MATH 101A: ALGEBRA I PART A: GROUP THEORY 39

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1 MAT 101A: ALEBRA I PART A: ROUP TEORY Categorical limits (Two lectures by Ivan orozov. Notes by Andrew ainer and Roger Lipsett. [Comments by Kiyoshi].) We first note that what topologists call the limit, algebraists call the inverse limit and denote by lim. Likewise, what topologists call the colimit, algebraists call the direct limit and denote lim. Example Take the inverse system C[x]/(x n ) C[x]/(x 3 ) C[x]/(x 2 ) C[x]/(x) Let f lim C[x]/(x n ) = C[[x]]. Then f = n 0 a nx n is a formal power series in x over C. We note that Z p = lim n Z/p n [is the inverse limit of] Z/p n+1 Z/p n Z/p colimits in the category of sets. Definition Let X α be sets indexed by α I and let f α,β : X α X β be functions with α, β I. [Only for pairs (α, β) so that α < β in some partial ordering of I.] Then {f α.β, X α } α I is a directed system of sets if, for every pair of composable morphisms f α,β : X α X β, f β,γ : X β X γ [i.e., wherever α < β < γ in I], the following diagram commutes f α,β X α X β f α,γ X γ and, for every α, β I there exists a γ I for which there are maps f α,γ and f β,γ [i.e., α, β γ] as such: f β,γ X α f α,γ X γ X f β β,γ One can think of a directed system as a graph with sets as points and arrows as edges. We can now define the direct limit on directed systems of sets by lim X α = X α / α I α where, for each f α,β and for all x X α, we set x f α,β (x). Informally then, the direct limit is the set of equivalence classes induced by all functions f α,β. That is an equivalence relation follows from the.

2 40 MAT 101A: ALEBRA I PART A: ROUP TEORY two properties illustrated diagramatically above. [The colimit of any diagram of sets exists. The assumption of being directed implies that any two elements of the colimit, represented by say x X α, y X β, are equivalent to elements of the same set X γ.] pull-back. In the following diagram, is a pull-back : β K The pull-back here is a subgroup (or subset) of given by K = {(g, h) α(g) = β(h)} universal property of pull-back. If is a group such that α (g, h) K commutes then there exists a unique map K such that h β K α commutes. K K push-forward [push-out] of groups. K K

3 MAT 101A: ALEBRA I PART A: ROUP TEORY 41 In this diagram, if K = {e, } then is the amalgamated free product of and given by = {g 1 h 1 g 2 h 2 g n h n g i, h i }. Note that (g 1 hg 2 ) 1 = g2 1 h 1 g1 1. [So, the set of such products is closed under the operation of taking inverse. So, is a group.] More generally, K is the quotient group / where and (gα(k)) (β(k) 1 h ) gh hg (hβ(k) (α(k) 1 g ). Exercise. Compute (Z/2Z) (Z/2Z) and explain the computation direct limit of groups. In order to take the direct limit of groups, we require a directed system of groups: Definition { α, f α,β } is a directed system of groups if i) f α,β : α β, f β,γ : β γ are homomorphism then [there is a homomorphism f α,γ : α γ in the system and] f αβ α β f αβ γ f βγ [i.e. the diagram commutes.] ii) For every α, β I there exists γ I such that f α,γ, f β.γ defined and α f αγ γ f β βγ We then define a new object [the weak product] α I α α I α: Definition For x = {x α } α we let x α I α if x has only finitely many x α coordinates which are not e α α. [The direct limit of a directed system of groups is then the same set as the direct limit of sets: lim α = α / α I α I where, x α α is equivalent to x β β if and only if f α,γ (x α ) = f β,γ (x β ) are

4 42 MAT 101A: ALEBRA I PART A: ROUP TEORY for some γ I, with the additional structure that the product of x α α, x β β is defined to be the product of their images in γ.] In practice, one works with lim α α in the following way: [ere orozov says that each element of the direct limit is represented by a single element of a single group α. I wrote that as the definition.] universal property of the direct limit of groups. If { α, f α,β } is a directed system of groups and g α : α are homomorphisms such that g α α f αβ g β β commutes, then there exists a unique homomorphism g : lim α such that α h α lim α!g g α commutes for all α I. [Any element x of the direct limit is represented by some element of some group x α α. Then we let g(x) = g α (x α ). If x β β is another representative of the same equivalence class x then, by definition, f α,γ (x α ) = f β,γ (x β ) = x γ γ for some γ I. But then g α (x α ) = g γ (x γ ) = g β (x β ). So, g is well-defined.] free groups. Let X be a set. The free group on X, F (X), is defined by the following universal property: given any group and set map f : X, there is a unique g : F (X) that is a group homomorphism such that the diagram f X i!g F (X) commutes. It remains to define F (X) and say what the map i is. Suppose X = {x 1, x 2, } (note that X need not be countable; we use this subscript notation simply for ease of use). Then the words in X, w, are all finite sequences chosen from the set X X 1, where X 1 = {x 1 1, x 1 2, }

5 MAT 101A: ALEBRA I PART A: ROUP TEORY 43 (here the 1 notation is purely formal). If W = {w} is the set of such word then F (X) = W/ where is the smallest possible relation so that we get a group: i.e., that for all i, both x i x 1 i and x 1 i x i are trivial. Then each word in F (X) has a unique reduced form, in which no further simplification induced by the above relation are possible, and F (X) thus consists of all the reduced words an important example. P SL 2 (Z) = SL 2 (Z)/ ± I. This group acts on the upper half-plane = {z C Iz > 0}: A SL 2 (Z) gives a map z az + b cz + d = az b cz d. ( ) ( ) Thus, for example, translates by 1, while is inversion It turns out that P SL 2 (Z) = Z/2Z Z/3Z This has something to do with fixed points in C under this action. Similarly, SL 2 (Z) = Z/4Z Z/2Z Z/6Z

6 44 MAT 101A: ALEBRA I PART A: ROUP TEORY 12. More about free products I decided to explain the example of P SL 2 (Z) more thoroughly Amalgamated products again. The free product formulas for P SL 2 (Z) and SL 2 (Z) are an example of the following theorem. Theorem Suppose that and have isomorphic normal subgroups N 1, N 2. N 1 = N2 = N. Then N N and N N = /N /N. Proof. N is the push-out (colimit) of the following diagram. N /N is also given by a universal property: It is the pushout of the diagram N {e} = 1 /N To show that N N = /N /N we need to show that: For any group X and homomorphism N X which is trivial on N, there exists a unique homomorphism /N /N X making the appropriate diagram commute. This condition is equivalent to the commuting diagram N X 1 owever, this commuting diagram includes the diagram N 1 X

7 MAT 101A: ALEBRA I PART A: ROUP TEORY 45 So, there is an induced map /N X and similarly, there is an induced map /N X as indicated in the following diagram. N X 1 /N /N The two morphism /N X, /N X induce a morphism from the free product /N /N X: N 1 /N This proves the theorem. X /N /N /N I restated the theorem in terms of specific 2 2 matrices A, B ( ) ( ) A = A = = I ( ) ( ) ( ) B = B = B = = I So, A 4 = I 2 = B 6. The statement is SL 2 (Z) = A I2 B = Z/4 Z/2 Z/6 This means that every element of SL 2 (Z) has a unique expansion of one of the following two forms (1) A m B n 1 AB n 2 A AB n k (2) A m B n 1 AB n 2 A AB n k A where, in both cases, each n i = 1 or 2 and m = 0, 1, 2 or 3.

8 46 MAT 101A: ALEBRA I PART A: ROUP TEORY Proof. By definition, elements of the amalgamated product A I2 B have the form A m 1 B n 1 A m 2 B n 2 A m3 A m k B n k subject to the condition that A 4 = B 6 = I and A 2 = B 3. This last condition implies that, if any of the powers of A are 2 or more or any of the powers of B are 3 or more, then we can convert it into a power of the other letter and move it to the left. For example, ABAB 2 AB 4 = ABAB 2 AB(B 3 ) = ABAB 2 (A 2 )AB = ABA(B 3 )B 2 AB = = (A 2 )ABAB 2 AB The only question is whether the last letter is A or B free group as adjoint functor. If S is a set, let (S) be the free group generated by S. This is the set of all sequences (reduced words) w = s n 1 1 s n 2 2 s n k k, k 0 where n i Z, n i 0 and s i s i+1 S. The length of w is l(w) = ni. Theorem : Ens ps is adjoint to the forgetful functor F. I.e., om Ens (S, F ) = om ps (S, ) Proof. The bijection sends the mapping f : S to the group homomorphism f : S given by f(s n 1 1 s n 2 2 s n k k ) = f(s 1) n 1 f(s 2 ) n2 f(s k ) n k. The inverse is the restriction map; 1 (f : S ) = f S actions and free products. As Ivan orozov pointed out, we can tell that P SL 2 (Z) is a free product from the way that it acts on upper half-space. The following theorem, which is Exercise 54 on p.81, explains how this works. Theorem Suppose that 1, 2,, n are subgroups of which generate. Suppose that acts on a set S. Suppose there are subsets S 1, S 2,, S n S and an element s S\ S i in the complement of the sets S i with the following property. For all g i, g i e, (1) g(s j ) S i for all j i and (2) g(s) S i.

9 MAT 101A: ALEBRA I PART A: ROUP TEORY 47 Then is the free product of the groups i : = 1 2 n Proof. By the universal property of the free product, there is a homomorphism φ : 1 2 n which is the inclusion map on each i. Since the groups i generate, this homomorphism is onto. Thus, it suffices to show that the kernel of φ is trivial. So, suppose that there is an element in the kernel of φ. This has the form g 1 g 2 g k where g j e is an element of ij and i j i j+1. Suppose for example that this element is g 1 g 2 g 3 where g 1 5, g 2 9, g 3 4. Then g 1 g 2 g 3 (s) 5 since g 3 (s) S 4, g 2 (g 3 (s)) g 2 (S 4 ) S 9, g 1 (g 2 g 3 (s)) g 1 (S 4 ) S 5. Therefore, g 1 g 2 g 3 e. And in general, g 1 g k e which implies that φ has a trivial kernel and is thus an isomorphism. Corollary P SL 2 (Z) = Z/2 Z/3. ( Proof. We ) apply the theorem to = P( SL 2 (Z), ) 1 = A where A = and = B where B =. Let S = be the 1 1 upper half plane. Let S 1 = {z = x + iy x < 0} Then Az = 1/z. So, A reverses the sign of the real part of z. Therefore, A sends s = 1//4 + i and the set {z = x + iy x > 0} into S 1. Let S 2 = X Y where X = {z z 1 < 1 and z 1} Y = {z = x + iy x 1/2 and z > 1} Then B(X) = Y, Bs X and B(S 1 ) X. Therefore, the conditions of the theorem are satisfied and we conclude that P SL 2 (Z) is the free product of the subgroups generated by A and B.

10 48 MAT 101A: ALEBRA I PART A: ROUP TEORY