Math 901 Notes 3 January of 33

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1 Math 901 Notes 3 January of 33 Math 901 Fall 2011 Course Notes Information: These are class notes for the second year graduate level algebra course at UNL (Math 901) as taken in class and later typed by Kat Shultis. The notes are from the Fall of 2011, and that semester the course was taught by Brian Harbourne. During the fall semester we covered the following topics: Category Theory, Functors, Natural Transformations Group Theory: Free (Abelian) Groups, Group Actions, Semi-Direct Products, Sylow Theorems, Nilpotent Groups, Solvable Groups Field Theory: Algebraic Extensions, Separable Extensions, Separable Degree, Galois Theory, Galois Extensions, Finite Fields, Transcendence Degree, Transcendence Bases For each class day, I ve indicated the topic at the top of that day s notes. Disclaimer: I created these notes in order to help me study, and so I ve expanded proofs where I find it useful, shortened things I was comfortable with before this course, and changed at least one proof to one that I like better than the one presented in class. These notes are not meant to be a substitute for your own notes. They have been proof-read, but are not guaranteed to be without errors. If you find errors, please Kat at the following address: s-kshulti1 at math.unl.edu 22 August 2011 Categories Comment: The purpose of categories and functors is to describe structure that is common in many parts of mathematics. Comment: Intuitively, a category is a structure consisting of objects and arrows (morphisms), and the arrows are maps between the objects. Examples: Category Objects Arrows Typical Notation Sets sets set maps S Groups groups group homomorphisms G Abelian Groups abelian groups group homomorphisms Ab Rings rings ring homomorphisms Vector Spaces finite dimensional R-vector spaces linear transformations V Manifolds topological manifolds continuous maps differentiable manifolds differentiable maps Group, G elements of G Definition: A category, A, consists of a collection of objects, Ob(A), and for each pair of objects, A, B Ob(A), there is a set of morphisms, Mor(A, B), and a composition law defined for morphisms Mor(A, B) and ψ Mor(B, C) when A, B, C Ob(A), such that ψ Mor(A, C), which satisfies the following: 1. Composition is associative when defined. 2. For all A Ob(A), there exists a unique identity morphism, which is a left and right identity whenever composition is defined. That is, there is a morphism 1 A Mor(A, A) such that if Mor(A, B) and ψ Mor(B, A), then 1 A = and 1 A ψ = ψ. 3. Mor(A, B) Mor(A, B ) =, unless A = A and B = B, 1 in which case Mor(A, B) = Mor(A, B ). 24 August 2011 More Categories & Universal Properties Example: Let X be a topological space. We ll use T X as the collection of open sets in X. We can turn this into a category, by setting Ob(X) = T X, and if U, V T X, we have either Mor(U, V ) = when U V or Mor(U, V ) = {U V } when U V. Definition: A morphism f : A B in a category C is an isomorphism if and only if there is a morphism g : B A such that g f = id A and f g = id B. Comment: We discussed the idea of direct products, and how a direct product is really just a function. This was a bit confusing, but the general idea was to generalize n-tuples. I ll only put it here in the most general context, not in the more specific cases we discussed. 1 These are actually equalities and not isomorphisms. In the category of groups, for example, my copy of Z is isomorphic, but not necessarily equal to, Sarah s copy of Z, and the morphisms from my copy to Caitlyn s copy are hence fundamentally different from the morphisms from Sarah s copy to Caitlyn s copy.

2 Math 901 Notes 3 January of 33 Definition: Let {A i : i I} be a family of objects in a category C. We say P Ob(C) is a product of the objects A i, sometimes written i I A i, if P was given together with morphisms π i : P A i called projections such that they satisfy the universal property. That is, given any B Ob(C) and morphisms f i : B A i there is a unique morphism f i = Mor(B, P ) such that the diagram below commutes for each i I. i I B P f i π i Aside: Suppose P and P are both products, that is we have projections π i : P A i and π i : P A i. Then by the universal property we have morphisms f : P P and g : P P, and these are inverse morphisms because the diagram below commutes. A i f g P P P π i π i π i A i In other words π i g f = π i and so we must have that g f = 1 P and similarly 1 P = f g. Hence P and P are isomorphic. However, note that we can actually only use this diagram for one direction, that is we can only show that g f = 1 P. However, a similar argument will show the other direction. Once things are more formal, we ll see that objects satisfying universal properties are unique August 2011 Monomorphisms & Epimorphisms Review: From last time, we discussed the categorical product, and we ve discussed the category of a topological space, X. The product of U and V in this category is their intersection. For homework, we ll ask more sophisticated questions about the product. Example: Let C be the category whose objects are all the elements in the set R [0, 1], and Mor(x, y) = {, if x > y {x y}, if x y.. Let {x i : i I} be a collection of objects in this category, then the categorical product is p = inf{x i : i I}. This is easily seen to be what you want if you do the case of two objects and their product, so that the product of x and y is min(x, y). Definition: Let C be a category and let A, B Ob(C), with Mor(A, B). We say is a monomorphism if given any γ, δ Mor(C, A) with γ = δ then γ = δ. Proposition: In the category of sets, is a monomorphism if and only if it is injective. Definition: (Again, using pictures) A map is a monomorphism if whenever we have γ C A B δ and it commutes, then γ = δ. Comment: An epimorphism is the categorical dual of a monomorphism, and also a generalization of a surjective map. Definition: A map is an epimorphism if whenever we have and it commutes, then γ = δ. 2 Also, the argument is frequently almost identical to this one. γ C δ A B

3 Math 901 Notes 3 January of 33 Example: Back to the order relations. Here, all morphisms are both monomorphisms and epimorphisms, but only the identity morphisms are isomorphisms. This indicates that the mono- and epi-morphisms are related to injective and surjective maps, but do not always act the same. Comment: The categorical dual of a product is a coproduct. Also, in most cases, categorical duals can be defined by simply reversing the directions of the arrows. 29 August 2011 Coproducts; Initial & Terminal Objects Comment: We ll repeat the definition of a product here, so that we can compare it to the dual, the coproduct. Definition: Let {A i : i I} be a family of objects in a category C. The product (if it exists) is an object A i with a family of morphisms π j : A i A j such that given any object i I i I B and family of morphisms p j : B A j, there is a unique morphism : B A i such that the diagram below i I commutes for each j I. B i I A i p j π j A j The coproduct (if it exists) is an object A i with a family of morphisms π j : A j A i such that given any i I i I object B and family of morphisms p j : A j B, there is a unique morphism : A i B such that the diagram i I below commutes for each j I. B i I A i p j π j A j Example: The category of abelian groups. In this category, the coproduct is the direct sum. Example: The category of sets. In this example, the coproduct is the disjoint union. Let s make that concept a bit more rigorous: Let {A i : i I} be our collection of sets. Then set Ãi = {i} A i I i I A i, so that Ãi Ãj = if i j. Then P = i I Ã i works and π j : A j P is given by x (j, x). Definition: An object, I in a category C is said to be initial (aka universally repelling) if there is a unique morphism I A for every object A Ob(C). Definition: An object T in a category C is said to be terminal (aka universally attracting) if there is a unique morphism A T for every object A Ob(C). Example: The initial object in the category of abelian groups is the trivial group, I = (0). In the category corresponding to a topological space, the empty set is an initial object, and the whole space is a terminal object. Theorem: If A and B are initial (respectively terminal) objects in a category C, then there is a unique isomorphism : A B. 31 August 2011 Topology Review; (co)products as (Initial)Terminal Objects Comment: We discussed a variety of possible definitions of continuity, and how these arose in history. We also discussed the idea that topology was in some ways a natural abstraction of other areas of mathematics. We then proceeded to define a topology (defined using open sets), closed sets, the closure and interior, the topological definition of continuous, and a basis. We then proceeded to continue our discussion of initial and terminal objects. The purpose

4 Math 901 Notes 3 January of 33 of this discussion was to introduce ideas that show up in the first homework assignment due Friday. I omit the details here as I was already comfortable with these definitions. Definition: Let C be a category, and let F = {A i : i I} be a family of objects in C. We define a new category C F. In this category, the objects are the objects of C together with a family of morphisms f i Mor(C, A i ) for each i I. We ll denote this object as C {fi}. Given two objects in this category C {fi} and D {gi}, a morphism Mor(C {fi}, D {gi}) is a morphism Mor(C, D) in the original category such that the diagram (existing in the category C) below commutes for all i I. C D f i g i Comment: The product in C is precisely a terminal object in C F. 3 Definition: Let C be a category, and let F = {A i : i I} be a family of objects in C. We define a new category C F. In this category, the objects are the objects of C together with a family of morphisms f i Mor(A i, C) for each i I. We ll denote this object as C {fi}. Given two objects in this category C {fi} and D {gi}, a morphism Mor(D {gi}, C {fi} ) is a morphism Mor(D, C) in the original category such that the diagram (existing in the category C) below commutes for all i I. C A i D f i g i Comment: The coproduct in C is precisely an initial object in C F. 4 A i 2 September 2011 Fiber Products Comment: For intuition, we discuss the fiber product in the category of sets. Given X, Y, Z sets with f : X Z and g : Y Z the fiber product is the product of the fibers, that is, X Z Y = {f 1 (z) g 1 (z) : z Z}, or more traditionally though of as X Z Y = {(x, y) X Y : f(x) = g(y)}. Definition: Let A, B Ob(C) for C some category, and let f Mor(A, C) and g Mor(B, C) be morphisms. We say an object F together with morphisms α Mor(F, A) and β Mor(F, B) is a fiber product of f and g if the diagram below commutes: α F β A B f C g and whenever we have morphisms Mor(G, A) and ψ Mor(G, B) such that the diagram below commutes G ψ A B f C g 3 This is obvious if you go through the definitions of a product and a terminal object. 4 This is obvious if you go through the definitions of a product and a terminal object.

5 Math 901 Notes 3 January of 33 then there is a unique morphism θ Mor(G, F ) such that the third diagram below commutes. G A α F θ β ψ B f Comment: The categorical fiber product of f Mor(A, C) and g Mor(B, C) in a category C is just a categorical product of objects A {f} and B {g} in the category C {C}. 5 Comment: We next started to think about functors, and discussed the idea of a subcategory, but did not complete the definition of a functor. Definition: We say a category C is a subcategory of a category D if Ob(C) Ob(D) and if given A, B Ob(C), then Mor C (A, B) Mor D (A, B) and if moreover an identity in C is an identity in D, and the composition law in C agrees with the composition law in D. Example: The category of abelian groups is a subcategory of the category of groups. The inclusion of the category of abelian groups in the category of groups is an example of a covariant functor. C g 7 September 2011 Functors Definition: Let C, D be categories. A covariant functor F : C D is an assignment of an object F (A) in D for each object A in C, and for each morphism : A B of objects in C, a morphism F () : F (A) F (B) such that F (1 A ) = 1 F (A), and if we have morphisms f : A B and g : B C in C, then F (g f) = F (g) F (f). A contravariant functor F : C D is an assignment of an object F (A) in D for each object A in C, and for each morphism : A B of objects in C, a morphism F () : F (B) F (A) such that F (1 A ) = 1 F (A), and if we have morphisms f : A B and g : B C in C, then F (g f) = F (f) F (g). Examples: The inclusion of abelian groups into groups is a covariant functor. The forgetful functor is a covariant functor. For example, we could have this functor from the category of groups to the category of sets, where the functor sends a group to its underlying set and a group homomorphism is sent to itself (as a set map). Comment: The next few items here are intended as a review of things necessary to understand the double dual as a covariant functor, and the dual as a contravariant functor. Definition: Given a vector space V, the dual of V is V = Mor(V, R). Definition: Given vector spaces V, W and a linear map f : V W, we define the dual of the map as f : W V via f () = f where : W R is any linear transformation. Recall: If V is a real finite dimensional vector space, then Mor(V, R) is as well; moreover, they have the same dimension, so there exists an isomorphism between V and V (and with V but we care significantly less about that here). Definition: Let V be the category whose objects are real finite dimensional vector spaces, with morphisms as linear transformations. Example: We now define the functor F : V V as follows. Given an object V Ob(V), then F (V ) = V. Also, if V, W Ob(V), with f Mor(V, W ), then F (f) is a map from F (V ) to F (W ), that is from V to W given by F (f) = (f ) = f. Lemma: Given f : V W and g : W Z, where V, W, Z are finite dimensional real vector spaces, and f, g are linear transformations, then (g f) = f g. Proof. Let Z, that is : Z R. Then it is clear that (g f) () = (g f). Also, we have that (f g )() = f (g ()) = f ( g) = ( g) f. Now, as composition of morphisms is associative, we have that the action of (g f) and of f g is the same, and hence, they are the same function. 5 See 31 August 2011 for a reminder of what this category is.

6 Math 901 Notes 3 January of 33 Proposition: As defined F : V V is a covariant functor. Proof. First, we ll check that F (1 V ) = 1 F (V ) or equivalently, that (1 V ) = 1 V. So let V so that : Mor(V, R) R. Then clearly, 1 V () =. Also, 1 V () = 1 V. Both of these are maps from V to V, and we want to check they are the same map. So let λ V. Then 1 V ()(λ) = λ. Also, 1 V ()(λ) = ( 1 V )(λ) = (1 V (λ)) = λ 1 V = λ. Hence, we have F (1 V ) = 1 F (V ). Now, let f : V W, and g : W Z where V, W, Z Ob(V) and f, g are morphisms. Then we ll check that F (g f) = F (g) F (f). From the Lemma, we know that (g f) = f g. So applying the lemma to f g = (g f) then gives that F (g f) = ((g f) ) = (f g ) = g f = F (g) F (f) as required. 9 September 2011 More on Functors Last Time: The double dual gave us a covariant functor. Note here that we didn t need finite dimensional or the field we were working over to be R to get a covariant functor, but we ll discuss this functor again later, and then we ll make use of those properties. Example: We can get a contravariant functor D : V V by taking duals. On objects, the functor is defined by D(W ) = W = {λ : W R λ is a linear transformation}. On morphisms, f : V W, D(f) = f known as the pull back morphism and is obtained by composing with f. 6 We checked in class that this was a contravariant functor, but Sarah and I did that work already, and the explanation is already incorporated into the notes for the previous class (7 Sept 2011). Comment: We get the covariant functor from last time by composing D with itself, or equivalently, using push forwards. That is, we can define F : V V by F (W ) = W and F (f) = f. We now describe this push forward. Definition: Given f : V W we want to define f : V W. Let V. We have a morphism, : V R and we know what f is, and so we define f () = f. Alternate Viewpoint: Given f : V W and : V R, we want f () to be an element of W, so we should be able to evaluate f () at elements λ W. We then define (f ())(λ) = (λ f). 7 Definition: Let F be a covariant functor F : C D. Then for any two objects C 1, C 2 in C, there is a map Mor C (C 1, C 2 ) Mor D (F (C 1 ), F (C 2 )) which is given by f F (f). If the map is injective for all C 1, C 2 in C, then we say F is faithful. If the map is surjective for all C 1, C 2 in C, then we say F is full. If the map is bijective for all C 1, C 2 in C, then we say F is fully faithful. Examples: The inclusion functor from the category of abelian groups to the category of groups is fully faithful. The forgetful functor from the category of groups to the category of sets is faithful but not full. The functor from the category of sets to the category with a single object, and single morphism is full, but not faithful. 12 September 2011 Functor Categories & Representability of Functors Motivation: We discussed the motivation for functor categories. The basic idea was to stick in things that we want to have in our categories, and then maybe they were already there, but we have them now. We then defined two representation functors. Definition: Let A be a category, and let A be an object in A. Then we define a covariant functor, M A from A to the category of sets, S. On objects, this functor is M A (B) = Mor A (A, B), and if f Mor A (C, D), then M A (f) should be a morphism from Mor A (A, C) to Mor A (A, D), so we set M A (f) = f where f is the push forward map, f (g) = f g. Proposition: The above defines a covariant functor. 8 Definition: Similarly, we define a contravariant functor, M A from A to S. On objects, the functor is M A (B) = Mor A (B, A), and if f Mor A (C, D), then M A (f) should be a map from Mor A (D, A) to Mor A (C, A), so we set M A (f) = f where f is the pull back map, f (g) = g f. Proposition: The above defines a contravariant functor. 9 6 and was also defined on 7 Sept I m not sure how this helps, it may make more sense if we had other examples of a push forward, right now, it seems like a pull back still and to be honest, I m a bit confused. 8 The proof is not difficult, but it can be confusing to think about. 9 The proof is not difficult, but it can be confusing to think about.

7 Math 901 Notes 3 January of 33 Definition: Let C and D be categories. Also, let F and G be covariant functors from C to D. A natural transformation from F to G is a rule, Φ which assigns to every object C of C a morphism Φ C Mor D (F (C), G(C)) such that given any morphism f Mor(C 1, C 2 ) in C, the diagram below commutes. 10 F (f) F (C 1 ) F (C 2 ) Φ C1 Φ C2 G(f) G(C 1 ) G(C 2 ) Fact: Let C and D be categories. Let C D be the collection of covariant functors from C to D. This becomes a category if given two covariant functors F and G from C to D we define Mor C D(F, G) to be the natural transformations from F to G. 11 Definition: A natural equivalence is a natural transformation which is an isomorphism in the functor category. Definition: A covariant functor, F, from a category C to the category S is said to be representable if F is naturally equivalent to M A for some object A in C. A contravariant functor, G from C to the category S is representable if G is naturally equivalent to M A for some object A in C. Comment: We re now ready to discuss the vector space example again. Next time, we re going to show that the double dual functor is naturally equivalent to the identity functor on the category V as defined on 7 September September 2011 Natural Equivalence Example Example: We spent all of this day explaining why the identity functor on V, which we ll denote 1 V, is naturally equivalent to the double dual functor, F (as defined on 7 September 2011). Step 1: First we defined an evaluation map e W w : W R for any fixed object W in V and any fixed element w W, by λ λ(w). We showed this is a linear map, and hence is an element of W. Step 2: We next defined a map e W : W W for any finite dimensional real vector space, W via w e W w. We again showed 12 this is a linear map, and hence an object of Mor V (W, W ). Step 3: We next defined a natural transformation Φ from 1 V to F. For each object W of V we need a morphism Φ W : 1 V (W ) F (W ). Note that 1 V (W ) = W and F (W ) = W. We ll take Φ W = e W. We must show that this is natural. That is, given any morphism L : W 1 W 2 of real finite dimensional vector spaces, we must show that the following diagram commutes: L 1 V (W 1 ) 1 V (W 2 ) e W1 e W2 F (W 1 ) L = L F (W 2 ) We ll rewrite the diagram with easier notation: W 1 L W 2 e W1 e W2 L = L W1 W2 We checked that the diagram commutes in class, and it was not difficult. Aside: For a natural transformation Φ to be a natural equivalence, we just need for each object W, that Φ W is an isomorphism. We can then define an inverse natural transformation Ψ as Ψ W = (Φ W ) 1 and then we just need to check Ψ is a a natural transformation and that it is an inverse to Φ. However, I believe Brian Harbourne said this will ALWAYS happen if each Φ W is an isomorphism. 10 You can similarly define a natural transformation of contravariant functors by changing the direction of the horizontal arrows. 11 I believe that we call the category C D the functor category. 12 We actually just claimed this, and didn t show it; but it took me all of 60 seconds to write it down including finding a working marker.

8 Math 901 Notes 3 January of 33 Example: Back to our example, it is enough then to show that e W is an isomorphism for each W. We ll assume without proof that dim(w ) = dim(w ) = dim(w ). However, this fails unless dim(w ) <. We ll also use the fact that if dim(w ) = dim(w ) and e W is injective, then e W is an isomorphism. 13 We then made a clever choice in order to show that ker(e W ) = 0. Comment: People often say that W and W are canonically isomorphic. They re referring to the isomorphism e W because it defines a natural equivalence from 1 V to F. 16 September 2011 Examples of Representability Example: Let F be the forgetful functor from V to S. Note that F is a covariant functor. We showed that F is representable. That is, there is a vector space U such that F is naturally equivalent to M U = Mor V (U, ). Or restated again, we showed there is a vector space U such that for all objects W in V, there is a natural bijection between W = F (W ) and M U (W ) = Mor V (U, W ). In fact, we can say that R represents F, i.e., that F is naturally equivalent to M R. We define the natural transformation Φ : M R F via Φ W : Mor V (R, W ) W is given by L L(1). It ll be a homework problem to actually check that Φ W is a bijection for all objects W in V and also that Φ is natural. Example: Now, let W 1 and W 2 be fixed objects in V. We ll define F to be the contravariant functor from V to S which is given by F (W ) = Mor V (W, W 1 ) Mor V (W, W 2 ), and F (f) will be given by (λ 1, λ 2) (λ 1 f, λ 2 f). This functor is also representable. In this case, it is represented by U = W 1 W 2. Again, we didn t actually show the natural transformation, and I don t feel like typing the details of the setup. Example: Let F = {W i : i I} be a family of objects in V. We define a contravariant functor F from V to S similarly to the previous example. That is, on objects, F (W ) = i I Mor V(W, W i ) and given a morphism L : W W, we define F (f) to be the map from F (W ) to F (W ) which is given by λ i L : i I. i I λ i : i I i I However, this functor is not representable if i I dim W i =, as the vector space we would want it to be representable by (namely U = i I W i is not an object in V). Comment: Grothendieck had a solution to this issue of non-representability. The solution basically boils down to embedding the category into a larger category where the object you want exists. We ll start explaining this with the Yoneda Lemma next time. 19 September 2011 Yoneda Lemma Yoneda Lemma Setup: Let C be a category. We can define a covariant functor F from C to C S where C S is the category of contravariant functors C to S. 14 On objects, we define F (C) = M C. 15 Given a morphism f Mor C (C, D) we define F (f) to be the natural transformation F (f) : M C M D. In particular, if B is an arbitrary object in C, then we have (F (f))(b) : M C (B) M D (B) or equivalently (F (f))(b) : Mor C (B, C) Mor C (B, D) is given by g f g. Yoneda Lemma: F as defined above is a fully faithful functor. Proof. We need to check four things here. They are: 1. F (f) is a natural transformation: Let A, B be objects in C with a morphism g : B A. We must show the following diagram commutes: M C (A) M C (g) M C (B) F (f)(a) F (f)(b) M D (A) M D (g) M D (B) 13 This is because the dimensions are equal, so a basis must map to a basis to get injectivity. 14 See Fact on 12 September. 15 Recall that M C is a contravariant functor.

9 Math 901 Notes 3 January of 33 So let M C (A) = Mor C (A, C). Then we have (F (f)(b) M C (g))() = F (f)(b)( g) = f g. Similarly, we get (M D (g) F (f)(a))() = M D (g)(f ) = f g, so that the diagram commutes. However, this is obvious as M C (g) and M D (g) act by composing with g on the right, and F (f)(a) and F (f)(b) act by composing with f on the left, and composition is associative. 2. F is a covariant functor: Let C be an object in C, and let 1 C be the identity morphism on C in C. Then, F (1 C ) is a natural transformation from M C to M C. So let B Ob(C), then F (1 C )(B) : M C (B) M C (B) is given by g g 1 C = g. Thus, F (1 C ) = 1 F (C). We now need to look at compositions. So let f Mor C (A, B) and g Mor C (B, C). Then we want to show that F (g f) = F (g) F (f). For any object D in the category C, then F (g f)(d) : M A (D) M C (D) is given by composition with g f on the left, so we get that h (g f) h. Similarly, F (f)(d) : M B (D) M C (D) is given by composition with f on the left, so h f h, and F (g)(d) : M A (D) M B (D) is given by composition with g on the left, so h g h. Now, if we have h M A (D), then (F (g) F (f))(h) = F (g)(f h) = g (f h). However, composition is associative, and so we have that F (g f) = F (g) F (f). 3. F is faithful: Let C, D be objects in C and consider the map Mor C (C, D) Mor C S (M C, M D ) which is given by f F (f). We must show that this map is injective. So let g, h Mor C (C, D) such that g h. We must show that F (g) F (h). We ll show this on C. So consider F (g)(c), F (h)(c) : M C (C) M D (C). Then, F (g)(c)(1 C ) = g 1 C = g h = h 1 C = F (h)(c)(1 C ), so that F (g)(c) F (h)(c) and hence F (g) F (h). 4. F is full: Again, Let C, D be objects in C and consider the map Mor C (C, D) Mor C S (M C, M D ) which is given by f F (f). We must show that this map is surjective. So let η Mor C S (M C, M D ). We ll set α = η(c)(1 C ). Note first that η(c) : M C (C) M D (C) so that α Mor C (C, D). We now claim that F (α) = η. To show this, let A be an object in C and Mor C (A, C). Then as η is a natural transformation, we know the following diagram commutes: M C (C) η(c) M D (C) = M C () M D () = M C (A) η(a) M D (A) Now, tracing the identity morphism on C will give us what we want. Note here that the red equality is due to the definition of α, the blue equality is due to the diagram commuting, and the green equality is due to the definition of F (α). id C η(c)(id C )= α id C = η(a)()= α =F (α)(a)() Thus the diagram tells us that η(a)() = F (α)(a)() and so we have that η = F (α) as desired so that the map is surjective. 21 September 2011 Representability and Free Groups Example: Let X be a set. We ll define a covariant functor F X from Ab to S. On objects, F X (A) = Mor S (X, A) and on morphisms, if g : A B, then F X (g) : F (A) F (B) is given by g. We wonder now if F X is representable. It is, and I ll show how we got there in class. Proof. We started by assuming that F X is representable and seeing what that meant about the functor and about the object that represents it. So we ll assume that B represents F X, i.e. that F X is naturally equivalent to M B. Let Φ : M B F X be the natural transformation. Then in particular we d have Φ B : M B (B) F X (B) which is given by 1 B Φ B (1 B ) =, where is simply defined to be the image of Φ B applied to the identity on B. However, we really want to know what Φ is in general. So let D be an abelian group, and g : B D a group homomorphism. We create the following commutative diagram to help us figure out what Φ D (B) is.

10 Math 901 Notes 3 January of 33 M B (B) M B (g) = g M B (D) Φ B Φ D F X (B) F X (g) = g F X (D) Tracing the element 1 B tells us what we need, namely, that Φ D (g) = g = g (). See the below diagram where the red equality is what will make the diagram commute. 1 B g (1 B ) = g g () = g = Φ D (g) Claim: The pair (B, : X B) satisfies the following universal property: for any abelian group D and any set map ψ : X D there is a unique homomorphism h ψ : B D such that the following diagram commutes: B h ψ D X Sub-Proof. This proof was fairly simple and was based entirely on the fact that we re assuming that Φ is a natural equivalence, i.e., that Φ D is an isomorphism of sets (which is just a bijection). Comment: The pair (B, ) exists, and is referred to as the free abelian group on the set X. Comment: The above shows that the free abelian group on the set X represents the functor F X. Construction/Definition: A much more explicit construction of a free abelian group can also be given. Here, we did this as the free abelian group on the set X is B = x X Z, with the map : X B. More generally, if {B x : x X} is a family of abelian groups such that B x = Z for all x X, then B = x X B x and : X B is { 1, x = y given by x e x where e x : X Z is the map e x (y) = δ xy = 0, x y. With all this, we denote the canonical inclusions by π x : B x B. Claim: We claim now that B is a free abelian group, i.e. that it satisfies the appropriate universal property. The idea is to either: Show directly that (B, ) satisfies the universal property OR Use the universal property of B as a categorical coproduct (sum) to show that (B, ) satisfies the universal property of a free abelian group. ψ 23 September 2011 Free Groups & Coproducts Claim: (B, ) as defined in the last class is a free abelian group, that is, it satisfies the following universal property: given any set map Ψ : X D, where D is an abelian group, there is a unique homomorphism h Ψ : B D such that the following diagram commutes: h Ψ B D X Ψ Proof. We did most of the details of this in class. However, we used the fact that B is a categorical coproduct, and that basically gave us what we wanted, so I m going to not rewrite down the details since neither existence nor uniqueness seemed at all tricky.

11 Math 901 Notes 3 January of 33 Definition: Let X be a set, F a group, and : X F a set map. We say (F, ) is a free group on the set X if for every group G and every map Ψ : X G, there exists a unique homomorphism h Ψ : F G such that the diagram below commutes. h Ψ F G Fact: Coproducts and free groups exist in the category of groups. Proposition: Let A and B be groups with A B = (for simplicity). Define the free product A B = A B to be the set of sequences (a 1,..., a m ) for m 0 such that a i (A \ {1 A }) (B \ {1 B }) and such that if a i A, then a i+1 / A and if a i B then a i+1 / B. Then A B is a group under the following: 1 A B is the unique sequence of length 0, namely (), (a 1,..., a m ) 1 = (a 1 m,..., a 1 1 ), and (a 1,..., a m, b 1,..., b n ) if a m, b 1 are in different groups, A, B ab = (a 1,..., a m ) (b 1,..., b n ) = (a 1,..., a m b 1,..., b n ) if a m, b 1 A (or B), and a m b 1 1 A or 1 B. (a 1,..., a m 1 ) (b 2,..., b n ) if a m b 1 = 1 A or 1 B X Proof. We discussed the ideas behind this proof in the next class, but I don t think that any of them are really worth typing up. Basically, these things make sense. The only difficulty arose in showing associativity, 16 and a sort of clever induction argument made it not so bad. Ψ 26 September 2011 More on Coproducts & Free Groups Comments: A B = B A We have the obvious injective homomorphisms: 17 π A : A A B, and π B : B A B where x (x) for x A (or x B as is appropriate), and obviously in each case the identity maps to the identity in A B which is the sequence of length 0. Given any group homomorphisms, : G for = A, B, then we claim there is a unique 18 group homomorphism : A B G such that the following diagrams commute for = A, B: A B G π In fact, we define as follows: Clearly we set (()) = e G. Also, ((x)) = (x) where x and = A, B, and in general, if x = (x 1,..., x n ), then (x) = ((x 1 ))... ((x n )). There are a bunch of cases to check it is a homomorphism, but it seems clear enough without checking them. Also, the fact that the diagrams commute seems clear, so I won t write down any of the details of either. 19 Corollary: Let A and B be groups, which we regard as having disjoint sets of elements. Then (A B, π A, π B ) is a coproduct of A and B in the category of groups. Proof. It only remains to show uniqueness of the map. However, this is easy to show since the diagrams must commute and must be a homomorphism, so I ll omit the details from these typed notes. Examples: The coproduct of A = Z/2Z = {0, 1 A } and B = Z/2Z = {1 B, 1} contains the elements: A B = {(), (1 A ), ( 1), (1 A, 1), ( 1, 1 A ), (1 A, 1, 1 A ), ( 1, 1 A, 1),...}. Note here that there are exactly two elements of each length. However, in the category of abelian groups, the coproduct is the direct sum, namely A B = Z/2Z Z/2Z which is the Klein Vier Group. 20 Lemma: Let X 1, and X 2 be disjoint sets, and assume (F i, i ) is a free group on X i for i = 1, 2. Then F 1 F 2 is the free group on the set X = X 1 X 2 with respect to the map : X F 1 F 2 defined by x π i ( i (x)). 16 as always! 17 that they are injective and homomorphisms is clear. 18 I haven t asserted uniqueness, but it will be unique. It s the freeness that makes it unique. B.H. 19 I m aware that this silly notation sucks, but I m lazy and don t want to do everything twice for both A and B. 20 Vier is 4 in German.

12 Math 901 Notes 3 January of September 2011 Free (Abelian) Groups Lemma: If X 1, X 2 are disjoint sets, and (F i, i ) is a free group on X i for i = 1, 2, then F 1 F 2 is a free group on X = X 1 X 2. Proof. Let π i : F i F 1 F 2 be the coproduct canonical homomorphisms, and define : X F 1 F 2 via x π i ( i (x)) for x X i. We need to show that (F 1 F 2, ) is a free group on X. So let ψ : X G be any set map to any group G. We must show there is a unique homomorphism h ψ : F 1 F 2 G such that h ψ = ψ. In order to work the coproduct into the proof, we ll need homomorphisms Γ i : F i G. Let Γ i be the unique homomorphism guaranteed by the universal property of F i free from the set map γ i : X i G, where γ i = ψ Xi. The maps Γ i then induce the map Γ : F 1 F 2 G since F 1 F 2 is a coproduct. Set h ψ = Γ. We must show that h ψ = ψ and that this map is unique. So, let x X = X 1 X 2. Without loss of generality (WLOG), let x X 1. Then, (h ψ )(x) = Γ π i i (x) = Γ 1 1 (x) = γ 1 (x) = ψ X1 (x) = ψ(x). 21 Examples: 1. The free group on the empty set is the trivial group. 2. The free group on X = {1} is isomorphic to Z. 3. Inductively, using the lemma, we have the the free group on a set X = {1,..., n} is Z }. {{.. Z }. n times 4. It is possible to define a free group on an infinite set as well, and Lang describes this. It is similar in idea to what we ve done so far, but is a bit messier. 22 Lemma 1: Let (F, : X F ) be a free group. Then is injective, and (X) generates F. Proof. Notice that { the injectivity is trivial if X < 2, so we may let x 1, x 2 be distinct elements of X. Define ψ : X 1, x x1 Z/2Z as ψ(x) =. Then by the universal property, there is a unique homomorphism h ψ : F Z/2Z 0, x = x 1 such that h ψ = ψ. However, ψ(x 1 ) ψ(x 2 ), so we must have that (x 1 ) (x 2 ) due to the fact that h ψ = ψ. Next, let G be the subgroup of F generated by (X). We ll show G = F. Consider the diagram below: h i F G F X Notice that by definition of F as a free group, there is a unique homomorphism h : F F such that h =. As 1 F works for h, then we must have that 1 F = i h. Since 1 F is a bijection, then we must have that i is surjective, meaning that F G, and hence F = G. Lemma 2: Let (A, : X A) be a free abelian group. Then is injective and (X) generates A. Proof. The proof is the same as the proof of Lemma 1, mutatis mutandis September 2011 Free (Abelian) Groups Example: If X is a set, then A = x X Z together with e : X A given by x e x is a free abelian group, where e x : X Z is given by e x (y) = 0 for y x, and e x (x) = 1. Lemma 3: Let (A, : X A) be a free abelian group on X. Then every element a A has a unique expression a = x X m x (x), where m x Z. 21 This is obvious once you draw the pictures, but there are too many to be worth putting in here. 22 obviously! 23 See Hungerford page xv. This is latin for (roughly) by changing the things which (obviously) must be changed (in order that the argument will carry over and make sense in the present situation).

13 Math 901 Notes 3 January of 33 Proof. Notice that existence here( is due to) Lemma 2 since (X) generates A as an abelian group. So we really only need to show uniqueness. Since Z, e is a free abelian group, then we have homomorphisms h e and h such x X that h e = e and h e =. Now, suppose m x (x) = n x (x). We ll apply h e to get the following: x X x X mx e x = ( ( ) m x h e ((x)) = h e mx (x)) = h e nx (x) = n x h e ((x)) = n x e x. We can apply these functions to any element y X to get that m y = m x e x (y) = n x e x (y) = n y for all y X. Hence, the two expansions are equivalent, which gives us uniqueness. Definition: A subset, B, of an abelian group, A, is a basis for A if B generates A in a unique way. Definition: If B is a basis for A, then we say B is the rank of A. Lemma 4: If (F, ψ : X F ) is a free abelian group and we have a commutative diagram: F ψ then (G, ) is free abelian on X as well. Proposition: If B is a basis for an abelian group A, then (A, ) is a free abelian group where : B A is the inclusion of B into A. Proof. Let (F, ψ) be a free abelian group on the set B. Then we have the induced commutative diagram: X G F h A ψ B We now claim that in fact h is an isomorphism. We ll first do surjectivity. As B = (B) generates A, and (B) = ψ(h (B)), then h (F ) contains B, so that h (F ) = A since B is a basis for A. Next, we ll do injectivity. Let f ker(h ). Since ψ(b) generates F, then we get f = b B m b ψ(b) for some m b Z. Now, ( ) 0 = h (f) = h mb ψ(b) = m b h (ψ(b)) = m b (b), which means that m b = 0 for all b B, and hence f = 0, so that h is injective. Applying Lemma 4 now completes the proof. Lemma 5: If we have a free abelian group (A, : X A) and a set Y with a bijection β : X Y, then (A, β 1 ) is a free abelian group on Y. 24 Corollary: Let A be an abelian group. Let X be a set with a map : X A. then (A, ) is a free abelian group if and only if (X) is a basis for A and is injective. 25 Fact: If X is an infinite set, and Fin(X) is the set of finite subsets of X, then X = Fin(X). 3 October 2011 Free Groups, Group Actions Theorem: Let (A i, i : X i A i ) be free abelian groups for i = 1, 2. Then A 1 = A2 if and only if X 1 = X The proof here is easy if you draw the pictures. 25 The proof here is an application of the lemmas and the proposition.

14 Math 901 Notes 3 January of 33 Proof. ( ) Let ψ : X 1 X 2 be a bijection. Since (A 2, 2 ) is free abelian on X 2, then (A 2, 2 ψ) is free abelian on X 1 by Lemma 5. Thus, A 1 = A2 by the universal property of their both being free on X 1. ( ) Now, assume A 1 = A2, and let h : A 1 A 2 be an isomorphism. Set 2A i = {2a a A i } and note that 2A i A i. Set A i = A i /2A i. We claim now that h induces an isomorphism h : A 1 A 2. So set h(a) = h(a). We first show this is well defined. Suppose a = b A 1. Then a b 2A 1. Note that h(2a 1 ) = 2h(A 1 ) 2A 2 so that h(a) h(b) 2A 2 which gives h(a) = h(b). Next, we must show that h is an isomorphism. In fact, we have the following commutative diagram: h A 1 A 2 h A 1 A 2 h Since h and are surjective, then so is h. We claim then that ker( h) = 2A 1 and apply the first isomorphism theorem. 26 Note that ( h)(2a) = 2h(a) = 0 in A 2 so that 2A 1 ker( h). Now, suppose ( h)(a) = 0. Then we have h(a) = 0 so that h(a) 2A 2. So let h(a) = 2b for some b A 2. But h is an isomorphism, so there exists some β A 1 such that h(β) = b. Thus, h(a) = 2h(β) = h(2β) so that a = 2β 2A 1 since h is an isomorphism. Thus, h is an isomorphism. Now, for i = 1, 2 we have the following: Z Z A i x X i = 2 x X i = = Z/2Z, Z 2Z x X i x X i x X i which gives the isomorphism Z/2Z = Z/2Z. Now, if X 1 <, then we get x X 1 x X 2 2 X1 = Z/2Z = Z/2Z = 2 X2 x X 1 x X 2 which means that X 1 = X 2. If, on the other hand, X 1 (and hence also X 2 ) is infinite, then we have X 1 = Fin(X 1 ) = Z/2Z = Z/2Z = Fin(X 2) = X 2 x X 1 x X 2 which completes the proof. Comment: We re now going to move on from free groups and start discussing group actions. We ll also briefly cover the Sylow Theorems, and will probably cover semi-direct products. Definition: Let G be a group and X a set. An action of G on X is a map α : G X X (usually we write g x or simply gx for the action of g on the set element x) such that 1. 1 G x = x for all x X, and 2. g 1 (g 2 x) = (g 1 g 2 ) x for all x X and for all g 1, g 2 G. Definition: We say the orbit of an element x X is the set G x = {g x g G}. Definition: Let x X. Then the stabilizer of x is Stab G (x) = {g G g x = x}. Notice here that the stabilizer is a subgroup of G. Comment: Sometimes the notation G x is used for the stabilizer of x, but that notation sucks given our notation for the orbits! Definition: We say an action is transitive if it has a single orbit. Equivalently, if for some x X (but really it will be true for all x X), we have G x = X. 26 Apparently the isomorphism theorems are due to Emmy Noether?

15 Math 901 Notes 3 January of 33 Comment: Giving an action of G on X is equivalent to giving a homomorphism from G to the group of permutations on the set X. Note here that this group is really the group of bijections X X where the group law is composition. We ll call this group of permutations Perms(X). 5 October 2011 More Group Actions; Starting Semi-Direct Products Recall: We have two equivalent definitions of an action being transitive. The first is that there is a unique orbit. The second is that given any two elements x, y X, there is a group element g G such that g x = y. Example: We discussed the action of S 1 on C where S 1 acts on C via regular multiplication in C. Here, this action is not transitive, but each orbit is a circle centered about the origin. Theorem: (In particular a version of Cayley s Theorem) Let : G X X be an action of a group G on a set X, let x X, and H = Stab G (x). 1. Then induces a homomorphism Φ : G Perms(X) which is given by g π g where π g (y) = g y. 2. If the action is transitive, then ker Φ H and moreover, ker Φ = ghg 1 and lastly, ker Φ contains every subgroup N G such that N G and N H. Proof. We ll show each part: 1. We first need to show that π g is actually a permutation for every g G. So note that given g G, then (π g π g 1)(x) = π g (g 1 x) = g(g 1 x) = e G x = x and (π g 1 π g )(x) = g 1 (gx) = e G x = x so that π g is a bijection for each g and hence an element of Perms(X). Next, we need to show that Φ is a homomorphism. That is, we need to show that π g1g 2 = Φ(g 1 g 2 ) = Φ(g 1 ) Φ(g 2 ) = π g1 π g2. So let x X. We ll show that both permutations send x to the same place. Now, π g1g 2 (x) = (g 1 g 2 )x = g 1 (g 2 x) = π g1 (g 2 x) = π g1 (π g2 (x)) = π g1 π g2 (x). Thus, Φ is a homomorphism. 2. First, we ll show that ker Φ H. So let a ker Φ. Then π a = 1 X so that π a (y) = a y = y for any y X. Hence, a x = x so that a Stab G (x) = H. Next, we need to show that ker Φ ghg 1. So let a ker Φ and let g G. It is clearly enough to show that g G a ghg 1. If a ker(φ) then a(gx) = gx so that g 1 agx = x. this means that g 1 ag H and so a ghg 1. Next, we ll show the other inclusion ker Φ ghg 1. So let b ghg 1, and let y X. Then y = γx for g G some γ G since the action is transitive. Also, b = γhγ 1 for some h H by choice of b. Thus, multiplying both sides of these two equations together gives by = γhγ 1 γx. Simplifying the right hand side gives by = γhx = γx = y. Thus, Φ(b) = π b = 1 X so that b ker(φ). Lastly, we must show that ker(φ) is the largest normal subgroup of G which is contained in H. So let N G with N H, and choose any g G. Then N = gng 1 ghg 1. Thus, N ghg 1 ker(φ) which completes the proof. Corollary: If a group G acts transitively on a finite set X and if G > n! where n = X > 1, then G has a nontrivial normal subgroup, N. Proof. We didn t prove this in class, but I ll do it now. Since G > X! = Perms(X), then the homomorphism Φ from part 1 of the theorem cannot be injective. This means that ker(φ) {1 G }. By part 2 of the theorem above, we know that H = Stab G (x) contains ker(φ) for any x X. Now, since the action is transitive, and X has more than one element, then there must be at least one g G such that gx x. Hence, H G, so that ker(φ) is a proper subgroup of G as we now have {1 G } ker(φ) H G. Since the kernel of any group homomorphism is a normal subgroup, then we have that ker(φ) is a non-trivial normal subgroup. Semi-Direct Products: We can do this in a more general setting, but we ll give an example first to illustrate the idea. Example: Let G be a group, with subgroups N, H such that N H = {1 G } and N G. Then, NH = {nh : n N, h H} is a subgroup of G. We let H act on N by conjugation, and can refer to NH as an internal semi-direct product. g G g G g G

16 Math 901 Notes 3 January of 33 7 October 2011 Semi-Direct Products & Back to Group Actions Definition: Let N, H be groups and Φ : H Aut(N) where Aut(N) is the set of automorphisms of N. Note that Aut(N) Perms(N). We define the semi-direct product, N Φ H, as follows. As a set N H is just the set N H. We define the group operation 27 as ( (n 1, h 1 ) (n 2, h 2 ) = (n 1 Φ(h1 )(n 2 ) ) ), h 1 h 2. Claim: N = N {1 H } N H and H = {1 N } H N H. 28 Recall: If N = Z/pZ where p is a prime, then Aut(N) = Z/(p 1)Z, and in general, Aut(Z/nZ) is of order φ(n) where φ is the Euler φ function. Example: Let N = Z/7Z and H = Z/3Z. Then, Aut(N) = Z/6Z. Let Φ : H Aut(N) be an injective homomorphism given by [1] 3 [2] 6. Now, N H has order 21, but is not an abelian group. You can see this by considering the two elements ([1] 7, [0] 3 ) and ([0] 7, [1] 3 ) which do not commute since Φ([1] 3 ) is not the identity automorphism. On the other hand if we were to try and construct a semi-direct product of the groups in the other order, H N, then we must use the zero map Φ : N Z/2Z and so H N = H N. Lemma: Let G be a finite group, acting on a finite set X and choose any x X. Then G = G x G x. Proof. This is mostly just an application of Lagrange s theorem, and noting that there is a bijection between elements of the orbit G x and the left cosets of Stab G (x) = G x in G. Orbit Decomposition Theorem: Let G be a finite group acting on a finite set X. Let X 1,..., X r be the distinct orbits in X for the action. Choose x i X i for each i. Then X = r G/G xi = i=1 r [G : G xi ]. Proof. This proof is a simple application of the lemma and Lagrange s theorem. i=1 10 October 2011 More Group Actions Proposition: Let G be a finite group acting on a finite set X. Let n be the number of orbits. Let f : G Z be a map where f(g) is the number of fixed points of g, i.e. f(g) = {x X : gx = x}. Then n G = g G f(g). 29 Proof. Let X 1,..., X n be the distinct orbits, and set f i (g) = {x X i : gx = x}. Note then that f(g) = n f i (g). Let Γ i be a correspondence in G X i. In particular, let Γ i = {(g, x) G X i gx = x}. We then have the following diagram where π 1 and π 2 are the usual projection maps: i=1 Γ i G X i π 1 G π 2 X i Consider π 2 Γi : Γ i X i. Note that this map is surjective as (e G, x) Γ i for all x X i. Also, the fibers are (π 2 Γi ) 1 ({x}) = {(g, x) : gx = x} = Stab G (x) {x}, and (π 1 Γi ) 1 ({g}) = {(g, x) : x X i and gx = x}, 27 We didn t actually show this operation turned N H into a group, but it isn t hard to show, and some people even did it as homework. The only tricky part is determining that (n, h) 1 = (Φ(h 1 )(n 1 ), h 1 ). 28 This seems not too hard to show and part of it may have been done by some people for homework. 29 This was actually stated the previous day, but proved this day, so it s listed here.

17 Math 901 Notes 3 January of 33 which is the set of fixed points of G in X i and so (π 1 Γi ) 1 ({g}) = f i (g). The fibers of a map are always disjoint, and their union is the domain; hence, f i (g) = Γ i = Stab G (x). Recall that G = Stab G (x) G x. g G x X i Hence, f i (g) = Γ i = G / X i = G. Now, we add these up over i to get g G x X i n n n f i (g) = f i (g) = G = n G. g G f(g) = g G i=1 i=1 g G Corollary: Say G is a finite group acting transitively on a finite set X with X > 1. Then there exists an element g G which is fixed-point free, that is, so that gx x for all x X. In the language of the previous proposition, this says there is an element g G such that f(g) = 0. Proof. Suppose not, then for each g G there exists an x g X such that g x g = x g. Or equivalently, f(g) 1 for every g G. Since the action is transitive, G = g G f(g). This means that f(g) = 1 for every g G, but this is a contradiction since f(1 G ) = X 2. Corollary: Let G be a group, H a subgroup such that H G. If G is a finite group, then ghg 1 G. Proof. We know G acts on the left cosets of H by left multiplication. Set X = {gh} g G. Since H G, then we have that X > 1. Also, note that this action is transitive, and choose any a G. Then a ghg 1 for some g G if and only if g 1 ag H, which happens if and only if ag gh, or equivalently if and only if a is in the stabilizer of gh under the group action, i.e. a gh = gh. Now, let S = ghg 1. Then a S if and only if f(a) 1, where as before, f(a) is the number of fixed points of a under the action, or equivalently, f(a) is the number of conjugates ghg 1 which contain a. However, by the previous corollary, there is some b G such that f(b) = 0. Hence, b / S. g G i=1 g G 12 October 2011 More Group Actions & the Class Equation We now provide an alternate proof for the Corollary proved last time: Proof. Alternate Proof: Let G act on the set of conjugates of H by conjugation, and set X = {ghg 1 }. 30 Note that Stab G (H) = {g G : ghg 1 = H} = N G (H). 31 Recall that ghg 1 = H for all g G, and that 1 G ghg 1 for all g G. Let n be the number of conjugates of H, i.e. n = X = G / Stab G (H) = G / N G (H). If n = 1, then H G and for all g G, we have ghg 1 = H, so clearly ghg 1 G. So suppose n 1. We have the following computation: ghg 1 G n ( H 1) + 1 = N G (H) ( H 1) + 1 = G G H N G (H) N G (H) + 1. g G Now, N G (H) G, H / N G (H) 1, and G / N G (H) = n 2, and so from the computation we get that ghg 1 G = G 1 < G. g G Class Equation: Let G act on itself via conjugation g h = ghg 1. This gives us a homomorphism Φ : G Aut(G) with kernel the center of the group, which we recall is denoted Z(G). The image of Φ is a subgroup of Aut(G) which is known as the group of inner automorphisms, and is sometimes denoted Inn(G). For h G, we have that Stab G (h) is {g G : ghg 1 = h} = C G (h), the centralizer of h in G. More generally, the centralizer of a subset S G is C G (S) = {g G gs = sg for all s S}. Note here that C G (G) = Z(G). We now assume that G is a finite group, and that X 1,..., X n are the disjoint orbits of the group under the conjugation action. For each i, choose any x i X i. Then we have the class equation: 30 This IS an action. 31 Recall that N G (H) is the normalizer of H in G. G = n X i = i=1 n [G : C G (x i )]. i=1 g G