21-373 Algebraic Structure Notes

Home , Algebraic structure, Left and right (algebra), Pointwise

Instructor: Anton Bernshteyn Notes by: Lichen Zhang Carnegie Mellon University 21-373 Algebraic Structure

March 14, 2020

Contents

1 Sets, Functions and Equivalence Relations3 1.1 Basic deﬁnitions...... 3 1.2 Higher-order functions...... 5 1.3 Equivalence relations & quotients...... 6

2 Functors 8 2.1 Covariant functors...... 8 2.2 Universal elements...... 10 2.3 Contravariant functors...... 12

3 Groups 13 3.1 Concrete groups...... 13 3.2 Abstract groups...... 17 3.3 Abelian groups, subgroups, homomorphisms...... 21 3.4 Actions, cosets, orbits...... 23 3.5 Orbit-stabilizer theorem...... 28 3.6 Normal subgroups and kernels...... 32 3.7 Groups and counting...... 34

4 Category 36 4.1 Basic deﬁnitions & examples...... 36 4.2 Monomorphism, epimorphism...... 38

5 Advanced topics on Groups 41 5.1 Product & coproduct of groups...... 41 5.2 Free groups...... 43 5.3 Structure of ﬁnite groups...... 51

1 6 Rings and Modules 55 6.1 Basic deﬁnitions...... 55 6.2 Free modules, independent, spanning, and basis...... 57 6.3 Commutative rings and IBN...... 60 6.4 Polynomials...... 66

2 1 Sets, Functions and Equivalence Relations

1.1 Basic definitions We start this notes by reminding readers some basic definitions of sets, functions and equivalence relations, which, though fundamental, are critical in the study of various algebraic structures. Given two sets A, B, except for the standard notation A ∪ B,A ∩ B,A \ B,A ⊆ B,A ⊂ B, we introduce symmetric difference of two sets.

Definition 1.1. The symmetric difference of two sets A, B, denoted by A4B, is defined as (A \ B) ∪ (B \ A).

Deﬁnition 1.2. The set of all functions f : A → B is denoted as BA.

|A| Exercise 1.3. If both A, B are ﬁnite, show that BA = |B| . Proof. To count all functions from A to B, it is useful to think of it in a bucket-item manner: each element of A can be viewed as a bucket, while each element of B is item, count the number of functions is just counting the number of ways to put items into buckets. For each bucket, there are |B| items we can put into, moreover, these items can be put into as many buckets as we want, so there are |B||B| ... |B| = |B||A| functions in total. | {z } |A| times Deﬁnition 1.4. Let f : A → B be a function.

• f is injective, if for a1, a2 ∈ A, a1 6= a2 ⇒ f (a1) 6= f (a2). • f is surjective, if for any b ∈ B, b ∈ im (f), or equivalently, B = im (f).

• f is bijective if it is both injective and surjective.

Deﬁnition 1.5. Two functions f, g are equal, if dom (f) = dom (g), and for any x ∈ dom (f) , f (x) = g (x).

Example 1.6. 1. Let A be a set, {0} be a 1-element set, what is A{0}? It’s the set of all functions from {0}. It is clear that there is a bijection between A{0} and A by f 7→ f (0). In such scenario, we say A is isomorphic to A{0}.

Deﬁnition 1.7. Given two sets A, B, we say A is isomorphic to B, denoted by A =∼ B, if there exists a bijection from A to B.

2. A{0,1} =∼ A × A, the bijection is given by f 7→ (f (0) , f (1)).

3. What is A∅, ∅A? They are diﬀerent as long as A 6= ∅.

4. {0, 1}A =∼ P (A), the bijection is f 7→ {a ∈ A : f (a) = 1}.

Deﬁnition 1.8. Given three sets A, B, C, two functions f : B → C, g : A → B, deﬁne composition of functions f, g, denoted by f ◦ g, as follows:

• dom (f ◦ g) = {x ∈ dom (g): g (x) ∈ dom (f)}.

3 • For x ∈ dom (f ◦ g), deﬁne (f ◦ g)(x) = f (g (x)).

g A B

f f◦g

Exercise 1.9 (Important). Verify function composition is associative, i.e., given f, g, h three functions with well-behaved domains & codomains, verify f ◦ (g ◦ h) = (f ◦ g) ◦ h.

Example 1.10. Applying a function to an element can be “encoded” as a composition operation: let f : A → B be our function, consider f (a), we can replace a by function g : {0} → A by 0 7→ a, then f ◦ g : {0} → B : 0 7→ f (a).

Deﬁnition 1.11. Given a set A, the identity functions on A, denoted by idA : A → A, is the function a 7→ a.

Exercise 1.12. Let f : A → B, then f ◦ idA = idB ◦ f = f.

Deﬁnition 1.13. Let f : A → B, g : B → A, if f ◦ g = idB, then f is a left inverse of g and g is a right inverse of f. If f is both left and right inverse of g, then f is a two-sided inverse of g.

Exercise 1.14. Give an example such that f is a left inverse, but not a right inverse of g.

Exercise 1.15. Let A 6= ∅,B be two sets and f : A → B, g : B → A. Prove the following:

1. f has a left inverse g if and only if f is injective.

2. f has a right inverse g if and only if f is surjective.

The proof is not hard, for the ﬁrst part, one simply choose g to send everything to its preimage under f. For the second part, g can simply send element to one of its preimage under f.

Deﬁnition 1.16. Given f : A → B, the followings are equivalent:

1. f is bijective.

2. f has a right and left inverse.

3. f has a two-sided inverse.

In this case, we denote (two-sided) inverse of f as f −1. Also, f −1 is a bijection from B to A and f −1−1 = f.

Exercise 1.17. Suppose f : A → B, g : B → A are both bijective, then (g ◦ f)−1 = f −1 ◦ g−1.

4 1.2 Higher-order functions In this section, we study higher-order functions, which take functions as inputs and spits out functions. This idea has been heavily exploited in functional programming and the study of programming language. We motivate with an example.

Y Example 1.18. Let X,Y,A be sets, then we claim AX×Y =∼ AX . Given f ∈ AX×Y , i.e., f : X × Y → A, we associate f to the function F : Y → AX as follows: for any y ∈ Y, x ∈ X,(F (y)) (x) := f (x, y). This is called function currying. The inverse of this operation is as follows: to each F ∈ AX Y , assign f : X × Y → A by f (x, y) := (F (y)) (x).

Deﬁnition 1.19. Let X,Y,A be sets, to each function f : X → Y , corresponds a function f A : XA → Y A, given by f A (g) := f ◦ g. g A X f A(g) = f f◦g

Proposition 1.20. Let X,Y,Z be sets and f : X → Y, g : Y → Z, then (g ◦ f)A = gA ◦ f A.

Proof. First, let’s compare the domains.

• g ◦ f : X → Z, so (g ◦ f)A : XA → ZA

• gA : Y A → ZA, f A : XA → Y A, so gA ◦ f A : XA → ZA

Take and h ∈ XA, let’s do some computations.

• (g ◦ f)A (h) = (g ◦ f) ◦ h

• gA ◦ f A (h) = gA (f ◦ h) = g ◦ (f ◦ h)

By 1.9, we know that function composition is associative, therefore, they are equal.

Deﬁnition 1.21. Let X,Y,A be sets, to each function f : X → Y , corresponds a function Af : AY → AX by Af (g) = g ◦ f. Af (g) X Y

f g

Exercise 1.22. Let X,Y,Z be sets and f : X → Y, g : Y → Z, then Ag◦f = Af ◦ Ag.

5 1.3 Equivalence relations & quotients Deﬁnition 1.23. A (binary) relation on a set X is a subset R ⊆ X × X, usually, we write xRy instead of (x, y) ∈ R.

Deﬁnition 1.24. An equivalence relation E on set X is a relation such that

1. E is reﬂexive.

2. E is symmetric.

3. E is transitive.

Example 1.25. Consider the following relation on Z:

• xEy ⇔ x − y ∈ Z, this is an equivalence relation.

• xEy ⇔ x + y ∈ Z, this is not an equivalence relation, transitivity does not hold.

Deﬁnition 1.26. Given function f : X → Y , the equivalence kernel of f is the relation Ef on X, given by xEf y ⇔ f (x) = f (y), this Ef is an equivalence relation. Conversely, given an equivalence relation E on X, for x ∈ X, let [x]E := {y ∈ X : xEy}, deﬁned as the equivalence class of x.

Let X/E := {[x]E : x ∈ X} be the quotient of X by E, the map pE : X → X/E : x 7→ [x]E is called the quotient map.

Proposition 1.27. Let E be an equivalence relation on X, then pE : X → X/E is a surjection, with equivalence kernel E.

Proof. The first part is clear, any equivalence class [x]E has at least one canonical representation, namely x in it, this is given by E is reflexive. To see the second part, consider the following chain of equivalence: xEy ⇔ [x]E = [y]E ⇔ pE (x) = pE (y). Definition 1.28. Let E be an equivalence relation on set X, a function f : X → Y is E-invariant if xEy ⇒ f (x) = f (y), i.e., f is constant on equivalence classes.

Theorem 1.29. Let E be an equivalence relation on X, if f : X → Y is E-invariant, then there is a unique function g : X/E → Y such that f = g ◦ pE. Proof. The theorem is equivalent to the following diagram: p X E X/E

f ∃!g

Y Deﬁne function g as follows: for each C ∈ X/E, let g (C) := f (x), for any x ∈ C. Since f is E-invariant, this does not depend on the choice of x ∈ C, so g is well-deﬁned.

Moreover, (g ◦ pE)(x) = g (pE (x)) = g ([x]E) = f (x), this proves the existence of such an g. To see uniqueness, if f = g ◦ pE, then for any x,(g ◦ pE)(x) = g ([x]E) = f (x), therefore, it’s unique.

6 Exercise 1.30. Based on the construction of g above, prove the following:

1. g is injective if and only if the equivalence kernel of f is E.

2. g is surjective if and only if f is surjective.

3. If Ef = E and im (f) = Y , then g is a bijection. Corollary 1.31 (Canonical decomposition theorem). Let f : X → Y , then f can be decomposed p b X X/Ef im (f) as follows: f i Y where:

• p is the quotient map pE, which is surjective. • i is the inclusion map, which is injective.

• b is a bijective map, which is g from 1.29.

7 2 Functors

In this section, we study the ﬁrst algebraic structure in this course, which operates on sets and functions.

2.1 Covariant functors Deﬁnition 2.1. A functor (on sets to sets) is a rule F that assigns each set S a set F (S) and to each function f : S → T a function F (f): F (S) → F (T ) such that

1. For each set S, F (idS) := idF(S). 2. For any f : S → T , g : R → S, we have F (f ◦ g) = F (f) ◦ F (g).

One can think of a functor F consists of

• an “object map”: FO : sets → sets.

• a “mapping map”: FM : functions → functions. Let’s look at some examples of functors to form a better sense of the deﬁnition.

Example 2.2. 1. The identity functor, given by Id (S) = S, Id (f) = f.

2. Fix a set A, let F (S) := A. for any set S, and let F (f) := idA, for any f. Exercise 2.3. Are there any other functors with F (S), for any set S?

3. The powerset functor, given by F (S) = P (S), and for function f : S → T , we need to have F (f): P (S) → P (T ). For any subset A ⊆ S, let (F (f)) (A) := {f (a): a ∈ A}, i.e., the image of A under f, or f (A). Let’s verify it’s a functor.

Proof. Notice F (idS)(A) = idS (A) = A, and idF(S) (A) = A, so they are equal. For composition, we use f∗ as a shorthand notation for F (f), then

• (f ◦ g)∗ (A) = {(f ◦ g)(a): a ∈ A} = (f ◦ g)(A) = f (g (A)) • f∗ ◦ g∗ (A) = f∗ (g∗ (A)) = f∗ (g (A)) = f (g (A)) Again, the composition is preserved.

4. Fix a set A, deﬁne the functor to raise sets and functions to the power of A: S 7→ SA, f 7→ f A. This deﬁnes a functor, since in 1.20, we have shown that (f ◦ g)A = f A ◦ gA. Identity is easy to verify.

5. For each set S, let [S]<∞ be the set of all finite subsets of S, define a functor as follows: for each set S, F (S) := [S]<∞, and for each function f : S → T , F (f): F (S) → F (T ), given by (F (f)) (A) := {f (a): a ∈ A}. Since input subset A is finite, output is also finite, so this functor is well-defined.

Deﬁnition 2.4. A functor G is called a subfunctor of F if for any set S, G (S) ⊆ F (S), and for any f : S → T , G (f) = (F (f)) |G(S).

8 A quick observation: suppose F is a functor and let G0 be a mapping that assigns each set S a subset G0 (S) ⊆ F (S), then there is a unique subfunctor G of F whose object map is G0 if and only if im F (f) |G0(S) ⊆ G0 (T ), for any f : S → T . Example 2.5. Fix a set A, let F be the functor given by F (S) := SA, and for each f : S → T , de- A fine F (f) := f . Let E be an equivalence relation on A, let G0 (S) := {E-invariant functions f : A → S}. A This gives rise to a subfunctor of F: first, it’s clear that for any set S, we have G0 (S) ⊆ S , our goal A A is to show that f |G0(S) : G0 (S) → G0 (T ) is well-defined and well-behaved, i.e., f |G0(S) (g) = f ◦g, is this function E-invariant? Notice that by definition, g is E-invariant, so if a1Ea2 ⇒ g (a1) = g (a2) ⇒ f (g (a1)) = f (g (a2)).

9 2.2 Universal elements Deﬁnition 2.6. A universal element for a functor F is a pair (R, u), where R is a set, u ∈ F (R) such that for any set S and each s ∈ F (S), there exists a unique function h : R → S, satisfying (F (h)) (u) = s.

Example 2.7. Fix a set A and an equivalence relation E, we have a functor F that maps S to F (S) = {E-invariant functions g : A → S}, and each f : S → T to F (f) by g 7→ f ◦ g. What is the universal element of this functor? We need a set R, and u ∈ F (R), where u is an E-invariant function from A to R, our goal is to satisfy the universal property, i.e., for all set S and s : A → S which is E-invariant, there is a unique function h : R → S such that F (h)(u) = h◦u = s.

Let’s set R = A/E, and u = pE, set h ([x]E) = s (x), given such u, h must be unique, so (A/E, pE) is a universal element of functor F.

Example 2.8. Fix sets A, B, deﬁne F (S) = SA × SB, given f : S → T , deﬁne F (f) by (g, h) = (f ◦ g, f ◦ h).

Exercise 2.9. Prove it’s a functor.

What is the universal element? It’s called disjoint union of A and B. Let R be our target set, u ∈ F (R) = RA × RB = (i, j), for set S and s ∈ F (s) = (g, h), there is a unique function f : R → S such that F (f)(i, j) = (g, h) = (f ◦ i, f ◦ j). A i R B j g ∃!f h

S If A ∩ B = ∅, then we can simply set R = A ∪ B and i, j be inclusion maps. Set ( g (x) , if x ∈ A f (x) = h (x) , if x ∈ B

On the other hand, if A ∩ B 6= ∅, then let A0,B0 be A0 := A × {0} ,B0 = B × {1}, and i, j maps element with an extra bit. By this bit, f has the ability to decode the input. Set ( g i−1 (x) , if x ∈ A0 f (x) = h j−1 (x) , if x ∈ B0

This construction is referred as the disjoint union of A and B.

Proposition 2.10. Let (R, u) be a universal element of F, suppose b : R → R0 is a bijection, then (R0, u0) is also a universal element, where u0 = F (b)(u).

Proof. Take any set S and s ∈ F (S), we need to show that there exists a unique h0 : R0 → S such that F (h0)(u0) = S. We separately show the existence and uniqueness. Existence: since (R, u) is universal, there exists a unique h : R → S with F (h)(u) = s. Deﬁne

10 h0 = h ◦ b−1, then

F h0 u0 = F h0 (F (b)(u)) = F h0 ◦ F (b)(u) = F h0 ◦ b (u) = F h ◦ b−1 ◦ b (u) = F (h)(u) = s

Uniqueness: Notice that we must have F (h0 ◦ b)(u) = F (h)(u), by uniqueness of h, we must have h0 ◦ b = h, the only choice for h0 is h ◦ b−1.

Next theorem is the converse of above proposition.

Theorem 2.11 (Uniqueness of universal element). Let F be a functor, if (R, u) , (R0, u0) are universal for F, then there exists a unique b : R → R0 such that u0 = F (b)(u).

Proof. Since (R, u) is universal, there exists a unique b : R → R0 such that F (b)(u) = u0. Similarly, (R0, u0) is universal, so there exists a unique b0 such that F (b0)(u0) = u. It suﬃces to show b is a 0 0 0 bijection, and b is its inverse, i.e., b ◦ b = idR0 , b ◦ b = idR.

F b0 ◦ b (u) = F b0 ◦ F (b)(u) = F b0 u0 = u

Apply universal property on u itself, there exists a unique h : R → R such that F (h)(u) = u. First observe that idF(R) (u) = u, and F (idR) = idF(R), h must be idR. On the other hand, 0 0 F (b ◦ b )(u) = u, which implies b ◦ b = h = idR. The other identity is similar.

11 2.3 Contravariant functors

Deﬁnition 2.12. A contravariant functor C assigns to each set S a C (S) and to each f : S → T a function C (f): C (T ) → C (S) such that

• C (idS) = idC(S) • If f : S → T, g : R → S, then C (f ◦ g) = C (g) ◦ C (f)

Example 2.13 (Contravariant powerset functor). Deﬁne C as follows: S 7→ P (S), f 7→ f∗ by f∗ : P (T ) → P (S), for each set B ⊆ T , f∗ (B) = {a ∈ S : f (a) ∈ B}, i.e., the preimage pf B under f.

Exercise 2.14. Check it’s a contravariant functor.

Example 2.15. Fix a set A, the following is contravariant: S 7→ AS, f 7→ Af , i.e., f : S → T,Af : AT → AS, and Af (g) = g ◦ f.

Deﬁnition 2.16. Let C be a contravariant functor, a universal element for C is a pair (R, u), where R is a set and u ∈ C (R), such that for any S and s ∈ C (S), there exists a unique h : S → R such that C (h)(u) = s. Universal element for contravariant functors is very similar to universal element for covariant functors, except for the function h, now its domain is S and codomain is R.

S S Example 2.17. Fix sets A, B, deﬁne a contravariant functor C as follows: C (S) := A × B , for f : S → T , deﬁne C (f): C (T ) → C (S) by C (f)(g, h) := (g ◦ f, h ◦ f). R R What is the universal element? We want (R, u), u ∈ C (R), so u = (p, q) ∈ A × B , for any S and s = (g, h) ∈ S, there exists a unique f : S → R such that s = C (f)(p, q) = (p ◦ f, q ◦ f) = (g, h). Set R = A × B, p (a, b) = a, q (a, b) = b, and f (x) = (g (x) , h (x)). It’s not hard to check this is a universal element.

Similar to covariant functor, contravariant functor also has a uniqueness theorem for universal element.

0 0 Theorem 2.18. If (R, u) , (R , u ) are universal elements for contravariant functor C, then there 0 0 0 0 exists a unique bijection b : R → R such that u = C (b )(u).

12 3 Groups

In this section, we study group, one of the most famous algebraic structures, with a simple deﬁnition but surprisingly complicated dynamics.

3.1 Concrete groups Deﬁnition 3.1. A concrete group on a set X is a set G of bijections X :→ X, such that

1. idX ∈ G 2. if f ∈ G, then f −1 ∈ G 3. if f, g ∈ G, then f ◦ g ∈ G Example 3.2. 1. The set f ∈ XX : f is a bijection is a concrete group on X, we call it symmetric group on X, denoted by Sym (X). The elements of Sym (X) are called permutations.

2. {idX } is a concrete group.

3. I = {f ∈ Sym (R): ∀x, y ∈ R, x < y ⇒ f (x) < f (y)} is a concrete group on R.

(a) idR ∈ I −1 −1 −1 (b) f ∈ I ⇒ f ∈ I. Take x, y ∈ R with x < y, our goal is to show that f (x) < f (y). Suppose not, then f −1 (x) > f −1 (y), but f is increasing, so f f −1 (x) > f f −1 (y) ⇒ x > y, contradicts x < y. (c) f, g ∈ I ⇒ f ◦ g ∈ I, take x, y ∈ R with x < y, then g (x) < g (y), so f (g (x)) < f (g (y)). n 4. Let X ⊆ R , a bijection f : X → X is an isometry if for all x, y ∈ X, ||x − y||2 = ||f (x) − f (y) ||2. Iso (X) := {f ∈ Sym (X): f is an isometry} is a concrete group. n −1 5. Let X ∈ R , a bijection f : X → X is a homeomorphism if f and f are continuous. Homeo (X) := {f ∈ Sym (X): f is a homeomorphism} is a concrete group on X. Exercise 3.3. Show that {f ∈ Sym (X): f is continuous} might not be a concrete group.

6. Graphs: Let G = (V,E), an automorphism of G is a bijection f : V → V , such that for any x, y ∈ V with {x, y} ∈ E ⇔ {f (x) , f (y)} ∈ E. Aut (G) := {f ∈ Sym (V ): f is an automorphism} is a concrete group on V . Below are several graphs illustrated this idea:

13 The first graph is just a long path, there are only two automorphisms, namely, the identity and flip the entire path, notice no matter the length of this path, there are only these two kinds of automorphisms. The second graph is a triangle, and Aut (G) = Sym ({0, 1, 2}). Why? No matter how we per- mute the three vertices, it remains an automorphism, thus, the automorphisms are exactly all permutations. The last one is a 4-cycle, and |Aut (G) | = 8. To do the counting, we can first put vertex 0: 4 positions to put. Then there are 2 ways to put 1. After that, everything is fixed, so in total 4 × 2 = 8 automorphisms. Two quick definitions: Cn := cycle of length n, and D2n := Cn. Perhaps another name for D2n is much more famous: it is the dihedral group of order 2n, and |Aut (G) | = 2n, we typically use |D2n| to denote this number. 7. Let G be a concrete group on a set X, a function f ∈ Sym (G) is an automorphism of G if for any g, h ∈ G, f (g ◦ h) = f (g) ◦ f (h). It is not hard to show that if f is an automorphism of G, then f (idX ) = idX , therefore, Aut (G) := {f ∈ Sym (G): f is an automorphism of G} is a concrete group on G.

n 0 8. Let X be a set and f ∈ Sym (X), for n ∈ N, we write f := f ◦ f ◦ ... ◦ f, f := idX , | {zn } f n+1 := f ◦ f n. Also, let f −n := f −1n.

Exercise 3.4. For any n, m ∈ Z, show that: (a) f −n = (f n)−1 (b)( f n)m = f nm (c) f n ◦ f m = f n+m

n From this, it follows that hfi := {f : n ∈ Z} is a concrete group on X, this group is called the cyclic group generated by f.

Now let’s dive deeper into the study of dihedral group. It is useful to keep the following picture of dihedral group D2n in mind:

D2n is just all automorphisms on the cycle of length n, and for n ≥ 3, we have |D2n| = 2n. Consider the following 2 elements of D2n: • g :=

14 i.e., g rotates the cycle to the left, by 1.

• h :=

i.e., h ﬂips the cycle along the ﬁxed 0 axis.

0 1 2 n−1 0 1 2 Proposition 3.5. Elements of D2n are id = g , g , g , . . . , g , and h = g ◦ h, g ◦ h, g ◦ h, . . . , gn−1 ◦ h.

k k 0 −k Proof. Take any f ∈ D2n, let k := f (0), we wish to express f either as g or g ◦h, let f := g ◦f, clearly, f 0 (0) = g−k ◦ f (0) = g−k (k) = 0, so f 0 is either id or h. Moreover, we can write f as gk ◦ f 0, this completes the proof.

Exercise 3.6. For each k, `, compute gk ◦ h` (0) and gk ◦ h` (1), and conclude that 2n functions listed in the claim are distinct.

Proof Sketch. For different k, gk ◦ h` (0) must be different. For fixed k, gk (1) 6= gk ◦ h (1).

Observe that D2n is the smallest group containing g, h. Indeed, if G is a group containing g, h, k ` then for any k, `, g ◦ h ∈ G, so D2n ⊆ G. We say that D2n is generated by g, h. Lemma 3.7. Let X be a set, G be a set of concrete groups of G, deﬁne T G := {f ∈ Sym (X): ∀G ∈ G, f ∈ G}. Then T G is also a concrete group. T Proof. 1. idX ∈ G, since for any G, idX ∈ G. 2. if f ∈ T G, then f −1 ∈ T G, to see this, notice that for any G, f ∈ G, and G is a concrete group, so for any G, f −1 ∈ G ⇒ f −1 ∈ T G.

3. if f, g ∈ T G, then apply the same argument, f ◦ g ∈ T G.

15 Deﬁnition 3.8. If S ⊆ Sym (X), then hSi := T {G : G is a concrete group on X such that S ⊆ G} is a concrete group by 3.7, furthermore, S ⊆ hSi, so hSi is the smallest group containing S as a subset. hSi is called the group generated by S.

For example, if f ∈ Sym (X), then hfi = h{f}i, and more generally, we write h{f1, f2, . . . , fn}i as hf1, f2, . . . , fni, e.g., with g, h ∈ D2n, we have D2n = hg, hi. The next proposition will show that, we can write elements of hSi in a more structured way.

Proposition 3.9. Let X be a set, S ⊆ Sym (X), then

1 2 k hSi = g1 ◦ g2 ◦ ... ◦ gk : k ∈ N, g1, . . . , gk ∈ S, 1, . . . , k ∈ {−1, 1} Proof. Let H denote the set on RHS, clearly, H ⊆ hSi, and also, S ⊆ H. So to argue for the other inclusion ,it suﬃces to show H is a group, then we will have hSi ⊆ H automatically.

• Take k = 0, we have g0 = idX ∈ H.

1 k −1 −k −k−1 −2 −1 • Suppose f ∈ H = g1 ◦ . . . gk , then f = gk ◦ gk−1 ◦ . . . g2 ◦ g1 ∈ H. • If f, g ∈ H, then it’s clear f ◦ g ∈ H: we just express it by adding a ◦ between the representation of f and g.

Let’s look at some examples of groups generated by a set of elements.

Example 3.10. 1. Rubik’s cube: let X be a set of cardinality 48, and S := set of 6 individual rotations, viewed as permutations on X. The Rubik’s group is hSi.

2. The lamplighter group is deﬁned as follows: X = {0, 1}Z × Z, and there are two functions: • g : move to the next light, i.e., g (x, n) := (x, n + 1). • h : switch the light the person is currently standing by, i.e., h (x, n) := (x0, n), where ( 0 1 − xi, if i = n xi = xi, otherwise

We use hg, hi to denote the lamplighter group.

16 3.2 Abstract groups Deﬁnition 3.11. A binary operation on a set S is a function ? : S × S → S.

Example 3.12. 1. If G is a concrete group on X, then ◦ : G × G → G is a binary operation, where ◦ (f, g) = f ◦ g. Note: ? is always written as an inﬁx operator.

2. There are tons of binary operations on Z: + on Z, ∗ on Z, and Z × Z → Z :(n, m) 7→ n. Deﬁnition 3.13. A binary operation ? : S × S → S is associative if ∀a, b, c ∈ S :(a ? b) ? c = a ? (b ? c).

Deﬁnition 3.14. Let ? be a binary operation on S, then an element e ∈ S is an identity for ? if for all a ∈ S, a ? e = e ? a = a.

Lemma 3.15. Let ? be a binary operation on S, then ? has at most one identity.

Proof. Suppose both e and e0 are identities for ?, then e ? e0 = e = e0.

Deﬁnition 3.16. Let ? be a binary operation with identity element e on S, an inverse of an element a is an element b ∈ S such that a ? b = b ? a = e.

Exercise 3.17. Given an example of a binary operation ? with identity such that there is an element with more than one inverse.

You can intentionally assign two elements as inverse to some element, then force those two elements to be diﬀerent from each other.

Lemma 3.18. If ? is associative, with identity e, then for any a ∈ S, if it has inverse, then it has at most one inverse.

Proof. Suppose b, b0 are inverses of a, then compute b ? a ? b0 = (b ? a) ? b0 = e ? b0 = b0, on the other hand, b ? a ? b0 = b ? (a ? b0) = b ? e = b, by associativity, we have b = b0.

After setting up enough deﬁnitions, we are ready to deﬁne abstract groups.

Deﬁnition 3.19. An abstract group, or group for short, is a set G with a binary operation ? such that

• ? is associative.

• ? has an identity.

• each a ∈ G has an inverse with respect to ?.

The main example: if G is a concrete group on X, then (G, ◦) is a group. Some quick examples: (N, +) is not a group, but (Z, +) is a group. (R, ∗) is not a group, but (R \{0} , ∗) is a group. Exercise 3.20. Let X be a set, recall 4 is the symmetric diﬀerence operation on two sets, then (P (X) , 4) is a group.

Some more examples of groups:

17 Example 3.21. 1. Z/nZ: let n ≥ 2 be an integer, and Zn = {0, 1, . . . , n − 1}, +n : addition modulo n, i.e., for x, y ∈ Zn, deﬁne x +n y to be the remainder of x + y after divided by n. Then, (Zn, +n) is a group. Please verify this. The nasty part is to verify that +n is associative.

2 2. Elliptic curves: An elliptic curve is the set of points (x, y) ∈ R , satisfying an equation of the form y2 = x3 + ax + b, for example, y2 = x3 − x + 1. We denote the set by E. Let’s illustrate this with diagram:

Let ∗ be an extra element not in E, define an operation ⊕ on E ∪{∗} as follows: P ⊕Q ⇒ take tangent of the line PQ, find its third intersection with E, then flip that point with respect to x axis. What if PQ intersects E only on P and Q? For example, consider

Then deﬁne P ⊕ Q = ∗. For all P ∈ E, deﬁne ∗ ⊕ P = P ⊕ ∗ = P , and ∗ ⊕ ∗ = ∗, i.e., ∗ is the identity element. This turns E ∪ {∗} into a group.

3. Let G be a connected graph, ﬁx a vertex x ∈ V (G), deﬁne Π1 (G, x), the homotopy group of G, as follows:

Π1 (G, x) := {all non-backtracking closed walks in G, start and end at x}

Here, non-backtracking just means if you have used edge (u, v), you cannot immediately take the edge (v, u). Given two walks u and w, u · w is the walk obtained from u and w by concatenating them and removing all backtrack edges. For example,

18 This deﬁnes a group.

Theorem 3.22 (Cayley). Every abstract group is isomorphic to a concrete group, i.e., for all groups (G, ?), there exists a set X, a concrete group H on X, and a bijection ϕ : G → H, such that for all f, g ∈ G, we have ϕ (f ? g) = ϕ (f) ◦ ϕ (g).

Proof. Take X = G. For each g ∈ G, let ϕ (g): G → G be the function given by ϕ (g)(h) := g ? h. We prove a sequence of propositions, to show that this ϕ gives us the desired result. Proposition 3.23 (Closure of ϕ). For all f, g ∈ G, ϕ (f ? g) = ϕ (f) ◦ ϕ (g).

Proof of Closure. Take h ∈ G. ϕ (f ? g)(h) = (f ? g) ? h, while ϕ (f) ◦ ϕ (g)(h) = ϕ (f)(g ? h) = f ? (g ? h). By associativity of ?, the identity holds.

The next part will focus on proving ϕ is an isomorphism, i.e., it is a bijection. By 3.23, we need to show the following:

Proposition 3.24. Let e be the identity of G, then ϕ (e) = idG.

Proof of identity. Take h ∈ G, ϕ (e)(h) = e ? h = h, for all h, so ϕ (e) = idG. The next proposition will be useful in ﬁnding our concrete group H. Proposition 3.25. For all g ∈ G, ϕ (g) is a bijection.

Proof of ϕ (g) bijectivity. We claim that the inverse of ϕ (g) is ϕ g−1. −1 −1 ϕ (g) ◦ ϕ g = ϕ g ? g = ϕ (e) = idG −1 −1 ϕ g ◦ ϕ (g) = ϕ g ? g = ϕ (e) = idG So we can view ϕ as a function from G to Sym (G), where H := im (ϕ) = {ϕ (g): g ∈ G}. Proposition 3.26. H is a concrete group on G.

Proof of H is a concrete group. • idG ∈ H, since e ∈ G, ϕ (e) ∈ H. • If f ∈ H, then f −1 ∈ H: suppose f = ϕ (g) for some g ∈ G, then f −1 = ϕ g−1 ∈ H.

• If f1, f2 ∈ H, then f1 ◦ f2 ∈ H: since ϕ (g1) ◦ ϕ (g2) = ϕ (g1 ? g2) ∈ H.

Finally, we are ready to show ϕ is a bijection. Proposition 3.27. ϕ : G → H is a bijection.

Proof. Surjectivity comes from deﬁnition of H: it is im (ϕ). What about injectivity? If ϕ (g1) = ϕ (g2), then ϕ (g1)(e) = g1 ? e = g1 = ϕ (g2)(e) = g2 ? e = g2.

19 This completes the proof of Cayley’s theorem.

Let’s examine 3.22 on Dihedral group. Recall D2n = hg, hi, where g is rotate by 1, and h is ﬂip with respect to 0 axis. What is ϕ (g)? If we apply it to other elements, the result will be further shifted by 1. What about ϕ (h)? It will ﬂip all elements.

20 3.3 Abelian groups, subgroups, homomorphisms In this section, we extend more on structures of abstract groups. We introduce three important concepts, that will be used across all later sections. We ﬁrst introduce some conventions, to simplify the notation. Often, we just say “G is a group” without mentioning the operation. In such case, the operation is called “multiplication” and denoted by a · b or simply, ab. There’s a notable exception, i.e., groups with addition, such as (Z, +). Deﬁnition 3.28. A binary operation ? on set S is commutative if order does not matter, i.e., a ? b = b ? a, for any a, b ∈ S.

Deﬁnition 3.29. If (G, ?) is a group and ? is commutative, we say that (G, ?) is an abelian group. In this case, often the operation is denoted by “+”. We write −x for x−1, and 0 as identity. Also, nx is used to denote xn.

An exception to the above convention: (Q \{0} , ·) is abelian. Deﬁnition 3.30. Let (G, ?), (H, ·) be groups, a function ϕ : G → H is a homomorphism if ϕ (g ? h) = ϕ (g) · ϕ (h).

Exercise 3.31. If ϕ : G → H is a homomorphism, where G, H are groups with identities eG, eH , −1 −1 then ϕ (eG) = eH , and for all g ∈ G, ϕ g = (ϕ (g)) .

It’s not hard to see that, for any g ∈ G, ϕ (g) = ϕ (g · eG) = ϕ (g) · ϕ (eG), so ϕ (eG) = eH .A similar argument can be used to show the inverse part. Given the deﬁnition of homomorphism, it is natural to ask: if we have two groups G, H, does there exist a homomorphism from G to H? The answer is yes. Simply map every element of G to eH , then this is a valid homomorphism. Remarkably, one can show that this is the only valid homomorphism between certain groups.

Exercise 3.32. Consider (Zn, +) as a group, show that there is no non-trivial homomorphism from Z4 to Z7. We have seen the deﬁnition of isomorphism between two sets before, here we extend it to groups.

Deﬁnition 3.33. A bijective homomorphism from G to H is called an isomorphism. If an iso- morphsim from G to H exists, we say that G and H are isomorphic, denoted by G =∼ H.

Exercise 3.34. Show that the inverse of an isomorphism from G to H is an isomorphism from H to G.

Recall that in 3.22, our argument has showed that the map from G to Sym (G) by g 7→ (h 7→ gh) is an injective homomorphism. We abstract this intuition further.

Deﬁnition 3.35. Let (G, ?) be a group. Given a subset S ⊆ G, deﬁne ?S : S × S → G by a ?S b = a ? b. If ?S is a binary operation on S, i.e., im (?S) ⊆ S, and (S, ?S) is a group, then (S, ?S) or S is called a subgroup of G, denoted by S ≤ G.

Lemma 3.36. Let G be a group, S ⊆ G, then S ≤ G if and only if the following conditions hold:

1. e ∈ S

21 2. if g ∈ S, then g−1 ∈ S

3. if g, h ∈ S, then gh ∈ S

In particular, a concrete group on set X is the same as a subgroup of Sym (X).

Proof. (⇐) : By the three conditions, S is a group, and S is closed under group operation, so S ≤ G. (⇒) : Suppose S ≤ G, condition 3 holds by deﬁnition, it suﬃces to show the other two conditions.

2 1. We need to show eS = e. The ﬁrst observation is, for any element g ∈ G, if g = g, then 2 −1 2 g = e. Why? Given g = g, we can multiply both sides by g , get g = e. Indeed, eS = eS, so eS = e. 2. We need to show that, if g ∈ S and h is the inverse of g in S, then h = g−1. Indeed, −1 gh = hg = eS = e, and h = g , as desired. Proposition 3.37. Suppose ϕ : G → H is a homomorphism, then im (ϕ) ≤ H.

Proof. It suﬃces to check three conditions in 3.36.

• Since ϕ (eG) = eH , we have eH ∈ im (ϕ). • Suppose h ∈ im (ϕ), then there exists some g ∈ G such that ϕ (g) = h. Since (ϕ (g))−1 = ϕ g−1, h−1 ∈ im (ϕ).

• Suppose h1, h2 ∈ im (ϕ), then there exists g1, g2 ∈ G being mapped to them by ϕ. Since ϕ is a homomorphism, we have h1h2 = ϕ (g1g2), so h1h2 ∈ im (ϕ).

Observe that if ϕ is injective, then trivially, we have G =∼ im (ϕ).

22 3.4 Actions, cosets, orbits Groups are interesting structures with nice property we can leverage. Can we extend it to arbitrary set? We do so by deﬁning action.

Deﬁnition 3.38. Let G be a group, an action on a set X is a homorphism α : G → Sym (X).

For notations, we usually denote domain and codomain of α as α : G y X, and often, instead of α (g)(x), we write g ·α x or g · x if α is clear from context.

Example 3.39. 1. Consider α : G y G given by g · h = gh, this is called the left multiplication action.

−1 2. α : G y G given by g · h = hg , this is called the right multiplication action. Let’s verify it −1 −1 −1 −1 is an action. Suppose g1, g2, h ∈ G, then (g1g2) · h = h (g1g2) = hg2 g1 = g1 · hg2 = g1 · g2 · h. Moreover, the map G → G : h 7→ hg−1 is a bijection, we left to reader to check this.

3. The trivial action: G y X : g · x = x, for all g ∈ G, x ∈ X. Deﬁnition 3.40. A right action of G on X is a function β : G → Sym (X), such that for all g1, g2 ∈ G, β (g1g2) = β (g2) ◦ β (g1). We usually write it as β : X x G, and x ·β g or x · g for β (g)(x). With this notation, the requirement of β can be written as (x · g1) · g2 = x · (g1g2). So we can deﬁne right multiplication action via h · g = hg, also, h · g = g−1h.

Definition 3.41. The shift action σ : G y P (G), defined as g · S := gS := {gh : h ∈ S}. ∼ G G Notice that P (G) = {0, 1} , so we can view this action as G y {0, 1} . More generally, there G G G is a shift action G y X for any set G: take g ∈ G, f ∈ X , then g · f ∈ X is a function G → X, given by (g · f)(h) := f g−1h, for all h ∈ G. It is instrumental for the reader to verify this is in fact an action. Y Y Even more generally, given any action G y Y , we can “lift” it to an action G y X : for f ∈ X , define g · f : y 7→ f g−1 · y. An important type of action to study is actions of (Z, +). What does this kind of actions look like? n Let α : Z y X, then α (0) = idX , suppose α (1) := f ∈ Sym (X), then α (n) = f , for all n ∈ Z. This is true, since we can view Z as generated by 1, since α is a homomorphism, it must preserve such a structure. Thus, there is a natural bijection between {actions of Z y X} and Sym (X), given by α 7→ α (1).

3 4   5 Example 3.42. Consider α : Z y [6], given by α (1) =  . We write in a vector form, since 2   1 6 essentially, Sym ([6]) are permutations, so we just need to specify where does each number go. The

23 following diagram illustrates this action:

Deﬁnition 3.43. Let α : G y X, x ∈ X. The orbit of x under α is the set G ·α x = G · x := {g · x : g ∈ G}.

Deﬁnition 3.44. The orbit equivalence relation, Eα, is the binary relation on set X, given by xEαy ⇔ y ∈ G ·α x. Exercise 3.45. Verify this is an equivalence relation.

Exercise 3.46. For actions of (Z, +), orbits are the following type: single element orbit, two elements orbit, three elements orbit, etc., and another type: a long chain that does not end.

The next example introduces an important type of action, which plays a major role in a bunch of concepts in abstract groups.

−1 Example 3.47. The conjugation action, αconj : G y G : g · h := ghg . Let’s verify it −1 −1 −1 −1 is a homomorphism: αconj (g1g2)(h) = g1g2h (g1g2) = g1g2hg2 g1 = g1αconj (g2)(h) g1 = αconj (g1) ◦ αconj (g2)(h). Recall that Aut (G) = {automorphisms of G} ≤ Sym (G), and deﬁnition of automorphism can be found in7, i.e., functions that preserve compositions of elements. We can eﬀectively view αconj : G → Aut (G). What we need to argue is: for all g ∈ G, αconj (g): G → G is a homomorphism, i.e., for any h1, h2 ∈ G, αconj (g)(h1h2) = αconj (g)(h1) αconj (g)(h2).

−1 αconj (g)(h1h2) = gh1h2g −1 −1 = gh1g gh2g −1 −1 = gh1g gh2g

= αconj (g)(h1) αconj (g)(h2)

Let’s examine conjugation action on dihedral group, D2n. What is the orbit of h under conjugation? Clearly, hhh = h, and for any rotation gk, gkhg−k is as follows: first rotate 0 to k, and n − k to 0, then flips it with respect to fixed 0 axis, which results in k goes to n − k, which can be viewed as the original 0 goes to n − k. Finally, g−k rotates n − k to n − 2k and k to 0. It is a bit abstract to talk in this way, the following diagram illustrates the behavior:

24 One can view this as ﬂip with respect to n − k axis. Similar to Eα, we use Econj to denote the orbit equivalence relation under conjugation action, i.e., g and h are conjugate if and only if gEconjh. A quick observation on abelian group: G is abelian if and only if the conjugation action is trivial. For forward implication, notice that α (g)(h) = ghg−1 = gg−1h = h. For converse, ghg−1 = h → gh = hg.

Deﬁnition 3.48. Let Sub (G) := {H ⊆ G : H ≤ G}, instead of conjugating elements, we can extend the deﬁnition by conjugating the entire subgroup, i.e., αconj : G y Sub (G) by g · H = gHg−1 := ghg−1 : h ∈ H . Moreover, g · H is actually a subgroup of G, so we can further extend Econj := the orbit equivalence relation on Sub (G). We say two subgroups H1,H2 are conjugate, if −1 H1EconjH2 ⇒ gH1g = H2, for some g ∈ G. This even gives an isomorphism between this two ∼ subgroups, the mapping is h1 7→ gh1g−1, so H1EconjH2 ⇒ H1 = H2.

Deﬁnition 3.49. A subgroup H ≤ G is normal if G ·conj H = {H}, notation: H E G. Example 3.50. 1. {e}, the trivial subgroup.

2. If G is abelian, then every subgroup is normal.

3. G = D2n, the dihedral group. hhi = {h, e} is not a normal subgroup of D2n, deﬁne hi := ﬂip i with respect to axis i, then h is conjugate to hi by g . So hhi is conjugate to hhii, but they are not equal if i 6= 0. k On the other hand, hgi E D2n is a normal subgroup: any g act on it results in the same group, and h · gi = gn−i.

Deﬁnition 3.51. Let α : G y X, β : G y Y , a homomorphism from α to β is a function ϕ : X → Y such that for any g ∈ G, x ∈ X, we have ϕ (g ·α x) = g ·β ϕ (x). Another term to describe ϕ is, it is G-equivalent. A diagram:

25 α(g) X X

ϕ ϕ

β(g) Y Y If ϕ is bijective, then ϕ is an isomorphism. 3 4   Example 3.52. Consider the following two Z actions on [5], i.e., permutations on [5]: 5 and   2 1 2 1   4, these two Z actions are isomorphic. How to ﬁgure it out? Notice in the ﬁrst action, there are   5 3 two orbits: 1 − 3 − 5 and 2 − 4, while in the second action, there are also two orbits: 1 − 2 and 3 − 4 − 5, so we can build a bijection between this two orbits. More generally, α, β are isomorphic ⇔ there exists a bijection b : X/Eα → Y/Eβ, such that the type of each orbit in O ∈ X/Eα is the same type, or size of b (O), i.e., |O| = |b (O) |. Generalize this idea further, instead of saying the size of the orbit is the same, we formalize it using action: if there exists a bijection b : X/Eα → Y/Eβ such that for each orbit O ∈ X/Eα, the induced actions G y O and G y b (O) are isomorphic. This helps ti reduce the problem to studying actions with just one orbit, and such actions are called transitive.

In the next part, it is useful to recall 1.31, the canonical decomposition. Let α : G y X be an action, pick x ∈ X, deﬁne fx : G → X : g 7→ g · x. Apply 1.31 on fx, we get the following diagram: =∼ G X/Efx im (fx)

Where Efx is the equivalence kernel of fx: 1.26, i.e., g1Efx g2 ⇔ fx (g1) = fx (g2) ⇔ g1 ·x = g2 ·x ⇔ −1 g2 g1 · x = x. Let’s leverage this special structure by deﬁning stabilizer of x. Deﬁnition 3.53. The stabilizer of x is Stab (x) := {g ∈ G : g · x = x}.

Proposition 3.54. Stab (x) ≤ G.

Proof. Let’s check three properties in 3.36.

• First, e · x = x. To see this, notice any action is an homomorphism, so α (e) = idX , therefore, c · x = x.

• If g · x = x, then by property of homomorphism, g · g−1 · x = g−1 · g · x = g−1 · x = x, so g−1 ∈ Stab (x).

26 • If g, h are all ﬁx points, then (gh) · x = g · h · x = g · x = x, so gh ∈ Stab (x).

Deﬁnition 3.55. Let H ≤ G, the left coset equivalence relation of H, denoted by LCH on G, is −1 given by g1LCH g2 ⇔ g2 g1 ∈ H.

LCH is in fact an equivalence relation, and H is the orbit equivalence relation under the right action:G x H : g · h = gh. The orbits of this action are called left-cosets. We say that g1 ∈ −1 orbit of g2 ⇔ g2 g1 ∈ H ⇔ g1 ∈ g2H. Consider G/LCH = {gH : g ∈ G}, we use G/H as a shorthand notation for G/LCH . Similarly, right-cosets of H are the orbits of left action H y G by h · g = hg, notation: H\G = G/RCH = {Hg : g ∈ G}.

With this deﬁnition, we can write G/Efx as G/Stab (x). The next part in the diagram is im (fx). What is im (fx)? im (fx) = {fx (g): g ∈ G} = {g · x : g ∈ G} = G · x, i.e., orbit of x. So we can complete the diagram as follows: ∼ G X/Stab(x) = G · x

fx X This is the orbit-stabilizer theorem.

27 3.5 Orbit-stabilizer theorem

Theorem 3.56 (Orbit-Stabilizer Theorem). Let G be a group, G y X be an action, ﬁx x ∈ X, let fx : G → X : g 7→ g · x. The following diagram holds: ∼ G X/Stab(x) = G · x

fx X

Proof. Direct application of canonical decomposition theorem, with Efx = Stab (x) and im (fx) = G · x.

Let’s further explore this diagram. Suppose our action is transitive, i.e., there is only one orbit, then G · x = X. By orbit-stabilizer, it is isomorphic to G/Stab (x), i.e., G quotient a subgroup of G. This observation leads to the following theorem:

Theorem 3.57. Let α : G y X be a transitive action, then α is isomorphic to an action β : G y G/H, for a subgroup H ≤ G.

Proof. The action we consider is given by g 7→ (H 7→ gH), it is clearly an action: (g1g2) ·β H = g1g2H = g1 ·β g2H = g1 ·β g2 ·β H. ∼ Since α is transitive, G ·α x = X, and by 3.56, G ·α x = G/Stab (x). Let’s fix an x ∈ X, set H = Stab (x), define ϕ : G/H → X : gH 7→ g ·α x. We will show ϕ is an isomorphism by showing it’s a homomorphism and a bijection. But first of all, let’s show ϕ is a well-defined function.

• Suppose g1H = g2H, we need to show that g1 ·α x = g2 ·α x. Since g1H = g2H, we have −1 −1 g2 g1H = H, this means g2 g1 ∈ H. Why? Let’s prove a lemma for this. Lemma 3.58. Let H be a subgroup of G, and g ∈ G, then gH = H ⇒ g ∈ H.

Proof of Lemma. Let’s do the contrapositive: if g 6∈ H, then gH 6= H. This is clear: since H is a subgroup, e ∈ H, so ge = g ∈ gH, but g 6∈ H, so gH 6= H.

−1 −1 So g2 g1 ∈ H, recall that H = Stab (x), which means g2 g1 ·α x = x ⇒ g1 ·α x = g2 ·α x.

• Homomorphism: Let y ∈ X, our goal: ϕ (g1 ·β g2H) = g1 ·α ϕ (g2H). LHS is just ϕ (g1g2H) = (g1g2) ·α x, while RHS is g1 ·α g2 ·α x, since α is an action, the identity holds.

• Bijection: Define φ : X → G/H by y 7→ gH, where g ·α x = y. This is well-defined, since −1 −1 if g1 ·α x = g2 ·α x = y, then g2 g1 ·α x = x, which means g2 g1 ∈ Stab (x), so g1 ∈ g2H. Similarly, g2 ∈ g1H, this shows the mapping is well-defined. Moreover, it is clear that φ is the inverse of ϕ, so ϕ is a bijection.

Theorem 3.59. Given 2 subgroups H1,H2 ≤ G, the actions G y G/H1,G y G/H2 given by g 7→ (H 7→ gH) are isomorphic, if and only if H1,H2 are conjugate to each other. Proof. Exercise.

28 Combine 3.57 and 3.59, we obtain a 1-to-1 correspondence between

{isomorphism classes of transitive actions of G} and {conjugacy classes of subgroup of G}

Example 3.60. Consider G = (Z, +). In 3.46, we have studied orbits of (Z, +), now let’s establish connections between isomorphic classes of transitive actions or “the shape of orbits” and conjugacy classes of subgroups of Z.

One way to think of this is actions of Z is determined by α (1) = f, and corresponding subgroup we choose is Stab (x), i.e., the integer power k such that f k (x) = x. If the orbit only contains one k k element, then any f has the property that f (x) = x, so Stab (x) = Z. Similarly, if orbits contain 2 elements, then f 2k (x) = x, so all even powers work. If orbit is an inﬁnite chain, then any f k for k 6= 0 will not become idX , so only {0} works. Exercise 3.61. Recall: a group G is cyclic if G = hgi for some g ∈ G, i.e., G is generated by a single element. Show that every subgroup is cyclic.

For this exercise, pick the least positive integer power of g, prove it generates the entire subgroup. Next, we talk about some important applications and corollaries of orbit-stabilizer.

1. Let H ≤ G, H y G by left multiplication:h·g := hg. Take some g ∈ G, consider the diagram: ∼ H H/Stab(g) = H · g

G The orbit H ·g is the right-coset of H corresponding to g: Hg. Stab (g) = {h ∈ H : hg = g} = {e}, so we further have Hg =∼ H/Stab (g) =∼ H, i.e., every right-coset of H is in bijection with H.

29 Theorem 3.62 (Lagrange). If G is a ﬁnite group, H ≤ G, then |G| = |H||G/H|, where we usually use |G : H| to denote |G/H|, called the index of H in G.

Proof. By our above observation, each coset of H is in bijection with H, which means they all have cardinality |H|. Moreover, either left-coset or right-coset partitions G, and there are |G/H| or |HG| such cosets in total, so |G| = |H||G/H|.

Corollary 3.63. |H| | |G|.

Proof. By 3.62, |G| = |H||G/H|, and |G/H| is a positive integer.

For instance, if |G| is prime, then subgroups of G are either {e} or G.

Exercise 3.64. Show that the only group of size p a prime, up to isomorphism, is Zp.

One can show such group must be cyclic: suppose not, then you can construct another subgroup of size smaller than p, contradicts 3.62.

Corollary 3.65. Let G be a ﬁnite group, g ∈ G.

+ n (a) {n ∈ N : g = e}= 6 ∅ + n n (b) Let |g| := min {n ∈ N : g = e}, called the order of g, then g = e if and only if |g| | n. (c) |g| | |G|, i.e., g|G| = e.

Proof. (a) Since G is ﬁnite, we cannot keep generating new elements by keep powering g, + j i j−i so there must exists some positive i, j ∈ N with j > i, such that g = g ⇒ g = e. k (b)( ⇐) : If |g| | n, then n = k|g| for some positive integer k, so gn = gk|g| = g|g| = ek = e. (⇒) : Use division algorithm. Suppose |g| 6 |n, so n = k|g| + r, for some 0 < r < |g|, so gn = gk|g|+r = gk|g|gr = egr = gr = e, however, r < |g|, so contradicts the deﬁnition of order. (c) Consider the subgroup hgi, it has size |g|: every element in hgi can be expressed as gk, for 0 ≤ k < |g|. By 3.63, |g| | |G|, as desired.

Exercise 3.66. Let p be a prime number, consider G = (Zp, ·). (a) Show this is an abelian group. p (b) Conclude that for every a ∈ Z, a = a mod p, this is Fermat’s little theorem.

The ﬁrst part is a little bit painstaking to show. The second part is just apply 3.65 to G.

2. Let ϕ : G → H be a group homomorphism, it gives rise to an action αϕ : G y H, given by (αϕ (g)) (h) := ϕ (g)(h). Let eH be the identity of H, the orbit of eH is im (ϕ), since G ·α eH = {ϕ (g): g ∈ G}. Stab (eH ) = {g ∈ G : ϕ (g)(eH ) = eH } = {g ∈ G : ϕ (g) = eH } = preim (ϕ (eH )), we call it kernel of ϕ.

30 Theorem 3.67 (First Isomorphism Theorem). Let ϕ : G → H be a group homomorphism, then im (ϕ) =∼ G/ ker (ϕ).

Proof. Apply orbit-stabilizer, with element eH , so Stab (eH ) = ker (ϕ), and G · eH = im (ϕ).

31 3.6 Normal subgroups and kernels In this part we focus on normal subgroup and kernel of homomorphisms. Let ϕ : G → H be a homomorphism, recall 3.67, combine with 3.56, we have the following diagram: ∼ G G/ ker(ϕ) = im(ϕ)

H In this diagram, im (ϕ) is a subgroup of H, by the isomorphism between G/ ker (ϕ) and im (ϕ), we can “copy” the group structure and operation to G/ ker (ϕ). This makes all the arrows in the diagram group homomorphisms. Let K := ker (ϕ), what is the group structure of G/K? For each choice C1,C2 ∈ G/K, let’s do the most natural thing. Let C1 = g1K,C2 = g2K, then deﬁne C1C2 := (g1K)(g2K) = (g1g2) K. So the group operation on G/K is denoted by (g1K)(g2K) = (g1g2) K.

Theorem 3.68. Let K ≤ G, then there is a group operation on G/K deﬁned by (g1K)(g2K) = (g1g2) K if and only if K is a normal subgroup. Proof. (⇐) : Assume K is normal, we need to check that the group operation is well-deﬁned, i.e., 0 0 0 0 0 −1 0 if g1K = g1K, g2K = g2K, then (g1g2) K = (g1g2) K. Since g1K = g1K, we have g1 g1K = K, −1 0 0 −1 −1 0 0 −1 −1 so g1 g1 ∈ K, similarly, g2g2 ∈ K, and g1 g1g2g2 ∈ K. Since K is normal, K = g2Kg2 , −1 0 0 −1 −1 −1 −1 0 0 −1 0 0 so g1 g1g2g2 ∈ g2Kg2 ⇒ g2 g1 g1g2 ∈ K, which means (g1g2) (g1g2) ∈ K, therefore, −1 0 0 0 0 (g1g2) (g1g2) K = K ⇒ (g1g2) K = (g1g2) K, as desired. −1 (⇒) : Suppose G/K is a group, we want to show that K EG, this means for any g ∈ G, K = gKg . Take g ∈ G. 0 0 (⊆) : We can reuse some arguments in the converse part, i.e., suppose g1K = g1K, g2K = g2K, then 0 0 −1 0 0 −1 −1 0 0 −1 −1 0 0 (g1g2) K = (g1g2) K, this means that K = (g1g2) (g1g2) K = g2 g1 g1g2K, so g2 g1 g1g2 ∈ K, −1 0 0 −1 −1 −1 0 0 −1 −1 g1 g1g2g2 ∈ g2Kg2 . Moreover, if we have g1 g1 ∈ K, g2g2 ∈ K, then we show K ⊆ gKg . −1 −1 −1 −1 Now, let’s take g2 = g, for any k ∈ K, we can write k as e kgg , clearly, e k ∈ K, gg = e ∈ K, so we have k ∈ gKg−1. (⊇) : Since K ≤ G, K ⊆ gKg−1, we have Kg ⊆ gK ⇒ g−1Kg ⊆ K. Set h = g−1, so we have hKh−1 ⊆ K, as desired.

We abstract the argument in ⊇ part into a Lemma:

Lemma 3.69. If K ≤ G satisﬁes K ⊆ gKg−1, then K is normal.

In fact, if α : G y X and Y ⊆ X such that Y ⊆ g ·α Y , then g ·α Y = Y . Corollary 3.70. A subgroup K ≤ G is a kernel of some homomorphism if and only if K is normal in G. ∼ Proof. (⇒) : Let K = ker (ϕ) with ϕ : G → H, by 3.67, G/K = im (ϕ) as group. By 3.68, K E G. (⇐) : Suppose K is normal, by 3.68, G/K is a group and G G/K, the quotient map, is a homomorphim, we claim the kernel is K. What is the identity of G/K? It is K, since (gK)(eK) = (ge) K = gK. Moreover, what elements g ∈ G satisfy gK = K? Exactly the elements in K.

32 Let {∗} be an one-element trivial group, if K E G, consider the following sequence: {∗} K G G/K {∗} In this sequence, the image of every homomorphism is equal to the kernel of next.

• im ({∗} → K) = {eK }, ker (K,→ G) = {eK }

• im (K,→ G) = K, ker (G G/K) = K Deﬁnition 3.71. The sequence with above property is called exact sequence.

Exercise 3.72. Let K, G, H be groups, suppose we have a sequence of homomorphisms which is exact: {∗} K G H {∗} ∼ 0 ∼ 0 Show that K = K E G, H = G/K . Exercise 3.73 (important). Suppose ϕ : G → H is a homomorphism, then

1. ϕ is surjective ⇔ im (ϕ) = H

2. ϕ is injective ⇔ ker (ϕ) = {eG} Both parts are not hard to show, but the result is very useful.

Deﬁnition 3.74. A group G is simple if the only normal subgroups of G are G and {e}.

Exercise 3.75. (Zn, +n) is simple if and only if n is prime.

Notice (Zn, +n) is abelian, so every subgroup is normal. If n is not prime, then it has subgroups other than G and {e}, so it’s not simple.

33 3.7 Groups and counting In this section, we use structure of groups to facilitate counting, especially, we study the famous necklace problem.

Lemma 3.76 (Burnside). Let α : G y X be an action of a finite group G, on a finite set X. Then 1 X |X/Eα| = | {x ∈ X : g · x = x} | | {z } |G| | {z } number of orbits g∈G points fixed by g

Before proving lemma, let’s ﬁrst look at an example.

Example 3.77. Suppose α : G y X is an action such that for any x ∈ X, Stab (x) = {e}, such actions are called free. By orbit-stabilizer, every orbit is isomorphic to G/Stab (x) =∼ G, so |X| every orbit has size |G|, there are |G| orbits in total. For each g ∈ G, | {x ∈ X : g · x = x} | = ( 0, if g 6= e , so RHS is also |X| . |X|, if g = e |G|

Proof. Let’s do some calculations. X |X/Eα| = 1

O∈X/Eα X X 1 = |O| O∈X/Eα x∈O X 1 = orbits partition X |G · x| x∈X

By orbit-stabilizer, G · x =∼ G/Stab (x), so |G · x| = |G/Stab (x) | = |G|/|Stab (x) |, where the last equality is due to 3.62, so we can write |X/Eα| as:

X 1 |X/E | = α |G · x| x∈X X |Stab (x) | = |G| x∈X 1 X = |Stab (x) | |G| x∈X 1 X X = δ (g · x = x) δ (·) is the indicator function |G| x∈X g∈G 1 X X = δ (g · x = x) |G| g∈G x∈X 1 X = | {x ∈ X : g · x = x} | |G| g∈G

34 Example 3.78. How many diﬀerent types of necklaces with n gems of k kinds are there, if the necklaces are the same up to rotation? Let’s ﬁrst look at n = 4, k = 2:

Let G : all rotations, X : kn configurations of necklaces, so the problem reduces to count the number of orbits of the action G y X by rotating one configuration. In this setting, |G| = n: we can rotate one configuration clockwise up to n − 1 steps. |X| = kn, so in general, number of orbits = 1 P | {x ∈ X : s · x = x} |. Let’s examine our n = 4, k = 2 case. n s∈Zn • For s = 0, all 24 = 16 configurations are fixed.

• For s = 1, only 2 conﬁgurations, namely, all squares or all triangles are ﬁx points.

• For s = 2, diagonals must be the same, so 4 conﬁgurations.

• For s = 3, only 2 conﬁgurations.

1 By 3.76, the count is 4 (16 + 2 + 4 + 2) = 6. Exercise 3.79. Show that | {x ∈ X : s · x = x} | = kgcd(n,s).

Exercise 3.80. What is the count, if you can ﬂip the necklace?

For the second exercise, choose G = D2n.

Corollary 3.81 (Polya). Let G be a ﬁnite group, X,Y ﬁnite sets, and α : G y X be an action. For each g ∈ G, let c (g) be the number of cycles in the permutation α (g) on X, i.e., the number X of orbits in the action Z y X : 1 7→ α (g). The action α gives rises to an action G y Y by (g · f)(x) := f g−1 · x, i.e., the shift action. Then the number of orbits of this action is equal to 1 P c(g) |G| g∈G |Y | . Proof.

1 X number of orbits = | f ∈ Y X : g · f = f | |G| g∈G 1 X = | f ∈ Y X : f (x) = f g−1 · x , ∀x ∈ X | this means f is constant on cycles in α (g) |G| g∈G 1 X = |Y |c(g) |G| g∈G

35 4 Category

Category is a more abstract deﬁnition of “collections” of objects. Category is useful because the same deﬁnition can work in several categories.

4.1 Basic deﬁnitions & examples

Deﬁnition 4.1. A category C consists of • a class Obj (C) of objects

• for all A, B ∈ Obj (C), a set HomC (A, B), or simply Hom (A, B) of morphisms from A to B • for all A, B, C ∈ Obj (C), a function ◦ : Hom (B,C) × Hom (A, B) → Hom (A, C), called composition, such that:

1. composition is associative: Let A, B, C, D ∈ Obj (C), f ∈ Hom (A, B) , g ∈ Hom (B,C) , h ∈ Hom (C,D), then h ◦ (g ◦ f) = (h ◦ g) ◦ f

2. identities: For each A ∈ Obj (C), there is a 1A ∈ Hom (A, A), such that for all B ∈ Obj (C) , f ∈ Hom (A, B) , g ∈ Hom (B,A):

f ◦ 1A = f, 1A ◦ g = g

Notice for ﬁxed A, 1A is unique: duplicate the argument that proves identity of a binary operation is unique.

Example 4.2. 1. The most basic example is the category Set of sets:

• Obj (Set) = all sets

• HomSet (A, B) = functions from A to B •◦ = composition of functions

2. One can obtain diﬀerent categories by taking fewer morphisms:

(a) • Obj = all sets • Hom (A, B) = bijections from A to B •◦ = composition (b) • Obj = all sets ( ∅, if A 6⊆ B • Hom (A, B) = {the inclusion map A,→ B} if A ⊆ B •◦ = composition ( ∅, if n > m 3. Obj = Z, Hom (n, m) = one element set {fnm} , if n ≤ m For composition, we need ◦ : Hom (n, m) × Hom (k, n) → Hom (k, m).

• If k ≤ n ≤ m, fnm ◦ fkn := fkm.

36 • Otherwise, there’s nothing to deﬁne: simply ∅.

4. The category of graphs, Graph:

• Obj Graph = {all graphs} • Hom (G, H) = {all homomorphisms from G to H}. Let’s recall the deﬁnition of homomorphism of graphs: it is a function f : V (G) → V (H) such that if {x, y} ∈ E (G), then {f (x) , f (y)} ∈ E (H). •◦ = composition

5. The category of groups, Grp:

• Obj Grp = {all groups } • Hom (G, H) = {all group homomorphisms from G to H} •◦ = composition

37 4.2 Monomorphism, epimorphism In this section, we introduce some general concepts for diﬀerent categories, which are useful in understanding more about groups.

Deﬁnition 4.3. Let C be a category, and A, B ∈ Obj (C).

• f ∈ Hom (A, B) is a monomorphism if for all C ∈ Obj (C) and g, h ∈ Hom (C,A), f ◦ g = f ◦ h ⇒ g = h

• Similarly, f is an epimorphism if for all C ∈ Obj (C) and g, h ∈ Hom (B,C), g ◦ f = h ◦ f ⇒ g = h

In the category Set, it is not hard to show that

• monomorphism ⇔ injective

• epimorphism ⇔ surjective

What about in Grp? Clearly, injectivity implies monomorphism. What about monomorphism implies injectivity?

Proof. Suppose f : A → B is a group homomorphism, take a, a0 ∈ A with f (a) = f (a0). Then n 0 n there are homomorphisms g, h : Z → A given by g (n) := a , h (n) := (a ) . Let’s verify f and g, h satisfy the monomorphism condition:

(f ◦ g)(n) = f (g (n)) = f (an) = f (a)n since f is a homomorphism = f a0n = f a0n = f (h (n)) = (f ◦ h)(n)

So f ◦ g = f ◦ h, combined with f is a monomorphism, we have g = h ⇒ g (1) = h (1) ⇒ a = a0, therefore, f is injective.

So indeed, monomorphism in Grp ⇔ injective. Note: in this proof, we have used the property of integers: for every group A and a ∈ A. there is a unique homomorphism: Z → A sending 1 to a. What about epimorphisms? Surjectivity implies epimorphisms. How about the converse? In the category of Grp, the converse is true, however, it is a non-trivial task to show.

Theorem 4.4 (Schreier). If a group homomorphism ϕ : H → G is an epimorphism in the category Grp, then ϕ is surjective.

Notice it is not true in every category.

38 n Example 4.5. Consider the category of subsets of R , n ∈ N, with continuous functions as n morphisms. The embedding ι : Q ,→ R is, surprisingly an epimorphism. Given g, h : R → R as continuous functions, if g ◦ ι = h ◦ ι, since Q is dense in R, so a continuous function on R is determined by its values on rational points. Therefore, g = h, and ι is an epimorphism. But clearly, ι is not surjective. Proof of Schreier. We can ﬁrst decompose ϕ into the following diagram: H im (ϕ)

ϕ ι

G The claim is, if ϕ is an epimorphism, then so is ι. Let g, h : G → K be group homomorphisms, if g ◦ ι = h ◦ ι, then g ◦ ϕ = h ◦ ϕ: for any h ∈ H, let t = ϕ (h), since so g ◦ ι (t) = h ◦ ι (t) ⇒ g ◦ ι ◦ ϕ (h) = h ◦ ι ◦ ϕ (h) ⇒ g ◦ ϕ = h ◦ ϕ. So g = h since ϕ is an epimorphism. Therefore, it suffices to consider the case where H ≤ G, and ϕ : H,→ G is the inclusion map. Since ϕ is the inclusion map, ϕ is surjective if and only if H = G, and we will show this. Lemma 4.6. If H is normal in G, then H = G. Proof of Lemma. Consider the two homomorphisms from G to G/H: • the quotient homomorphism q : G G/H • the trivial homomorphism t : G → G/H : g 7→ H The claim is q ◦ ϕ = t ◦ ϕ, since for all h ∈ H, we have (q ◦ ϕ)(h) = q (h) = H, and (t ◦ ϕ)(h) = t (h) = H. Since ϕ is an epimorphism, we have q = t, i.e., the quotient map is nothing but the trivial map, this means G/H = {H}, i.e., G = H, as desired. Corollary 4.7. In such scenario, we can assume |G/H| ≥ 3. Proof of Corollary. If |G/H| = 1, then G = H. Otherwise, if |G/H| = 2, then there are exactly two cosets: H and gH, for some g ∈ G. Suppose H is not normal, i.e., there is some a ∈ G such that aHa−1 6= H, which means there is some h ∈ H, with aha−1 6∈ H. Clearly, a 6∈ H, so we can write a = g`, for some ` ∈ H, and aha−1 6∈ H implies that aha−1 = gh0, for some h0 ∈ H. Multiply both sides by g−1, we get g−1aha−1 = h0 ⇒ g−1g`h`−1g−1 = h0 ⇒ `h`−1g−1 = h0, since `, h, h0 ∈ H, it must be the case that g−1 ∈ H ⇒ g ∈ H, contradicts that gH and H partitions G. Thus, H must be normal. Towards a contradiction, let’s assume H 6= G, so H cannot be normal, by above corollary, it’s safe to say |G/H| ≥ 3. We will show that in this case, it’s possible to construct two homomorphisms α, β : G → Sym (G), such that α ◦ ϕ = β ◦ ϕ, but α 6= β, which contradicts ϕ is an epimorphism. Define α = left multiplication action, and β = α conjugated by “swapping” two of cosets of H. More specifically, since |G/H| ≥ 3, there are two elements u, v ∈ G such that H, Hu, Hv are all distinct. Let π : G → G be the bijection such that  hv, if g = hu for some h ∈ H  π (g) := hu, if g = hv for some h ∈ H  g, if g 6∈ Hu ∪ Hv

39 −1 Deﬁne an action β : G y G by β (g) := π ◦ α (g) ◦ π = π ◦ α (g) ◦ π. We need to check α, β agree on H. Take h, h0 ∈ H, then

• α (h)(h0) = hh0, β (h)(h0) = π ◦ α (h) ◦ π (h0) = π ◦ α (h)(h0) = π (hh0) = hh0

• α (h)(h0u) = hh0u, β (h)(h0u) = π ◦ α (h) ◦ π (h0u) = π ◦ α (h)(h0v) = π (hh0v) = hh0u

• α (h)(h0v) = hh0v, β (h)(h0v) = π ◦ α (h) ◦ π (h0v) = π ◦ α (h)(h0u) = π (hh0u) = hh0v

Thus, α|H = β|H ⇒ α ◦ ϕ = β ◦ ϕ, but α 6= β, in particular, α (u) 6= β (u): α (u)(e) = u, β (u)(e) = v. This completes the contradiction, and thus completes the proof of the theorem.

40 5 Advanced topics on Groups

In this section, we study more advanced topics on groups, including products & coproducts, free groups, and Sylow groups.

5.1 Product & coproduct of groups Before studying products & coproducts of groups, let’s first recall them on Sets. It is best to illustrate the definitions of products & coproducts with diagrams. The first is what we need for product of sets: p X × Y X

q f ∃!h

Y g A The second one is what we need for coproduct of sets, and coproduct is sometimes called disjoint union: X t Y X i

j ∃!h f

g Y A In Grp: Given groups G, H, we want a group G × H and homomorphisms p : G × H → G, q : G × H → H, such that p G × H G

q ϕ ∃!ξ

H F ψ As a set, G × H is just the Cartesian product of G and H, where operations are deﬁned as (g1, h1) · (g2, h2) := (g1g2, h1h2), i.e., pointwise multiplication. Deﬁne homomorphisms p, q as p (g, h) = g, q (g, h) = h. Given homomorphisms ϕ : F → G, ψ : F → H, there exists a unique ξ : F → G × H, such that p ◦ ξ = ϕ, q ◦ ξ = ψ, by ξ (f) = (ϕ (f) , ψ (f)).

Deﬁnition 5.1. G × H is called direct product of G and H.

Example 5.2. 1. G × trivial group =∼ G

2 2. Z × Z = Z ∼ 3. Z3 × Z7 = Z21 ∼ Exercise 5.3. Show that if n, m ≥ 2, gcd (m, n) = 1, then Zn × Zm = Znm. ∼ However, if gcd (m, n) 6= 1, we cannot say anything about Zn ×Zm: for example, Z2 ×Z2 =6 Z4, since Z4 is cyclic but the former is not.

41 Now let’s switch the gear to look at coproducts. In Grp, what we want is: G ∗ H G i ∃!ξ j ϕ

ψ H F Recall that in Sets: 2.8, we have shown that if im (i) ∩ im (j) = ∅, then we can simply take the union of G and H. However, im (i) ∩ im (j) 6= ∅, since i (eG) = j (eH ). The question is, is there any group K such that there are homomorphisms i : G → K, j : G → K, i, j are injective, and im (i) ∩ im (j) = {eK }? Consider K = G × H, and i : g 7→ (g, eH ) , j : h 7→ (eG, h). Proposition 5.4. In the category Ab, a.k.a. abelian groups, G × H is the coproduct of G and H.

Proof. Take abelian group F with homomorphisms ϕ : G → F, ψ : H → F , we want to show there exists a unique ξ : G × H → F such that the diagram holds. Simply deﬁne ξ (g, h) := ϕ (g) + ψ (h), this is a homomorphism, since

ξ ((g1, h1) + (g2, h2)) = ξ (g1 + g2, h1 + h2)

= ϕ (g1 + g2) + ψ (h1 + h2)

= ϕ (g1) + ϕ (g2) + ψ (h1) + ψ (h2)

= ϕ (g1) + ψ (h1) + ϕ (g2) + ψ (h2)

= ξ (g1, h1) + ξ (g2, h2)

We leave uniqueness part as an exercise to the reader.

So in abelian groups, we call coproduct direct sums. What about Grp? We need to introduce the notion of free groups.

42 5.2 Free groups What is free groups? We use a diagram to introduce the idea. Let X be a set and G be a group, then free group FX is the group that satisﬁes the following diagram: X FX

f ∃!ϕ

G ∼ n Let’s ﬁrst do the easy case, say X = {x}, then we claim FX = Z, and ϕ : Z → G : n 7→ (f (x)) , and for the map from {x} to FX , we simply map x to 1. Given such setting, ϕ is clearly unique, and is also a homomorphism. What if X = {x, y}, in such scenario, what should F{x,y} look like? To address this problem, we need to introduce the concept of reduced words.

Deﬁnition 5.5. Let X be a set (generators), deﬁne

X0 = x0 : x ∈ X is associated with x0, serve as inverse

The words over X ∪X0 is simply (X ∪ X0)∗, where A∗ is the kleene star, i.e., all words over alphabet A with ﬁnite length. The reduced words is a subset of words over X ∪X0, with no x and x0 adjacent to each other.

0 Definition 5.6. The free group over a set X, denoted by FX is defined as {all reduced words over X ∪ X }. The group operation is concatenation+reduction, i.e., for two reduced words w1, w2, we first con- catenate them, then cancel out all adjacent x and x0.

Exercise 5.7. Show that such operation is associative. In other words, the result of reducing a word is independent of orders of reductions.

For simplicity, we write x−1 for x0. Whenever G is a group, f : X → G, then there exists a unique homomorphism ϕ : FX → G that completes the diagram in 5.2: let w = x1x2 . . . xn be a reduced word, deﬁne ϕ (w) = −1 −1 f (x1) f (x2) . . . f (xn), and deﬁne f x := (f (x)) . What is the coproduct Z ∗ Z? ∗ ∼ Z Z = F{x,y} 17→x Z

∃!ξ 17→y 17→a

17→b Z G Since ϕ : Z → G, ψ : Z → G, we can abstract them as 1 7→ a, 1 7→ b, so ξ is deﬁned as ξ (x) = ϕ (1) = a, ξ (y) = ψ (1) = b.

n 2 Example 5.8. Consider the dihedral group, D2n = hg, hi, where g = e, h = e, hghg = e. Proposition 5.9. Suppose G is a group, x, y ∈ G such that xn = y2 = yxyx = e, then there exists a unique homomorphism: ϕ : D2n → G, such that ϕ (g) = x, ϕ (h) = y.

43 m ` m ` Proof. Since D2n is generated by g, h, every element of D2n is of the form g 1 h 1 . . . g k h k , where m ` m ` m ` m ` mi, `i ∈ Z. Then ϕ g 1 h 1 . . . g k h k = x 1 y 1 . . . x k y k , this proves uniqueness. m1 `1 mk `k We need to show ϕ is in fact a function, and it suﬃces to check if g h . . . g h = eD2n , then m ` m ` x 1 y 1 . . . x k y k = eG. We proceed by cleaning up the exponents. 1. Get rid of negative powers of g by gn−1 = g−1, xn−1 = x−1

2. Get rid of negative powers of h by h−1 = h, y−1 = y

3. hghg = e ⇒ ghg = h ⇒ hg = gn−1h, yx = xn−1y, so we can move all g in front of h, and all x in front of y

This results in a form of gmh`, where 0 ≤ m < n, 0 ≤ ` ≤ 1, and we have xmy` as well, since m ` 0 0 g h = eD2n , we must have m = ` = 0 ⇒ x y = eG, as desired.

If w = x1 . . . xn is a word, let f˜(w) := f (x1) . . . f (xn), so our diagram becomes: X FX

f ∃!f˜

Deﬁnition 5.10. Let R be a set of reduced words over X, i.e., R ⊆ FX , call the elements of R ˜ relations. Given a group G and f : X → G, say (G, f) satisﬁes R if f (w) = eG, for all w ∈ R.

Example 5.11. Let X = {x, y} ,G = D2n, f : X → D2n :(x 7→ g, y 7→ h), then (D2n, f) satisﬁes xn, y2, xyxy .

Note that, if (G, f) satisfies R and ϕ : G → H is a homomorphism, then (H, ϕ ◦ f) also satisfies ˜ R, since ϕ]◦ f (w) = ϕ f (w) = ϕ (eG) = eH . ∼ Definition 5.12. Let G be a group, X be a set, f : X → G, R ⊆ FX , we write (G, f) = hX | Ri, if 1.( G, f) satisfies R.

2. If there exists other (H, g) satisﬁes R, then there exists a unique homomorphism ϕ : G → H, such that g = ϕ ◦ f.

We call hX | Ri a presentation of G.

Exercise 5.13. If (G, f) , (H, g) =∼ hX | Ri, show that there exists a unique isomorphism ϕ : G → H, such that g = ϕ ◦ f.

As a hint, use the homomorphism given by deﬁnition of (G, f) =∼ hX | Ri, and show the homomorphism given by (H, g) =∼ hX | Ri is its inverse. ∼ Example 5.14. 1. FX = hX | ∅i = hX |i n n ∼ 2. h{x} | {x }i = hx | x i = Zn Exercise 5.15. If (G, f) =∼ hX | Ri, then G = h{f (x): x ∈ X}i.

44 3. As a shorthand notation, below we will write G =∼ hX | Ri to denote that there exists f : ∼ ∼ n 2 X → G such that (G, f) = hX | Ri. With this notation, we have D2n = x, y | x , y , xyxy . 4. Free abelian group over X: x | x−1y−1xy : x, y ∈ X . ∼ 5. Let G be an arbitrary group, X = G, idG : G → G, then (G, idG) = hG | RGi, where

RG = {all reduced words w ∈ FG that are satisﬁed in (G, idG), i.e., w = g1 . . . gn such that g1 . . . gn = eG }

Corollary 5.16. If G is finite, then G is finitely represented, i.e., it has a presentation with finitely many relations. ∼ Proof. Since G is finite, there are only finitely many elements that can be eG, using G = hG | RGi gives us the desired result.

Theorem 5.17. Let X be a set, R ⊆ FX , then there exists a group G with a function f : X → G such that (G, f) =∼ hX | Ri.

We need a Lemma/deﬁnition in order to prove this theorem. T Lemma 5.18. Let G be a group, R ⊆ G, then deﬁne N (R) := {H E G : R ⊆ H}, it is a normal subgroup of G, called the normal closure of R. N (R) is the smallest normal subgroup of G containing R.

Proof. Take g ∈ G, we want to show that N (R) is normal, i.e., gN (R) g−1 = N (R). Take h ∈ N (R), then ghg−1 ∈ N (R), since h is in every normal group that contains R. This proves gN (R) g−1 ⊆ N (R). Recall in 3.69, we have shown that one side of containment suﬃces to show a group is normal. Thus, N (R) is normal, as desired.

Exercise 5.19. Let G be a group, R ⊆ G, show that N (R) = grg−1 : g ∈ G, r ∈ R .

Lemma 5.20. Let G, H1,H2 be groups, ϕ1 : G → H1, ϕ2 : G → H2 be homomorphisms such that im (ϕ1) = H1 and ker (ϕ1) ⊆ ker (ϕ2), then there exists a unique homomorphism ψ : H1 → H2 such that ψ ◦ ϕ1 = ϕ2. ∼ Proof. Let K1 := ker (ϕ1) ,K2 := (ϕ2), by 3.67, im (ϕ1) = G/K1, since im (ϕ1) = H1, we can assume H1 = G/K1, so one can view ϕ1 as a homomorphism from G to G/K1, i.e., the quotient map. Deﬁne ψ : G/K1 → H2 by ψ (gK1) = ψ (ϕ1 (g)) = ϕ2 (g), we need to check that ψ is 0 0 well-deﬁned, i.e., if gK1 = g K1, then ϕ2 (g) = ϕ2 (g ). Notice that

0 gK1 = g K1 −1 0 ⇒ g g K1 = K1 −1 0 ⇒ g g ∈ K1 −1 0 ⇒ g g ∈ K2 since K1 ⊆ K2 −1 0 ⇒ ϕ2 g g = eH2 0 ⇒ ϕ2 (g) = ϕ2 g

45 Uniqueness is left as an exercise to the reader. This gives us the following diagram: ϕ1 ∼ G H1 = G/ ker (ϕ1)

ϕ2 ∃!ψ

Now we are ready to prove the theorem.

Proof of Theorem. Take G := FX /N (R), consider the following diagram: p X FX FX /N (R)

f f˜ ∃!ϕ

H where (H, f) satisfies R. What we need to show is two things: 1). G satisfies R, and 2). there exists a unique homomorphism from G to H. For the first part, notice that R ⊆ N (R), if we let ι denote the embedding of X into FX , then p ◦ ι will satisfy R. For the second part, we apply ˜ 5.20 to FX − FX /N (R) − H, where we need to check that p is surjective, and ker (p) ⊆ ker f . The first part is clear, since p is the quotient map, for the second part, notice ker (p) = N (R). By our assumption, (H, f) satisfies R, so R ⊆ ker f˜ , and due to 3.70, ker f˜ is normal, so ˜ N (R) ⊆ ker f . Thus, by 5.20, there exists a unique homomorphism ϕ : FX /N (R) → H, as desired. ∼ ∼ Example 5.21. Coproducts of groups. Let G1,G2 be groups, say G1 = hX1 | R1i ,G2 = hX2 | R2i, assume X1 ∩ X2 = ∅.

Proposition 5.22. hX1 ∪ X2 | R1 ∪ R2i = G1 ∗ G2 is a coproduct of G1 and G2, it is also called the free product.

Proof. Consider the following diagram: X1 X1 ∪ X2 X2

G1 G1 ∗ G2 G2

∃!

H By 5.2, H both satisﬁes R1 and R2, so there exists a unique homomorphism from G1 ∗G2 to H.

2 This has some interesting dynamics: consider Z2 ∗ Z2, we can represent the ﬁrst with x | x 2 2 2 and the second with y | y , by 5.22, we have Z2 ∗ Z2 = x, y | x , y , this is an inﬁnite group: we can repeatedly form xy, xyxy, xyxyxy, . . ., i.e., (xy)∗.

46 Example 5.23. Consider x, y | xn, y2, x−1y−1x2y , where the last relation can be interpreted as x2y = yx, by this relation, we can swap all x in front of y, so every element can be written xmy`, for 0 ≤ m < n, 0 ≤ ` ≤ 1. The size of this group is thus at most 2n. Let’s do some more computations.

x = y2x = yx2y = x2yxy = x4y2 = x4

3 n 3 ∼ So x = e. If gcd (n, 3) = 1, then x = x = e ⇒ x = e, the group is then {e, y} = Z2. Exercise 5.24. What if n is divisible by 3?

Example 5.25. A famous computational problem, the word problem: given a ﬁnite presentation hX | Ri and w ∈ FX , is w the identity in hX | Ri? Though seems the problem is not so hard to solve, but it is in fact undecidable.

Let’s try to understand more structures about free groups. ∼ ∼ ∼ F0 is the trivial group, F1 = Z, F2 = 2d tree, we know that F1 =6 F2, since F1 is abelian, but F2 is ∼ not. One can prove F3 =6 F2 via abelianization or some counting argument on homomorphisms, but it is not obvious. Moreover, F3 is isomorphic to a subgroup of F3, making the structure even more mysterious. Even more, there exists an injective homomorphism from F∞ to F2. Still, something can be said as “general property” for free groups.

Theorem 5.26 (Nelsen-Schreir). Every subgroup of a free group is free.

The type of argument we are gonna use is new to this course, it comes from a specific field called geometric group theory, one can view it as a general type of graphs, where multi-edges and self-loops are allowed, and for each edge, we think of it as two arcs with different orientations. We formally re-define the notion of graph we are gonna use.

Deﬁnition 5.27. A graph G consists of the followings:

• a set V of vertices

• a set E of edges

• a pair of functions, start, end : E → V

• a bijection flip : E → E, such that

– flip (flip (e)) = e – start (flip (e)) = end (e) – end (flip (e)) = start (e)

Deﬁnition 5.28. For x, y ∈ V , an x-y walk is a sequence of edges ekek−1 . . . e2e1(possibly empty), where:

47 • start (ei+1) = end (ei)

• start (e1) = x, end (ek) = y Given a graph G, we form a subset E+ ⊆ E by picking an arbitrary edge from each pair {e, flip (e)}. Let E− := E \ E+. For an xy-walk p and a yz-walk q, one can deﬁne an xz-walk qp by concatenating the sequences, this gives rise to a category: • Objects are vertices

• Morphisms from x to y: xy-walks

• Composition: concatenation of walks

Deﬁnition 5.29. A walk ekek−1 . . . e1 is reduced if there is no i such that ei+1 = flip (ei).

For every xy-walk p, if p is not reduced, say there exists some i with ei+1 = flip (ei), then we can apply a reduction operation to p, by removing both ei and ei+1 from the sequence. After ﬁnitely many reductions, we have a reduced walk, denoted by Red (p). Deﬁne an operation on reduced walk p from x to y, reduced walk q from y to z, by qp := Red (concat (qp)). Proposition 5.30. 1. Reduced walk Red (p) does not depend on the order of reductions taken.

2. If p is a reduced xy-walk, q is a reduced yz-walk, r is a reduced zw-walk, then (rq) p = r (qp).

+ − Proof. Consider the free group FE+ . Notice walks are directly relating to words over E ∪ E , and reduced walks are essentially a subset of reduced words. Concatenation with reduction of reduced walks coincides with multiplication in FE+ . By using the property of FE+ , we have completed the proof of the proposition.

This gives rise to a new category: • Objects are vertices

• Morphisms from x to y: reduced xy-walks

• Composition: free group multiplication In this category, every morphism has an inverse: let p be a walk from x to y, then flip all edges along p gives p−1, a walk from y to x. For x ∈ V , let π (x, G) be the group of all reduced xx-walks. Notice π (x, G) ≤ FE+ . Proposition 5.31. If G is connected, then π (x, G) =∼ π (y, G), for any x, y ∈ V . So we can write π (G) without mentioning any vertex, this group is called the fundamental group of G. Proof. Since G is connected, there exists a walk p from x to y. Define the isomorphism as follows: given an xx-walk q, maps it to pqp−1, i.e., first travels from y to x, take an xx-walk, then travel −1 −1 −1 back to y. To see this is a homomorphism, observe that pq1q2p = pq1p pq2p , and the inverse of this mapping is, given a yy-walk o, map it to p−1op.

Theorem 5.32. Let X be a set, FX be the free group on X, and H ≤ FX , then H = π (G) for some graph G.

48 Proof. Look at FX /H. Let α : FX y FX /H be the left multiplication action. Deﬁne a graph G as follows: V = FX /H, E = {(C, x): C ∈ FX /H, x ∈ X}, i.e., vertices are cosets of FX /H, and edges compromise two bits of information: the start of the edge, C, and the end point xC. So flip (C, x) = xC, x−1, flip C, x−1 = x−1C, x, start (C, x) = C, end (C, x) = xC. Now our goal is to show π (G) =∼ H. First, π (G) =∼ π (H,G), what are elements of π (H,G)? That’s −1 all reduced HH-walks, i.e., sequences (Ck, xk) ... (C2, x2)(C1, x1), where x1, . . . , xk ∈ X ∪ X , C1 = Ck = H, and Ci+1 = xiCi, so we can write the sequence as xkxk−1 . . . x2x1H = H ⇒ xk . . . x2x1 ∈ H, a.k.a., each reduced word corresponds to exactly one reduced-walk in π (G, H). This completes the proof.

Deﬁnition 5.33. Let G be a graph, e = {u, v} be an edge, the edge contraction of e is the operation that removes e from E, and make one super vertex w, connects all neighbors of u and v to w. Notice edge contraction might yield a graph with self-loops, if the original graph admits multi-edges. We use G/e as a notation for G contracts edge e.

Proposition 5.34. If e ∈ E (G) and e is not a self-loop, then π (G) =∼ π (G/e).

Proof. Let e = {u, v}, and let w be the super vertex of u and v, after contraction. We will show π (u, G) =∼ π (w, G/e). We classify uu-walks as follows:

Type 1 Walks that do not use edge e. For this kind of walk, we simply map it to the same sequence of edges in G/e, with all u and v substituted by w.

Type 2 Walks that use edge e. This means at some time of the walk, it takes edge e to go from u to v, since this is essentially a uu-walk, it then must take some walk p to go from v to u, so it is safe to simply remove all occurrences of edge e in the original walk, the result is still a ww-walk, and it is not hard to see that this mapping is bijective.

Thus, π (u, G) =∼ π (w, G/e).

Deﬁnition 5.35. A spanning tree in a connected graph G is a set T ⊆ E such that

1. for all e ∈ T , flip (e) ∈ T

2. for all x, y ∈ V , there is an xy-walk with edges in T

3. there are no non-trivial reduced closed walks in T , i.e., T does not contain cycles

Proposition 5.36. If G is a connected graph, then G has a spanning tree.

Proof. A simple procedure will work: keep adding edges that do not form a cycle, until all vertices have been covered.

Exercise 5.37. If T ⊆ E is a spanning tree in G, then for all x, y ∈ V , there is a unique reduced xy-walk in T .

Proof. Suppose it’s not the case, i.e., there are at least two reduced xy-walk in T , then one can use the first walk to travel from x to y, and use the second to travel back from y to x: this is doable since by definition of spanning tree, both e and flip (e) are in T . This gives us a non-trivial closed walk, contradicts the definition of spanning tree.

49 Theorem 5.38. For every connected graph G, π (G) is a free group. ∼ Proof. Pick a vertex x ∈ V , let T ⊆ E be a spanning tree in G. We claim that π (G) = FE+\T , −1 where we interpret flip (e) as e . First of all, it is clear that π (x, G) ⊆ FE+ , deﬁne an embedding ( e, if e 6∈ T ϕ : FE+ → FE+\T , by e 7→ The goal is to show ϕ|π (x, G) is an isomorphism id, otherwise from π (x, G) to FE+\T , i.e., for every ek . . . e2e1 ∈ FE+\T , there is a unique p ∈ π (x, G) such that ϕ (p) = ek . . . e1. Let’s say we write p = pkekpk−1ek−1 . . . p2e2p1e1p0, where p0, p1, . . . , pk are walks using tree edges from T . Notice that p0 is a (x, start (e1))-walk, for each 1 ≤ i < k, pi is a (end (ei) , start (ei+1))-walk, and pk is a (end (ek) , x)-walk. These are all walks using purely T edges, so by 5.37, these are all unique. This proves ϕ|π (x, G) is a bijection, and therefore, ∼ π (G) = FE+\T , π (G) is a free group. With all these in hand, we are ready to prove 5.26. ∼ Proof of Nelsen-Schreir. Let H ≤ FX , then by 5.32, H = π (G) for some graph G, and by 5.38, π (G) is a free group, therefore, H is also a free group, as desired.

50 5.3 Structure of finite groups Groups, with a simple definition but complicated internal dynamics, so many people try to look at finite simple groups, and classify them based on certain criteria. In this section, we will try to do so.

Theorem 5.39 (Cauchy). If G is a ﬁnite group, p is a prime number such that p | |G|, then G has an element of order p.

p Proof. Consider the following set S := {(a1, a2, . . . , ap) ∈ G : a1a2 . . . ap = e}. What is the cardinality of S? The claim is |S| = |G|p−1, the idea is we can freely pick p − 1 elements of the −1 sequence, while the last element is fixed as (a1a2 . . . ap−1) . The sequence a1a2 . . . ap exhibits −1 −1 rotational invariance: a1a2 . . . ap = e ⇒ a1 a1a2 . . . apa1 = a1 a1 = e ⇒ a2 . . . apa1 = e, so let’s define an action α : Zp y S by α (1) (a1, . . . , ap) := (a2, . . . , ap, a1), and α (m)(a1, . . . , ao) := ∼ (am+1, . . . , a1, a2, . . . , am), i.e., shifts the sequence to the left by m. By orbit-stabilizer, Zp/Stab (s) = Zp · s, since |Zp| = p and Stab (s) ≤ Zp, |Zp/Stab (s) | is either 1 or p, which also indicated every orbit of α has size either 1 or p. Since orbits partition S, we can classify S into fixed points, i.e., elements with orbit size 1, let Z be the set of these points. Since (e, e, . . . , e) ∈ Z, |Z| ≥ 1. Moreover, |S| = |Z| + |S \ Z|p, taking modulo p, we have |S| = |Z| mod p, since |S| = |G|p−1 and p | |G|, |Z| = 0 mod p, so |Z| ≥ p ≥ 2. So let (a1, . . . , ap) ∈ Z 6= (e, e, . . . , e), and by definition of p fixed points, (a1, a2, . . . , ap) = (a2, a3, . . . , ap, a1) ⇒ a1 = a2 = a3 = ... = ap = a 6= e ⇒ a = e, as desired.

Exercise 5.40. Let G be a ﬁnite group, p is a prime number such that p | |G|, then number of cyclic subgroups of G, of order p, equals 1 mod p.

Proof. We make use of the ﬁxed points set Z in 5.39. Since |Z| = 0 mod p, we can write |Z| = kp for some positive integer k, get rid of (e, e, . . . , e), there are kp − 1 sequences remaining. For the same a, the sequence (a, a, . . . , a) is counted p − 1 times, so number of cyclic groups is essentially kp−1 the number of unique sequence (a, a, . . . , a), thus, the count is p−1 = 1 mod p. Proposition 5.41. If G is a ﬁnite group, |G| = mp where p is a prime number and 1 < m < p, then G is not simple.

Proof. The claim is G has a normal subgroup of order p, suppose this is not the case, then G has ≥ 2 subgroups of order p, by 5.40, number of subgroups of order p is 1 mod p, so the number must be ≥ p + 1. Note that if both H1,H2 ≤ G with |H1| = |H2| = p, H1 6= H2 ⇒ H1 ∩ H2 = {e}, so |G| ≥ 1 + (p + 1) (p − 1) = p2, contradicts |G| = mp < p2.

Example 5.42. Suppose we have a group G with |G| = 2019, notice 2019 = 673 × 3, and 673 is a prime, so there are no simple groups of order 2019.

Theorem 5.43 (Sylow’s First Theorem). Let G be a ﬁnite group, p be a prime number, and k ∈ N. If pk | |G|, then G has a subgroup of order pk.

Proof. By induction on |G|. If k = 0, then it’s trivial. Assume k ≥ 1, in particular, pk | |G|. If G has a subgroup H such that H 6= G and pk | |H|, then by induction hypothesis, we are done.

51 Suppose G has no such subgroups, we claim that G has a normal subgroup of order p. We will make use of the class formula of G: X |G| = |Z (G) | + |G : Z (a) | a∈A where Z (G) is the center of G, all elements in Z (G) can commute with any elements in G, and Z (a) is the centralizer of element a, which contains elements that can commute with a. Finally, A is a subset of G formed by picking an element from each non-trivial conjugacy class. To give a short proof of class formula, consider the conjugation action αconj : G y G, notice the stabilizer of a is exactly Z (a), and by orbit-stabilizer, each orbit has size |G : Z (a) |. Since orbits partition G, we get the class formula. Notice that each Z (a) is a subgroup of G, by our assumption, pk 6 ||Z (a) |, which means p | |G|/|Z (a) | ⇒ p | |G : Z (a) |, so we can rewrite class formula as

k X mp = |Z (G) | + p Ca a∈A where we write |G : Z (a) | as Cap. Taking modulo p on both sides, we get 0 = |Z (G) | + 0 mod p, so |Z (G) | = 0 mod p. Moreover, e ∈ Z (G), so |Z (G) | ≥ 1 ⇒ p | |Z (G) |. By 5.39, there exists a subgroup of Z (G) with order p, which is also normal: this comes from the fact that Z (G) is the center, so any subgroup of Z (G) is normal. k−1 Let N E G be the normal subgroup of G with order p, consider G/N. Notice that p | |G/N|, so there is a subgroup P ≤ G/N of order pk−1, by induction hypothesis. Let Q be the preimage of P , under the quotient map G G/N, one can verify that Q is a subgroup of G which contains N, so ∼ k−1 k N EQ, we can write P = Q/N ⇒ |Q| = |P ||N| = p p = p , and we get our desired subgroup. Deﬁnition 5.44. A p-group is a group of size equal to a power of p.A Sylow p-subgroup of a ﬁnite group G is a subgroup P ≤ G such that |P | = the largest power of p dividing |G|.

Remark 5.45. One can interpret 5.43 as saying that every ﬁnite group G has a Sylow p-subgroup, if p | |G|.

Theorem 5.46 (Sylow’s Second Theorem). Let G be a ﬁnite group, p be a prime number, H ≤ G be a p-group and P ≤ G be a Sylow p-subgroup, then H is contained in a conjugate of P , i.e., there exists g ∈ G such that H ⊆ gP g−1.

Before proving the theorem, let’s ﬁrst prove a handy Lemma.

Lemma 5.47. Let G be a p-group, S be a ﬁnite set, and Z denote the set of ﬁxed points under any action G y S, then |Z| = |S| mod p.

Proof. Given an element s ∈ S, let Os denote the orbit contains s, by orbit-stabilizer, |Os| = k k |G/Stab (s) | = |G|/|Stab (s) | ⇒ |Os| | |G|, since G is a p-group, let p := |G|, so |Os| | p . Since P orbits partition S, we can write |S| = |Z| + s∈A |Os|, where A is a subset of S formed by picking an element from each non-trivial orbit. For each non-trivial orbit, |Os| ≥ 2 ⇒ p | |Os|, so if we take modulo p on both sides, we get |Z| = |S| mod p, as desired.

52 Proof of Sylow’s Second Theorem. Let H y G/P by left multiplication, by orbit-stabilizer, every orbit has size equal to |H/Stab (gP ) |, so every orbit has size a divisor of |H|, hence a power of p. But |G/P | = |G|/|P |= 6 0 mod p, since P is Sylow p-subgroup, so at least one orbit has size p0 = 1, which means there is some g ∈ G, such that for any h ∈ H, hgP = gP ⇒ g−1hgP = P ⇒ g−1hg ∈ P ⇒ H ⊆ gP g−1.

Theorem 5.48 (Sylow’s Third Theorem). Let G be a ﬁnite group with |G| = mpk, m 6= 0 mod p, where p is a prime number. Then number of Sylow p-subgroups of G divides m, and is 1 mod p. −1 Deﬁnition 5.49. Let H ≤ G, the normalizer of H in G is NG (H) := g ∈ G : gHg = H , i.e., the stabilizer of H under conjugation action G y Sub (G), several properties of NG (H):

• NG (H) ≤ G

• H E NG (H) • Orbit-stabilizer: number of subgroups of G that are conjugate to H, i.e., size of orbit containing H, is equal to |G : NG (H) | Proof of Sylow’s Third Theorem. We ﬁrst prove the number of Sylow p-subgroups of G divides m. Let P be any Sylow p-subgroup of G, we use nP := number of subgroups conjugate to P = k k |G : NG (P ) | = |G|/|NG (P ) |, notice |G| = mp , and |NG (P ) | is divisible by |P | = p , suppose k mpk m |NG (P ) | = `p , then nP = `pk = ` ⇒ nP ` = m, nP | m. Next, we prove nP = 1 mod p. Consider the following: m = |G : P |

= |G : NG (P ) ||NG (P ): P |

= nP |NG (P ): P |

Lemma 5.50. Let G be a ﬁnite group, p be a prime number, H ≤ G be a p-group, then |NG (H): H| = |G : H| mod p.

Proof of Lemma. Consider the action H y G/H by left multiplication. Orbits have sizes which are powers of p. By 5.47, |G : H| = |G/H| = (number of fixed points of this action) mod p, so it suffices to show |NG (H): H| equals to the number of fixed points, modulo p. Consider the following chain of logical equivalence:

gH is a ﬁxed point ⇔ ∀h ∈ H, hgH = gH ⇔ g−1hgH = H ⇔ ghg−1 ∈ H ⇔ H = gHg−1

⇔ g ∈ NG (H)

So gH is a ﬁxed point if and only if g ∈ NG (H), i.e., it corresponds to a coset of NG (H) /H. Thus, we have the identity.

53 Proposition 5.51. There are no simple groups of order 12.

Proof. 12 = 22 × 3. Number of Sylow 2-subgroups: 1 or 3, number of Sylow 3-subgroups: 1 or 4. Assume for a contradiction that there is a simple group of order 12, then there must be 3 Sylow 2-subgroups and 4 Sylow 3-subgroups, so there are 3 subgroups of order 4, 4 subgroups of order 3. How many elements are there of order 3? At most 2 × 4 = 8 elements, so it remains 12 − 8 = 4 elements, and we cannot ﬁt 3 subgroups of order 4 into 4 elements.

Proposition 5.52. There are no simple groups of order 24.

Proof. 24 = 23 × 3. Number of Sylow 3-subgroups: 1 or 4, number of Sylow 2-subgroups: 1 or 3, again assume for a contradiction that there is a simple group of order 24, so number of Sylow 3-subgroups is 4, and number of Sylow 2-subgroups is 3. Let G y {Sylow 2-subgroups} by conjugation, we can view it as α : G → Sym ({1, 2, 3}). Now let’s examine ker (α), since |G| = 24, |Sym ({1, 2, 3}) | = 6, α is not injective, so ker (α}) 6= {e}, on the other hand, since α is conjugation, clearly ker (α) 6= G, since we have assumed G is simple, so all Sylow 2-subgroups conjugate with each other. Thus, we have found a non-trivial normal subgroup of G, which is ker (α).

54 6 Rings and Modules

6.1 Basic definitions Rings and modules are built upon groups, they are less “abstract”, in the sense that they allow more operations, so there are real life examples and applications of this concept. Definition 6.1. Let A be an abelian group. An endomorphism of A is a homomorphism f : A → A. Example 6.2. Some examples of endomorphisms: • identity map, 1 • zero map, i.e., map all elements to 0, 0 • inverse map: a 7→ −a, denoted by −1, this uses abelian: −1 (a + b) = −b − a = −a − b = −1 (a) + −1 (b) Definition 6.3. Let End (A) := {all endomorphisms of A}, except for standard function composition, we also equip End (A) with pointwise addition: for f, g ∈ End (A), define f + g : A → A by (f + g)(a) = f (a) + g (a). Proposition 6.4. f + g is an endomorphism. Proof.

(f + g)(a + b) = f (a + b) + g (a + b) = f (a) + f (b) + g (a) + g (b) = f (a) + g (a) + f (b) + g (b) = (f + g)(a) + (f + g)(b)

Exercise 6.5. Check that (End (A) , +) is abelian. Deﬁnition 6.6. A concrete ring is a subgroup R ≤ (End (A) , +), such that for all f, g ∈ R, f ◦ g ∈ R. If idA = 1 ∈ R, then R is called a unital ring. Example 6.7. 1. End (A), analogous to Sym (A), called the endomorphism ring of A. 2. {0} ≤ End (A) is a ring.

3. What is End (Z)? For f ∈ End (Z), all its values are determined by f (1), for eacn n ∈ Z, ∃!f ∈ End (Z) with f (1) = n, f (m) = nm. Let fn denote the function that f (1) = n, so ∼ End (Z) = {fn : n ∈ Z}, and fn + fk = fn+k, fn ◦ fk = fnk, so (End (Z) , +, ◦) = (Z, +, ·).

4. {fr : r ∈ R} ⊆ End (R), for each r ∈ R, fr : R → R is given by fr (s) = rs, then fr ∈ End (R), fr + fs = fr+s, fr ◦ fs = frs, so {fr : r ∈ R} is a concrete ring on R, and it is isomorphic to (R, +, ·).√ However, it is not the entire End (R): there exists f ∈ End (R) such that f (1) = π, f 2 = e. Deﬁnition 6.8. An (abstract) ring is a set R with two binary operations +, ·, such that (R, +) is an abelian group with identity 0 ∈ R. · is associative and for all a, b, c ∈ R, a · (b + c) = a · b + a · c. If · has an identity element 1, then R is called unital. If · is commutative, then R is called a commutative ring.

55 Analogous to concrete group and group, if R ⊆ End (A) is a concrete ring, f, g, h ∈ R, then (f ◦ (g + h)) (a) = f ◦ (g + h)(a) = f (g (a) + h (a)) = f (g (a)) + f (h (a)) = f ◦ g (a) + f ◦ h (a)

Theorem 6.9. Let (R, +, ·) be a ring, for each r ∈ R, let fr : R → R : a 7→ ra, then fr ∈ End (R, +), R˜ := {fr : r ∈ R} is a concrete ring on (R, +) and the map R → R˜ : r 7→ fr is an isomorphism from (R, +, ·) to R,˜ +, ◦ .

Proof. One can easily duplicate the argument from 3.22. We won’t repeat it here.

2 2 2 Example 6.10. End Z . Let f : Z → Z be an endomorphism, suppose f (1, 0) = (a, b) , f (0, 1) = (c, d), then f (x, y) = xf (1, 0) + yf (0, 1) = (xa, xb) + (yc, yd) = (xa + yc, xb + yd). It is best to write it in matrix notation: a c x f (x, y) = b d y 2 So End Z is all 2 × 2 integer matrices. 2 Exercise 6.11. Verify that in End Z , ring addition is matrix addition, and function composition is matrix multiplication. It’s worth noticing that this ring is not commutative. Deﬁnition 6.12. Let R,S be rings, a homomorphism from R to S is a function f : R → S such that

• f is a homomorphism of abelian groups (R, +R) , (S, +S) •∀ a, b ∈ R, f (ab) = f (a) f (b)

If R,S are unital, then a homomorphism f : R → S is called unital if f (1R) = 1S. Deﬁnition 6.13. An R-module or a module over R is an abelian group M, with a homomorphism α : R → End (M). If R is unital, then R-module α : R → End (M) is unital, if α (1R) = idM . Often, we say “M is an R-module” and write ra for α (r)(a). Example 6.14. 1. Let R be a ring, then R → End (R): r 7→ (a 7→ ra) is a homomorphism, so R is a module of itself. 2. Let R be a ring, M be any abelian group, the trivial homomorphism R → End (M): r 7→ 0.

3. Z-module: Let M be an abelian group, consider the following homomorphism: Z → End (M): 0 7→ 0, 1 7→→ idM , n 7→ n · idM = idM + idM + ... + idM , i.e., (nidM )(a) = a + a + ... + a = | {z } | {zn } n na, ∀n ∈ Z, a ∈ M. Exercise 6.15. Show that this is a ring homomorphism, i.e., it preserves multiplication.

Remark 6.16. This establishes a relation: unital Z-modules correspond to abelian groups. 4. Let R be a ring, Rn(as abelian group) can be equipped with an R-module structure as follows: r (a1, a2, . . . , an) = (ra1, . . . , ran).

56 6.2 Free modules, independent, spanning, and basis Deﬁnition 6.17. Let R be a unital ring and X be a set, a free R-module over X is a (unital) R-module MX , together with the function ι : X → MX , such that for all (unital) R-modules M and functions f : X → M, there exists a unique R-module homomorphism ϕ : MX → M, such that ϕ ◦ ι = f.

Remark 6.18. This is almost the same diagram as free group: ι X MX

f ∃!ϕ

Deﬁnition 6.19. Let M,M 0 be R-modules, a function ϕ : M → M 0 is an R-module homomorphism if it is a group homomorphism and for all r ∈ R, a ∈ M, ϕ (ra) = rϕ (a).

∼ n Let’s now discover the structure of MX . If X is finite, then MX = R , where n = |X|, The idea is to think of the embedding function ι as map each element of X to an indicator vector, or more ( 1R, if y = x formally, define MX = {ω : X → R} and ι : X → MX : x 7→ ωx, where ωx (y) = 0R, otherwise On the other hand, if X is infinite, we just change the definition of MX a little bit: MX = {ω : X → R : there are finitely many x ∈ X such that ω (x) 6= 0R}. For each r ∈ R, ω ∈ MX , define r · ω ∈ MX by (r · ω)(y) = r (ω (y)). Then the definition of ϕ P becomes obvious, let’s first write elements of MX as x∈X axω (x), where ax ∈ R, so ϕ (ω) := P P ϕ x∈X axωx = x∈X axf (x). Exercise 6.20. Check this is a homomorphism, and it is unique.

We will extend on ϕ in free modules, to introduce some concepts the reader might recall from linear algebra.

Deﬁnition 6.21. Let X ⊆ M.

• X is independent if ϕ is injective: ϕ is injective → ker (ϕ) = {0} → for any coeﬃcients ax, P x∈X ax · x = 0 implies ax = 0, ∀x ∈ X.

• X is spanning if ϕ is surjective: ϕ is surjective → ∀m ∈ M, ∃ coeﬃcients ax, x ∈ X, only P ﬁnitely many of which are non-zero, such that x∈X ax · x = m. We write span (X) as the span of X.

• X is a basis, if ϕ is bijective. In this case, M is a free R-module.

Before looking at free modules, let’s ﬁrst check out some examples on non-free modules.

Deﬁnition 6.22. Let G be a group, G is torsion-free, if for any g ∈ G, g has inﬁnite order, i.e., k for k ∈ Z \{0} , g 6= 0.

57 Example 6.23. 1. Zn is a Z-module that has no basis, there are no non-empty independent sets. ∀x ∈ Zn, n · x = 0 ⇒ {x} is not independent. Note that (Zn, +, ·) is a ring, and as a Zn-module, Zn is free, since {1} is a basis. Observe that any free Z-module is torsion-free, i.e., any element in module M has inﬁnite order: ∀r ∈ Z \{0} and x ∈ M \{0}, r · x 6= 0. P For basis, this is trivially true, so suppose x is not a basis, we can write it as b∈B ab · b, P so r · x = b∈B rab · b, since r is non-zero, there is at least one coeﬃcient is non- zero, so r · x 6= 0.

2.( Q, +) is a torsion-free abelian group. As a Z-module, it has no basis: suppose X ⊆ Q is a 1 P basis for Q as a Z-module, pick x ∈ X, x 6= 0, then x = y∈X cy · y, with only finitely many P 2 non-zero ⇒ 1 · x = y∈X (2cy) · y, the coefficient of x is odd, but coefficients on RHS are even, so X cannot be a basis. P P Fact: If X is a basis, then x∈X ax · x = x∈X bx · x, then ax = bx, ∀x ∈ X. X 3. If X is infinite, then Z is not a free Z-module. Definition 6.24. A division ring is a unital ring R in which every non-zero element has multiplicative inverse. A non-zero commutative division ring is called a field. √ Example 6.25. , , , where p is a prime number, a + b 2 : a, b ∈ ⊂ . Let’s verify the R Q C Zp √ Q √ R last one for the multiplicative inverse part: 1√ = a−b 2 = a − b 2. a+b 2 a2−2b2 a2−2b2 a2−2b2 In the next part, we will try to prove that every module over division ring is free.

Exercise 6.26. If every R-module is free, then R is a division ring.

Lemma 6.27. Let R be a division ring, M be an R-module, if I ⊆ M is an inclusion maximal independent set, i.e., if I ⊆ J and J is independent, then I = J. Then I is a basis.

Proof. We need to show span (I) = M. Take x ∈ M \ I, we want x ∈ span (I). Notice by definition I∪{x} of I, I ∪ {x} is not independent, so for some (cy : y ∈ I ∪ {x}) ∈ R , only finitely many are P P non-zero, we have y∈I∪{x} cy · y = 0 ⇒ cx · x + y∈I cy · y = 0, since I is independent, cx 6= 0, so P P −1 cx · x = y∈I −cy · y ⇒ x = y∈I − cx cy · y. Definition 6.28. Let P be a set, be a binary relation on P , we say that is a partially-ordered relation if

• is reﬂexive

• is antisymmetric, i.e., for x, y ∈ P , if x y and y x ⇒ x = y

• is transitive

We say (P, ) is a partially-ordered set, or poset for short.

Deﬁnition 6.29. Let (P, ) be a poset, a subset C ⊆ P is a chain if for every x, y ∈ C, x y or y x. An upper bound for C is an element x ∈ P such that ∀y ∈ C, y x.

Lemma 6.30 (Zorn). If (P, ) is a poset such that every chain has an upper bound, then P has a maximal element.

58 Proof. We refer readers any online resources for the proof, since it is out of the scope of this course.

Proposition 6.31. Let R be a ring, and R be a module of itself by left multiplication. If R is unital, then α : R → End (R) is injective.

Proof. Let r, s ∈ R with r 6= s, then α (r) (1) = r · 1 = r, α (s) (1) = s · 1 = s, so α (r) 6= α (s).

Theorem 6.32. All modules over a division ring are free.

Proof. Let R be a division ring, M be an R-module, let P = {I ⊆ M : I is independent}, consider the poset (P, ⊆), if (P, ⊆) has a maximal element, then by 6.27, M has a basis. Let C ⊆ P be S S a chain, we want to show that C has an upper bound, the claim is C = I∈C I is the upper S P bound we want, we need to show that C is independent. Suppose not, then x∈S C cx · x = 0 is a nontrivial linear combination, we write it as c1 · x1 + c2 · x2 + ... + ck · xk = 0, let xi ∈ Ii ⊆ C, reorder them as I1 ⊆ I2 ⊆ ... ⊆ Ik, then x1, . . . , xk ∈ Ik, but Ik is independent, a contradiction. By 6.30,(P, ⊆) has a maximal element.

Next, we want to deﬁne rank (Rn) to be n, but what if Rn =∼ Rm for n 6= m?

Deﬁnition 6.33. A unital ring R has the Invariant Basis Number, or IBN for short, if for all + n ∼ m n, m ∈ N ,R = R ⇒ n = m. Theorem 6.34. Every division ring has IBN.

Lemma 6.35 (Exchange property). Let R be a division ring, let M be an R-module, let I ⊆ M be independent, B ⊆ M a basis. Then for all x ∈ I \ B, there exists y ∈ B \ I such that (I \{x}) ∪ {y} is independent.

Proof of Exchange property. Suppose no such y, this means for any y ∈ B \ I,(I \{x}) ∪ {y} is not P independent, i.e., a∈(I\{x})∪{y} ca · a = 0, for coeﬃcients ca non-zero. By an argument analogous to 6.27, we have y ∈ span (I \{x}), so B \ I ⊆ span (I \{x}), also B ∩ I ⊆ I \{x} ⊆ span (I \{x}), so B ⊆ span (I \{x}), since span (B) = M, so I \{x} spans M. However, x ∈ span (I \{x}), contradicts I is independent.

Proof of Theorem. Suppose for a contradiction that Rn =∼ Rm but n 6= m, without loss of generality, n m assume n < m. Let Bn be a basis for R , and Bm be a basis for R , so |Bn| = n, |Bm| = m. We can then apply exchange property, to “swap” elements of Bn out, replace with elements in Bm. This 0 0 0 results in Bn ⊂ Bm, and Bn is a basis, however, |Bm| > |Bn|, this means Bm is not independent, contradicts Bm is a basis.

59 6.3 Commutative rings and IBN In this section, we will prove an important theorem

Theorem 6.36. Every commutative unital ring R has IBN.

Example 6.37. Z has the IBN. However, as a Z-module, it has maximal independent set, say {2}, but it is not a basis. So the idea is to choose a correct module. We provide two proofs.

m Proof #1. Notice Z ⊆ Q, and Q is a ﬁeld. Let Bn := {x1, . . . , xn} be a basis of Z of size n, so span (B ) = m, and span (B ) = m. Let’s show this. Let y = p1 ,..., pm ∈ m, let Z n Z Q n Q q1 qm Q y˜ = (q . . . q ) y ∈ m, soy ˜ = c x + . . . c x ⇒ y = c1 x + ... + cn x . However, since 1 m Z 1 1 n n q1...qm 1 q1...qm n Q m m is a division ring, by 6.34, rank (Q ) = m, contradicts Q can be spanned by Bn.

Proof #2. Let p be a prime number, πp : Z → Zp : k 7→ k mod p, again let Bn = {x1, . . . , xn} be a m m m m basis of Z with n elements, define πp : Z → Zp :(a1, a2, . . . , am) 7→ (πp (a1) , πp (a2) , . . . , πp (am)), m m m m so span πp (Bn) = Zp (here we overload πp to apply on each element of Bn): take y ∈ Zp , let m m m y˜ ∈ Z be the element such that πp (˜y) = y, writey ˜ = c1x1 + ... + cnxn, so y = πp (c1) πp (x1) + m ... + πp (cn) πp (xn), however, Zp is a field, so this contradicts Zp has IBN. n We can extend from Z , to show something about Fn, the free group. ∼ Corollary 6.38. Let n, m ∈ N, then Fn = Fm ⇒ n = m. n ∼ n ∼ m Proof. The abelianization of Fn is Z , then Fn = Fm ⇒ Z = Z ⇒ n = m. For the definition and property of abelianization, we refer reader to any online resources. Let R be a non-zero, commutative, unital ring. For 6.37 to work, we need to embed R into a field, but this might not work: for example, Z6 cannot be embedded into a field, since 2 × 3 = 0 mod 6, −1 −1 suppose we can embed Z6 into some field K, then in K, 3 has inverse 3 , so 2 × 3 × 3 = 2×3 × 3−1 = 2 = (2×)×3−1 = 0, which is not possible. For this to work, one requires additional structure on R.

Deﬁnition 6.39. A commutative, non-zero, unital ring R is called integral domain if ∀a, b ∈ R, a · b = 0 ⇒ a = 0 ∨ b = 0.

Theorem 6.40. A commutative, non-zero, unital ring R can be embedded into a ﬁeld if and only if R is integral domain.

Proof. (⇒): We have already shown the contrapositive. (⇐): Given an integral domain R, we shall construct a field Frac (R), called field of fractions of R. a Every element of Frac (R) is a fraction of the form b , a, b ∈ R, b 6= 0. a c ad+bc a b ab ˜ Idea: we want b + d = bd , c · d = cd . Let’s try to do so. Define +,˜· be binary operations on R × (R \{0}) via (a, b) +˜ (c, d) = (ad + bc, bd) , (a, b)˜· (c, d) = (ac, bd).

• +˜ ,˜· are commutative, by deﬁnition of R

60 • +˜ ,˜· are associative:

(a, b) +˜ (c, d) +˜ (e, f) = (ad + bc, bd) +˜ (e, f) = (adf + bcf + bde, bdf) (a, b) +˜ (c, d) +˜ (e, f) = (a, b) +˜ (cf + de, df) = (adf + bcf + bde, bdf)

One can follow the same strategy to show ˜· is associative.

• identity for +˜ : (0, 1), for ˜· : (1, 1)

• distributivity, however, does not hold:

(a, b)˜· (c, d) +˜ (e, f) = (a, b)˜· (cf + de, df) = (acf + ade, bdf) [(a, b)˜· (c, d)] +˜ [(a, b)˜· (e, f)] = (ac, bd) +˜ (ae, bf) = acbf + aebd, b2df

a c The problem is diﬀerent pairs may represent the same fraction, i.e., b = d ⇔ ad = bc. Deﬁne a binary relation ∼ on R × (R \{0}) via (a, b) ∼ (c, d) ⇔ ad = bc. Proposition 6.41. ∼ is an equivalence relation.

Proof. Reﬂexive and symmetric is obvious, let’s focus on transitivity. Suppose (a, b) ∼ (c, d) , (c, d) ∼ (e, f), then ad = bc, cf = de, we want to show: af = be.

ad = bc adef = bcef afde = becf afde = bede afde − bede = 0 (af − be) de = 0

Since R is an integral domain, either af − be is 0 or de is 0. In the former case, we are done, so let’s check for the latter case. If de is 0, then e must be 0, since cf = de, c must also be 0, and again, since ad = bc, a must be 0, so af = 0 = be, as desired.

a Let’s now formally define Frac (R) := R × (R \{0}) / ∼, notation: we write b for the ∼ class of the pair (a, b). a c ad+bc a c ac Define b + d = bd , b · d = bd , one can verify this is well-defined. Now +, · on Frac (R) are associative, commutative, distributive, and have identities:

a −a a −a 0 ab+b(−a) 0 2 0 0 • Additive inverse: − b = b : b + b = 1 , b2 = b2 , but 0 · 1 = 0 · b , so b2 = 1 . a −1 b 1 1 0 • Multiplicative inverse: b = a , the multiplicative identity is 1 , 1 6= 1 since 0 6= 1 in R. a So Frac (R) is a ﬁeld. Deﬁne ι : R → Frac (R): a 7→ 1 , this is an injective unital ring homomorphism.

61 Remark 6.42. Frac (R) is the smallest ﬁeld containing R, we always have the following diagram: R ι Frac (R)

f ∃!embedding

K where f is an unital embedding, K is a ﬁeld. The embedding from Frac (R) to K is given by a −1 b 7→ f (a) f (b) . Corollary 6.43. Integral domains have IBN.

Proof. Repeat the proof for Z using embedding Z ,→ Q for arbitrary integral domain to its fractional ﬁeld.

To achieve our goal, i.e., prove commutative ring has IBN, we need to adapt the idea in Z Zp, i.e., ﬁnd a surjective homomorphism R F , where F is a ﬁeld. Let R be a ring, M,N be R-modules, ϕ : M → N be an R-module homomorphism, we have the following diagram: ∼ M N/ ker (ϕ) = im (ϕ) N This is called the isomorphism theorem for R-module. In fact, im (ϕ) , ker (ϕ) and N/ ker (ϕ) are all R-modules. • im (ϕ) is a submodule of N: if x ∈ im (ϕ) , r ∈ R, then there exists some y ∈ M with x = ϕ (y), so r · x = ϕ (r · y) ∈ im (ϕ).

• ker (ϕ) is a submodule of M: if x ∈ ker (ϕ) , r ∈ R, then ϕ (x) = 0, ϕ (r · x) = r · ϕ (x) = r · 0 = 0, so r · x ∈ ker (ϕ). Generally speaking, if M 0 ⊆ M is a submodule, then M/M 0 becomes an R-module under r · (x + M 0) := (r · x) + M 0. This is well-deﬁned: if x + M 0 = y + M 0 ⇒ x − y ∈ M 0 ⇒ r · (x − y) ∈ M 0 ⇒ r · (x − y) + M 0 = M 0 ⇒ (r · x) + M 0 = (r · y) + M 0. Exercise 6.44. Check this makes M/M 0 an R-module. Remark 6.45. If R is a unital ring, M,N be unital R-modules, i.e., 1 · x = x, ϕ : M → N be a ∼ module homomorphism, then M N/ ker (ϕ) = im (ϕ) ,→ N, all intermediate modules are also unital. Deﬁnition 6.46. Let R be a ring, R is an R-module under left multiplication, submodules of R, viewed as an R-module, are called (left) ideals. In other words, I ⊆ R is a (left) ideal if 1. I is a subgroup of (R, +)

2. for all x ∈ I, r ∈ R, we have r · x ∈ I Note that an ideal is a subring. Example 6.47. • For every ring R: R, {0}.

• Z: {0} , Z, nZ, i.e., every subgroup of Z is an ideal, but not necessarily unital, for example, 2Z is not unital.

62 Proposition 6.48. Let K be a ﬁeld, the only ideals in K are {0} ,K. Proof. Suppose I ⊆ K is an ideal, {0} ⊂ I ⊂ K, take any x ∈ I \{0}, then for all y ∈ K, y = yx−1 x ∈ I, so I = K.

Proposition 6.49. Let R be a non-zero commutative, unital ring. R is a ﬁeld if and only if the only ideals are R and {0}. Proof. (⇒): 6.48. (⇐): Suppose R is not a ﬁeld, take x 6= 0 with no multiplicative inverse, consider I := {rx : r ∈ R}, I 6= R, since 1 6∈ I, and it is also non-zero, since x ∈ I. Let’s check I is an ideal: 1.0 ∈ I, rx + sx = (r + s) x ∈ I, − (rx) = (−r) x ∈ I.

2. s · rx = (sr) x ∈ I.

Proposition 6.50. Let I be an ideal in a unital ring R, then I 6= R if and only if 1 6∈ I, i.e., I is a proper ideal. Proof. (⇒) : Clear. (⇐) : If 1 ∈ I, then for any r ∈ R, r · 1 = r ∈ I.

Let R,Q be rings, ϕ : R → Q be a ring homomorphism, consider the diagram: ∼ R R/ ker (ϕ) = im (ϕ) Q where ker (ϕ) is an ideal of R: it’s a subgroup, and for x ∈ ker (ϕ) , ϕ (x) = 0 ⇒ ϕ (rx) = rϕ (x) = r ·0 = 0, so rx ∈ ker (α). im (ϕ) is a subring of Q: for x, y ∈ im (ϕ) , ϕ (a) = x, ϕ (b) = y ⇒ ϕ (ab) = xy, so xy ∈ im (ϕ). Remember the group situation: ϕ : G → H, then ∼ G G/ ker (ϕ) = im (ϕ) H where ker (ϕ) is normal, so G/ ker (ϕ) is a group, and im (ϕ) is a subgroup of H. In our ring analogy, ker (ϕ) is not only a left ideal, but also a right ideal: for x ∈ I, r ∈ R, xr ∈ I. In fact, ker (ϕ) is a two-sided ideal, both left and right ideal. Let R be a ring, I ⊆ R be a two-sided ideal, then R/I becomes a ring under multiplication: (x + I)(y + I) := (xy) + I, let’s check it’s well-defined: suppose x + I = x0 + I, y + I = y0 + I, want to show that (xy) + I = (x0y0) + I. Notice x − x0, y − y0 ∈ I, so (x − x0) y = xy − x0y ∈ I, x0 (y − y0) = x0y − x0y0 ∈ I ⇒ xy − x0y + x0y − x0y0 = xy − x0y0 ∈ I ⇒ (xy) + I = (x0y0) + I. Exercise 6.51. Check R/I is a ring. Theorem 6.52. Let R be a non-zero, commutative, unital ring, then there exists a surjective homomorphism R K, where K is a field. For what ideals I ⊆ R is R/I a field?

Example 6.53. R = Z, then Z/I is a ﬁeld for I = pZ, where p is a prime number. R/I 6= {0} ⇔ I 6= R, so I is a proper ideal. Lemma 6.54. Let R be a commutative, unital ring, I ⊆ R be an ideal ,then there is a one-to-one correspondence between {J ⊆ R/I, an ideal} and {I0 ⊆ R, an ideal,I ⊆ I0}, the bijective mapping is given by J 7→ {x ∈ R : x + I ∈ J} = S J, and the inverse is I0 7→ {x + I : x ∈ I0}.

63 Proof. We’ll verify the direction I0 7→ {x + I : x ∈ I0}, let I ⊆ I0 ⊆ R, and J := {x + I : x ∈ I0}, we need to check that J is an ideal of R/I. This is the case, since J is the image of I0 under quotient map R ⇒ R/I, so J is a subgroup of R/I. To see it’s an ideal, take x + I ∈ J, x ∈ I0, and y + I ∈ R/I, then (y + I)(x + I) = (yx) + I ∈ J, since yx ∈ I0. Now let’s consider the inverse, let I00 = {y ∈ R : y + I ∈ J}, the goal is to show I0 = I00. Since x ∈ I0 ⇒ x + I ∈ J, so x ∈ I00 ⇒ I0 ⊆ I00. For the other inclusion, suppose y + I ∈ J, then there exists x ∈ I0 such that x + I = y + I ⇒ x − y ∈ I ⊆ I0, so y ∈ I0 ⇒ I00 ⊆ I0. Thus, this is a valid bijection. The other direction is left as an exercise to the reader.

Let I ⊆ R be a proper ideal, consider the following chain of logical equivalence:

R/I is a ﬁeld ⇔ the only ideals in R/I are {0} , R/I ⇔ the only ideals in R containing I are I and R ⇔ the only proper ideal containing I is I ⇔ I is inclusion maximal among all proper ideals

Theorem 6.55 (Krull). Let R be a commutative, unital, non-zero ring, then R has a maximal proper ideal.

Proof. We will use 6.30. Let P := {I ⊂ R : I is a proper ideal}, consider the poset (P, ⊆), we want to show every chain has an upper bound in this poset. Take C ⊆ P be a chain, if C = ∅, then {0} is an upper bound, otherwise, C 6= ∅, then S C is an upper bound for C. Let’s check S C is an ideal, S S S S take x ∈ C, r ∈ R, suppose x ∈ Ii, then r · x ∈ Ii ⊆ C. Take x, y ∈ C, x ∈ I1 ⊆ C, y ∈ I2, S so I2 ⊆ I1 or I1 ⊆ I2, this means x + y belongs to the larger ideal, which is a subset of C. So S C is an ideal. Is S C proper? Notice 1 6∈ S C, this is true because for any I ∈ C, 1 6∈ I. So every chain in (P, ⊆) has an upper bound, by Zorn, it has a maximal element.

Corollary 6.56. If I is a proper ideal, then there is a maximal ideal containing I, i.e., I ⊆ I0.

Proof. Apply 6.55 to R/I, obtaining a maximal proper ideal in R/I, call it J, using the bijective mapping from R/I to I in 6.54, map J to S J, clearly, I ⊆ S J, so it suﬃces to show S J is a maximal proper ideal in R. Suppose it is not maximal, and adding y to S J still keeps it an ideal, then y + I is not in J, again using the mapping in 6.54, we can map S J ∪ {y} to J ∪ {y + I}, but the latter is a larger ideal, contradicts J is maximal. S J is proper comes from J is proper.

Let’s now prove 6.52.

Proof. Use 6.55 to get a maximal proper ideal I ⊂ R, consider R/I, it is a ﬁeld by our above reasoning. The surjective embedding: R R/I is then given by the quotient map. Now we have enough tools to prove 6.36.

m ∼ n m Proof. Suppose for a contradiction that R = R but n < m, take a basis x1, . . . , xn ∈ R , fix m m m a surjective homomorphism π : R K with K a field, set π : R → K :(a1, a2, . . . , am) 7→ m m m (π (a1) , π (a2) , . . . , π (am)), then the set {π (x1) , . . . , π (xn)} spans K , but n < m, K is a field, so it’s impossible.

64 For the spanning part, take any y ∈ Km, π is surjective, so y = πm (z) for some z ∈ Rm, where m m z = a1x1 + ... + anxn, for some a1, . . . , an ∈ R, therefore y = π (a1) π (x1) + ... + π (an) π (xn), so the set is spanning.

65 6.4 Polynomials We have studies free groups and free modules, so it’s time to consider what is free (unital, commutative) rings? The diagram always looks similar: X ι R (X)

f ∃!ϕ

Q where R (X) is a commutative unital ring, ϕ is a unital ring homomorphism, and so is Q. ∼ Let’s ﬁrst consider X = ∅, then R (∅) = Z, since the only restriction has become ϕ to be a unique unital ring homomorphism, and we achieve so by 0 7→ 0Q, 1 7→ 1Q. What if X = {x}? We will ∼ show R ({x}) = Z [x], the polynomial ring in one variable over Z.

Deﬁnition 6.57. A polynomial over Z in one variable x is a sequence p = (a0, a1,...) ∈ ZN, such 2 P∞ i that only ﬁnitely many ai’s are nonzero, usually, we write a0 + a1x + a2x + ... = i=0 aix = (a0, a1,...).

The set of all such polynomials is denoted by Z [x]. Z [x] is an abelian group under +, defined by (a , a ,...) + (b , b ,...) = (a + b , a + b ,...). As an abelian group, [x] ∼ L , we turn 0 1 0 1 0 0 1 1 Z = N Z n Z [x] into a commutative, unital ring by defining a product. For notation, let [x ](a0, a1,...) := 2 2 8 an, for example, x 1 + 3x − x + 7x = −1. The product is defined as follows: for p, q ∈ [x], [xn](pq) := P xi (p) xj (q) , or we can write it as P a xi P b xj = Z i+j=n,i,j∈N i i j j P n P n cnx , where cn = i+j=n aibj. Multiplicative identity: (1, 0, 0, 0,...), and x 7→ (0, 1, 0, 0,...). Observe that (0, 1, 0, 0,...)n = ! 0, 0, 0,..., 0, 1 , 0,... . |{z}n Exercise 6.58. Check Z [x] is indeed a commutative ring. We can now complete the diagram: x7→x {x} Z [x]

f7→a ∃!ϕ

Q 2 For p ∈ (c0, c1, c2,...) ∈ Z [x], define ϕ (p) = c0 + c1a + c2a + ..., with only finitely many ci’s are nonzero. By our definition, ϕ (p) is essentially p (a), polynomial p evaluates at point a. Z [x] is free abelian group, generated by (1, 0, 0,...) , (0, 1, 0,...) , (0, 0, 1, 0,...) ,.... What if X = {x, y}? We can extend to Z [x, y], integer polynomials over 2 variables. Definition 6.59. A polynomial over Z in two variables x,y is denoted by Z [x, y], defined as the 2 set functions c : N × N → Z, such that (i, j) ∈ N : c (i, j) 6= 0 is finite . Addition is defined pointwise, so [x, y] ∼ L , and multiplication is defined as follows: Z = N×N Z for p, q ∈ [x, y], [xnym](pq) := P xayb (p) xcyd (q) . Z a,b,c,d∈N,(a,b)+(c,d)=(n,m) P n m n m Define the evaluation of p on point (a,b) by, for p ∈ [x, y] , a, b ∈ Q, p (a, b) := 2 ([x y ]) a b . Z (n,m)∈N

66 Exercise 6.60. Check the map Z [x, y] → Q : p 7→ p (a, b) is a unital ring homomorphism.

t0 t1 t2 If X is infinite, say X = {x0, x1, x2,...}, then we can characterize them as x0 x1 x2 ..., where t0, t1, t2 ∈ N, finitely many of them are nonzero. Definition 6.61. Let R be a commutative, unital ring, and X be a set, a polynomial over R in variable X, denoted by R [X], is a polynomial with coefficients in R, with variable set X, and finitely many coefficients are nonzero.

Deﬁnition 6.62. Let M be an R-module, where R is a commutative, unital ring. An R-Endomorphism of M is a homomorphism f : M → M of R-modules, i.e.

• f is a homomorphism of groups (M, +) → (M, +).

• for all r ∈ R, x ∈ M, f (r · x) = r · f (x).

One can interpret R-Endomorphism as a standard endomorphism, with additional constraint that f is linear in r · x. Let EndR (M) := {all R-endomorphisms of M} ⊆ End (M, +). • It is a ring: closed under addition, additive inverses, contains 0, and closed under composition.

• It is an R-module: deﬁne for r ∈ R and f ∈ EndR (M), r · f : M → M via (r · f)(x) = r · (f (x)).

–

(r · f)(x + y) = r · (f (x + y)) = r · (f (x) + f (y)) = r · f (x) + r · f (y) = (r · f)(x) + (r · f)(y)

–

(r · f)(s · x) = r · (f (s · x)) = r · (s · f (x)) = r · s · f (x) = s · r · f (x) = s · (r · f)(x)

Exercise 6.63. Finish the check that EndR (M) is an R-module.

Deﬁnition 6.64. A concrete R-algebra is a subset A ⊆ EndR (M) that is both a subring and submodule of EndR (M). An (abstract) R-algebra is a set A together with a ring structure (A, +, ·) and a (unital) R-module structure such that for all r, s ∈ R, x, y ∈ A, (r · x) · (s · y) = (rs) · (xy).

So End 2 ∼ 2 × 2 real matrices, since any f ∈ End 2 must preserve linear scaling, we R R = √ R R won’t have something like f (1) = π, f 2 = e.

67 2 2 Proposition 6.65. EndZ R = EndQ R . Proof. (⊇) : Clear, if f is linear on all rationals, then it must be linear on all integers. 2 2 (⊆) : Take f ∈ EndZ (R), we want to show: f ∈ EndQ R , i.e., for all q ∈ Q, x ∈ R , f (qx) = s s s qf (x). Write q = t for s, t ∈ Z, t 6= 0, so the question is whether t f (x) = f t x . Notice that s s s s s s f (sx) = f t tx = sf (x), and f t tx = tf t x , so sf (x) = tf t x ⇒ t f (x) = f t x . Exercise 6.66. Show that R as a Q-vector space is inﬁnitely dimensional, in fact, the following inﬁnite subset of R is Q-linearly independent: √ √ √ √ • (Tricky): 1, 2, 3, 5, 7,...

• (Easy): ln 2, ln 3, ln 5, ln 7,...

• (Median): Give an example of an uncountable such set. ∼ Notice that EndR (R) = R, but EndQ (R) is much bigger.

Analogous to Z [X] as the free ring over set X, R [X] is a free commutative, unital R-algebra: X ι R [X]

f ∃!ϕ

A Here A is an R-algebra, and ϕ is a unital R-algebra homomorphism. From now on, we focus on the ring R [x].

Deﬁnition 6.67. Let R be a ring, r ∈ R is called unit if it has multiplicative inverse.

In R [x], units are all nonzero constants, whose coeﬃcients are units. Recall the fundamental theorem of arithmetic: every natural number admits a unique prime factorization. In the proof, we ﬁrst use a strong induction to show every natural number has a prime factorization, then we use a contradicting argument to show it’s unique. We remind reader the proof:

Theorem 6.68 (Fundamental Theorem of Arithmetic). Every natural number has a unique prime factorization.

Proof. Existence: Induction on n. For n = 0, 1, it’s clear. Otherwise, either n is a prime number, then we are done, or n is a composite, let p be one of its prime divisor, then by induction hypothesis, n/p has a prime factorization, plus p, we get a prime factorization for n. Uniqueness: Suppose otherwise, n has two prime factorizations: n = p1 . . . ps = q1 . . . qt, let pi be the ﬁrst uncommon prime divisor, then pi | n ⇒ pi | q1 . . . qt. Recall Euclid Lemma: if p is a prime, p | ab ⇒ p | a or p | b, so pi | q1 or p1 | q2 . . . qt, since pi 6= q1, it must be the latter case, we iteratively apply this argument to the product part, ﬁnally we will have pi | qj, for any 1 ≤ j ≤ t, but pi does not exist in q1 . . . qt, a contradiction. To this end, we will adapt the same argument to show that any polynomial in R [x] admits a unique “prime” factorization.

68 Deﬁnition 6.69. Let R be a ring, a nonunit a ∈ R is irreducible if it cannot be written as a = bc, for nonunits b, c.

Deﬁnition 6.70. A nonunit a ∈ R is prime if ∀b, c ∈ R, a | bc ⇒ a | b or a | c.

Deﬁnition 6.71. Let a, b ∈ R, the greatest common divisors of a,b, denoted by gcd (a, b), is the set {g : g | a ∧ g | b, ∀h, h | a ∧ h | b ⇒ h | g}.

Proposition 6.72. Any two gcd’s are related by units.

Proof. Suppose g1, g2 ∈ gcd (a, b), then g1 | g2, g2 | g1, so there exists k, `, g1 = kg2, g2 = `g1 ⇒ g2 = k`g2, which means k` = 1 ⇒ k, ` are units.

In our familiar Z, we have gcd (a, b) 6= ∅, gcd (a, b) = gcd (a − kb, b) , gcd (a, 0) = {±a}. More- over, for any n ∈ Z, n is prime ⇔ n is irreducible. We prove so by using Bezout identity.

Lemma 6.73 (Bezout identity). For any a, b ∈ Z, there exists x, y ∈ Z, such that ax+by = g, ∀g ∈ gcd (a, b).

Lemma 6.74 (Euclid Lemma). In Z, prime and irreducible are equivalent.

Proof. Suppose p is irreducible in Z and p | ab, without loss of generality, assume p 6 |a, then by 6.73, gcd (a, p) = 1, so there exists x, y such that px + ay = 1 ⇒ pbx + aby = b, since p | ab, ab = kp for some k, so bxp + kyp = b ⇒ p | b.

How to prove Bezout identity? We will make use of Euclidean algorithm, which relies on division algorithm, or more concretely, division with remainders. Consider R = F [x], where F is some ﬁeld, we need to deﬁne division with remainders for polynomials.

Lemma 6.75 (Division algorithm). For any a (x) , b (x) ∈ F [x], b (x) 6= 0, there exists q (x) , r (x) such that a (x) = q (x) b (x) + r (x), such that deg (r) < deg (b) or r = 0.

Proof. Let n := deg (a) , m := deg (b), we prove by induction on n. Consider two cases.

Case 1 m > n, then q (x) = 0, r (x) = a (x).

2 n 2 Case 2 m ≤ n, write a (x) = a0 +a1x+a2x +...+anx , with an 6= 0, and b (x) = b0 +b1x+b2x + ...+b xm, consider a0 (x) := a (x)− an xn−mb (x), so deg (a0) < n, by induction hypothesis, m bm a0 (x) = q0 (x) b (x) + r (x), take q (x) = q0 (x) + an xn−m, then a (x) = q (x) b (x) + r (x). bm

The proof also gives an algorithm, which is referred as the division algorithm.

69 Algorithm Division: input: a(x), b(x) != 0; output: q(x), r(x); n = deg(a); m = deg(b); // Base case if n < m: q(x) = 0; r(x) = a(x); else: a’(x) = a(x) - (a_n / b_m)x^(n-m) b(x); // Degree of a’ < a (q’(x), r(x)) = Division(a’(x), b(x)); q(x) = q’(x) + (a_n / b_m)x^(n-m); return (q(x), r(x));

Let’s now prove Euclidean algorithm works in F [x]. Lemma 6.76 (Euclidean algorithm). For any a (x) , b (x) , q (x), gcd (a (x) , b (x)) = gcd (a (x) − q (x) b (x) , b (x)).

Proof. (⊆) : Given g ∈ gcd (a, b), g | a, b ⇒ g | qb, ∀q, so g | a − qb. (⊇): g | a − qb, g | b ⇒ g | a.

Now we are ready to prove Bezout identity for polynomials.

Lemma 6.77 (General Bezout identity). For any a (x) , b (x) ∈ F [x], there exists w (x) , z (x), such that a (x) w (x) + b (x) z (x) = g (x), where g (x) ∈ gcd (a (x) , b (x)).

Proof. We will perform induction on degree. Using 6.75, we write a (x) = q (x) b (x) + r (x), by 6.76, gcd (a (x) , b (x)) = gcd (r (x) , b (x)), so by induction hypothesis,

g (x) = w (x) r (x) + z (x) b (x) = w (x)(a (x) − q (x) b (x)) + z (x) b (x) = w (x) a (x) − w (x) q (x) b (x) + z (x) b (x) = w (x) a (x) + (z (x) − w (x) q (x)) b (x)

With Bezout in hand, we get similar result for integers: irreducible polynomials are primes, i.e., p (x) | a (x) b (x) ⇒ p (x) | a (x) or p (x) | b (x).

Theorem 6.78. Let F be a ﬁeld, then for any f (x) ∈ F [x], f (x) admits a unique prime factorization.

In the last part of this notes, we will take a look at roots or zeros of polynomials.

Theorem 6.79. Let K be a field, p ∈ K [x] be a monic polynomial of degree ≥ 1, where monic means the coefficient of highest degree is 1. Then there is a field L such that K ⊆ L, p has a root in L.

70 Proof. Let’s introduce a new variable t, and consider K [t]. Let I be the ideal in K [t], generated by p (t), i.e., I := {q (t) p (t): q (t) ∈ K [t]}. Look at K [t] /I, deﬁne ϕ (t) = t + I ∈ K [t] /I, so

2 d p (t + I) = a0 + a1ϕ (t) + a2ϕ (t) + ... + ϕ (t) 2 d = a0 + a1ϕ (t) + a2ϕ t + ... + ϕ t since the quotient map is a homomorphism

2 d = a0 + ϕ (a1t) + ϕ a2t + ... + ϕ t

2 d = ϕ a0 + a1t + a2t + ... + t = ϕ (p (t)) = p (t) + I = I since p (t) ∈ I

= 0K[t]/I

2 2 Example 6.80. Consider p (x) = x + 1 ∈ R [x], then the ﬁeld we consider is R [t] / t + 1 , elements are of the form q (t)+I, and p (t + I) = (t + I)2 +(1 + I) = t2 +I +1+I = t2 + 1+I = I. It remains to show K [t] /I is a ﬁeld.

• K [t] /I is nonzero, especially, 1 6∈ I. Notice deg(p(t)) ≥ 1, so for any q (t) ∈ K [t], deg (q (t) p (t)) ≥ 1, but 1 has degree 0, so 1 6= q (t) p (t). Thus, 1 6∈ I, and 1 + I ∈ K [t] /I.

• K [t] /I, in fact, might fail to be a ﬁeld, unless p (x) is irreducible. More formally, if p (x) is a irreducible polynomial with degree ≥ 1, then the ideal in K [t] generated by p (t) is maximal. Suppose not, let J be a proper ideal of K [t] such that hp (t)i ⊂ J, take any q (t) ∈ J \ hp (t)i, let g (t) be a gcd of p (t) , q (t), then g (t) 6= p (t), since q (t) 6= hp (t)i, so g (t) = 1, but by 6.77, g (t) = s (t) p (t) + f (t) q (t) ∈ hp (t) , q (t)i ⊆ J, which means 1 ∈ J, J is no longer proper, a contradiction.

By 6.78, we factor p (x) = p1 (x) p2 (x) . . . pk (x), where all of them are irreducible, then pick one of pi (x) to generate hpi (t)i, and take a quotient, we get our desired result. Exercise 6.81. Every ideal in K [t] is generated by one element.

We can even embed K into K [t] /I. Let p be a irreducible polynomial, then K [t] / hp (t)i is a ﬁeld, moreover, the embedding K,→ K [t] / hp (t)i : a 7→ (a, 0, 0,...)+I, is injective. The injectivity comes from the fact that, except for 0, all other polynomials of degree 0 are not in hp (t)i.

Theorem 6.82. Let K be a ﬁeld, there is a ﬁeld L with K ⊆ L, such that L is algebraically closed, i.e., every polynomial p ∈ L [x] of degree ≥ 1 has a root in L.

Proof. Iteratively apply 6.79, until every polynomial has a root.

Finally, we are ready to state the Fundamental Theorem of Algebra, though it is not fundamental, and even not formally a theorem, nor does it closely connect to algebra.

Theorem 6.83 (Fundamental Theorem of Algebra). C is algebraically closed.