18 MATH 101A: ALGEBRA I PART A: GROUP THEORY7. Category theory and productsI want to prove the theorem that a finite group is nilpotent if andonly if it is the product of its Sylow subgroups. For this we first haveto go over the product of groups. And this looks like a good time tointroduce category theory.7.1. categories. I gave the definition of a category and two examplesto illustrate the definition: Gps is the category of groups and Ens isthe category of sets.Definition 7.1. A category C consists of four things: C = (Ob(C), Mor, ◦, id)where(1) Ob(C) is a collection of objects. This collection is usually nota set. For example, Ob(Gps) is the collection of all groups andOb(Ens) is the collection of all sets.(2) For any two objects X, Y ∈ Ob(C) there is a set of morphismsMorC(X, Y )which are written f : X → Y . For example, in MorGps(G, H)is the set of homomorphisms φ : G → H and MorEns(S, T ) isthe set of all mappings f : S → T .(3) For any three objects X, Y, Z, we have a composition law:MorC(Y, Z) × MorC(X, Y ) → MorC(X, Z)sending (g, f ) to g ◦ f. Composition must be associative.(4) Every object X ∈ O b(C) has an identity idX∈ MorC(X, X) sothat idY◦ f = f = f ◦ idXfor any f : X → Y .Note that there are only two assumptions about the structure. Namely,associativity of composition and the existence of units.The idea of category theory is to extract elementary concepts outof difficult mathematics. We look only at composition of morphismsand forget the rest of the structure. Then we can ask: What arethe properties that can be expressed only in terms of c omposition ofmorphisms? One of these is the product.7.2. product of groups.Definition 7.2. If G, H are groups, then the product G ×H is definedto be the cartesian product of setsG × H = {(g, h) | g ∈ G, h ∈ H}MATH 101A: ALGEBRA I PART A: GROUP THEORY 19with the group law given coordinate-wise by(g1, h1)(g2, h2) = (g1g2, h1h2).These are several things to notice about this definition. The first isthat G × H contains a copy of G, H which commute. By this I meanthat there are monomorphisms (1 − 1 homomorphisms):j1: G → G × H, j2: H → G × Hgiven by j1(g) = (g, e), j2(h) = (e, h). These inclusion maps havecommuting images sincej1(g)j2(h) = (g, e)(e, h) = (g, h) = (e, h)(g, e) = j2(h)j1(g)7.2.1. internal direct product.Lemma 7.3. If φ : G → K, ψ : H → K are homomorphisms withcommuting images then there is a unique homomorphism f : G × H →K so that f ◦ j1= φ and f ◦ j2= ψ.Proof. This is obvious. f must be given by f(g, h) = j1(g)j2(h). Thisis a homomorphism since [j1(G), j2(H)] = {e}. !Theorem 7.4. Suppose that G contains normal subgroups H, K so thatH ∩ K = {e}, [H, K] = {e} and HK = G. Then the homomorphismf : H × K → G given by the inclusion maps H #→ G, K #→ G is anisomorphism.We say that G = H × K is the internal direct product in this case.Proof. The map is given by f(h, k) = hk. This is surjective sinceHK = G. It is 1 − 1 since H ∩ K = {e}. It is a homomorphism since[H, K] = {e}. !7.2.2. universal property. The product G ×H has two other projectionhomomorphismsp1: G × H → G, p2: G × H → Hgiven by p1(g, h) = g, p2(g, h) = h. These satisfy the following “uni-versal” prop e rty which is obvious (obviously true) and which I alsoexplained in categorical terms.Theorem 7.5. Suppose that G, H, K are groups and φ : K → G, ψ :K → H are homomorphisms. Then there exists a unique homomor-phism f : K → G × H so that p1◦ f = φ and p2◦ f = ψ.The unique homomorphism is f(x) = (φ(x), ψ(x)) and it is writtenf = φ × ψ.20 MATH 101A: ALGEBRA I PART A: GROUP THEORY7.3. categorical product. The last theorem is categorical since itinvolves only composition of homomorphism. It says that G × H is acategorical product.Definition 7.6. Suppose that X, Y are objects of a category C. ThenZ ∈ C is the product of X and Y if there are morphisms p1: Z →X, p2: Z → Y so that for any other object W and any morphismsφ : W → X, ψ : W → Y there is a unique morphism f : W → Z sothat p1◦ f = φ and p2◦ f = ψ.The condition can be written as a commuting diagram:XWφ!!!!!!!!!!ψ""""""""""∃!f##Zp1$$#######p2%%$$$$$$$YWe say that Z is the product of X, Y in the category C and we writeZ = X × Y . We also call Z the categorical product of X and Y .Theorem 7.5 was written in such a way that it is obvious that theproduct of groups is the categorical product.The next point I made was that the definition of a product definesZ = X × Y uniquely up to isomorphism.The concept of an isomorphism is categorical:Definition 7.7. Two objects X, Y in any category C are isomorphicand we write X∼=Y if there are morphisms f : X → Y and g : Y → Xso that f ◦ g = idYand g ◦ f = idX.The definition of product is by a “universal condition” which forcesthe object Z to be unique up to isomorphism if it exists. (If the productdoes not exist, it suggests that the category is not large enough andperhaps we should add more objects.)Theorem 7.8. The product Z = X × Y is unique up to isomorphismassuming it exists.Proof. Suppose that Z"is another product. This means what we havemorphisms p"1: Z → X, p"2: Z → Y so that for any W , such asW = Z, any morphisms W → X, W → Y (such as p1, p2) there is aunique morphism g so that p"i◦ g = pifor i = 1, 2. In other words, theMATH 101A: ALGEBRA I PART A: GROUP THEORY 21following diagram commutes.XZ"p!1!!%%%%%%%%p!2""&&&&&&&&Z∃!g&&p1$$#######p2%%$$$$$$$YSimilarly, since Z is the product, there is a unique morphism f : Z"→Z so that pi◦ f = p"ifor i = 1, 2. Now, take Z and Z. We have twomorphisms Z → Z making the following diagram commute:XZp1!!$$$$$$$p2""#######f◦gidZ##Zp1$$#######p2%%$$$$$$$YBy the uniqueness clause in the definition of the product, we must havef ◦ g = idZ.Similarly, g ◦ f = idZ!. So, Z∼=Z". !7.4. products of nilpotent groups. Finally, I proved the followingtheorem which we need. I used three lemmas without proof. But, I amgiving the proofs here after the proof of the theorem (and retroactivelyin Corollary 5.3).Theorem 7.9. If G, H are nilpotent groups of nilpotency class c1, c2resp. then G × H is nilpotent of class c = max(c1, c2).The proof is by induction on c using the following lemma.Lemma 7.10. The center of G × H is Z(G) × Z(H).Proof. This is obvious. A n element (x, y) ∈ G × H …