zdt ⊕ Zr,
∼= G2 = Zd01 ⊕ ... ⊕ Zd0
t0 ⊕ Zr0
Then clearly, G1tor ∼= G2tor and also G1/G1tor= Zr ∼= G2/G2tor = Zr0. Hence in particular r = r0.
4. It remains to show that t = t0 and di = d0i.
To see this, it’s equivalent to show the uniqueness of elementary divisors of finite abelian groups. So now we assume that G is finite abelian group. Also note that if G1 ∼= G2, then G1(p) ∼= G2(p). Thus we may even assume that G is a finite abelian p-group.
Suppose now that
G1 := Zpa1 ⊕ ... ⊕ Zpat
∼= G2 := Zpb1 ⊕ ... ⊕ Zpbs, with a1 ≤ a2 ≤ ... ≤ at, b1 ≤ b2 ≤ ... ≤ bs.
Then we have pG1 ∼= pG2 and G1/pG1 ∼= G2/pG2. Note that G1/pG1 ∼= Zcp1, with c1 = {i|ai > 1}. It follows that c1(G1) = c1(G2).
Similarly, we can define ck:= {i|ai > k} and ck(G1) = ck(G2).
Moreover, G1[p] ∼= Ztp ∼= G2[p] ∼= Zsp. Hence t = s.
Since t, c1(G1), c2(G1)... determine a1, ..., atuniquely and s, c1(G2), c2(G2)...
determine b1, ..., bs uniquely. It follows that t = s and ai = bi for all i.
2.8. Nilpotent groups, solvable groups. Given a group G, if G has a normal subgroup N, then we have a quotient group G/N. One can expect that knowing N and G/N would give some information on G.
In this section, we are going to introduce the general technique of this idea.
Let G be a group. If G has no non-trivial normal subgroup, then G is said to be simple.
In general, there are two natural way to produce normal subgroups.
The first one is the the center Z(G). It is a normal subgroup of G.
And we have the canonical projection G → G/Z(G). Let C2(G) be the inverse image of Z(G/Z(G)) in G. By the correspondence theo- rem, Z(G/Z(G)) is a normal subgroup of G/Z(G) hence C2(G) is a normal subgroup of G. And then let C3(G) to be the inverse image of Z(G/C2(G)). By doing this inductively, one has an ascending chain of normal subgroups
{e} < C1(G) := Z(G) < C2(G) < ...
Notice that by the construction, each Ci(G) C G and Ci+1(G)/Ci(G) is abelian.
Definition 2.8.1. G is nilpotent if Cn(G) = G for some n.
Proposition 2.8.2. A finite p-group is nilpotent.
Proof. We use the fact that a finite p-group has non-trivial center. Thus one has Ci Ci+1. The group G has finite order thus the ascending chain must terminates, say at Cn. If Cn 6= G, then G/Cnhas non-trivial center. One has Cn Cn+1 which is impossible. Hence Cn= G. ¤ Theorem 2.8.3. If H, K are nilpotent, so is H × K.
Proof. The key observation is that Z(H × K) = Z(H) × Z(K). Then inductively, one proves that Ci(H × K) = Ci(H) × Ci(K). If Cn(H) = H, Cm(K) = K then Cl(H × K) for l = max(m, n). ¤ Lemma 2.8.4. Let G be a nilpotent group and H G be a proper subgroup. Then H NG(H).
Proof. Let C0(G) = {e}. Let k be the largest index such that Ck(G) <
H. Then Ck+1(G) 6< H. Pick a ∈ Ck+1− H, then for every h ∈ H, we have Ckha = CkhCka = CkaCkh = Ckah for Ck+1/Ck = Z(G/Ck(G)).
Thus aha−1 ∈ Ckh ⊂ H for all h ∈ H. That is a ∈ NG(H) − H. ¤ Then we are ready to prove the following:
Theorem 2.8.5. A finite group is nilpotent if and only if it’s a direct product of Sylow p-subgroups.
Proof. By the previous two results, it’s clear that a direct product of Sylow p-subgroups is nilpotent.
Conversely, if G is nilpotent, then we will prove that every Sylow p-subgroup is a normal subgroup of G. By checking the decomposition criterion, one has the required decomposition.
It remains to show that if P is Sylow p-subgroup, then P C G.
To this end, it suffices to prove that NG(P ) = G. By applying this Claim to NG(P ), then it says that NG(P ) can’t be a proper subgroup of G since NG(NG(P )) = NG(P ). Thus it follows that NG(P ) = G. ¤ Example 2.8.6.
Let G = D12 = {xiyj|x6 = y2 = e, xy = yx5}. One of it’s Sylow 2- subgroup is {e, x3, y, x3y} isomorphic to V4 and it’s Sylow 3-subgroup is {e, x2, x4} ∼= Z3.
However Z(G) = {e, x3} and G/Z(G) ∼= D6 ∼= S3 and Z(S3) = {e}.
Thus G is not nilpotent. And therefore, D12 6∼= V4× Z3. ¤ We have seen that we have a series of subgroup by taking centers.
Another natural construction is to take commutators.
Definition 2.8.7. Let G be a group. The commutator of G, denoted G0 is the subgroup generated by the subset {aba−1b−1|a, b ∈ G}.
Roughly speaking, the subgroup G0measures the non-commutativity of a group. More precisely, G0 = {e}, if and only G is abelian. The smaller G0, the more commutative it is.
Proposition 2.8.8. We have:
1. G0C G,
2. and G/G0 is ableian.
3. if N C G, then G/N is abelian if and only if G0 < N .
Proof. 1.) for all g ∈ G, g(aba−1b−1)g−1 ∈ G0, hence gG0g < G0. So G0C G.
2.) aG0bG0 = abG0 = ab(b−1a−1ba)G0 = baG0 = bG0aG0.
3.) Consider π : G → G/N. If G/N is abelian, then π(aba−1b−1) = e, hence G0 < N . Conversely, if G0 < N , we have a surjective homomor- phism G/G0 → G/N. G/G0 is abelian, hence so is it homomorphic
image G/N. ¤
Definition 2.8.9. We can define the the commutator inductively, i.e.
G(2) := (G0)0, etc. The G(i) is called the i-th derived subgroup of G. It’s clear that G > G0 > G(2) > ....
A group is solvable is G(n)= {e} for some n.
Example 2.8.10.
Take G = S4. The commutator is the smallest subgroup that G/G0 is abelian. Since the only non-trivial normal subgroups of S4 are V, A4. It’s clear that G0 = A4 (Or one can prove this by hand). Similarly, one finds that G(2) = A04 = V , and G(3) = {e}. Hence S4 is solvable. ¤ Another useful description of solvable groups is the groups with solv- able series.
Definition 2.8.11. A groups G has a subnormal series if there is a series of subgroups of G
G = H0 > H1 > H2 > ... > Hn, such that HiC Hi−1 for all 1 ≤ i ≤ n.
A subnormal series is a solvable series if Hn = {e} and Hi−1/Hi is abelian for all 1 ≤ i ≤ n.
A subnormal series is a normal series if all Hi are normal subgroups of G.
Theorem 2.8.12. A group is solvable if and only it has a solvable series.
Proof. It’s clear that G > G0 > ...G(n) = {e} is a solvable series. It suffices to prove that a group with a solvable series is solvable. Suppose now that G has a sovable series {e} = Hn < ... < H0 = G. First observe that G0 < H1 since G/H1 is abelian. We claim that G(i) < Hi
for all i inductively. Which can be proved by the observation that the intersection of the series {e} = Hn < ... < H0 = G with G(i) gives a
solvable series of G(i). ¤
Example 2.8.13.
A finite p-group has a solvable series, hence is solvable.
Moreover, a nilpotent group is solvable. To see this, let G be a nilpotent group. Then there exist a series
{e} < C1(G) := Z(G) < C2(G) < ... < Cn(G) = G.
Notice that Ci+1(G)/Ci(G) = Z(G/Ci(G)) is abelian. Therefore this
is a solvable series. ¤
Oct. 27, 2006 (Fri.)
Proposition 2.8.14. Let H be a subgroup of a solvable group G, then H is solvable.
Let N be a normal subgroup of G. Then G is solvable if and only if both N and G/N are solvable.
Sketch. G has a solvable series, intersecting the series with H gives a solvable series of H.
If N C G, then we have π : G → G/N. Projecting the solvable series of G to G/N gives a solvable series of G/N.
Finally, if N and G/N are solvable, they have solvable series respec- tively. Apply π−1 to the solvable series of G/N gives a series from N to G. Combine this series with the serious of H gives a solvable series
of G. ¤
Example 2.8.15.
We will prove in the coming subsection that A5 is not solvable, hence
so is Sn for n ≥ 5. ¤
2.9. normal and subnormal series. We turning back to series a little bit more. A subnormal series is called a composition series if every quotient is a simple group.
Definition 2.9.1. For a subnormal series, {e} = Hn < ... < H0 = G, the factors of the series are the quotient groups Hi−1/Hi and the length is the number of non-trivial factors. A refinement is a series obtained by finite steps of one-step refinement which is {e} = Hn < . < K <
.. < H0 = G.
Definition 2.9.2. Two series are said to be equivalent if there is a one-to-one correspondence between the non-trivial factors. And the corresponding factors groups are isomorphism.
It’s clear that this defines an equivalent relation on subnormal series.
The main theorems are
Theorem 2.9.3 (Schreier). Any two subnormal (resp. normal) series of a group G have a subnormal (resp. normal) refinement that are equivalent.
An immediate corollary is the famous Jordan-H¨older theorem.
Theorem 2.9.4 (Jordan-H¨older). Any two composition series of a group are equivalent.
The main technique is the Zassenhaus Lemma, or sometimes called butterfly Lemma.
Lemma 2.9.5 (Zassenhaus). Let A∗C A and B∗C B be subgroups of G. Then
