Notas de Aula Grupos Profinitos. Martino Garonzi. Universidade de Brasília. Primeiro semestre 2018

Notas de Aula Grupos Profinitos Martino Garonzi Universidade de Brasília Primeiro semestre 2018 1

Le risposte uccidono le domande. 2

Contents 1 Topology 4 2 Profinite spaces 6 3 Topological groups 10 4 Profinite groups 14 5 The Tychonoff Theorem 17 6 Inverse limits 18 7 Pro-C groups 22 8 Completion 26 9 Exercises - list 1 31 10 Lagrange and Sylow 37 11 Pronilpotent and pro-p groups 40 12 A theorem of Serre about finitely generated pro-p groups 44 13 Finitely generated profinite groups 48 14 A dense subgroup of finite index 51 15 The automorphism group 53 16 Hopfian groups 56 17 Exercises - list 2 59 18 Exercises - list 3 62 19 Infinite Galois Theory 65 20 The fundamental theorem of Galois Theory 68 21 Free profinite groups 71 22 Projective profinite groups 73 23 Positively Finitely Generated groups 76 24 PMSG groups 80 3

1 Topology A topological space is a set X endowed with a family τ of subsets of X (called the open subsets of X, and τ is called the topology of X) with the following properties. 1. If {U i : i I} is an arbitrary family of open subsets of X then i I U i is open (τ is closed under arbitrary unions). 2. If U, V are open subsets of X then U V is open (τ is closed under finite intersections). 3. and X are open (, X τ). We may indicate such topological space also as a pair (X, τ) Let (X, τ) be a topological space. A subset F of X is called closed if X F is open. It is easy to see that the family of closed subsets contains and X, it is closed under finite unions and arbitrary intersections. If Y is a subset of X then Y has the structure of topological space whose open subsets are of the form U Y where U is an open subset of X. This is called topological subspace. Whenever we will treat a subset of a topological space as a topological space we will mean it is endowed with the subspace topology. Example: The discrete topology. If X is any set let P (X) denote the set of all subsets of X. Then P (X) is (clearly) a topology, called the discrete topology on X. A discrete space is a topological space X whose topology is precisely the discrete topology. Example: The trivial topology. If X is any set then {, X} is (clearly) a topology on X, called the trivial topology. Example: If X is a topological space with exactly one element (we call such space point ) then there is only one topology on X, it is {, X} (which in this unique case is both discrete and trivial). Example: The real line R has the usual topology, its members are all the arbitrary unions of open intervals of the form (a, b) = {x R : a < x < b} where a < b. This topology is not discrete (for example {0} is not open). Exercise: Prove that in this space the subset (0, 1] = {x R : 0 < x 1} is not open and not closed. Exercise: Prove that Z (as a topological subspace of R) is discrete. Example: The unique possible topologies on the set {1, 2} are the trivial topology, the discrete topology and the two topologies τ 1 = {, {1, 2}, {1}} and 4

τ 2 = {, {1, 2}, {2}}. With the topology τ 1 the point {2} is closed, the point {1} is not closed; with the topology τ 2 the point {1} is closed, the point {2} is not closed. We also recall some other notions. A base (or basis) for the topology τ is a collection of open subsets of X (i.e. elements of τ) (U λ ) λ such that every non-empty open subset of X (i.e. element of τ) can be written as a union of some of the U λ. For example a base of the real line R (with the usual topology) is given by the open intervals (a, b) with a < b. X is said to be disconnected if it can be written as a disjoint union of two proper non empty open subsets, and connected otherwise. For example a discrete space with more than one element is clearly disconnected. For example the space {1, 2} with the topology {, {1, 2}, {1}} is connected. X is said to be totally disconnected if everytime a subset of X is connected, it has exactly one element. For example any discrete space is (clearly) totally disconnected. X is said to be compact if any open cover of X admits a finite subcover. In other words everytime {U i : i I} is a family of open subsets of X with the property that X = i I U i there exists J I, a finite subset of I, such that X = j J U j. Equivalently, any family of closed subsets with empty intersection has a finite subfamily with empty intersection. Equivalently if a family of closed subsets has the property that every finite subfamily has non-empty intersection then the family itself has non-empty intersection. Exercise: Let X be a discrete space. Show that X is compact if and only if it is finite. Exercise: Show that in the real line R with the usual topology the open intervals (a, b) are not compact, and the closed intervals [a, b] = {x R : a x b} are compact. Exercise: Let X be a topological space and let B be a base for the topology of X. Show that X is compact if and only if any open cover of X consisting of elements of B admits a finite subcover. If C X is closed and X is compact then C is compact. [If C is a union of some U i C where each U i is open in X then X is the union (X C) i U i hence there is a finite set I of indeces with X = (X C) i I U i therefore i I U i C = C.] A topological space X is called Hausdorff if for any two points x, y of X there are open subsets U, V of X such that x U, y V and U V =. Exercise: show that in a Hausdorff topological space the points are always closed. 5

If X is Hausdorff and C X is compact then C is closed. [Let x X C, for every c C let U cx, V cx be open in X with c U cx, x V cx and U cx V cx =, then C = c C U cx C so there are finitely many c i (x) C with C i U c i(x)x. Let V x := i V c i(x)x, it is open (finite intersection of opens) and contains x, also V x C =. Therefore X C = x X C V x is open.] A function f : X Y between two topological spaces is called continuous if f 1 (V ) is open in X whenever V is open in Y. Recall that f 1 (V ) := {x X : f(x) V } (pre-image of V ). Exercise: Let f : X Y be a continous map between two topological spaces. Show that if X is compact then f(x) (the image of f) is compact. Exercise: Let f : X Y be a continuous map between two topological spaces. Show that if X is connected then f(x) (the image of f) is connected. Exercise: Show that if f : X Y is a continous bijection, X is compact and Y is Hausdorff, then f is a homeomorphism (in other words the inverse f 1 is continuous). Exercise: Find an example of a continuous bijection f : X Y whose inverse is not continuous. A subset of a topological space is called clopen if it is both open and closed. For example if U and V are disjoint open subsets of a space X with U V = X then U and V are clopen (their complements are V and U respectively). In the next class we will prove the following theorem. Theorem 1. Let X be a Hausdorff compact and totally disconnected topological space. Then every open subset of X is a union of clopen subsets of X, that is, X admits a basis consisting of clopens. 2 Profinite spaces Lemma 1. Let X be a compact and Hausdorff topological space. Let C, D be closed and disjoint subsets of X. Then there exist U, V, open subsets of X such that C U, D V and U V =. Proof. First assume C = {c} is a single point. In this case for each d D there are disjoint open sets U cd, V cd with c U cd and d V cd. Observe that D = d D V cd D so since D is compact (because it is closed in a compact) there is a finite family d i of points of D with D V = i V cd i, now let U = i U cd i, then U is open (finite intersection of opens) and c U, moreover U is disjoint from V by construction. Now we consider the general case. Let c C. By the above paragraph there exist two disjoint open subsets U c, V c of X such that c U c and D V c. 6

Now the union of the U c covers C, so we can extract a finite subcovering, say U c1,..., U ck (C is closed in a compact, so it is compact). Take as U the union U c1... U ck and as V the intersection V c1... V ck. Exercise: If X is a topological space and F G X are subspaces, with F closed in G and G closed in X, show that F is closed in X. Lemma 2. Let X be a compact and Hausdorff topological space, and let x X. Let {C λ } λ Λ be the family of all the clopens of X that contain x. Then A := λ Λ C λ is connected. Proof. Let by contradiction be A = C D where C, D are non-empty open disjoint subsets of A. A is an intersection of closed, so it is closed, hence compact (because X is compact). C and D are closed in X (being closed in A, which is closed), so they are compact in X. By the previous lemma, there exist U and V open in X such that C U, D V and U V =. Then clearly A (X (U V )) =. That is, ( C λ ) (X (U V )) = λ (X (U V ))) = λ Λ λ Λ(C We have found a family of closed subsets of the compact space X with empty intersection, so it admits a finite subfamily with empty intersection: there exists J Λ finite such that C j U V j J Now, j J C j is clopen, being a finite intersection of clopens. Moreover, using the previous formula it is clear that A j J C j = (U j J C j ) (V j J C j ) Since U and V are disjoint, this says that V ( j C j) and U ( j C j) are closed in j C j which is closed in X, so that they are closed in X. It follows that U ( j C j) and V ( j C j) are clopens of X. Since x A, either x U ( j C j) or x V ( j C j). Without loss of generality assume that x U ( j C j). Then U ( j C j) is one of the C λ, being a clopen that contains x. It follows that it contains A (by definition of A), therefore A U. Since A = C D and D is disjoint from U it follows that A C U so D =. Contradiction. A topological space is called profinite if it is Hausdorff, compact and totally disconnected. For example any finite discrete space is (clearly) profinite. 7

Theorem 2. Let X be a profinite space. Then every open subset of X is a union of clopen subsets of X, that is, X admits a basis consisting of clopens. Proof. Let A be an open non-empty subset of X, and let x A. It suffices now to find a clopen C of X such that x C A. Now, by the previous lemma the intersection of all the clopens of X containing x is connected, so it has exactly one element (X is totally disconnected), that clearly has to be x. So if X y x there exists a clopen F y such that x F y and y F y. Now we have X = A ( (X F y )) x y X This is an open covering of X, so it admits a finite subcovering (in which A has to show up because x X F y for any x y X): X = A (X F y1 )... (X F yn ) = A (X (F y1... F yn )) So that n i=1 F y i is a clopen contained in A and containing x. Let X be a topological space and let A be a subset of X. The closure of A is the intersection of the closed subsets of X containing A. It is the smallest closed containing A. For example in the real line the closure of an open interval (a, b) is the closed interval [a, b]. The closure of A is denoted A. The subset A of the space X is called dense (in X) if A = X, equivalently for every non-empty open subset U of X we have A U (prove this equivalence as an exercise). For example Q is dense in R. Proposition 1. If X is a topological space, the closures of the points of X are connected. In particular if X is totally disconnected the points of X are closed. Proof. Given x X we need to show that Y := {x} is connected. Suppose that Y = A B where A and B are non-empty disjoint open subsets of Y. Then without loss of generality we have x A. Writing B = U Y with U open in X we have that x X U so X U is a closed subset of X containing x, hence B Y X U, contradicting the fact that B U. Usually the property of being Hausdorff is called T2. A space is called T0 if at most one of its points is dense. A space is called T1 if its points are closed. It is easy to show that T2 implies T1 and that T1 implies T0 (exercise). It is also easy to show (by means of counterexamples) that T0 does not imply T1 and that T1 does not imply T2 (exercise). T0, T1 and T2 are examples of separation axioms (there are more: T3, T4,...). Trivial example of profinite space: any finite discrete space is profinite. We want a non-trivial example of profinite space. Cartesian product. If I is a set of indeces and X i is a set for every i I define X := i I X i to be the set of functions f : I i I X i with the property 8

that f(i) X i for all i I. To such a function we associate the well-known I-tuple (x i ) i I identifying f(i) with x i, in other words we think of f as the sequence of its values. A finite cartesian product (i.e. a cartesian product with finitely many indeces) will be also denoted X 1... X n, and its points will be denoted (x 1,..., x n ). A direct power is a set of the form X n = X... X (n times). Geometrically for example R 2 = R R corresponds to the real plane. Product topology. Let X i be a topological space for every i in a set of indeces I. We want to give X = i I X i the structure of topological space. For every i I let π i : X X i be given by π i (x) := x i (canonical i-th projection). The product topology is the topology on X with base the family of finite intersections π 1 i 1 (U i1 )... π 1 i n (U in ) where U i1,..., U in are open subsets of X i1,..., X in respectively. In other words an open subset of X is by definition an arbitrary union of finite intersections of this type. Clearly when X is given the product topology all the projections π i are continuous. It is an easy exercise to show that the product topology on X is the intersection of all the topologies τ of X with the property that the projections π i are all continuous when X is given the topology τ. When we treat a cartesian product of topological spaces as topological space we will always mean that it is endowed with the product topology. Fact 1. A product of Hausdorff spaces is Hausdorff. To see this let x y in X = i I X i, then there exists an index i I such that x i y i. Since X i is Hausdorff there are two open subsets U, V of X i with x i U, y i V, U V =. Clearly x i = π i (x) U, y i = π i (y) V, in other words x π 1 i (U), y π 1 i (V ), and π 1 i (U) π 1 i (V ) is empty because if it contained an element α then π i (α) would belong to U V. We managed to separate x and y with open disjoint subsets of X (they are open because they are pre-images of open subsets of X i via the i-th canonical projection). Fact 2. A product of totally disconnected spaces is totally disconnected. To see this let A be a connected subspace of X = i I X i. We need to prove that A = 1. We know that the projections π i are continuous, and that a continuous function takes connected sets to connected sets. Therefore for every i I the set π i (A) is connected in X i, which is totally disconnected, therefore π i (A) = {a i }. It follows that setting a = (a i ) i I in the product, A = {a}. Fact 3. A product of compact spaces is compact. This is the Tychonoff theorem and I will not prove it now. It follows that a cartesian product of profinite spaces is a profinite space. In particular a cartesian product of finite discrete spaces is a profinite space (this is the first non-trivial example of a profinite space). Warning: it is not discrete in general! Exercise: a cartesian product of discrete spaces is not discrete in general. 9

Exercise: Let Y be a topological space and let = {(y, y) : y Y } Y Y (the diagonal ). Prove that Y is Hausdorff if and only if is closed in Y Y (where Y Y as usual is given the product topology). Exercise: Prove that a closed subset of a profinite space is profinite. Exercise: Let τ 1, τ 2 be two topologies of a space X. Prove that the identity (X, τ 1 ) (X, τ 2 ) is continuous if and only if τ 2 τ 1. Prove that the identity (X, τ 1 ) (X, τ 2 ) is a homeomorphism if and only if τ 1 = τ 2. 3 Topological groups Exercise. A composition of continuous functions is continuous. Exercise. Let X = i I X i be a product of topological spaces (with the product topology). A map f : Y X is continuous if and only if the compositions π i f are continuous (where π i : X X i are the canonical projections). Exercise. Let X = X 1 X 2 the product of two topological spaces (with the product topology). Show that a subset U of X is open in X if and only if for every (a, b) X there exist an open A of X 1 and an open B of X 2 with a A, b B and A B U. Let G be a group (multiplicative notation) and a topological space. G is called a topological group if v = v G : G G G, (x, y) xy 1 is continuous. For example the additive group R with the usual topology is a topological group. If G is a topological group then: φ : G G, g g 1 is continuous because it is a composition of continuous mappings (look at the compositions with the projections and use the exercises above): G G G v G g (1, g) g 1 µ : G G G, (x, y) xy is continuous because it s a composition of continuous mappings: G G G G G (x, y) (x, y 1 ) xy v 10

If x G the map G G, g gx is an homeomorphism. It s continuous because it is the composition G G G µ G g (g, x) gx The inverse is G G, g gx 1, and it is continuous for similar reasons. It follows that if g G and U G is open (resp. closed) then gu, Ug, U 1 are all open (resp. closed). Lemma 3. Let G be a topological group, and let H G. 1. If H is open then H is closed. 2. If H is closed and G : H is finite then H is open. 3. If H is open and G is compact, then G : H is finite. Proof. Let us define the equivalence relation in G by g 1 g 2 iff g 2 g1 1 H. The equivalence classes are the cosets Hg. Let {g λ } λ Λ be a set of representatives of different cosets. Then clearly G is the disjoint union of the Hg λ. We have G H = Hg λ Hg λ H We obtain that if H is open, G H is a union of cosets Hg, which are open, so G H is also open, i.e. H is closed. If H is closed and G : H is finite, G H is a finite union of closed subsets of G, so it is closed, i.e. H is open. If H is open and G is compact then we can extract from the open covering {Hg λ } λ Λ a finite one, so by disjointness G : H is finite (a covering that is also a partition does not have proper subcoverings). Let X be a topological space, and let P be an equivalence relation on X. X/P is a topological space with the so called quotient topology, defined using the canonical projection π : X X/P by saying that U X/P is open in X/P if π 1 (U) is open in X. The main example of this is the case of a quotient group G/N where N is a normal subgroup of the topological group G. Whenever we treat a quotient group G/N as a topological group we will mean that it is endowed with the quotient topology. If f : X Y is a continuous mapping of topological spaces, we can construct the equivalence relation on X defined by x y iff f(x) = f(y). There exists a unique continuous mapping f : X/P Y such that the diagram commutes. X π f X/P f Y 11

Let now G be a topological group. H G with the subspace topology is a topological group (this is because a restriction of a continuous function is always continuous). Recall that a function f : X Y between topological spaces is said to be open if f(u) is open in Y whenever U is open in X. Observe that a continuous function does not need to be open (for example sin : R R has the property that sin(r) = [ 1, 1]). Proposition 2. If N G then the canonical projection π : G G/N is open and G/N is a topological group. Proof. We prove that π is open. Let U G be open. We want to prove that π(u) is open in G/N, i.e. (by definition of quotient topology) that π 1 (π(u)) is open in G. But it is easy to see that π 1 (π(u)) = UN, so it is open, being the union of the cosets Un with n N. Now let v G : G G G and v G/N : G G G be both defined by the rule (x, y) xy 1. Since G is a topological group, v G is continuous. We want to prove that v G/N is continuous. Consider the commutative diagram G G v G G (π,π) G/N G/N v G/N π G/N where v G and v G/N are both defined by the rule (x, y) xy 1. Now, since π is open and surjective, f = (π, π) is also open and surjective (exercise). This implies that v G/N is continuous (which is what we need to prove). Indeed if U is open in G/N to show that v 1 G/N v 1 G/N (U) = f(f 1 (v 1 G/N f 1 (v 1 G/N (U) is open observe that since f is surjective (U))) and since f is open it is then enough to prove that (U)) is open; now apply the commutativity of the above diagram. We remark that in a Hausdorff topological space X the points are closed. The converse in general is false, as the following example shows. Let X be an infinite set and let τ be the cofinite topology on X, defined by saying that U X is open if and only if X U is finite (in other words F X is closed if and only if it is finite). In this topology points are closed (they are finite sets) but this topology is not Hausdorff (exercise). In the case of topological groups we have: Theorem 3. Let G be a topological group. G is Hausdorff if and only if {1} is closed in G. 12

Proof. Let {1} be closed, and let a b be elements of G. We have to show that there exist two disjoint open subsets of G containing respectively a and b. Clearly a 1 b 1. Moreover {a 1 b} = {1}a 1 b, hence {a 1 b} is closed. So that U := G {a 1 b} is open and 1 U. Now v : G G G is continuous, and v(1, 1) = 1 U, so that v 1 (U) is open and contains (1, 1). There exists an open subset Z of the form V W for some V, W open of G, such that V W v 1 (U) and 1 V W. Since v(z) U, clearly a 1 b v(z) = V W 1, i.e. av bw =. av and bw are disjoint and open, a av and b bw. Using the exercise of the previous lecture, another way of proving the theorem is the following: if {1} is closed then v 1 ({1}) is closed in G G (being v continuous), but v 1 ({1}) = {(g, g) : g G} (the diagonal) and we now that the diagonal being closed is equivalent to G being Hausdorff. It follows that for a topological group G, Let K G and consider G/K with the quotient topology. Then G/K is Hausdorff if and only if K is closed in G. If G is totally disconnected, it is Hausdorff because points are closed. Lemma 4. If the topological group G is compact and Hausdorff, and C, D are closed subsets of G, then CD is closed. Proof. C and D are compact because they are closed in the compact G, so that by Tychonoff s theorem C D is compact, hence µ(c D) = CD is compact, µ being continuous. Since G is Hausdorff, CD is closed. Observe that if A is a subset of a topological space X then a point x X belongs to the closure A if and only if every open of X containing x has a non-empty intersection with A (exercise). Proposition 3. Let H be a subgroup of a topological group G. Then the closure H is a subgroup of G. Proof. Let a, b H. It is enough to prove that ab 1 H. If by contradiction this is not the case then ab 1 U = G H. Observe that U is open. Since the map f : G G G taking (x, y) to xy 1 is continuous and (a, b) f 1 (U), there exist two open subsets A and B of G such that a A, b B and A B f 1 (U) (by definition of product topology), in other words f(a B) U. Since a, b H there exist x A H and y B H. Therefore xy 1 f(a H B H) f(a B) U, on the other hand clearly xy 1 H (being x, y H and being H a subgroup) hence U H, a contradiction (because H H and U = G H). Exercise. If f, g : X Y are continuous and Y is Hausdorff, the set F = {x X f(x) = g(x)} is closed in X, or equivalently the set U = 13

X F = {x X f(x) g(x)} is open in X. (Hint: recall that the diagonal = {(y, y) : y Y } is closed in Y Y ). Exercise. Let N be a normal subgroup of a topological group G. Prove that G/N is Hausdorff if and only if N is closed in G, and that G/N is discrete if and only if N is open in G. Exercise: do the exercises of Chapter 0 of the book of Wilson. 4 Profinite groups Lemma 5. Let G be a compact topological group. Take Y G closed. Let {X λ } λ Λ be a family of closed subsets of G with the property that for any λ 1, λ 2 Λ there exists µ Λ such that X µ X λ1 X λ2. Then ( X λ )Y = λ Y ) λ Λ λ Λ(X Proof. The inclusion ( ) is trivial because for any λ 0 Λ we have ( λ X λ )Y X λ0 Y. Assume now that g ( λ X λ )Y, so that gy 1 ( λ X λ ) =. Then {gy 1 } {X λ } λ is a family of closed subsets of G of empty intersection. Since G is compact, there exist λ 1,..., λ r Λ such that gy 1 X λ1... X λr =. Now, there exists µ Λ such that X µ X λ1... X λr, so that gy 1 X µ =, that is, g X µ Y. Lemma 6. Let G be a compact topological group, and let C G be a clopen such that 1 C. Then C contains an open normal subgroup of G. Proof. Let x C and W x := Cx 1. W x is open and contains 1. Since µ : G G G is continous, there exist L x, R x open subsets of G containing 1, such that µ(l x R x ) = L x R x W x, that is, (1, 1) L x R x µ 1 (W x ). S x := L x R x is open and contains 1, and S x S x W x. Now since 1 W x for all x C we have C S x x x C This is an open covering of C, which is closed, hence compact being G compact, so there exist x 1,..., x n C such that C n i=1 S x i x i. Now, S := n i=1 S x i is open and contains 1. Moreover n n n n n SC S S xi x i = SS xi x i S xi S xi x i W xi x i = C = C i=1 i=1 i=1 We deduce SC C. Since 1 C, this implies S C. Let now T := S S 1. It is open and 1 T. Let H := n N T n i=1 i=1 14

It is the subgroup of G generated by T. H is open because it is a union of sets of the kind T x for some x, which are open (specifically T n = x T n 1 T x). We claim that H C. Clearly T S C. If n N then T n+1 = T T n SC C by induction. This proves that T n C for any n N, so H C. Now let N = g G Hg where H g = g 1 Hg is the conjugate of H by g. We have N G (exercise) and N H C so to conclude it is enough to show that N is open in G. For this it is enough to show that H has finitely many conjugates in G, because then N is a finite intersection of conjugates of H, which are open because H is open, hence N is open as well. Now since H is open and G is compact, G : H is finite, therefore H has finitely many cosets, say they are Hg 1,..., Hg n. Let H g be an arbitrary conjugate of H, and let i {1,..., n} be such that g Hg i (such i exists because the cosets of a subgroup of G form a partition of G), then there is some h H with g = hg i hence H g = H hgi = H gi. This shows that any conjugate of H is one of H g1,..., H gn hence H has finitely many conjugates in G. Definition 1 (Profinite groups). A topological group G is called a profinite group if it is compact and totally disconnected. Every profinite group is Hausdorff. Indeed, as we have seen, in a totally disconnected space points are closed and in a topological group this implies the Hausdorff property. Therefore a profinite group is in particular a profinite space. Theorem 4. Let G be a profinite group. Then any open subset of G is a union of cosets of open normal subgroups of G. Proof. Let U G be open, and x U. Then since G is a profinite space, it admits a base consisting of clopen subsets, so x V U for some V clopen, and 1 V x 1. So V x 1 is clopen and contains 1. By the lemma above V x 1 contains a normal open subgroup N of G. Hence x Nx V. It follows that a basis for a profinite group G is {Ng g G, N G open}. This is indeed a base for the topology (see below). Corollary 1. Let G be a profinite group and let X G. Then X = {NX N G, N open} Proof. Let N G be open. Then NX is a union of cosets of N, and it is a finite union because N has finite index. Each of these cosets is a clopen, so it is closed. Hence NX is closed so X NX by definition of closure. ( ) is proved. Now let y G and y X. Then since y G \ X is open, by the theorem above there exists N G open such that Ny X =. Then y NX, and ( ) is proved. 15

In particular if G is a profinite group, the intersection of the open normal subgroups of G is equal to {1} (to see this just apply the corollary to the case X = {1} and recall that in a profinite group points are closed). Since the open subgroups have finite index, this implies in particular that profinite groups are residually finite, meaning that the intersection of the subgroups of finite index is {1}. Observe that this is a purely algebraic property. In particular, if a group is not residually finite, then there is no hope to construct a topology that would make it a profinite group. Exercise: Let X be a set and let B be a family of subsets of X with the property that B B B = X and for all A, B B and for all x A B there exists U B with x U A B. Then B is a base of a topology on X, in other words the family of all unions of members of B (including the empty union, that is ) is a topology on X. Proposition 4 (Costruction of topological groups). Let G be an abstract group (i.e. a group without a topology). Let L be a family of normal subgroups of G with the property that for any K 1, K 2 L there exists K 3 L such that K 3 K 1 K 2. We define a topology on G in this way: a basis of open subsets is given by the cosets Kg, where K L and g G. In this way G is a topological group. Proof. Observe that the given family of cosets is indeed a base for a topology. They clearly cover G, and if g 1, g 2 G and K 1, K 2 L then if K 1 g 1 K 2 g 2, taken x K 1 g 1 K 2 g 2 we have K 1 g 1 K 2 g 2 = K 1 x K 2 x = (K 1 K 2 )x = y K1 K 2 K 3 y where L K 3 K 1 K 2. So K 1 g 1 K 2 g 2 is open. Consider v : G G G, (g, h) gh 1. We have to show that v is continous. Suppose that g 1, g 2 G and g 1 g2 1 W for some open subset W of G. There exists N L such that Ng 1 g2 1 W (a coset of N containing g 1 g2 1 is equal to Ng 1 g2 1 ). Let U := Ng 1 and V := Ng 2. Then v(u V ) = UV 1 W. For example we can consider: (Profinite topology) L is the set of the normal subgroups of G of finite index. Careful: the construction in this case does not necessarily make G a profinite group. (Pro-π topology) L is the set of the normal subgroups N of G of finite index and such that G/N is a π-group, where π is a fixed set of prime numbers (a π-group is a finite group whose order factorizes in prime numbers that are element of π). We have to show that if N 1 and N 2 are element of L there exists N 3 L contained in the intersection N 1 N 2. If N 1, N 2 L then we have the homomorphism G G/N 1 G/N 2, g (gn 1, gn 2 ). The kernel is N 1 N 2. So G/(N 1 N 2 ) is isomorphic to a subgroup of G/N 1 G/N 2, which is a π-group, so that G/(N 1 N 2 ) is a π-group, and we can take N 3 := N 1 N 2. 16

In the case of a profinite group G we can reconstruct the topology via the above construction choosing L to be the family of all open normal subgroups. Indeed if K 1, K 2 are open normal subgroups then K 1 K 2 is also an open normal subgroup. 5 The Tychonoff Theorem Definition 2 (Filter). Let X be a set, L P (X) be a non empty family of subsets of X, closed under finite unions and intersections. A filter F on L is a subset of L such that: 1. If A, B F then A B F. 2. If A, B L and F A B then B F. 3. F. Lemma 7. Every filter F on L is contained in a maximal filter on F, said ultrafilter on L. Proof. Easy application of Zorn s lemma. Theorem 5. Let F be an ultrafilter on L. Let A, B L such that A B F. Then A F or B F. Proof. Suppose A F. Let D := {D L A M D M F} Clearly F D and A D (note that A (A B) A and A B F). In particular D is not a filter because it contains properly the ultrafilter F. But D is closed under finite intersections and unions, and if D L and D C D then D D. It follows that D because otherwise D would be a filter. Then there exists M F such that A M =. Now, B = B (A M) = (A B) (M B) F because A B and M B are elements of F. So B F. Exercise: Let X be a topological space and let L be the set of the closed subsets of X. Then X is compact if and only if the intersection of every ultrafilter of L is non empty. Theorem 6 (Tychonoff). Let X λ be a compact topological space, for every λ Λ. Then X := λ Λ X λ with the product topology is compact. Proof. Let L be the set of the closed subsets of X, and let F be an ultrafilter on L. By the exercise, it is enough to show that F. For every λ Λ consider D λ := {D X λ D closed in X λ, π λ (F ) D F F} 17

Clearly D λ is closed in X λ. To show that it is non empty it suffices to show that the intersection of any given finite subfamily is non empty (because X λ is compact). So consider r i=1 D λ i r i=1 π λ(f i ) for some F i F. If it is empty then r r = π λ (F i ) π λ ( F i ) i=1 so r i=1 F i = F because F is closed under finite intersections. Contradiction because F is a filter, so F. Now let a λ D λ for every λ Λ, and consider a := (a λ ) λ Λ X. We want to show that a F for every F F. Suppose by contradiction that a F for some suitable F F. Then a X F, which is open, so a j J i=1 π 1 j (U j ) X F for some suitable J Λ finite, U i X i open in X i. This implies clearly that π j (a) U j for every j J. Now F i J (X π 1 j (U j )) F But since F is an ultrafilter, one of the sets in this finite union belongs to F, say X π 1 j (U j ) F for a suitable j J. π j (X π 1 j (U j )) X j U j, so by definition of D j, X j U j D j. This implies that a j X j U j, i.e. a j U j, contradiction. 6 Inverse limits Recall that a partial order on a set I is a relation on I that is reflexive, transitive and antisymmetric. Usually partial orders are denoted by. Definition 3 (Directed sets). A directed set is a poset (i.e. a partially ordered set) I such that for any a, b I there exists c I such that a c and b c. Definition 4 (Inverse limit). Let I be a directed set. An inverse system is a family {X i } i I of sets with a map ϕ ij : X j X i for any i j, such that ϕ ii = id Xi for any i I, and if i j k then ϕ ij ϕ jk = ϕ ik. A set of compatible maps on the inverse system {X i, ϕ ij } i,j is a set of maps {ϕ i : Y X i } i such that whenever i j the diagram commutes. ϕ j Y ϕ i X j ϕ ij X i 18

An inverse limit on the inverse system {X i, ϕ ij } i,j is a set X together with a set of compatible maps α i : X X i, with the following universal property: for any set of compatible maps β i : Y X i there exists a unique map σ : Y X such that the diagram commutes. Y σ β i X i To help imagination observe that if all maps are inclusions then the inverse limit of a family of sets coincides with their intersection. Replacing set by group, topological space, topological group and map by homomorphism, continuous map and continuous homomorphism respectively we get analogous definitions in the category of groups, topological spaces and topological groups respectively. Recall that in such categories the word isomorphism means group isomorphism, homeomorphism and homeomorphic group isomorphism. In the following proposition isomorphism means morphism (map, homomorphism, continuous function...) f : X Y that admits an inverse, that is a morphism g : Y X such that f g is the identity of Y and g f is the identity of X. Proposition 5. If it exists, the inverse limit is unique up to isomorphism. Proof. Let (X, α i ) and (Y, β i ) be two inverse limits of the direct system {X i, ϕ ij } i,j. By definition of inverse limit there are unique maps σ, τ such that the following diagrams commute for all i I. Y σ X β i α i X i τ β i Y α i X X τ Y α i β i X i This implies that the following diagrams commute too. σ α i X Y τ σ Y X σ τ X β i α i β i α i X i X i The identity Y Y clearly makes the first diagram commute as well. By unicity of the compatible map we deduce that τ σ is the identity of Y. The same argument applied to the second diagram shows that σ τ is the identity of X. Therefore σ and τ are isomorphisms. 19

The above proof is purely categorical. Existence of inverse limit for sets, groups, topological spaces, topological groups, etc. Let {X i, ϕ ij } i,j be an inverse system and let X := {(x i ) i I i I X i j i, ϕ ij (x j ) = x i } and consider the maps α i := π i X, the restrictions of the projections. X with the α i is an inverse limit for the inverse system {X i, ϕ ij } i,j. Indeed if β i : Y X i is a set of compatible maps then there is a unique map σ : Y X satisfying α i σ = β i for all i, indeed such compatibility condition is saying precisely that if y Y then π i (σ(y)) = β i (y) in other words σ(y) is the tuple (β i (y)) i. This defines uniquely σ, and observe that the universality works categorically because if each β i is a homomorphism then σ is also a homomorphism, and if each β i is a continuous map then σ is also a continuous map. When talking about inverse limits, if not specified we will refer to this construction. Again, if we work in the category of groups, X is a group. The same holds for many objects you could consider (topological spaces, topological groups, rings...). Now let {X i, ϕ ij } i,j be an inverse system of topological spaces (the ϕ ij are continuous mappings), and let X be its inverse limit. We have: If the X i are Hausdorff then X, as a subspace of a cartesian product of Hausdorff topological spaces, is Hausdorff. If the X i are totally disconnected then X is totally disconnected (because the cartesian product is totally disconnected). If the X i are Hausdorff then X is closed in i I X i: if j i consider D ij := {c l X l ϕ ij (π j (c)) = π i (c)}. Since X i is Hausdorff and all the considered maps are continuous, D ij is closed. Since X is the intersection of the D ij, X is closed. If the X i are Hausdorff and compact, then X is compact, being a closed subspace of a compact Hausdorff topological space. It follows that if each X i is a finite discrete set then the corresponding inverse limit is a profinite space, and if each X i is a finite discrete group then the corresponding inverse limit is a profinite group. We conclude with a motivation. Proposition 6 (Density criterion). Let {X i, ϕ ij } i,j be an inverse system of Hausdorff topological spaces, with the set of indeces being a directed set, and with X (together with the maps ϕ i ) its inverse limit. Assume that Y X is such that ϕ i (Y ) = X i for any i I. Then Y is a dense subset of X. 20

Proof. We will prove this next time. Theorem 7. A topological space is a profinite space if and only if it is homeomorphic to an inverse limit of finite discrete spaces. Proof. We saw above that an inverse limit of finite discrete spaces is a profinite spaces. We now prove the converse. Let X be a profinite space, that is a compact Hausdorff totally disconnected topological space. Let Y be a discrete space (a set with the discrete topology). Let α : X Y be surjective and continuous. Since X is compact and α is surjective, Y is finite (being discrete and compact). Since the points are clopen in Y, their preimages are a finite number of clopens of X that yield a partition of X. So that the set of discrete images of X corresponds to the partitions of X in a finite number of clopens. Call Σ the set of partitions σ of X in a finite number of clopens, and for any σ Σ define X σ := X/σ (recall that to give a partition of a set is equivalent to give an equivalence relation on the same set), and give it the quotient topology. Then clearly every X σ is finite and discrete. Consider β σ : X X σ, the natural projection. It is continuous and epimorphic. The set Σ is a poset with the following partial order: σ 1 σ 2 if σ 2 is a refinement of σ 1, i.e. if any clopen of the partition σ 1 is a union of suitable clopens of the partition σ 2. It is easy to show that with this partial order Σ is a directed set. Define moreover, whenever σ 1 σ 2, ϕ σ1σ 2 : X σ2 X σ1, [x] σ2 [x] σ1. Here [x] σ denotes the clopen of the partition σ containing the point x. In this way {X σ, ϕ σ2σ 1 } is an inverse system (this is easy to show). Denote by (Y, ϕ σ ) its inverse limit. Recall that Y is a topological subspace of σ Σ X σ. By universality of the inverse limit Y, there exists a unique continuous map β : X Y such that the diagram commutes for any σ Σ. X β β σ X σ β is surjective: let Z := β(x). Then everytime σ Σ ϕ σ ϕ σ (Z) = ϕ σ (β(x)) = β σ (X) = X σ By the density criterion, this says that Z is dense in Y. But since X is compact, its image by the continuous β, that is Z, is compact. It follows that Z is closed in Y because Y is Hausdorff and Z is compact. Z is dense and closed in Y, so Z = Y, that is, β is surjective. β is injective: take x 1, x 2 X and assume that β(x 1 ) = β(x 2 ). Then β σ (x 1 ) = ϕ σ (β(x 1 )) = ϕ σ (β(x 2 )) = β σ (x 2 ) Y 21

for any σ Σ. This means exactly, by definition of β σ, that every clopen of X that contains x 1 contains also x 2, and conversely (because we can always consider the partition σ = {C, X C} where C is a clopen). We have seen that in a compact and Hausdorff space the intersection of all the clopens that contain a point is connected. Now since X is also totally disconnected, every point is the intersection of all the clopens that contain the point. It follows that x 1 = x 2. Moreover, β is a homeomorphism because it is bijective and continuous from a compact to a Hausdorff. More precisely, a profinite space is the inverse limit of its finite discrete quotient spaces. Exercises. 1. Wilson exercises 1, 2, 3, 4, 5, 6 of Chapter 1. 2. Ribes-Zalesski exercise 1.1.14. 3. Compute the inverse limit of an inverse system {X i, ϕ ij } when the partial order on the set of indices is equality (the trivial order). 7 Pro-C groups Proposition 7. Let {X i, ϕ ij } i,j be an inverse system of Hausdorff topological spaces on a directed set I, and let X (together with the ϕ i ) be its inverse limit. A basis for the topology of X is given by the following subsets: ϕ 1 k (U) where k I and U is open in X k. Proof. Since X is a subspace of the product of the X i, a basis for the topology of X is given by the subsets of the kind P := {π 1 j (U j ) j J I finite, U j X j open j J} X We need to prove that for any a P there exists k I and U X k open such that a ϕ 1 k (U) P (recall that ϕ j = π j X ). Take k I such that j k for any j J, and consider ϕ 1 jk (U j). It is an open subset of X k and it contains a k. Let U := ϕ 1 jk (U j) j J We claim that ϕ 1 k (U) P. In fact, assume that b = (b i) i I ϕ 1 k (U). Then b k U, so for any j J, That is, b P. ϕ j (b) = ϕ jk (ϕ k (b)) = ϕ jk (b k ) U j 22

Proposition 8 (Density criterion). Let {X i, ϕ ij } i,j be an inverse system of Hausdorff topological spaces on a directed set I, with X (together with the ϕ i ) its inverse limit. Assume that Y X is such that ϕ i (Y ) = X i for any i I. Then Y is a dense subset of X. Proof. It suffices to show that Y has non empty intersection with any non empty element of a basis of open subsets for X. Take a basis open subset, of the kind ϕ 1 k (U), where k I and U is open in X k. Then Y ϕ 1 k (U) because ϕ k (Y ) = X k U. Lemma 8. Let {X i, ϕ ij } i,j be an inverse limit of Hausdorff topological spaces on a directed set I, let Y be a topological space and let α : Y X be a map. α is continuous if and only if ϕ i α : Y X i is continuous for any i I. Proof. One direction is trivial: a composition of continuous mappings is continuous. Conversely if α ϕ i is continuous for any i I then (ϕ i α) 1 (U) = α 1 (ϕ 1 i (U)) is open for any i I and for any open U of X i. This yields the continuity of α because ϕ 1 i (U) is a generic open basis subset. Lemma 9. Let G be an inverse limit of Hausdorff topological groups G i, and take L G open. Then there exists i I such that ker(ϕ i ) L. Proof. Since 1 L, there exists an open basis subset ϕ 1 i (U), where U is an open subset of G i, such that 1 ϕ 1 i (U) L. It follows that ker(ϕ i ) L because 1 U (being ϕ i (1) = 1). Definition 5 (Filtering bases). Let G be a group. A filtering basis for G is a family I of subgroups of G such that for any K 1, K 2 I there exists K 3 I such that K 3 K 1 K 2. Theorem 8. Let G be a topological group. Let I be a filtering basis of normal closed subgroups of G. We can put an order on I: say that K 1 < K 2 if K 2 K 1. Consider the maps ϕ K1K 2 : G/K 2 G/K 1, gk 2 gk 1. I is a directed set and {ϕ KiK j } i,j is a set of compatible maps. Consider the inverse limit Ĝ := lim G/K G/K K I K I Let θ : G K G/K be the natural map g (gk) K. Then: 1. θ is a continuous homomorphism. [all the projections are continuous] 2. θ(g) Ĝ. [If K i < K j, ϕ KiK j (gk j ) = gk i ] 3. θ(g) is dense in Ĝ. [density criterion] 4. ker(θ) = K I K. 5. If G is compact then θ(g) = Ĝ. [θ(g) is compact in a Hausdorff, so it is closed] 23

6. If G is compact and K I K = {1} then G and Ĝ are homeomorphic and isomorphic. [θ is continuous and bijective from the compact G to the Hausdorff K G/K] Definition 6 (C-groups, pro-c groups). Let C be a class of finite discrete (i.e. with the discrete topology) groups closed under subgroups, epimorphic images (quotients) and finite cartesian products. A C-group is an element of this class. A pro-c group is a topological group that can be obtained as an inverse limit on an inverse system of C-groups. For example we can take as C The class of all finite groups. The class of all finite p-groups, where p is a prime. The class of all finite π-groups, where π is a set of primes. The set of all finite abelian groups. The set of all finite solvable groups. The set of all finite nilpotent groups. Notice that every C-group is a pro-c group. Theorem 9. Let G be a topological group, and let C := {G λ λ Λ} as before. The following are equivalent: 1. G is a pro-c group. 2. G is homeomorphically isomorphic to a closed subgroup of λ G λ. 3. G is compact and {N G N open, G/N C} = {1}. 4. G is compact, totally disconnected and G/N C for any open N G. Proof. (1) (2). G is an inverse limit, so it is closed in the cartesian product of some G λ s, therefore it can be obtained as a closed subspace of the cartesian product of all the G λ s (simply intersecting with pre-images of {1}), and the cartesian product λ G λ is Hausdorff since the G λ are Hausdorff (being finite and discrete). (2) (3). G is compact because it is closed in the product, which is compact (being the G λ compact). Now recall that the ϕ λ are the restrictions of the projections. N λ := ker(ϕ λ ) is closed of finite index (because G/N λ is isomorphic to a subgroup of G λ, which is finite) so it is open and G/N λ belongs to C because it is an isomorphic to a subgroup of G λ C. So the intersection of the N λ s contains the intersection in the statement. But λ N λ = {1}. 24

(3) (1). We use theorem 8, recalling that I := {N G open G/N C} is a filtering basis (in fact if N 1, N 2 I then N 1 N 2 I because G/(N 1 N 2 ) = G/N 1 G/N 2 ). Let Y be the inverse limit of the class {G/N N G, G/N C}. The continuous homomorphism θ : G Y is an isomorphic homeomorphism because its kernel is {1}. (1) (4). If G is a pro-c group, for any open normal subgroup N of G there exists λ Λ such that N λ := ker(ϕ λ ) N (lemma 9). Then G/N is isomorphic to a quotient of G/N λ, which is isomorphic to a subgroup of G λ. So G/N C. G is compact and totally disconnected because it is a closed subspace of a product of compact and totally disconnected topological spaces. (4) (3). Since G is totally disconnected, {1} is the intersection of all the clopen subsets of G that contain 1. Since a basis for the open subsets of G is given by the cosets of the open normal subgroups of G we clearly have that {N G N open} = {1}. By assumption, this is exactly what (3) says. In particular, by (2) is clear that a closed subgroup of a pro-c group is a pro-c group. A profinite group is a pro-c group, where C is the class of all finite groups. In other words the above implies that profinite = pro-finite. More precisely, Corollary 2. The following are equivalent for the topological group G: 1. G is a profinite group (i.e. compact and totally disconnected). 2. G is a closed subgroup of a cartesian product of finite groups. 3. G is compact and the intersection of all the open normal subgroups of G is {1}. 4. G is (homeomorphically isomorphic to) an inverse limit of finite discrete groups. Proposition 9. Let G be a pro-c group, and let K be a closed normal subgroup of G. Then G/K is a pro-c group. Proof. If G is a pro-c group, N = {1} where N := {N G N open, G/N C}. Now let π be the natural projection G G/K. Then if N N, π(n) = KN/K is open (because π is open) and normal in G/K. Moreover G/KN C being a quotient of G/N. G/K is compact because it is the image of G by π which is continuous. By item 3 of the previous theorem, we are left to check that the intersection of the open normal subgroups M/K of G/K such that G/M C is the trivial subgroup, {K}. This intersection is clearly contained in 25

N N (KN/K), because KN/K is open and normal in G/K and G/KN C. It follows from the distributivity lemma that (KN/K) = (KN)/K = (K N)/K = K/K = {K} N N N N N N This concludes the proof. Exercises. 1. Exercises 7, 8, 9, 10, 11, 12, 13 of Chapter 1 of Wilson s book. 2. Prove that pro-abelian implies abelian, pro-solvable does not imply solvable and pro-nilpotent does not imply nilpotent. 3. Determine all the open subgroups and the closed subgroups of R (usual topology). 8 Completion Definition 7 (Completion). Let G be an abstract group. Let I be a filtering basis of normal subgroups of finite index. We define a topology on G with basis open subsets given by the cosets Kx where K I and x G. The completion of G with respect to I is a pair (Ĝ, j) where Ĝ is a profinite group and j : G Ĝ is continuous, such that the following universal property is satisfied: for any finite discrete group H and any continuous homomorphism θ : G H, there exists a unique ˆθ : Ĝ H continuous homomorphism such that θ = ˆθ j. G j Ĝ H θ!ˆθ Existence: Let Ĝ be the inverse limit of the G/K on I. It is a profinite group. Let j : G Ĝ, g (Kg) g G. It is a map from G to the product of the G/K. We have to check the universal property. Let H be a discrete finite group, and θ : G H be a continuous homomorphism. Then ker(θ) is open (θ is continuous and {1} is open in H), so there exists K I such that K ker(θ) (by definition of the topology on G). Consider the canonical map θ K : G/K H. We get θ = θ K π K. Define ˆθ = θ K ϕ K where ϕ K is the canonical map coming from the inverse limit. The following diagram commutes. Ĝ ϕ K G/K θ K H j π K 26 G θ

We have to show that ˆθ is unique. So let ˆθ 1 and ˆθ 2 be two such maps. By the density criterion, j(g) is dense in Ĝ. Since H is discrete, it is Hausdorff, so the subset of Ĝ in which ˆθ 1 = ˆθ 2 is closed, and it contains j(g) because ˆθ 1 (j(g)) = ˆθ 2 (j(g)) for all g G (by compatibility). Since j(g) is dense, ˆθ 1 = ˆθ 2. Generalization to a profinite H. We show that the universal property holds even if H is profinite. Let H be profinite and let θ : G H be a continuous homomorphism. Take M, open normal subgroup of H. By the universal property of the finite case, since M has finite index in H, there exists a unique map θ ˆ M : Ĝ H/M such that the diagram G H θ j π M θ Ĝ ˆ θ M π M H/M commutes. By universal property of the inverse limit H, since { θ ˆ M } M is a compatible set of maps Ĝ H/M, there exists a unique ˆθ : Ĝ H such that ˆθπ M = θ ˆ M for any open normal subgroup M of H. The compatibility implies that π M (ˆθ(j(g))) = π M (θ(g)) for any M open and normal in H, and for any g G. Then ˆθ(j(g)) = θ(g) for any g G because, equivalently, they are congruent modulo every open normal subgroup M of the profinite group H (the intersection of the open normal subgroups of H is trivial). To check the unicity of ˆθ just use the argument of the previous point. Unicity of the completion. Standard argument using the previous point. Let Ĝ 1, Ĝ 2 be two different choices. Then by universality there exist α : Ĝ 1 Ĝ2, β : Ĝ 2 Ĝ1 such that αβ = idĝ1 and βα = idĝ2. G G Ĝ 1 j 1 j 2 α Ĝ 2 β j 1 Ĝ 1 Ĝ 2 j 2 β j 1 j 2 Ĝ 1 α Ĝ 2 id ˆ G1 id ˆ G2 Definition 8 (Pro-C completion of an abstract group). Let G be an abstract group. Let C be a set of finite discrete groups closed under subgroups, epimorphic image and finite cartesian product. Let I be the set of normal subgroups N of G such that G/N C. The inverse limit G C of the G/N on I is called the pro-c completion of G. The image of the canonical map j : G G C is dense in the pro-c completion and ker(j) = G/N C N. 27