Notas De Aula Grupos Profinitos Martino Garonzi Universidade De

Notas de Aula Grupos Proﬁnitos

Martino Garonzi

Universidade de Bras´ılia

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. S 1. If {Ui : i ∈ I} is an arbitrary family of open subsets of X then i∈I Ui 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 τ1 = {∅, {1, 2}, {1}} and

4 τ2 = {∅, {1, 2}, {2}}. With the topology τ1 the point {2} is closed, the point {1} is not closed; with the topology τ2 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)(Uλ)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 Uλ. 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 {Ui : i ∈ I} is a family of open subsets of X with S the property that X = Ui there exists J ⊆ I, a finite subset of I, such S i∈I that X = j∈J Uj. 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 Ui ∩ C where each Ui is open in X then X is S the union (X − C) ∪ Ui hence there is a finite set I of indeces with S i S X = (X − C) ∪ i∈I Ui therefore i∈I Ui ∩ 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 Ucx, Vcx be open in X with c ∈ Ucx, S x ∈ Vcx and Ucx ∩ Vcx = ∅, then C = Ucx ∩ C so there are finitely S c∈C T many ci(x) ∈ C with C ⊆ i Uci(x)x. Let Vx := i Vci(x)x, it is open (finite intersection of opens) and contains x, also Vx ∩ C = ∅. Therefore S X − C = x∈X−C Vx 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 Ucd, Vcd with c ∈ Ucd and d ∈ Vcd. Observe that S D = Vcd∩D so since D is compact (because it is closed in a compact) there d∈D S T is a finite family di of points of D with D ⊆ V = i Vcdi , now let U = i Ucdi , 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 Uc, Vc of X such that c ∈ Uc and D ⊆ Vc.

6 Now the union of the Uc covers C, so we can extract a ﬁnite subcovering, say

Uc1 , ..., Uck (C is closed in a compact, so it is compact). Take as U the union

Uc1 ∪ ... ∪ Uck and as V the intersection Vc1 ∩ ... ∩ Vck . 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 {Cλ}λ∈Λ be the family of all the clopens of X that contain x. Then \ A := Cλ λ∈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, \ \ ( Cλ) ∩ (X − (U ∪ V )) = (Cλ ∩ (X − (U ∪ V ))) = ∅ λ∈Λ λ∈Λ

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 \ Cj ⊆ U ∪ V j∈J T Now, j∈J Cj is clopen, being a finite intersection of clopens. Moreover, using the previous formula it is clear that \ \ \ A ⊆ Cj = (U ∩ Cj) ∪ (V ∩ Cj) j∈J j∈J j∈J T T Since U and V are disjoint, this says that V ∩ ( Cj) and U ∩ ( Cj) are T j j closed in Cj which is closed in X, so that they are closed in X. It follows Tj T that U ∩ ( Cj) and V ∩ ( Cj) are clopens of X. Since x ∈ A, either T j Tj x ∈ U ∩ ( Cj) or x ∈ V ∩ ( Cj). Without loss of generality assume that T j T j x ∈ U ∩( j Cj). Then U ∩( j Cj) is one of the Cλ, 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 3 y 6= x there exists a clopen Fy such that x ∈ Fy and y 6∈ Fy. Now we have [ X = A ∪ ( (X − Fy)) x6=y∈X This is an open covering of X, so it admits a finite subcovering (in which A has to show up because x 6∈ X − Fy for any x 6= y ∈ X):

X = A ∪ (X − Fy1 ) ∪ ... ∪ (X − Fyn ) = A ∪ (X − (Fy1 ∩ ... ∩ Fyn )) Tn So that i=1 Fyi 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 6= ∅ (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 ∅= 6 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 Xi is a set for every i ∈ I Q S deﬁne X := i∈I Xi to be the set of functions f : I → i∈I Xi with the property

8 that f(i) ∈ Xi for all i ∈ I. To such a function we associate the well-known “I-tuple” (xi)i∈I identifying f(i) with xi, 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 X1 × ... × Xn, and its points will be n denoted (x1, . . . , xn). A direct power is a set of the form X = X × ... × X (n times). Geometrically for example R2 = R × R corresponds to the real plane.

Product topology. Let Xi be a topological space for every i in a set of Q indeces I. We want to give X = i∈I Xi the structure of topological space. For every i ∈ I let πi : X → Xi be given by πi(x) := xi (canonical i-th projection). The product topology is the topology on X with base the family of finite intersections π−1(U ) ∩ ... ∩ π−1(U ) where U ,...,U are open i1 i1 in in i1 in subsets of Xi1 ,...,Xin 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 πi 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 πi 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 Q x 6= y in X = i∈I Xi, then there exists an index i ∈ I such that xi 6= yi. Since Xi is Hausdorff there are two open subsets U, V of Xi with xi ∈ U, yi ∈ V , −1 U ∩ V = ∅. Clearly xi = πi(x) ∈ U, yi = πi(y) ∈ V , in other words x ∈ πi (U), −1 −1 −1 y ∈ πi (V ), and πi (U) ∩ πi (V ) is empty because if it contained an element α then πi(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 Xi via the i-th canonical projection).

Fact 2. A product of totally disconnected spaces is totally discon- Q nected. To see this let A be a connected subspace of X = i∈I Xi. We need to prove that |A| = 1. We know that the projections πi are continuous, and that a continuous function takes connected sets to connected sets. Therefore for every i ∈ I the set πi(A) is connected in Xi, which is totally disconnected, therefore πi(A) = {ai}. It follows that setting a = (ai)i∈I in the product, A = {a}.

Fact 3. A product of compact spaces is compact. This is the Ty- chonoﬀ 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 τ1, τ2 be two topologies of a space X. Prove that the identity (X, τ1) → (X, τ2) is continuous if and only if τ2 ⊆ τ1. Prove that the identity (X, τ1) → (X, τ2) is a homeomorphism if and only if τ1 = τ2.

3 Topological groups

Exercise. A composition of continuous functions is continuous.

Q Exercise. Let X = i∈I Xi be a product of topological spaces (with the product topology). A map f : Y → X is continuous if and only if the compositions πi ◦ f are continuous (where πi : X → Xi are the canonical projections).

Exercise. Let X = X1 × X2 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 X1 and an open B of X2 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

−1 v = vG : G × G → G, (x, y) 7→ xy 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 7→ 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 7→ (1, g) 7→ g−1

• µ : G × G → G, (x, y) 7→ xy is continuous because it’s a composition of continuous mappings:

G × G → G × G →v G (x, y) 7→ (x, y−1) 7→ xy

10 • If x ∈ G the map G → G, g 7→ gx is an homeomorphism. It’s continuous because it is the composition

µ G → G × G → G g 7→ (g, x) 7→ gx

The inverse is G → G, g 7→ 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. −1 Proof. Let us define the equivalence relation ∼ in G by g1 ∼ g2 iff g2g1 ∈ H. The equivalence classes are the cosets Hg. Let {gλ}λ∈Λ be a set of representatives of different cosets. Then clearly G is the disjoint union of the Hgλ. We have [ G − H = Hgλ

Hgλ6=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 {Hgλ}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 f X / Y = π f˜ ! X/P commutes.

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 E 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 vG : G × G → G and vG/N : G × G → G be both deﬁned by the −1 rule (x, y) 7→ xy . Since G is a topological group, vG is continuous. We want to prove that vG/N is continuous. Consider the commutative diagram

v G × G G / G

(π,π) π

vG/N G/N × G/N / G/N

−1 where vG and vG/N are both deﬁned by the rule (x, y) 7→ xy . Now, since π is open and surjective, f = (π, π) is also open and surjective (exercise). This implies that vG/N is continuous (which is what we need to prove). Indeed if U −1 is open in G/N to show that vG/N (U) is open observe that since f is surjective −1 −1 −1 vG/N (U) = f(f (vG/N (U))) and since f is open it is then enough to prove that −1 −1 f (vG/N (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 6= 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−1b 6= 1. Moreover {a−1b} = {1}a−1b, hence {a−1b} is closed. So that U := G − {a−1b} 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−1b 6∈ v(Z) = VW −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 E 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 6= ∅, 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) 6= 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 {Xλ}λ∈Λ be a family of closed subsets of G with the property that for any λ1, λ2 ∈

Λ there exists µ ∈ Λ such that Xµ ⊆ Xλ1 ∩ Xλ2 . Then \ \ ( Xλ)Y = (XλY ) λ∈Λ λ∈Λ

Proof. The inclusion (⊆) is trivial because for any λ0 ∈ Λ we have (∩λXλ)Y ⊆ −1 Xλ0 Y . Assume now that g 6∈ (∩λXλ)Y , so that gY ∩ (∩λXλ) = ∅. Then −1 {gY } ∪ {Xλ}λ is a family of closed subsets of G of empty intersection. Since −1 G is compact, there exist λ1, ..., λr ∈ Λ such that gY ∩ Xλ1 ∩ ... ∩ Xλr = ∅. −1 Now, there exists µ ∈ Λ such that Xµ ⊆ Xλ1 ∩ ... ∩ Xλr , so that gY ∩ Xµ = ∅, that is, g 6∈ XµY . 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. −1 Proof. Let x ∈ C and Wx := Cx . Wx is open and contains 1. Since µ : G × G → G is continous, there exist Lx,Rx open subsets of G containing 1, −1 such that µ(Lx × Rx) = LxRx ⊆ Wx, that is, (1, 1) ∈ Lx × Rx ⊆ µ (Wx). Sx := Lx ∩ Rx is open and contains 1, and SxSx ⊆ Wx. Now since 1 ∈ Wx for all x ∈ C we have [ C ⊆ Sxx x∈C This is an open covering of C, which is closed, hence compact being G compact, n n so there exist x1, ..., xn ∈ C such that C ⊆ ∪i=1Sxi xi. Now, S := ∩i=1Sxi is open and contains 1. Moreover n n n n n [ [ [ [ [ SC ⊆ S Sxi xi = SSxi xi ⊆ Sxi Sxi xi ⊆ Wxi xi = C = C i=1 i=1 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 := T n n∈N

14 It is the subgroup of G generated by T . H is open because it is a union of sets n S of the kind T x for some x, which are open (speciﬁcally T = x∈T n−1 T x). We claim that H ⊆ C. Clearly T ⊆ S ⊆ C. If n ∈ N then T n+1 = TT n ⊆ SC ⊆ C by induction. This proves that T n ⊆ C for any n ∈ N, so H ⊆ C.

T g g −1 Now let N = g∈G H where H = g Hg is the conjugate of H by g. We have N E 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 g Hg1, . . . , Hgn. Let H be an arbitrary conjugate of H, and let i ∈ {1, . . . , n} be such that g ∈ Hgi (such i exists because the cosets of a subgroup of G form a g hgi gi partition of G), then there is some h ∈ H with g = hgi hence H = H = H . This shows that any conjugate of H is one of Hg1 ,...,Hgn 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 E 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 E G, N open}

Proof. Let N E 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 6∈ X. Then since y ∈ G \ X is open, by the theorem above there exists N E G open such that Ny ∩ X = ∅. Then y 6∈ 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 S 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 K1,K2 ∈ L there exists K3 ∈ L such that K3 ⊆ K1 ∩ K2. 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 g1, g2 ∈ G and K1,K2 ∈ L then if K1g1 ∩K2g2 6= ∅, taken x ∈ K1g1 ∩ K2g2 we have K1g1 ∩ K2g2 = K1x ∩ K2x = (K1 ∩ K2)x =

∪y∈K1∩K2 K3y where L 3 K3 ⊆ K1 ∩ K2. So K1g1 ∩ K2g2 is open.

Consider v : G×G → G,(g, h) 7→ gh−1. We have to show that v is continous. −1 Suppose that g1, g2 ∈ G and g1g2 ∈ W for some open subset W of G. There −1 −1 exists N ∈ L such that Ng1g2 ⊆ W (a coset of N containing g1g2 is equal to −1 −1 Ng1g2 ). Let U := Ng1 and V := Ng2. Then v(U × V ) = UV ⊆ 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 N1 and N2 are element of L there exists N3 ∈ L contained in the intersection N1 ∩ N2. If N1,N2 ∈ L then we have the homomorphism G → G/N1 × G/N2, g 7→ (gN1, gN2). The kernel is N1 ∩ N2. So G/(N1 ∩ N2) is isomorphic to a subgroup of G/N1 × G/N2, which is a π-group, so that G/(N1 ∩ N2) is a π-group, and we can take N3 := N1 ∩ N2.

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 K1,K2 are open normal subgroups then K1 ∩K2 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 3 A ⊆ B then B ∈ F. 3. ∅ 6∈ 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 6∈ 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 3 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 Xλ be a compact topological space, for every Q λ ∈ Λ. Then X := λ∈Λ Xλ 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 6= ∅. For every λ ∈ Λ consider

Dλ := {D ⊆ Xλ | D closed in Xλ, πλ(F ) ⊆ D ∃F ∈ F}

17 Clearly ∩Dλ is closed in Xλ. To show that it is non empty it suffices to show that the intersection of any given finite subfamily is non empty (because Xλ is r r compact). So consider ∩i=1Dλi ⊇ ∩i=1πλ(Fi) for some Fi ∈ F. If it is empty then r r \ \ ∅ = πλ(Fi) ⊇ πλ( Fi) i=1 i=1 r so ∩i=1Fi = ∅ ∈ F because F is closed under finite intersections. Contradiction because F is a filter, so ∅ 6∈ F. Now let aλ ∈ ∩Dλ for every λ ∈ Λ, and consider a := (aλ)λ∈Λ ∈ X. We want to show that a ∈ F for every F ∈ F. Suppose by contradiction that a 6∈ F for some suitable F ∈ F. Then a ∈ X − F , which is open, so

\ −1 a ∈ πj (Uj) ⊆ X − F j∈J for some suitable J ⊆ Λ ﬁnite, Ui ⊆ Xi open in Xi. This implies clearly that πj(a) ∈ Uj for every j ∈ J. Now

[ −1 F ⊆ (X − πj (Uj)) ∈ F i∈J But since F is an ultrafilter, one of the sets in this finite union belongs to F, −1 −1 say X − πj (Uj) ∈ F for a suitable j ∈ J. πj(X − πj (Uj)) ⊆ Xj − Uj, so by definition of Dj, Xj − Uj ∈ Dj. This implies that aj ∈ Xj − Uj, i.e. aj 6∈ Uj, 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 {Xi}i∈I of sets with a map ϕij : Xj → Xi

for any i ≤ j, such that ϕii = idXi for any i ∈ I, and if i ≤ j ≤ k then ϕij ◦ ϕjk = ϕik.

• A set of compatible maps on the inverse system {Xi, ϕij}i,j is a set of maps {ϕi : Y → Xi}i such that whenever i ≤ j the diagram Y

ϕi

ϕj ~ ϕij Xj / Xi commutes.

18 • An inverse limit on the inverse system {Xi, ϕij}i,j is a set X together with a set of compatible maps αi : X → Xi, with the following universal property: for any set of compatible maps βi : Y → Xi there exists a unique map σ : Y → X such that the diagram

σ Y / X

βi

αi ~ Xi commutes. 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, αi) and (Y, βi) be two inverse limits of the direct system {Xi, ϕij}i,j. By deﬁnition of inverse limit there are unique maps σ, τ such that the following diagrams commute for all i ∈ I.

σ τ τ σ Y / X / Y X / Y / X

βi αi αi βi β αi ~ i ~ Xi Xi This implies that the following diagrams commute too.

τ◦σ σ◦τ Y / Y X / X

βi αi β αi ~ i ~ Xi Xi 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 {Xi, ϕij}i,j be an inverse system and let Y X := {(xi)i∈I ∈ Xi | ∀j ≥ i, ϕij(xj) = xi} i∈I and consider the maps αi := πi|X , the restrictions of the projections. X with the αi is an inverse limit for the inverse system {Xi, ϕij}i,j. Indeed if βi : Y → Xi is a set of compatible maps then there is a unique map σ : Y → X satisfying αi ◦ σ = βi for all i, indeed such compatibility condition is saying precisely that if y ∈ Y then πi(σ(y)) = βi(y) in other words σ(y) is the tuple (βi(y))i. This deﬁnes uniquely σ, and observe that the universality works “categorically” because if each βi is a homomorphism then σ is also a homomorphism, and if each βi 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 {Xi, ϕij}i,j be an inverse system of topological spaces (the ϕij are continuous mappings), and let X be its inverse limit. We have:

• If the Xi are Hausdorff then X, as a subspace of a cartesian product of Hausdorff topological spaces, is Hausdorff.

• If the Xi are totally disconnected then X is totally disconnected (because the cartesian product is totally disconnected). Q • If the Xi are Hausdorﬀ then X is closed in Xi: if j ≥ i consider Q i∈I Dij := {c ∈ l Xl | ϕij(πj(c)) = πi(c)}. Since Xi is Hausdorﬀ and all the considered maps are continuous, Dij is closed. Since X is the intersection of the Dij, X is closed.

• If the Xi are Hausdorﬀ and compact, then X is compact, being a closed subspace of a compact Hausdorﬀ topological space.

It follows that if each Xi is a finite discrete set then the corresponding inverse limit is a profinite space, and if each Xi is a finite discrete group then the corresponding inverse limit is a profinite group.

We conclude with a motivation.

Proposition 6 (Density criterion). Let {Xi, ϕij}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 ϕi) its inverse limit. Assume that Y ⊆ X is such that ϕi(Y ) = Xi 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 Xσ := 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 Xσ is finite and discrete. Consider βσ : X → Xσ, the natural projection. It is continuous and epimorphic. The set Σ is a poset with the following partial order: σ1 ≤ σ2 if σ2 is a refinement of σ1, i.e. if any clopen of the partition σ1 is a union of suitable clopens of the partition σ2. It is easy to show that with this partial order Σ is a directed set. Define moreover, whenever

σ1 ≤ σ2, ϕσ1σ2 : Xσ2 → Xσ1 ,[x]σ2 7→ [x]σ1 . Here [x]σ denotes the clopen of the partition σ containing the point x. In this way {Xσ, ϕσ2σ1 } is an inverse system (this is easy to show). Denote by (Y, ϕσ) its inverse limit. Q Recall that Y is a topological subspace of σ∈Σ Xσ. By universality of the inverse limit Y , there exists a unique continuous map β : X → Y such that the diagram β X / Y

βσ

ϕσ ~ Xσ commutes for any σ ∈ Σ. • β is surjective: let Z := β(X). Then everytime σ ∈ Σ

ϕσ(Z) = ϕσ(β(X)) = βσ(X) = Xσ

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 x1, x2 ∈ X and assume that β(x1) = β(x2). Then

βσ(x1) = ϕσ(β(x1)) = ϕσ(β(x2)) = βσ(x2)

21 for any σ ∈ Σ. This means exactly, by definition of βσ, that every clopen of X that contains x1 contains also x2, 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 x1 = x2. 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 {Xi, ϕij} when the partial order on the set of indices is equality (the “trivial” order).

7 Pro-C groups

Proposition 7. Let {Xi, ϕij}i,j be an inverse system of Hausdorﬀ topological spaces on a directed set I, and let X (together with the ϕi) be its inverse limit. −1 A basis for the topology of X is given by the following subsets: ϕk (U) where k ∈ I and U is open in Xk.

Proof. Since X is a subspace of the product of the Xi, a basis for the topology of X is given by the subsets of the kind

\ −1 P := {πj (Uj) | j ∈ J ⊆ I finite, Uj ⊆ Xj open ∀j ∈ J} ∩ X

We need to prove that for any a ∈ P there exists k ∈ I and U ⊆ Xk open such −1 that a ∈ ϕk (U) ⊆ P (recall that ϕj = πj|X ). Take k ∈ I such that j ≤ k for −1 any j ∈ J, and consider ϕjk (Uj). It is an open subset of Xk and it contains ak. Let \ −1 U := ϕjk (Uj) j∈J −1 −1 We claim that ϕk (U) ⊆ P . In fact, assume that b = (bi)i∈I ∈ ϕk (U). Then bk ∈ U, so for any j ∈ J,

ϕj(b) = ϕjk(ϕk(b)) = ϕjk(bk) ∈ Uj

That is, b ∈ P .

22 Proposition 8 (Density criterion). Let {Xi, ϕij}i,j be an inverse system of Hausdorﬀ topological spaces on a directed set I, with X (together with the ϕi) its inverse limit. Assume that Y ⊆ X is such that ϕi(Y ) = Xi 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 −1 −1 ϕk (U), where k ∈ I and U 6= ∅ is open in Xk. Then Y ∩ ϕk (U) 6= ∅ because ϕk(Y ) = Xk ⊇ U.

Lemma 8. Let {Xi, ϕij}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 ϕi ◦ α : Y → Xi is continuous for any i ∈ I. Proof. One direction is trivial: a composition of continuous mappings is con- −1 tinuous. Conversely if α ◦ ϕi is continuous for any i ∈ I then (ϕi ◦ α) (U) = −1 −1 α (ϕi (U)) is open for any i ∈ I and for any open U of Xi. This yields the −1 continuity of α because ϕi (U) is a generic open basis subset.

Lemma 9. Let G be an inverse limit of Hausdorff topological groups Gi, and take L E G open. Then there exists i ∈ I such that ker(ϕi) ⊆ L. −1 Proof. Since 1 ∈ L, there exists an open basis subset ϕi (U), where U is an −1 open subset of Gi, such that 1 ∈ ϕi (U) ⊆ L. It follows that ker(ϕi) ⊆ L because 1 ∈ U (being ϕi(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 K1,K2 ∈ I there exists K3 ∈ I such that K3 ⊆ K1 ∩ K2. 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 K1 <∗ K2 if

K2 ≤ K1. Consider the maps ϕK1K2 : G/K2 → G/K1, gK2 7→ gK1. I is a directed set and {ϕKiKj }i,j is a set of compatible maps. Consider the inverse limit Y Gˆ := lim G/K ⊆ G/K ←− K∈I K∈I Q Let θ : G → K G/K be the natural map g 7→ (gK)K . Then: 1. θ is a continuous homomorphism. [all the projections are continuous] ˆ 2. θ(G) ≤ G. [If Ki <∗ Kj, ϕKiKj (gKj) = gKi]

3. θ(G) is dense in Gˆ. [density criterion] T 4. ker(θ) = K∈I K. 5. If G is compact then θ(G) = Gˆ.[θ(G) is compact in a Hausdorﬀ, so it is closed]

23 T ˆ 6. If G is compact and K∈I K = {1} then G and G are homeomorphic and isomorphic. [θ is continuous and bijective from the compact G to the Q 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 := {Gλ | λ ∈ Λ} as before. The following are equivalent: 1. G is a pro-C group. Q 2. G is homeomorphically isomorphic to a closed subgroup of λ Gλ.

3. G is compact and ∩{N E G | N open, G/N ∈ C} = {1}. 4. G is compact, totally disconnected and G/N ∈ C for any open N E G. Proof. (1) ⇒ (2). G is an inverse limit, so it is closed in the cartesian product of some Gλ’s, therefore it can be obtained as a closed subspace of the cartesian product of all the Gλ’s (simply intersecting with pre-images of {1}), and the Q cartesian product λ Gλ is Hausdorff since the Gλ are Hausdorff (being finite and discrete).

(2) ⇒ (3). G is compact because it is closed in the product, which is compact (being the Gλ compact). Now recall that the ϕλ are the restrictions of the projections. Nλ := ker(ϕλ) is closed of ﬁnite index (because G/Nλ is isomorphic to a subgroup of Gλ, which is ﬁnite) so it is open and G/Nλ belongs to C because it is an isomorphic to a subgroup of Gλ ∈ C. So the intersection of the Nλ’s contains the intersection in the statement. But ∩λNλ = {1}.

24 (3) ⇒ (1). We use theorem 8, recalling that I := {N EG open | G/N ∈ C} is ∼ a ﬁltering basis (in fact if N1,N2 ∈ I then N1 ∩N2 ∈ I because G/(N1 ∩N2) =≤ G/N1 ×G/N2). Let Y be the inverse limit of the class {G/N | N EG, 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 Nλ := ker(ϕλ) ≤ N (lemma 9). Then G/N is isomorphic to a quotient of G/Nλ, which is isomorphic to a subgroup of Gλ. 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 E 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, T N = {1} where

N := {N E 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 T 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 (G,ˆ j) where Gˆ is a profinite group and j : G → Gˆ 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 θˆ : Gˆ → H continuous homomorphism such that θ = θˆ ◦ j.

j G / Gˆ

θ ˆ ∃!θ H

• Existence: Let Gˆ be the inverse limit of the G/K on I. It is a profinite group. Let j : G → Gˆ, g 7→ (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 θK : G/K → H. We get θ = θK ◦ πK . Define ˆ θ = θK ◦ϕK where ϕK is the canonical map coming from the inverse limit. The following diagram commutes.

ϕK θ Gˆ / G/K K / H a O = θ πK j G

26 ˆ ˆ ˆ We have to show that θ is unique. So let θ1 and θ2 be two such maps. By the density criterion, j(G) is dense in Gˆ. Since H is discrete, it is ˆ ˆ Hausdorff, so the subset of Gˆ in which θ1 = θ2 is closed, and it contains ˆ ˆ j(G) because θ1(j(g)) = θ2(j(g)) for all g ∈ G (by compatibility). Since ˆ ˆ j(G) is dense, θ1 = θ2. • 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 ˆ exists a unique map θM : Gˆ → H/M such that the diagram

j G / Gˆ

πM ◦θ θ θˆM

πM ! H / H/M

ˆ commutes. By universal property of the inverse limit H, since {θM }M is a compatible set of maps Gˆ → H/M, there exists a unique θˆ : Gˆ → H such ˆ ˆ that θπM = θM for any open normal subgroup M of H. The compatibility ˆ implies that πM (θ(j(g))) = πM (θ(g)) for any M open and normal in H, and for any g ∈ G. Then θˆ(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 θˆ just use the argument of the previous point.

• Unicity of the completion. Standard argument using the previous point. Let Gˆ1, Gˆ2 be two diﬀerent choices. Then by universality there exist α : Gˆ → Gˆ , β : Gˆ → Gˆ such that αβ = id and βα = id . 1 2 2 1 Gˆ1 Gˆ2

G G

j1 j2 j2 j1 j1 j2 ~ β ~ β Gˆ α Gˆ Gˆ Gˆ Gˆ α Gˆ 1 / 2 /6 1 2 / 1 /6 2

id ˆ id ˆ G1 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 GC of the G/N on I is called the pro-C completion of G. The image of the canonical map j : G → GC is dense in the T pro-C completion and ker(j) = G/N∈C N.

27 Similarly as above it is possible to show (exercise) that the pro-C completion of an abstract group G is a pro-C group with the following universal property: for any pro-C group H and any continuous homomorphism θ : G → H, there ˆ ˆ exists a unique θ : GC → H continuous homomorphism such that θ = θ ◦ j. A group G is called residually C if the intersection of the normal subgroups N such that G/N ∈ C is trivial. In this case the canonical map j from G to its pro-C completion is injective. Therefore a residually C group can be thinked as a dense subgroup of a pro-C group.

The p-adic integers.

Let p be a prime. We denote by Zp the pro-p completion of Z, by deﬁnition the inverse limit of the Z/N on I, where I is the set of the open normal subgroups of Z such that |Z/N| is ﬁnite of order equal to a power of p. Now, Z/N has n n n order p if and only if N = p Z, so that Zp is the inverse limit of the Z/p Z, where {Z/pnZ | n ∈ N} is an inverse system: if m ≥ n we get Z/pmZ → Z/pnZ.

In this way, x ∈ Zp is a sequence

2 Y n x = (x1 + pZ, x2 + p Z, ...) ∈ Z/p Z n∈N m such that if n ≥ m then xn ≡ xm mod(p ). Let x1 + pZ = a0 + pZ with 2 2 2 0 ≤ a0 ≤ p−1. Let x2+p Z = y2+p Z with 0 ≤ y2 ≤ p −1, so that y2 = b0+b1p with 0 ≤ b0, b1 ≤ p − 1. It follows that b0 ≡ a0 mod(p), that is, b0 = a0. Going 2 on this way we ﬁnd that x0 = a0, x1 = a0 + a1p, x2 = a0 + a1p + a2p , ...

2 x = (a0, a0 + a1p, a0 + a1p + a2p , ...) P i We can think of a p-adic integer just as a formal inﬁnite sum aip for i∈N unique integers ai such that 0 ≤ ai ≤ p − 1 for all i.

More precisely: call Zp the pro-p completion of Z and call S the set of formal P∞ j j inﬁnite sums j=0 ajp where 0 ≤ aj < p for all j ≥ 1. Consider the reduction i i mod p as a map ϕi : S → Z/p Z and let θ : S → Zp given by s 7→ (ϕi(s))i. The injectivity of θ follows from the unicity of the representation of an integer i i in base p. For the surjectivity let x = (xi + p Z) ∈ Zp where 0 ≤ xi < p for all i i. Choose a0 = x1. By deﬁnition we have xi+1 − xi = aip for some integer ai i+1 P∞ j and 0 ≤ ai < p being xi+1 < p . It follows that x = θ( j=0 ajp ).

Observe that for example

∞ X −1 = (p − 1)pj, (p − 1)(1 + p + p2 + p3 + ...) = −1. j=0

Zp is a topological ring (it is an inverse limit of rings). In particular it is an additive abelian group. We now study its topological properties as a ring and

28 as an additive group. Let G := Zp. Observe that since G contains a subring isomorphic to Z (speciﬁcally j(Z)), Zp has characteristic zero. The embedding Pn j j : Z → Zp is given by z 7→ j=0 ajp where the power sum is the representation of z in base p.

• G/pnG =∼ Z/pnZ. n The map πn : G → Z/p Z is continuous, its codomain is ﬁnite and dis- n crete, so ker(πn) is open. If for some n ∈ N we have xn ≡ 0 mod (p ) i n then clearly xi ≡ 0 mod(p ) for any i = 0, ..., n, that is, z = p z¯ for a suit- n n ∼ n ablez ¯ ∈ G. It follows that ker(πn) = p G. Therefore G/p G = Z/p Z. • The open subgroups of G are precisely G ⊃ pG ⊃ p2G ⊃ p3G ⊃ .... Let H ≤ G be open. Then G/H is a ﬁnite p-group, with |G/H| = pm (for some m ≥ 0). Then clearly pmG ≤ H, so since |G : H| = pm equals |G : pmG| (by the previous point), we have H = pmG. • The closed subgroups of G are of the type pnG (with n ≥ 0) or {0}. If K is a closed subgroup of G, K is an intersection of open subgroups, so it is an intersection of some of the pmG (because in general the closure of a subset Y is the intersection of all the YN where N varies in the family of the open normal subgroups of G). So either K = pmG or K = {0}. In particular {0} is the only closed non-open subgroup of G.

• Zp is an integral domain. Let 0 6= x ∈ Zp. Consider the map αx : Zp → Zp, z 7→ zx. It is continuous. ker(αx) = {z ∈ Zp | zx = 0} is the preimage of 0, so it is closed. So it is either {0} or one of the pmG. But if x 6= 0 then pmx 6= 0 for any m m ∈ N (think of the inﬁnite sum representation), so ker(αx) 6= p Zp for any m ∈ N. So ker(αx) = {0}, that is, Zp is a domain.

• Zp is a local principal ideal domain. Let xZp be a non trivial principal ideal. xZp = αx(Zp) is closed in Zp m because it is compact and G is Hausdorff. So it is of the form p Zp for some m ∈ N. A non trivial ideal of Zp is generated by principal ideals. But the principal ideals are totally ordered by inclusion, so any ideal is principal. In particular the unique maximal ideal of Zp is pZp ∼ (it is maximal because Zp/pZp = Z/pZ is a field). So Zp is local and ∼ Zp/pZp = Z/pZ (a field). It follows that z ∈ Zp is invertible if and only P i if z 6∈ pZp, and i aip is invertible if and only if a0 6= 0. For example i the inverse of 2 in Z3 is given by the sequence of the inverses of 2 mod 3 (which equals (3i + 1)/2) for all i, which can be written as infinite sum 2−1 = −1 − 3 − 32 − 33 − ...

• The additive group Zp is torsion-free. k If x ∈ Zp and m ∈ N is such that mx = 0 then writing m = ap we have k that a is invertible in Zp and Zp is a domain, so p x = 0 implying x = 0.

29 Now we want to study the proﬁnite completion of Z. We want to prove the following: ˆ Q Theorem 10. Z is homeomorphic to p prime Zp.

We will interpret this result in the Sylow theory language, Zp being the (unique) Sylow p-subgroup of Zˆ.

We have the canonical map θp : Z → Zp, which is injective because Z is residually p, and whose image is dense in Zp. Let Y G := Zp p prime

It is a proﬁnite group (being a cartesian product of proﬁnite groups). The θp’s induce an embedding Z ,→ G, z 7→ (θp(z))p. We will write Z ⊆ G, by meaning that we identify Z with θ(Z). Lemma 10. For any 0 6= n ∈ N we have nG + Z = G.

Proof. Let π := {p1, ..., pr} be the set of the prime factors of n:

α1 αr n = p1 ...pr = q1...qr

αi where qi = pi for i = 1, ..., r. Take x = (xp)p ∈ G. We want to build z ∈ Z such that x − z ∈ nG. That is, xp − z ∈ nZp for any prime p. If p 6∈ π, n is invertible mod(p), so nZp = Zp and we have nothing to prove. Take now p = pi ∈ π. qiZp + Z is open, closed and dense in Zp (it is open because it is a ﬁnite union of cosets of subgroups of Zp, it is closed because it is open; for the density recall that Z is dense in Zp), so it equals Zp. So there exists zi ∈ Z such that xpi − zi ∈ qiZpi . Take z ∈ Z such that z ≡ zi mod qi for every i = 1, ..., r (it exists by the chinese remainder theorem). For i = 1, ..., r, xpi − z ∈ qiZpi = nZpi (recall: mZp = Zp if p does not divide m). Lemma 11. G/nG =∼ Z/nZ. Proof. Remark that

G/nG = (nG + Z)/nG =∼ Z/(nG ∩ Z) is a quotient of Z/nZ, hence |G/nG| ≤ n. We have

πi ϕqi G / Zpi / Z/qiZ 4 µi

The maps µi are surjective because πi and ϕqi are surjective. The µi’s induce a surjective µ, from G to the product: Y µ : G → Z/qiZ i

30 g 7→ (µi(g))i Q ∼ Moreover, i Z/qiZ = Z/nZ by the chinese remainder theorem. Now ker(µ) ≥ nG, so G/nG has a quotient isomorphic to Z/nZ. Since |G/nG| ≤ n we obtain that G/nG =∼ Z/nZ. Lemma 12. If H is an open subgroup of G of ﬁnite index n then H = nG. Proof. It follows from nG ⊆ H and |G/nG| = n. Now being G proﬁnite G = lim G/H = lim G/nG = lim /n = ˆ ←− ←− ←− Z Z Z H≤G open n n Exercise 2.1.9 of Ribes-Zalesski. Exercises 14 and 15 of Chapter 1 of Wilson’s book.

9 Exercises - list 1

1. A compact discrete space is ﬁnite. This is because if X is discrete the sets {x} with x ∈ X form an open cover of X. 2. A product of discrete spaces is not discrete in general.

For example if Xi is a finite space with |Xi| ≥ 2 for all i ∈ I and I is infinite Q then i∈I Xi is infinite and compact (by the Tychonoff theorem, being Xi compact for all i) hence it is not discrete (by the previous exercise). 3. If X is an infinite set with the cofinite topology then the points are closed but X is not Hausdorff. Indeed the points are finite subsets so they are closed (recall that in the cofinite topology closed means finite) however such a space cannot be Hausdorff because if U and V are open and nonempty then U ∩ V 6= ∅ (there are no disjoint nonempty open subsets). 4. If f : X → Y is continuous and X is compact then f(X) is compact. S S −1 Let f(X) = i∈I Ui be an open cover of f(X), then X = i∈I f (Ui) is an open cover of X. Since X is compact there is a finite set J ⊆ I with S −1 S X = j∈J f (Uj). Hence f(X) = j∈J Uj is a finite subcover. 5. If f : X → Y is continuous and bijective, X is compact and Y is Hausdorff then f is a homeomorphism. We need to prove that if U is open in X then f(U) is open in Y . Let F = X − U. Since f is bijective f(F ) = f(X − U) = Y − f(U) so f(U) is open if f(F ) is closed. Hence it is enough to show that f(F ) is closed in Y . But F is closed in X, which is compact, so F is compact, so f(F ) is compact (being f continuous) hence f(F ) is closed in Y (being Y Hausdorff).

31 6. A topological space Y is Hausdorﬀ if and only if ∆ = {(y, y): y ∈ Y } is closed in Y × Y . Let x, y ∈ Y with x 6= y. Let U and V be open subsets of Y . Saying that x ∈ U, y ∈ V and U ∩ V = ∅ is equivalent to saying that

(x, y) ∈ U × V ⊆ Y × Y − ∆

Using this (and the definition of product topology) the result is immediate. 7. Let f, g : X → Y be two continuous functions and Y Hausdorff. Show that the set {x ∈ X : f(x) = g(x)} is closed in X. It is the preimage via γ : X → X × Y , γ(x) = (f(x), g(x)) of the closed subset {(y, y): y ∈ Y } (Y is Hausdorff). Observe that γ is continuous because the two components of γ are.

8. Proabelian implies abelian, prosolvable does not imply solvable, pronilpotent does not imply nilpotent. A pro-C group is a closed subgroup of a cartesian product of C-groups, so a proabelian group is always abelian (it is a subgroup of a product of abelian groups). However if Gi is a ﬁnite solvable group for every i ∈ N, Q with Gi of derived length i, then Gi is not solvable. Similarly if Gi i∈N is a ﬁnite nilpotent group for every i ∈ N, with Gi of nilpotency class i, Q then Gi is not nilpotent. i∈N

9. Wilson, chapter 0, exercise 2. Let G be a Hausdorﬀ topological group. For S ⊆ G let CG(S) = {g ∈ G : gs = sg ∀s ∈ S} (centralizer −1 −1 of S) and for H ≤ G let NG(H) = {g ∈ G : g hg ∈ H, ghg ∈ H ∀h ∈ H} (normalizer of H).

• If S ⊆ G then CG(S) is closed. This is because CG(S) is the intersection of the Cs = {g ∈ G : gs = sg} and each Cs is closed being the “equality set” of two continuous functions (left and right multiplication by s) and G being Hausdorﬀ.

• If H is a closed subgroup of G then NG(H) is closed. It is the intersection of sets of the form {g ∈ G : g−1hg ∈ H} and {g ∈ G : ghg−1 ∈ H} (for ﬁxed h) which are closed being H closed and multiplication and inverse continuous operations.

• Each closed abelian subgroup of G is contained in a maximal closed abelian subgroup of G.

We use Zorn lemma. Let {Ai : i ∈ I} be a chain of closed abelian subgroups of G, we need to ﬁnd a closed abelian subgroup of G

32 S containing all of the Ai’s. Let A := i∈I Ai. Since the Ai’s form a chain, A is clearly an abelian subgroup of G, however in general it is not closed. We claim that the closure A is an abelian closed subgroup of G containing all of the Ai’s. It is clear that A is a closed subgroup of G containing all of the Ai’s (it contains A and we have seen that the closure of a subgroup of a topological group is a subgroup).

Hence it is enough to show that if A is an abelian subgroup of G then the closed subgroup A is also abelian (observe that this is false in general if G is not Hausdorff: as a counterexample take any nonabelian group G with the trivial topology and A = {1}, then G is a topological group, A is abelian and A = G). For this consider F = {(x, y) ∈ A × A : xy = yx}. We need to show that F = A×A. Since A×A is closed in G×G, F is a closed subset of G×G (being defined by the equation xyx−1y−1 = 1 and G being Hausdorff) and clearly F ⊇ A×A (because A is abelian) so F ⊇ A × A. If we show that A × A = A × A we will deduce F = A × A which is what we want.

We show that A × A = A×A. The inclusion ⊆ is clear because A×A is closed and it contains A × A. Now assume the other inclusion ⊇ does not hold (by contradiction) and let (a, b) ∈ A × A such that (a, b) 6∈ A × A. Let T = A × A and W = G × G − T . Since T is closed, W is open, and (a, b) ∈ W . By deﬁnition of product topology there are two open subsets U, V of G with (a, b) ∈ U ×V ⊆ W . Since a ∈ U ∩ A and b ∈ V ∩ A we deduce A ∩ U 6= ∅ and A ∩ V 6= ∅, hence (A×A)∩(U ×V ) 6= ∅, and this contradicts the fact that T is disjoint from U × V (recalling that T ⊇ A × A).

• If {1} = G0 C G1 C G2 C ... C Gn = G is such that Gi/Gi−1 is abelian for all i = 1, . . . , n then there is such series consisting of closed subgroups (take closures).

Let Hi be the closure of Gi for all 0 ≤ i ≤ n. Observe that the closure of a normal subgroup of a topological group is a normal subgroup (if H is normal in G and g ∈ G deﬁne γ : G → G to be the conjugation by g, then H is contained in the preimage γ−1(H) which is closed so γ−1(H) ⊇ H). So we indeed have a series of closed subgroups, each normal in the next, {1} = H0 C H1 C H2 C ... C Hn = G. The fact that Hi/Hi−1 is abelian is equivalent to the fact that [Hi,Hi] ⊆ Hi−1, −1 equivalently Hi × Hi ⊆ f (Hi−1) where f : G × G → G is given by −1 −1 −1 −1 f(x, y) = xyx y . But clearly Gi × Gi ⊆ f (Gi−1) ⊆ f (Hi−1) −1 being Gi−1 ⊆ Hi−1, and since f (Hi−1) is closed and it contains Gi × Gi, it contains Gi × Gi = Gi × Gi = Hi × Hi.

33 10. Wilson, chapter 1, exercise 4. Let I be a directed set. A subset J of I is called coﬁnal if for all i ∈ I there is some j ∈ J with j ≥ i.

• If I is countable then it has a coﬁnal subset isomorphic (as an ordered set) to N. Since I is countable we may write its elements as i1, i2, i3,... Let j1 = i1. For every n ≥ 2 choose a jn with the property that jn ≥ jn−1, in (it exists because I is a directed set). Then J = {j1, j2, j3,...} is coﬁnal (because if in ∈ I then jn ≥ in) and it is isomorphic to N because j1 ≤ j2 ≤ j3 ≤ ....

• Let J be a coﬁnal subset of I. Show that if G = lim G ←−i∈I i then G =∼ lim G . ←−j∈J j Whenever we say “map” we mean morphism (in the case of topological groups, a morphism is a continuous homomorphism). Let L = lim G . We have canonical maps ψ : L → G for all j ∈ I. If ←−j∈J j j j i ∈ I then choose j ∈ J with j ≥ i, composing ψj with the canonical map ϕij gives a map αi : L → Gi. It is clear that the maps αi form a compatible family of maps. We now show that L together with the αi is the inverse limit of the Gi’s (over I). For this it is enough to prove the universal property. Let βi : Y → Gi be a family of compatible maps. We need to show that there is a unique compatible map γ : Y → L. ∃!γ Y / L

βj ψ j βi Gj

α ϕij i ~ Gi q

The family {βj : j ∈ J} is compatible so by the universal property of L there is a unique map γ : Y → L compatible with the βj’s, so we are left to show that if i ∈ I then αi ◦ γ = βi. Recalling that αi = ϕij ◦ ψj for some (and therefore any) j ≥ i and that ψj ◦ γ = βj we obtain that

αi ◦ γ = ϕij ◦ ψj ◦ γ = ϕij ◦ βj = βi

being the βi’s compatible.

34 11. Wilson, chapter 1, exercise 6. Let θ : G → H be a homomorphism between two proﬁnite groups. Then θ is continuous if and only if whenever A is an open subgroup of H, θ−1(A) is an open subgroup of G. The implication ⇒ is clear, now let us prove ⇐. Since a basis of open subsets of H is given by the cosets hN where h ∈ H and N is an open normal subgroup of H (being H proﬁnite) it is enough to show that θ−1(hN) is open in G. If θ−1(hN) is empty then it is open, now assume θ−1(hN) contains some element x. Then θ(x) ∈ hN so hN = θ(x)N hence

θ−1(hN) = {g ∈ G : θ(g) ∈ hN} = {g ∈ G : θ(g) ∈ θ(x)N} = {g ∈ G : θ(x−1g) ∈ N} = {g ∈ G : x−1g ∈ θ−1(N)} = {g ∈ G : g ∈ xθ−1(N)} = xθ−1(N).

Now xθ−1(N) is open because G is a topological group and θ−1(N) is open in G being N an open subgroup of H.

12. Wilson, chapter 1, exercise 10. Let C be a class of groups closed for subgroups, quotients and ﬁnite direct products. Given a group G denote by Gˆ the pro-C completion of G.

• Each homomorphism f : A → B induces a continuous homomorphism fˆ : Aˆ → Bˆ. Observe that f is continuous (when A and B are given the pro-C topology). For this observe that if K E B is such that B/K ∈ C then A/f −1(K) ∈ C being isomorphic to a subgroup of B/K (by the isomorphism theorem) hence f −1(K) is open in A whenever K is an open normal subgroup of B. Now the argument of the previous exercise shows that f is continuous. Composing f : A → B with the canonical map B → Bˆ gives a continuous map A → Bˆ hence we obtain a unique compatible continuous homomorphism fˆ : Aˆ → Bˆ given by the universal property of the pro-C completion of A.

• Suppose f surjective. Show that fˆ is surjective (completion is right exact) and that ker(fˆ) is the closure of j(ker(f)) in Gˆ where j : G → Gˆ is the canonical map. For every B/K ∈ C the composition A → B → B/K is surjective (composition of surjective maps) and such maps A → B/K induce exactly the map A → Bˆ given by the composition A → Aˆ → Bˆ. Therefore fˆ(Aˆ) projects surjectively onto each B/K hence fˆ(Aˆ) is dense in Bˆ by the density criterion. On the other hand Aˆ is compact, so fˆ(Aˆ) is compact, so it is closed in Bˆ. It follows that fˆ(Aˆ) is closed and dense in Bˆ, so it equals Bˆ.

35 Suppose A and B are residually C. Identify A with its image in Aˆ. Let H = ker(fˆ), then H ∩ A is dense in H (because A is dense in Aˆ) so H ∩ A = H. It is then enough to show that H ∩ A = ker(f). The inclusion ⊇ is clear, we prove the other inclusion. An element a ∈ H ∩A veriﬁes f(a) ∈ K whenever B/K ∈ C. Since B is residually C we deduce f(a) = 1.

• Show that the inclusion Z → Q induces the zero map Zˆ → Qˆ (completion is not left exact). This is because Qˆ = {0} because Q is a divisible group, so it does not have proper subgroups of ﬁnite index. Indeed if H ≤ Q has index n then nQ ⊆ H, on the other hand nQ = Q.

∼ • Show that G\1 × G2 = Gˆ1 × Gˆ2. The idea is to check that the universal property of the completion of G1 × G2 is satisﬁed by Gˆ1 × Gˆ2.

13. Wilson, chapter 1, exercise 12. Let G be a residually finite group and let Gˆ be its profinite completion. Let j : G → Gˆ be the canonical map, so that G =∼ j(G). Let x, y be elements of G (identified with j(x) and j(y)). Prove that the following are equivalent. • x, y are conjugated in Gˆ. • xK and yK are conjugated in G/K for all K E G such that G/K is finite. Let A = {g ∈ Gˆ : g−1xg = y}. We need to prove that A 6= ∅. Observe that g−1xg = y is equivalent to g−1xgy−1 = 1, hence A is the intersection −1 −1 of the sets AN = {g ∈ Gˆ : g xgy ∈ N} where N varies in the family of open normal subgroups of Gˆ. Since Gˆ is compact (being profinite) and each AN is closed (being N closed), if by contradiction A = ∅ then there exist N1,...,Nk open normal subgroups of Gˆ with the property that Tk ˆ i=1 ANi = ∅. Let N := N1 ∩...∩Nk. It is an open normal subgroup of G. There is no g ∈ G that conjugates x into y modulo N, indeed if there was such a g then applying the compositions G → Gˆ → G/Nˆ → G/Nˆ i (which equal the canonical maps G → G/Nˆ i by compatibility) we would obtain that x and y are conjugate modulo Ni for all i = 1, . . . , k, contradicting the Tk fact that i=1 ANi = ∅. Let K = N ∩ G. By one of the exercises N = K. If we show that G/Nˆ =∼ G/K (canonically), since this group is finite (because any open normal subgroup of a profinite group has finite index) we can deduce that x and y are not conjugated in G/K, a contradiction. Observe that K is equal to the kernel of G → Gˆ → G/Nˆ because K ⊆ N, hence we have a canonical map G/K → G/Nˆ , we are left to show that

36 |G/K| = |G/Nˆ |. Since G is dense in Gˆ we have GN = Gˆ, so there is a left transversal t1, . . . , tn of N in Gˆ with ti ∈ G for all i = 1, . . . , n, now we have a disjoint union Gˆ = t1N ∪ ... ∪ tnN so

G = (t1N ∪...∪tnN)∩G = t1(N ∩G)∪...∪tn(N ∩G) = t1K ∪...∪tnK. We deduce |G/Nˆ | = |G/K|. [cf. Page 81 of Ribes-Zalesski]

10 Lagrange and Sylow

A Steinitz number (or supernatural number) is a formal infinite product Y n = pn(p) p prime where n(p) ∈ N (where N 3 0) or n(p) = ∞. Clearly every natural number is also a supernatural number (defined by the fact that n(p) 6= ∞ for every p and n(p) 6= 0 only for finitely many p), for example the number 1 is defined Q n(p) Q m(p) by 1(p) = 0 for all prime p. Let n = p p , m = p p be supernatural numbers. Q n(p)+m(p) 1. Product: n · m = p p . 2. n divides m means that n(p) ≤ m(p) for all p.

Q min(n(p),m(p)) 3. The gcd (greatest common divisor) between n and m is p p . We can deﬁne analogously the gcd of an inﬁnite family of supernatural numbers.

Q max(n(p),m(p)) 4. The lcm (least common multiple) between n and m is p p . We can define analogously the lcm of an infinite family of supernatural numbers. 5. n and m are said to be coprime if gcd(n, m) = 1, equivalently lcm(n, m) = nm. This means that for every prime p we have min(n(p), m(p)) = 0. Definition 9 (Index of a closed subgroup). Let G be a profinite group, H ≤ G a closed subgroup. The index of H, |G : H|, is the least common multiple of the indices of the open subgroups of G containing H. The order of G is |G| := |G : {1}| and the order of x ∈ G is the order of hxi. Equivalently, the index of H ≤ G closed is

|G : H| = lcm(|G : NH| | N Eo G) because every U ≤ G open containing H contains some open normal subgroup N of G.

Note: In particular, the pro-p groups are the proﬁnite groups of order pn with n ≤ ∞.

37 Lemma 13. Let H be a subgroup of a proﬁnite group G. Then a subgroup K ≤ H is open in H if and only if there is an open normal subgroup N of G With N ∩ H ≤ K. S Proof. If there is such N then K is open in H because K = k∈K k(N ∩ H) is a union of open subsets of H. Conversely if K is open in H then K = U ∩ H with U open in G, so there is an open normal subgroup N of G with 1 ∈ N ⊆ U hence N ∩ H ⊆ U ∩ H = K. Theorem 11 (Lagrange). Let G be a proﬁnite group, K ≤ H ≤ G closed. Then |G : K| = |G : H| · |H : K|.

Proof. If N E G is open then N ∩ H E H is open in H and |G : NK| = |G : NH| · |NH : NK| = |G : NH| · |H :(N ∩ H)K|

Now observe that if ai, bj are supernatural numbers for every i ∈ I, j ∈ J then

lcm(ai | i ∈ I) · lcm(bj | j ∈ J) = lcm(aibj | i ∈ I, j ∈ J) It follows that |G : K| divides |G : H| · |H : K|.

Conversely, let N1 EG be open, N2 EH be open. Then N2 is the intersection with H of some open of G, so that there exists M E G open such that M ∩ H ≤ N2. Let N := M ∩ N1. Then N E G is open in G, and |G : N1H| · |H : N2K| divides |G : NH| · |H :(N ∩ H)K| = |G : NK|. So that |G : H| · |H : K| divides |G : K|.

Lemma 14. Let (Hi | i ∈ I) be a family of closed subgroups of the proﬁnite group G such that for every i, j ∈ I there exists k ∈ I such that Hk ≤ Hi ∩ Hj. Then \ |G : Hi| = lcm(|G : Hi| | i ∈ I) i T Proof. lcm(|G : Hi| | i ∈ I) divides |G : i Hi| by Lagrange’s theorem: \ \ |G : Hi| = |G : Hi0 | · |Hi0 : Hi| i i T for every i0 ∈ I. Let U ≤ G be open such that i Hi ⊆ U. Then \ (Hi ∩ (G \ U)) = ∅ i

By compactness there exist i1, ..., ir ∈ I such that r r \ \ (Hλi ∩ (G \ U)) = ∅ ⇒ Hiλ ⊆ U λ=1 λ=1

Now, if Hk ≤ Hi for every λ = 1, ..., r then Hk ≤ U, so |G : U| divides |G : Hk|. Tλ Therefore |G : i Hi| divides lcm(|G : Hi| | i ∈ I).

38 Observe that if H is a closed subgroup of a profinite group G and Hg denotes g−1Hg for g ∈ G, if H ⊆ Hg then Hg = H (warning: this is false in general). Indeed if N is an open normal subgroup of G then HN/N is finite and HN/N ⊆ HgN/N = (HN/N)gN hence HgN = HN by finiteness. Since Hg is (clearly) closed in G it follows that \ \ Hg = HgN = HN = H.

NEoG NEoG Exercise: |G : Hg| = |G : H| for all H ≤ G closed, g ∈ G. Definition 10 (Sylow p-subgroups). Let G be a profinite group. If p is a prime, a Sylow p-subgroup of G is a closed subgroup P ≤ G of order pn (with n ≤ ∞) such that |G : P | and p are coprime. Theorem 12 (Sylow’s theorem, profinite version). Let G be a profinite group, p a prime. Then: 1. G has Sylow p-subgroups. 2. If P is a Sylow p-subgroup of G and T is a closed pro-p subgroup of G then T ⊆ P g for some g ∈ G. 3. Every closed pro-p subgroup of G is contained in a Sylow subgroup of G. 4. Any two Sylow p-subgroups of G are conjugate. Proof. (1). Let I be the set of all closed subgroups of G of index coprime with p. Clearly G ∈ I (because G has index 1), so I 6= ∅, and I is partially ordered by inclusion. We can apply Zorn lemma because if J is a chain in I then T{H | H ∈ J} ∈ I by the lemma above; let P ∈ I be minimal in I. |G : P | is coprime with p. If P is not a pro-p group then there exists M E P open in P such that P/M is not a p-group. Since P/M is finite, we can use Sylow’s theory, finite case: there exists Q/M < P/M, Sylow p-subgroup. Q is closed in P because it is a union of finitely many cosets of M, which is open in P . Since P is closed in G, Q is closed in G. By Lagrange theorem, |G : Q| = |G : P |·|P : Q|. Since |P : Q| = |(P/M)/(Q/M)| is coprime with p, |G : Q| is coprime with p. This implies Q ∈ I, and contradicts the minimality of P .

∼ (2). Let N E G be open. Then N ∩ P E P is open in P and NP/N = P/(N ∩ P ), so NP/N is a p-group. Moreover |G : NP | divides |G : P |, so NP/N is a Sylow p-subgroup of G/N. NT/N =∼ T/(N ∩ T ) is a p-subgroup of G/N. We can apply Sylow’s theory of the ﬁnite case: there exists g ∈ G such that (NT/N)g ⊆ NP/N, or equivalently T g ⊆ NP . Let g RN := {g ∈ G | T ⊆ NP }= 6 ∅

Then RN is closed, because it is a union of cosets of N (if g ∈ RN and n ∈ N then T gn ⊆ (NP )n ⊆ NPN = NP ). If M ≤ N are open normal subgroups then RM ⊆ RN , so the intersection of ﬁnitely many RN is non-empty:

RN1 ∩ ... ∩ RNk ⊇ RN1∩...∩Nk 6= ∅.

39 T By compactness of G, the whole intersection is non empty. Let g ∈ N RN . Then T g ⊆ NP for every open normal subgroup N of G. So \ \ T g ⊆ (NP ) = ( N)P = P N N by the usual lemma.

(3), (4). They follow immediately from the following note: the conjugate of a Sylow p-subgroup of G is again a Sylow p-subgroup of G. In particular for (4): if P and T are Sylow p-subgroups of G then T ⊆ P g ⊆ T h for some g, h ∈ G, so T = P g = T h (see the remark above). Exercises: Wilson, chapter 1 exercises 16, 17, 18, 19; chapter 2 exercises 1,2,3,6.

11 Pronilpotent and pro-p groups

Proposition 10. Let G be a proﬁnite group and let p be a prime number. 1. If G is a pro-p group and H is a closed subgroup of G then H is a pro-p group.

2. If K is a closed normal subgroup of G then G is a pro-p group if and only if both K and G/K are pro-p groups. Proof. To prove (1) let L be an open normal subgroup of H, and let N be an open normal subgroup of G with N ∩ H ≤ L, then H/N ∩ H =∼ HN/N ≤ G/N hence H/N ∩ H is a finite p-group (being G/N a finite p-group), however H/L is a quotient of H/N ∩ H so H/L is also a finite p-group. To prove (2) let K be a closed normal subgroup of G. Observe that if G is a pro-p group then K is a pro-p group by the previous point and we already know that G/K is a pro-p group. Now assume that both K and G/K are pro- p groups. Let N be an open normal subgroup of G. Then both G/NK and K/N ∩ K =∼ NK/N are finite p-groups, so G/N is also a finite p-group. Proposition 11. Let G be a profinite group, K a normal closed subgroup and P a Sylow p-subgroup of G. Then 1. K ∩ P is a Sylow p-subgroup of K. 2. KP/K is a Sylow p-subgroup of G/K.

3. G = NG(Q)K for each Sylow p-subgroup Q of K.

4. H = NG(H) whenever H is a closed subgroup of G which contains NG(Q) for some Sylow p-subgroup Q of K.

40 Proof. For (1) and (2) observe that both K ∩ P and KP/K are pro-p groups, since the latter is isomorphic (as a topological group) to P/K∩P . Since |G : KP | divides |G : P |, (2) follows. We have K ∩ P ≤ Q for some Sylow p-subgroup Q of K and g−1Qg ≤ P for some g ∈ G. Thus

K ∩ P ≤ Q ≤ K ∩ (gP g−1) = g(K ∩ P )g−1 hence K ∩ P = Q = g(K ∩ P )g−1. (3) Let g ∈ G. Then g−1Qg is a Sylow p-subgroup of K and so k−1(g−1Qg)k = −1 Q for some k ∈ K. Therefore x = gk ∈ NG(Q) and g = xk ∈ NG(Q)K. −1 (4) Let u ∈ NG(H). Now Q is a Sylow p-subgroup of K ∩H and so u Qu is a Sylow p-subgroup of u−1(K ∩H)u = K ∩H. Thus h−1(u−1Qu)h = Q for some h ∈ K ∩ H so that uh ∈ NG(Q) ≤ H. It follows that u ∈ H as required.

Lemma 15. Let Hi (i ∈ I) be a family of normal closed subgroups of a proﬁnite group G. Suppose that G is the closure of the abstract subgroup generated by the union of all the Hi’s and let Ki be the closure of the abstract group generated S T by j6=i Hj. Suppose Ki ∩ Hi = {1} for all i ∈ I. If i∈I Ki = {1} then G is isomorphic (as a proﬁnite group) to the cartesian product of all the Hi’s.

Proof. We clearly have Ki E G for all i ∈ I, moreover Ki ∩ Hi = {1} and KiHi S is a closed subgroup containing the union j∈I Hj so that G = KiHi. Hence ∼ Q G/Ki = Hi for each i. Define ϕ : G → G/Ki by g 7→ (Kig)i. We have T i∈I ker(ϕ) = i∈I Ki = {1} and ϕ is continuous, since its composition with each projection map is continuous. We shall prove that ϕ(G) is dense in the cartesian product. Since ϕ(G) is also compact, hence closed, it will follow that ϕ is surjective and therefore is an isomorphism of profinite groups. Now ϕ(Hj) consists of all elements (ci)i Q of i G/Ki such that ci is trivial for all i 6= j, and so ϕ(G) contains the set D consisting of the c = (ci)i such that ci = 1 for all but finitely many i (if J denotes the set of j with cj 6= 1 choose hj ∈ Hj with ϕ(hj) = cj for all j ∈ J and let h be the product of the hj’s, then ϕ(h) = c). Therefore every non-empty Q open subset of i G/Ki intersects D (this follows from the definition of product topology). Hence ϕ(G) is dense. We now characterize pronilpotent groups as those profinite groups that are the cartesian products of their Sylow subgroups. Observe that if X is a subset of a profinite group G then X = G if and only if NX = G for every open normal subgroup N of G. This follows from X = T NX. NEoG Theorem 13. Let G be a profinite group. The following conditions are equivalent.

1. G is pronilpotent. 2. Each Sylow subgroup of G is normal in G. 3. G is isomorphic to the cartesian product of its Sylow subgroups.

41 4. NG(U) 6= U for each proper open subgroup U of G. Proof. (1) ⇒ (4). Let U be a proper open subgroup of G and let N be the normal core of U in G. Then G/N is nilpotent, hence NG/N (U/N) 6= U/N hence NG(U) 6= U. (4) ⇒ (2). Let P be a Sylow subgroup of G and let U be an open subgroup containing NG(P ). Then U = NG(U) by the lemma. By (4) we deduce U = G. This means that G is the only open subgroup of G containing NG(P ). Since P is closed, NG(P ) is closed hence NG(P ) = G (it is the intersection of the open subgroups containing it). (2) ⇒ (3). Let I be the set of prime divisors of |G|, let {Pi : i ∈ I} be the set of Sylow subgroups of G, and for each i let Ki be the closure of the abstract S group generated by j6=i Pj. Note that |Ki| is coprime to i. Indeed if the prime p divides |Ki| then p divides |Ki/M| for some open normal subgroup M of Ki. Only finitely many of the groups MPj/M (j 6= i) are non-trivial, and since they are normal Sylow subgroups of Ki/M, the finite group Ki/M is the direct product of them. We conclude that p divides |Pj| for some j 6= i hence p = j 6= i. T T It follows that Ki ∩ Pi = {1} for each i. Also, i Ki = {1}, indeed if C = i Ki was not trivial then given a prime divisor p of |C| we have |Kp| = |Kp : C| · |C| hence p divides |Kp|, a contradiction. Every profinite group is the closure of the abstract group generated by all of its Sylow subgroups, since this closure clearly has index 1 (it is coprime to every prime divisor of |G|). Applying the lemma the result follows. (3) ⇒ (1). Because finite p-groups are nilpotent, pro-p groups are pronilpotent and any cartesian product of pronilpotent groups is pronilpotent. The Frattini subgroup Φ(G) of a profinite group G is the intersection of all maximal open subgroups of G. It is a normal closed subgroup of G. Proposition 12. Let G be a profinite group, let H and K be closed subgroups of G, with K normal in G. 1. If HΦ(G) = G then H = G. 2. KΦ(G)/K ≤ Φ(G/K). 3. If K ≤ Φ(H) then K ≤ Φ(G).

4. If Φ(G) ≤ K E G and K/Φ(G) is pronilpotent then K is pronilpotent. In particular Φ(G) is pronilpotent. 5. G is pronilpotent if and only if G/Φ(G) is abelian. Proof. (1) H is the intersection of the open subgroups containing H, so if H 6= G then there is a maximal open subgroup M containing H. This implies HΦ(G) ≤ M hence HΦ(G) 6= G. (2) Write Φ(G/K) = T/K. Then T is the intersection of the maximal open subgroups of G containing K, so T contains Φ(G).

42 (3) Suppose that K is not contained in Φ(G). Then there is a maximal open subgroup M of G that does not contain K. We have G = KM so H = K(M ∩H) since K ≤ H. We also have K ≤ Φ(H) hence H = M ∩ H by (1). Therefore K ≤ H ≤ M and G = KM = M, a contradiction. (4) We write Φ = Φ(G). Let Q be a Sylow subgroup of K. Then ΦQ/ΦEK/Φ since K/Φ is pronilpotent, and we deduce

G/Φ = (K/Φ)NG/Φ(ΦQ/Φ) = NG/Φ(ΦQ/Φ) so ΦQEG. Since Q is a Sylow subgroup of K and Φ ≤ K, Q is a Sylow subgroup of ΦQ, hence G = (ΦQ)NG(Q) thus G = QNG(Q) = NG(Q) by (1). This shows that all Sylow subgroups of K are normal hence K is pronilpotent. (5) If G/Φ(G) is abelian then G is pronilpotent by (4) choosing K = G. Suppose G is pronilpotent, then NG(U) 6= U for each proper open subgroup U, so for each maximal open subgroup M we have M E G and G/M is cyclic of prime order. Therefore if x, y ∈ G then x−1y−1xy ∈ M for each M, so that x−1y−1xy ∈ T M = Φ(G) and G/Φ(G) is abelian. Since we have seen that any pronilpotent group is a direct product of pro-p groups, we now discuss pro-p groups. Proposition 13. Let G be a pro-p group and let Gp = hgp : g ∈ Gi. Then

Φ(G) = Gp[G, G].

Proof. Let K := Gp[G, G], a normal closed subgroup of G. We want to prove that Φ(G) = K. If M is a maximal open subgroup of G let N Eo G with N ≤ M, then M/N is a maximal subgroup of the finite p-group G/N hence M is normal ∼ in G and |G : M| = p, in particular G/M = Cp is abelian. This implies that Gp[G, G] ≤ M hence K ≤ M being M closed, so K ≤ Φ(G). To prove the other inclusion define Q := G/K. It is a pro-p group (being a quotient of a pro-p group over a closed normal subgroup) so if N Eo Q then Q/N is a finite p-group, hence it is elementary abelian (by definition of K) so NΦ(Q)/N ≤ Φ(Q/N) = {1}. Therefore Φ(Q) ≤ T N = {1} hence being NEoQ K ≤ Φ(G),

{1} = Φ(Q) = Φ(G/K) ≥ KΦ(G)/K = Φ(G)/K and we deduce Φ(G) = K. A profinite group G is said to be “topologically finitely generated” or simply “finitely generated” if there are finitely many elements g1, . . . , gn ∈ G such that hg1, . . . , gni = G, equivalently hg1, . . . , gniN = G whenever N Eo G.

Exercise: G is procyclic if and only if it is topologically generated by one ˆ element. So for example Zp and Z are topologically generated by one element.

43 Theorem 14. Let G be a pro-p group. Then G is ﬁnitely generated if and only if Φ(G) is open in G.

Proof. If Φ(G) is open then G/Φ(G) is finite, let g1, . . . , gn be elements of G that map to generators of the finite group G/Φ(G), then hg1, . . . , gniΦ(G) = G hence hg1, . . . , gniΦ(G) = G and we deduce hg1, . . . , gni = G. Now suppose G is finitely generated. Let X be a finite subset of G with |X| = d and hXi = G. Since Φ(G) = Gp[G, G], if N is an open normal subgroup of G containing Φ(G) then G/N is a finite elementary abelian p- group (because Gp and [G, G] are contained in N) which can be generated by d elements (because being G/N finite and discrete, X maps to a set of abstract generators of G/N) so |G/N| ≤ pd. This implies that the open normal subgroups containing Φ(G) have bounded index. Let N0 be an open normal subgroup of G containing Φ(G) with |G : N0| as large as possible. Such N0 is then contained in all the open normal subgroups of G containing Φ(G), so N0 equals the intersection of the open normal subgroups of G containing Φ(G). But since Φ(G) is normal and closed, it is the intersection of the open normal subgroups containing it (exercise), hence Φ(G) = N0 is open.

12 A theorem of Serre about ﬁnitely generated pro-p groups

Recall that a profinite group G is said to be topologically finitely generated (or simply finitely generated) if there is a finite subset X of G with the property that hXi = G, in other words G has a dense subgroup that is finitely generated in the abstract sense. For example if G is a finitely generated abstract group then its profinite completion is a finitely generated profinite group. Recall that T Y = YN, therefore hXi = G if and only if hXiN = G for every N o G, NEoG E equivalently XN/N = {xN : x ∈ X} generates G/N abstractly for every N Eo G.

Observe that an infinite profinite group cannot be finitely generated in the abstract sense because a finitely generated abstract group is at most countable, and we have the following observation. Proposition 14. Let G be a profinite group. If G is infinite then G is not countable. Proof. By contradiction, let G be a countable not finite profinite group. Then there exists a bijection N → G, n 7→ gn. Moreover G admits a basis of clopens, because it is totally disconnected and compact. Any open normal subgroup of G has finite index, so it is infinite. There exists a non-empty clopen U1 contained in G − {g1}, infinite because it contains a coset of an open normal subgroup of

G. Let r1 = 1. Let r2 be the smallest index such that gr2 ∈ U1 and gr2 6= gr1 .

There exists a non-empty clopen U2 contained in U1 − {gr2 }. Let r3 be the smallest index such that gr3 ∈ U2 and gr3 6= gr2 . There exists a non-empty

44 clopen U3 contained in U2 − {gr3 }. Going on this way, we obtain a strictly descending chain of non-empty clopens {Ui}i∈N, with g1, g2, g3, . . . , gri 6∈ Ui. It follows that this chain has empty intersection. Since G is compact and the Ui’s are closed, there exists an empty element in the chain: Uk = ∅ for a suitable k ∈ N. Contradiction. The following is a very important result. It depends on the classification of finite simple groups. Theorem 15 (Nikolov, Segal 2003). Let G be a finitely generated profinite group. Then every subgroup of G of finite index is open. The pro-p version of this is due to Serre and we will prove it below. Observe that this theorem says that in a finitely generated profinite group the algebraic structure alone completely determines the topology. In particular it implies that if θ : G → H is an abstract group homomorphism, where G is a finitely generated profinite group and H is any profinite group, then θ is continuous. Indeed if K is an open normal subgroup of H then θ−1(K) is a normal subgroup of G and G/θ−1(K) embeds into H/K which is finite, hence θ−1(K) has finite index in G therefore it is open by the Nikolov-Segal theorem. Proposition 15. If G is a 1-generated profinite group then every subgroup of G of finite index is open. Proof. Let G be topologically generated by g ∈ G (meaning that hgi = G). The z ˆ map Z → G, z 7→ g can be uniquely extended to a continuous αg : Z → G ˆ (universal property of the completion). Now αg(Z) is closed and it contains g, ˆ z ˆ ˆ so αg(Z) = G, αg is surjective and G = hg | z ∈ Zi. So G is a quotient of Z. We have G =∼ Zˆ/N for some closed normal subgroup N of Zˆ. Let H =∼ K/N be a subgroup of G of index n. Then Zˆ/K is an abelian group of order n, so nZˆ ⊆ K. Since |Zˆ : nZˆ| = n, K = nZˆ is open, so K/N is open. Theorem 16. Let G be a profinite group, H ≤ G of finite index, say n (H not necessarily closed in G). Then n divides |G|.

Proof. The intersection of the conjugates of H in G, HG (the normal core of H in G), has ﬁnite index (at most n!), and if |G : HG| divides |G|, also |G : H| divides |G|. This implies that we can suppose H to be normal in G.

Let us show that if a prime p divides n then it divides |G|. Now, p divides |G/H|, so there exists gH ∈ G/H of order p. Let K := hgi. Since K/(H ∩K) =∼ KH/H, |K : H ∩ K| is ﬁnite. By the previous, since K is 1-generated, H ∩ K is open in K, so it is closed in K and hence in G. By Lagrange theorem,

|G : H ∩ K| = |G : K| · |K : H ∩ K|

Now g ∈ K − H, and g has order p in G/H so gp ∈ H. So g(H ∩ K) has order p in K/(H ∩ K), and therefore p divides |K : K ∩ H|, which divides |G| = |G : K| · |K : H ∩ K| · |H ∩ K : {1}|.

45 Now let pα be the largest p-power dividing n, and let pβ be the largest p- power dividing |G| (it is possible that β = ∞). We have to show that α ≤ β. β By definition of |G|, there exists U E G open such that p divides |G/U| but pβ+1 does not divide |G/U|, and (|U|, p) = 1. U and H have finite index in G, so H ∩ U has finite index, too. Since p does not divide |U|, p does not divide |U : H ∩ U| (by the first part of this proof: U is profinite and H ∩ U is a subgroup of finite index!), so |G : H ∩ U| is not divisible by pβ+1, but it is divisible by |G : H|, divisible by pα. This clearly implies that α ≤ β.

We prove the following theorem. Theorem 17 (Serre 1975). Let G be a finitely generated pro-p group. Then every subgroup of G of finite index is open. Recall that if G is a group and A, B are subgroups of G the commutator [A, B] is the subgroup of G generated by the commutators [a, b] = a−1b−1ab where a ∈ A and b ∈ B. The lower central series of G is the series of subgroups γi(G) defined by induction setting

γ1(G) = G, γi+1(G) = [γi(G),G] and the nilpotency class c of the nilpotent group G is the smallest positive integer such that γc+1(G) = {1}. For example G is abelian if and only if G has nilpotency class 1. Observe that γc(G) is central in G. Lemma 16. Let x, y, z be elements of a group G and let n be a positive integer. Recall that [x, y] = x−1y−1xy (the commutator of x and y). xy := y−1xy. 1. [xy, z] = [x, z]y[y, z]. 2. [x, yz] = [x, z][x, y]z.

n−1 n−2 3. [xn, y] = [x, y]x [x, y]x ··· [x, y]x[x, y].

n−1 4. [x, yn] = [x, y][x, y]y ··· [x, y]y . Proof. Omitted (it is just a computation). Lemma 17. Let G be a ﬁnitely generated pro-p group. Then the derived subgroup [G, G] is closed in G.

Proof. First we prove that if H = ha1, . . . , adi is a nilpotent group then every element of [H,H] is equal to a product of the form [x1, a1] ··· [xd, ad] with x1, . . . , xd ∈ H. We prove this by induction on the nilpotency class c of H. If c = 1 then H is abelian so [H,H] = {1} and there is nothing to prove. Assume c ≥ 2. If u ∈ γc−1(H) then since γc(H) is central in H we have

ei ei ei ei e e [u, a 1 ··· a k ] = [u, a 1 ] ··· [u, a k ] = [u, a ] i1 ··· [u, a ] ik i1 ik i1 ik i1 ik

ei1 eik = [u , ai1 ] ··· [u , aik ].

46 If u1, . . . , ud, v1, . . . , vd ∈ γc−1(H) then since γc(H) is central in H we have

[u1, a1] ··· [ud, ad][v1, a1] ··· [vd, ad] = [u1v1, a1] ··· [udvd, ad].

This implies that every element of γc(H) can be written in the form w = [w1, a1] ··· [wd, ad] with w1, . . . , wd ∈ γc−1(H). Now let g ∈ [H,H]. By induction we may write g = [y1, a1] ··· [yd, ad]w with y1, . . . , yd ∈ H and w ∈ γc(H). Writing w = [w1, a1] ··· [wd, ad] with w1, . . . , wd ∈ γc−1(H) we obtain

g = [y1, a1] ··· [yd, ad][w1, a1] ··· [wd, ad] = [w1y1, a1] ··· [wdyd, ad] using that each [wi, ai] is central. Choose xi = wiyi for i = 1, . . . , d.

Now let G be a pro-p group generated topologically by {a1, . . . , ad}. Let

X = {[g1, a1] ··· [gd, ad]: g1, . . . , gd ∈ G}.

Then X is closed in G being the image of a continuous map Gd → G. Let N be an open normal subgroup of G, then G/N is a finite p-group generated by a1N, . . . , adN. By the first part of this proof we have [G/N, G/N] = XN/N so T T [G, G]N = XN. It follows that [G, G] ⊆ N [G, G]N = N XN = X. On the other hand it is clear that X ⊆ [G, G] hence [G, G] = X is closed. Now we proceed to the proof of the theorem. Let G be a finitely generated pro-p group. Write G{p} = {gp : g ∈ G}. This is the image of the map g 7→ gp hence it is closed in G (because it is compact). Since G/[G, G] is abelian, Gp[G, G] = G{p}[G, G] is closed (because [G, G] is closed) so it equals Φ(G). Since G is finitely generated, it follows that Gp[G, G] = Φ(G) is open.

Let K be a proper normal subgroup of G of ﬁnite index. We prove that K is open by induction on the index |G : K| (which is ﬁnite). If |G : K| = 1 then K = G is open, so now assume |G : K| > 1. Let

M = Φ(G)K = Gp[G, G]K.

Since the index of a ﬁnite index subgroup always divides |G| (as we have shown above), |G : K| divides |G|, which is a power of p, so G/K is a ﬁnite p-group. Let H/K a subgroup of G/K of index p, then Gp ⊆ H because H is a normal subgroup of G of index p,[G, G] ⊆ H because G/H is abelian (having order p), therefore M ⊆ H.

[NB: observe that if we had [G, G] instead of [G, G] we could not deduce that [G, G] ⊆ H only from the fact that G/H is abelian!]

This proves that M 6= G, and since Φ(G) = Gp[G, G] is open in G, M is a proper open subgroup of G. Since |M : K| < |G : K| we have that K is open in M by induction. Hence K is open in G.

47 If K is an arbitrary subgroup of G of ﬁnite index then its normal core

\ g KG = K E G g∈G has also ﬁnite index: calling n = |G : K| we know that G/KG embeds into the symmetric group of degree n (because KG is the kernel of the action of G on the set of right cosets of K given by right multiplication) hence |G : KG| ≤ n!. So KG is open (by the discussion above) hence K is open too (being a union of cosets of KG).

13 Finitely generated proﬁnite groups

Definition 11 (Finitely generated profinite groups). The profinite group G is said to be a finitely generated profinite group if there exists X ⊆ G finite such that the subgroup generated by X is dense in G, i.e. hXi = G. Note: X ⊆ G generates a dense subgroup if and only if hXiN = G for every N E G open. This is because \ hXi = hXiN.

NEoG

Lemma 18. Let G be a proﬁnite group topologically generated by X := {x1, ..., xd}. Then for any r ∈ N there are only ﬁnitely many open normal subgroups N of index r in G. In particular G is countably based. Proof. Each such N is the kernel of a continuous surjective group homomorphism θ from G to a discrete group of order r. The number of the isomorphism classes of the images θ(G) is at most the number of the isomorphism classes of 2 a group of order r, which is at most rr (count the multiplication tables). Since G = NhXi, θ is completely determined by θ(X) = {θ(x1), ..., θ(xd)}, so there are at most rd choices for θ when the codomain is given. It follows that the number of possible θ, surjective homomorphism from G to a group of order r is 2 at most rr rd.

N Example: let F2 = Z/2Z, and G = F2 . G has only countably many open subgroups of index 2: let Ni be the preimage of 1 along the i-th component; if N is an open subgroup of G of index 2 then it contains a finite intersection of Ni’s so it is determined up to finitely many choices by a choice of a finite subset of N, and N has only countably many finite subsets. Observe that for all natural n number n there is a discrete quotient of G isomorphic to F2 , hence G cannot be d+1 finitely generated! (if d is any number, G is not d-generated because it has F2 d+1 as a finite discrete quotient, and F2 is not d-generated). Observe that seen as a vector space over the field with 2 elements G has uncountable dimension (! this is not trivial and you could try to prove it) hence it has uncountably many subgroups of index 2 (take the ones spanned by B − {b} where B is a base of the space and b ∈ B). So in G not every subgroup of finite index is open!

48 Proposition 16. A proﬁnite group G can be generated by d elements if and only if for any N E G open, G/N can be generated by d elements. Proof. (⇒). Say that G is topologically generated by X ⊆ G, with |X| = d. Then for every N E G open, hXiN = G, so G/N is generated by the images in G/N of the elements of X.

(⇐). Assume that for any N open and normal subgroup of G, G/N can be generated by d elements. For any N E G open deﬁne

d d SN := {(g1, ..., gd) ∈ G | hN, g1, ..., gdi = G} ⊆ G

Then for N E G open we have:

• SN 6= ∅. Take g1, ..., gd ∈ G, representatives of d generators of G/N. Then (g1, ..., gd) ∈ SN .

d d d • SN is closed in G . SN is a union of cosets of N in G . In fact if (g1, ..., gd) ∈ SN then for every n1, ..., nd ∈ N,(g1n1, ..., gdnd) ∈ SN . Now d d since N has finite index in G, N has finite index in G , so SN is in fact a finite union of cosets of N d in Gd, which are closed because N d is closed. Then SN is closed.

Now assume that {N1, ..., Nt} is a family of open normal subgroups of G. Let M be their intersection. Then M is again an open normal subgroup of G, and T SM ⊆ SN for every i = 1, ..., t. So SN ⊇ SM 6= ∅, and since G is compact, i T i i the intersection N SN where N varies in the set of all open normal subgroups of G is non empty (here we are using that the SN are closed!). Pick then (g1, ..., gd) in this intersection, and let X := {g1, ..., gd}. Then hXiN = G for every open normal subgroup N of G, i.e. hXi is dense in G. Since X has exactly d elements, we are done. We have these three possible properties for a proﬁnite group G: 1. G is ﬁnitely generated.

2. For every n ∈ N, G has only finitely many open subgroups of index n. 3. G has countably many open normal subgroups (i.e. G is a countably based profinite group, by the next lemma). We have the implications (1) ⇒ (2) ⇒ (3). Lemma 19. The following are equivalent for a profinite group G:

1. G is countably based. 2. G has at most countably many open normal subgroups.

49 Proof. The implication (2) ⇒ (1) is clear because every open normal subgroup of G has finitely many cosets (i.e. finite index!) so the basis of G consisting of the cosets of its open normal subgroups is countable. now we prove (1) ⇒ (2). Let {Bi | i ∈ N} be a countable basis for G. Since G is a profinite group, the cosets of its open normal subgroups form a basis, so for every i ∈ N there exists an open normal subgroup Hi and gi ∈ G such that giHi ⊆ Bi. For every i ∈ N define Fi = {N Eo G : Hi ≤ N}. S Since each Hi has finite index, each Fi is finite, so Fi is countable. It i∈N is then enough to show that for each N Eo G there is some i ∈ N such that Hi ≤ N. Let N be an open normal subgroup of G. Let i ∈ N be such that Bi ⊆ N. Then giHi ⊆ Bi ⊆ N hence gi = gi · 1 ∈ N therefore Hi ⊆ N.

Proposition 17. Let {Gi | i ∈ I} be a family of ﬁnite discrete groups, and let Q H be a closed subgroup of C := i∈I Gi. For every i ∈ I let πi : C → Gi be the projection. Then H = C if and only if for every {i1, ..., ir} ⊆ I the map

H → Gi1 × ... × Gir

h 7→ (πi1 (h), ..., πir (h)) is surjective. Suppose in addition that if i, j ∈ I and i 6= j then Gi and Gj have no isomorphic composition factors. Then H = C if and only if for every i ∈ I the restriction πi|H is surjective.

Proof. Take a = (ai)i∈I ∈ C − H. Since C − H is open, there exists a (basis) open neighbourhood of a in C, disjoint from H, say V := π−1(a )∩...∩π−1(a ), j1 j1 js js with a ∈ V ⊆ C − H. Then (aj1 , ..., ajs ) is not in the image of the canonical map H → Gj1 × ... × Gjs .

Now suppose that for every i 6= j the groups Gi and Gj have no isomorphic ˜ composition factors. Let i1, ..., ir ∈ I, and let H be the image of H in Gi1 ×

... × Gir . Suppose that the restrictions πi|H are all surjective. Then clearly the ˜ induced map H → Gij is surjective for every j = 1, ..., r. Since Gi1 , ..., Gir have ˜ ˜ no isomorphic composition factors, |H| ≥ |Gi1 |...|Gir |, so that H = Gi1 × ... ×

Gir . The result follows by the ﬁrst statement.

In particular if {Gi}i∈I is a family of ﬁnite groups that have pairwise no common composition factor and each Gi can be generated with d elements then Q the proﬁnite group i Gi can be generated with d elements.

Suggestion: try to prove that if A is any non-trivial finite group then the profinite group AN is not finitely generated.

Theorem 18. Let An be the alternating group of degree n and let Y H := An. n≥5

50 Then H is topologically 2-generated as a proﬁnite group, and every countably based proﬁnite group G can be embedded in H as a closed subgroup.

Proof. For n ≥ 5 let cn := (1 2 3), dn := (1 ... n) if n is odd, dn := (2 ... n) if n is even. Then An is generated by cn, dn for every n ≥ 5 and if c := (cn)n, d := (dn)n then c, d ∈ H and H is topologically generated by c, d by the previous proposition because An is simple for every n ≥ 5, so that every such An has only one composition factor, An itself.

Let {Ni | i ∈ N} be the set of the open normal subgroups of G (countable, because G is countably based). Consider the embedding Y G,→ G/Ni, g 7→ (gNi)i∈N i∈N

By Cayley’s theorem, every G/Ni embeds in An for every n ≥ |G/Ni| + 2, because Sn embeds in An+2 for every n ≥ 1: send every even permutation to itself and every odd permutation ρ to (n − 1 n)ρ. Let (ni)i∈N be a strictly increasing sequence of integers greater than or equal to 5 such that G/Ni embeds in Ani for every i ∈ N. Then since the ni’s are pairwise distinct Y Y Y G,→ G/Ni ,→ Ani ,→ An i∈N i∈N n≥5 as we want.

14 A dense subgroup of ﬁnite index

We want to give an example of a proper subgroup of a profinite group which is dense of finite index, in particular not open (taken from [2]). Q Let T be a finite non trivial group. Let G := λ∈Λ Tλ where Λ is an infinite ∼ set and Tλ = T for every λ ∈ Λ. Let F be the filter of cofinite subsets of Λ, i.e.

F := {A ⊆ Λ | Λ − A is finite}

(It is a filter: Λ is infinite, so ∅ 6∈ F ). Let U ⊆ P (Λ) be an ultrafilter containing F . For every g = (gλ)λ∈Λ ∈ G consider

Ω(g) := {λ ∈ Λ | gλ = 1}.

H := {g ∈ G | Ω(g) ∈ U }. We have:

• H is a normal subgroup of G.

Notice that Ω(1) = Λ, so 1 ∈ H. If g1, g2 ∈ H then Ω(g1) ∩ Ω(g2) ⊆ −1 Ω(g1g2), so Ω(g1g2) ∈ U , i.e. g1g2 ∈ H. If g ∈ H then Ω(g) = Ω(g ), so

51 g−1 ∈ H. This says that H ≤ G. If g ∈ H and x ∈ G then Ω(gx) = Ω(g), i.e. gx ∈ H, so H is also normal in G.

• H is dense in G.

For J = {λ1, . . . , λm} a ﬁnite subset of Λ, recall that it is enough to check

that the “composite” projection πJ : H → Tλ1 × · · · × Tλm is surjective. We have

πJ (H) = {(gλ1 , . . . , gλm ): g ∈ H} = {(gλ1 , . . . , gλm ) : Ω(g) ∈ U }

⊇ {(gλ1 , . . . , gλm ) : Ω(g) ∈ F } = Tλ1 × ... × Tλm

∼ m by definition of F , because Tλ1 ×...×Tλm = T is finite (being T finite).

• The index of H in G is |T |.

For any t ∈ T let t¯denote the element of G such that t¯λ = t for any λ ∈ Λ. Clearly if t 6= 1 then t¯ 6∈ H because Ω(t¯) = ∅ 6∈ U . Fix g ∈ G. For any t ∈ T consider Λt,g := {λ ∈ Λ | gλ = t} Clearly for ﬁxed g ∈ G we have [ Λ = Λt,g t∈T

and the union is disjoint. But Λ ∈ U , so we have that a finite union of subsets of Λ belong to the ultrafilter U . By lemma 5, there exists t ∈ T −1 −1 such that Λt,g ∈ U . We have Ω(gt¯ ) = Λt,g ∈ U , so gt¯ ∈ H. That is, g ∈ Ht¯. This implies that [ G = Ht¯ t∈T In particular H has finite index in G. To show that the index is indeed |T |, i.e. this last union is disjoint, it suffices to show that the elements −1 t¯ are not congruent mod H: if t1, t2 ∈ T are distinct then Ω(t¯1t¯2 ) = −1 −1 Ω(t1t2 ) = ∅ 6∈ U because t1t2 6= 1. • G/H =∼ T .

With the notation above the map T → G/H deﬁned by t 7→ Ht is a group isomorphism, being {t : t ∈ T } a transversal of H in G.

52 15 The automorphism group

Let G be a proﬁnite group. Consider Aut(G), the group of continuous automorphisms of G. Observe that an element of Aut(G) is precisely a group isomorphism that is also a homeomorphism (this is because in the proﬁnite context homeomorphism is the same thing as “bijective and continuous”). It is a topological group with the congruence subgroup topology: for any N Eo G set −1 AG(N) := {α ∈ Aut(G) | α(g)g ∈ N ∀g ∈ G}

It is easy to see that AG(N) is a subgroup of Aut(G). The congruence subgroup topology is deﬁned by setting {AG(N): N Eo G} to be a basis of open neighbourhoods of 1 (in other words their cosets form a base of the topology). This makes sense because AG(G) = Aut(G) and AG(N)∩AG(M) = AG(N ∩M) whenever N,M Eo G.

In general if G is a profinite group the automorphism group Aut(G) with the congruence subgroup topology is not a profinite group. For example consider N the profinite group C2 , described as follows.

∼ Y Y −1 Gi = hcii = C2 ∀i ≥ 1,G := Gi,U := Gi = π1 ({1}) Eo G. i≥1 i≥2

For ci ∈ Gi let ci be the N-ple which has ci in the place i and 1 anywhere else. For any j ≥ 2 deﬁne the automorphism

fj : G → G, c1 7→ cj, cj 7→ c1, ci 7→ ci ∀i 6= 1, j.

The fj are continuous automorphisms of G. If j1 6= j2 are larger than 1 then f −1(f (c )) · c −1 = c · c −1 6∈ U, therefore f −1f 6∈ A (U), and since the j2 j1 j2 j2 1 j2 j2 j1 G fj are infinitely many, AG(U) is an open subgroup of infinite index in Aut(G). So Aut(G) is not profinite.

In the following proof we use many times the following fact: if N is a characteristic subgroup of a group G (characteristic means that α(N) = N whenever α ∈ Aut(G)) then any automorphism α of G induces an automorphism of G/N in the following way: The composition of α with the canonical projection G → G/N has kernel α−1(N) = N, therefore it induces an isomorphismα ˜ : G/N → G/N. This induces a canonical homomorphism Aut(G) → Aut(G/N). Of course in the profinite context “characteristic” means preserved by the continuous automorphisms. Theorem 19. Aut(G) is profinite if the profinite group G satisfies (∗).

53 Proof. Assume then that G satisﬁes (∗) and consider \ Gn := {H ≤o G : |G : H| ≤ n} for every n ∈ N. Gn is open because the intersection is ﬁnite (here we use (∗)!), and it is normal because a conjugate of an open subgroup of index ≤ n is again an open subgroup of index ≤ n. Moreover \ \ Gn = H = {1}.

n H≤oG

Observe that G1 = G, and (Gn)n∈N is a descending chain of open (hence closed) normal subgroups whose intersection is {1}. In particular it is a coﬁnal family. Therefore we can realize G as the inverse limit of the Gn’s: G ∼ lim G/G . = ←− n n Pick α ∈ Aut(G). α is a homeomorphism, so it sends open subgroups to open subgroups. It also preserves the index of a subgroup, therefore α(Gn) ⊆ Gn, −1 and the same holds for α , hence α(Gn) = Gn for every n ≥ 1. In other words each Gn is a characteristic subgroup of G (where “characteristic” means preserved by continuous automorphisms). For n ∈ N deﬁne

αn : G/Gn → G/Gn

xGn 7→ α(x)Gn

Since α(Gn) = Gn, this αn is well deﬁned and it is an automorphism of G/Gn. Observe that in general we have a canonical map from the automorphism group of G to the automorphism group of a quotient of G by a characteristic subgroup. Therefore for n ≥ m we have a canonical map βmn : Aut(G/Gn) → Aut(G/Gm), and (Aut(G/Gn), βmn) is an inverse system. So we get the inverse limit Y L = lim Aut(G/G ) ⊆ Aut(G/G ). ←− n n n n Consider now Y β : Aut(G) → Aut(G/Gn) n∈N

α 7→ (αn)n∈N

Observe that β is continuous. To see this let πn be the canonical projection from the product to Aut(G/Gn) and compute

−1 −1 β (πn (1)) = {α ∈ Aut(G): αn = 1} =

= {α ∈ Aut(G): αn(x) ≡ x mod Gn} = AG(Gn) which is open in Aut(G) by deﬁnition of congruence subgroup topology. This −1 clearly implies that β is continuous because the open subgroups πn (1) form a

54 basis of open neighbourhoods of 1 (this also proves that the congruence subgroup topology is the smallest topology that makes β continuous).

Also, β is a group homomorphism, and since βnm(αm) = αn for m ≥ n we T have β(Aut(G)) ⊆ L. Also, β is injective because n Gn = {1} and

ker(β) = {γ ∈ Aut(G) | γ(x) ≡ x mod Gn ∀n} \ = {γ ∈ Aut(G) | γ(x) ≡ x mod Gn} = {1}. n It is then enough to prove the other inclusion, namely β(G) ⊇ L. Assume to have (γn)n∈N ∈ L (what this means is that whenever n ≥ m the map γm is precisely the one induced by γn modulo the characteristic subgroup Gm/Gn). Let γ : G = lim(G/G ) → G = lim(G/G ) ←− n ←− n n n

(xnGn)n∈N 7→ (γn(xn)Gn)n∈N

We need to check that γ is well deﬁned. If n ≥ m then ϕmn(γn(xn)Gn) = γm(xm)Gm where ϕmn denotes the reduction modulo Gm. But this is true because (γn)n ∈ L and xmGm = ϕmn(xnGn). In other words the following diagram commutes (by deﬁnition of the βmn’s).

γn G/Gn / G/Gn

γm G/Gm / G/Gm T Now γ is injective because n Gn = {1} and it is surjective by the density criterion. We proved that γ is an automorphism of G such that β(γ) = (γn)n∈N. This proves that β(Aut(G)) = L and concludes the proof. If you remember the Nikolov-Segal theorem (cf. a previous class), this implies in particular that if G is a ﬁnitely generated proﬁnite group then every automorphism is a topological automorphism, so Aut(G) is really the abstract automorphism group of G.

Example: Let G = Zp. We compute Aut(G).

We can apply the discussion above because Zp is topologically 1-generated. Observe that since the open subgroups are exactly the subgroups of the form n n p G, in the notation of the above proof Gpn = p G. Aut(G) ∼ lim(Aut(G/pnG)) = lim(Aut( /pn )) = ←− ←− Z Z n n Now Aut(Z/pnZ) is isomorphic to U(Z/pnZ). Assume p is odd. Then n n n−1 n−1 Aut(Z/p Z) =∼ Z/ϕ(p )Z = Z/(p − 1)p Z =∼ Z/(p − 1)Z × Z/p Z.

55 The isomorphisms are canonical, therefore there are induced inverse systems on the inverse systems {Z/(p − 1)Z} (a family of groups independent of n) and on {Z/pn−1Z : n ∈ N}. Therefore Aut(G) ∼ lim(Aut(G/pnG)) = lim(Aut( /pn )) = ←− ←− Z Z n n = lim( /(p − 1) × /pn−1 ) ←− Z Z Z Z n ∼ lim( /(p − 1) ) × lim( /pn−1 ) = C × = ←− Z Z ←− Z Z p−1 Zp n n If p = 2 then

Aut(G) ∼ lim(Aut(G/2nG)) = lim(Aut( /2n )) = ←− ←− Z Z n n = lim( /2 × /2n−1 ) ∼ C × ←− Z Z Z Z = 2 Z2 n

Observe that in both cases there is a nontrivial torsion part of Aut(Zp)(Cp−1 if p is odd, C2 if p = 2). Also, Aut(Zp) is a pro-p group if and only if p = 2.

Observe that U( ) is also canonically isomorphic to lim U( /pn ). This Zp ←−n Z Z is because a p-adic integer x = (x0, x1,...) is invertible if and only if each xn is invertible, and the U(Z/pnZ) naturally form an inverse system. Therefore Aut( ) ∼ lim Aut( /pn ) = lim U( /pn ) ∼ U( ). Zp = ←− Z Z ←− Z Z = Zp n n

Another way of saying this is that a continuous automorphism of Zp is determined by the image of 1 by continuity, and the image of 1 must be a topological generator of Zp hence it must be a unit u of the ring. Then the image of any x ∈ Zp is just xu. Indeed setting u = (un)n and given f ∈ Aut(Zp), x = (xn)n ∈ Zp, we have f(x) = (fn(x))n = (unx)n = ux.

16 Hopﬁan groups

For a profinite group G let (∗) be the following property: “For every non-zero n ∈ N there are only finitely many open subgroups of index n”. For example, we have seen that if G is finitely generated then it satisfies (∗). Definition 12 (Hopfian groups). A group G is called Hopfian if every surjective ∼ homomorphism α : G → G is injective, i.e. for every N E G, if G/N = G then N = {1}.

N N ∼ N Clearly every finite group is Hopfian. C2 is not Hopfian: C2 /C2 = C2 .

Example: There exist finitely generated abstract groups which are not Hopfian. Take the Baumslag-Solitar group ha, b | (b2)a = b3i. It is not Hopfian.

56 Definition 13 (Hopfian profinite groups). A profinite group G is called Hopfian ∼ if for every closed N E G, if G/N = G (as topological groups!) then N = {1}. Proposition 18. If the profinite group G satisfies (∗) then it is Hopfian. ∼ Proof. Assume that G/N = G for some closed N E G. Let Xn be the set of the open subgroups of G with index at most n, for every n ≥ 1. Consider Yn, the set of the open subgroups of G of index at most n, containing N. Clearly Yn ⊆ Xn. Since G satisfies (∗) the sets Xn and Yn are finite. There is a canonical bijection between the family of open subgroups of G containing N and the family of open subgroups of G/N, and such bijection preserves the index of the subgroups (exercise). Since G =∼ G/N, it follows that the set of the open subgroups of G/N with index at most n is equipotent to the set of the open subgroups of G which contain N and which have index at most ∼ n. This last set is precisely Yn, so since G = G/N, |Xn| = |Yn| hence Xn = Yn (because Yn ⊆ Xn and they are finite sets). This means that N is contained in every open subgroup of G, hence N = {1}.

Lemma 20. Assume that {Xi, ϕij}i,j∈I is an inverse system of non-empty topological spaces, with I a directed set, and that Xi is Hausdorff and compact for any i ∈ I. Then the inverse limit X is non-empty. Q Proof. Everytime j ≥ i consider Dij := {x ∈ Xl : ϕij(xj) = xi}. It Q l is a closed subset of Xi and X is the intersection of the Dji’s. Assume i Q by contradiction that X = ∅. Then since i Xi is compact, there exists a Tr finite set {(i1, j1), ..., (ir, jr)} such that ∅ = l=1 Djlil . Since I is directed, there exists k ∈ I such that jl ≤ k and il ≤ k for any l ∈ {1, ..., r}. Take an arbitrary xk ∈ Xk, and define x as follows: xj := ϕjk(xk) if j ≤ k, and arbitrary otherwise. We have that x ∈ Djl,il for any l ∈ {1, ..., r}, indeed

ϕiljl (xjl ) = ϕiljl (ϕjlk(xk)) = ϕilk(xk) = xil . Contradiction. Theorem 20. Let G, H be two profinite groups. Suppose that G satisfies (∗) ∼ and for any N E H open, there exists M E G open such that H/N = G/M. Then H is isomorphic to a quotient of G: there exists K E G closed such that G/K =∼ H (as topological groups!). Proof. For every integer r ≥ 1 define \ \ Ur := {M Eo G : |G : M| ≤ r},Vr := {N Eo H : |H : N| ≤ r}.

Since G satisﬁes (∗), Ur is an open normal subgroup of G, because it is a ﬁnite intersection of open normal subgroups.

We prove that Vr is open in H. Let F be the family of the open normal subgroups of H that can be obtained as a possibly inﬁnite intersection of open normal subgroups of H of index at most r. Let W ∈ F and write W = T i Hi where each Hi is an open normal subgroup of H of index at most r. By hypothesis, there exist K E G open and an isomorphism γ : H/W → G/K. We

57 have Hi ⊇ W for every i, and the γ(Hi) are open, normal and of index ≤ r T in G (here γ(Hi/W ) = γ(Hi)/K by abuse of notation); moreover γ(Hi) = T i γ( i Hi) = γ(W ) = K, so K is an intersection of open normal subgroups of G of index ≤ r. Clearly this implies that Ur ≤ K. Therefore H/W is an epimorphic image of G/Ur (because such is G/K). But Ur is open, so it has finite index: G/Ur is finite; then H/W is finite of order less than or equal to |G/Ur|. So that we can choose W ∈ F such that the order |H/W | is maximum. We claim that W = Vr, and therefore Vr is open. Indeed if it was W 6= Vr there would exist some Hi open normal in H of index at most r with Hi 6⊆ W hence W ∩ Hi would be an open normal subgroup of H of finite index larger than |H : W |, contradicting the maximality of |H : W |.

So Vr is open and H/Vr is an epimorphic image of G/Ur. Let

∅= 6 Λl := {epimorphisms G/Ul → H/Vl}, l ≥ 1.

It is a ﬁnite set, give it the discrete topology. If s ≥ r then Us ≤ Ur so using the fact that the composition G → G/Us → H/Vs → H/Vr (where the second arrow is an element of Λs) factors through Ur (because the preimage of a subgroup of index at most r is a subgroup of index at most r) we have the canonical map

ϕrs :Λs → Λr.

Now, (Λ , ϕ ) is an inverse system, so we can talk about Λ := lim Λ , the r rs ←−r r inverse limit. Call ϕr :Λ → Λr the canonical maps. Since the Λr are Hausdorff and compact, Λ 6= ∅ (by the lemma). Take then α ∈ Λ, and define αr := ϕr(α). Let πr denote the projection G → G/Ur or H → H/Vr. {Ur}r and {Vr}r are filtering bases of open normal subgroups of G and H respectively, with trivial intersection, so G and H can be realized as inverse limits of the G/Ur and H/Vr respectively, with the naturally induced maps as set of compatible maps. By the universal property of the inverse limit H, we obtain a unique compatible map β such that the following diagram commutes whenever s ≥ r.

πs G / G/Us / G/Ur

∃!β αs αr

πs H / H/Vs / H/Vr

We are left to check that it is surjective. Let then J := β(G). By the commutativity the diagram, since πr and αr are surjective, πr ◦ β is surjective for every r ≥ 1. So JVr/Vr = πr(J) = πr(β(G)) = H/Vr.

That is, JVr = H. Now J is closed because G, H are compact and β is continuous, and {Vr}r is a family of closed subgroups of H closed by intersection, T T so by our usual lemma H = (JVr) = J( Vr) = J. r∈N r∈N

58 Corollary 3. Let G, H be profinite groups such that G satisfies (∗) and G, H have exactly the same finite discrete images, that is, the following conditions are satisfied: ∼ • For every M E G open there exists N E H open such that H/N = G/M. ∼ • For every N E H open there exists M E G open such that G/M = H/N. Then G =∼ H. Proof. By the previous statement there exists α : G → H surjective. H satisfies (∗) because it is an epimorphic image of G, hence again from the previous statement, there exists β : H → G surjective. Consider β ◦ α : G → G. Clearly it is surjective. Since G is profinite and satisfies (∗), it is Hopfian, so β ◦α is also injective; in particular α is injective. So that α : G → H is an isomorphism. Now we give an example to show that if we drop property (∗) the theorem is no longer true. It is a general fact that if G is a residually finite group then iden- fifying it with its image in its profinite completion Gb, there is a index preserving bijection between the finite index subgroups of G and the open subgroups of Gb given by H 7→ H (whose inverse is K 7→ K ∩ G). Knowing this it is clear that N for G = C2 , since G has uncountably many subgroups of index 2, the profinite completion Gb has uncountably many open subgroups of index 2, however G only has countably many open subgroups of index 2, hence the profinite groups G and Gb are not isomorphic (as profinite groups), on the other hand they have the same finite discrete images. Using the Nikolov-Segal terminology, G is not strongly complete.

17 Exercises - list 2

1. Let G be a profinite group all of whose abstract subgroups are closed. Prove that G is finite. Suppose G is not finite. Let X be an infinite countable subset of G. Then H = hXi is a countable subgroup of G. To see this we can first assume that X contains the inverses of its elements (this does not affect countability) n and then observe that defining fn : X → H by fn(x1, . . . , xn) := x1 ··· xn the group H is a countable union of countable sets so it is countable: S n H = n fn(X ). However H is closed by assumption, so it is profinite, a contradiction (an infinite profinite group cannot be countable). 2. Separation axioms for a topological space. T0 = at most one dense point. T1 = all points are closed. T2 = Hausdorff. Prove that T2 implies T1 implies T0 but T0 does not imply T1 and T1 does not imply T2. If X is T2 then its points are closed because their complements are unions of open (separating it from the points in the complement), so X is T1, therefore X cannot have dense points unless |X| = 1 in which case there

59 is precisely one dense point, so X is T0. Any inﬁnite set with the coﬁnal topology is T1 but not T2 (as we have seen) and the space {1, 2} with the topology {∅, {1}, {1, 2}} is T0 but not T1, indeed {1} is not closed however it is the only dense point.

3. Find the open and the closed subgroups of R (usual topology). Let H be an open subgroup of G = R. The complement G − H is a union of cosets of H so it is open. Then G = H ∪ (G − H), however G is connected hence G − H = ∅, that is G = H. In other words the only open subgroup of G is G. Now let H be a closed subgroup of G, and assume that H 6= G. Let g be the inﬁmum of the positive real numbers belonging to H.

We claim that g ∈ H, indeed if not then let xn be a strictly decreasing sequence of points in H with the property that xn − g < 1/n for all n, then H 3 yn = xn − xn+1 < 1/n and yn > 0 so setting yn = xn − xn+1 we obtain subgroups ynZ of H whose union is clearly dense in G (because yn is arbitrarily small), however such union is contained in H, therefore H is dense, a contradiction because H is closed and H 6= G. We deduce that g ∈ H. A similar argument shows that g 6= 0, so g > 0. Now we claim that H = gZ. The inclusion ⊇ is obvious, now let h ∈ H. If (by contradiction) h 6∈ gZ we may choose z ∈ Z with h − gz > 0 minimum possible. Since H is a subgroup, h − gz ∈ H. We claim that h − gz < g, contradicting the minimality of g. Indeed it cannot be h−gz = g (because h 6∈ gZ) and if it was h − gz > g write h − gz = g(1 + a) with a > 0, then h−g(z+1) = ga > 0 (being both g and a positive) and h−g(z+1) < h−gz contradicting the minimality of h − gz. Therefore the only open subgroup of R is R and the closed subgroups of R are R and the subgroups of the form gZ where g ∈ R (the cyclic subgroups).

4. R is not profinite. This follows from the above because the only open subgroup of R is R, and in a profinite group the intersection of the open subgroups is trivial. However a more direct proof is the following: R is not even residually finite because the only subgroup of R of finite index is R, indeed if H is a subgroup of R of finite index n then nR ⊆ H (because the quotient group R/H has order n) on the other hand R = nR ⊆ H hence H = R. The same holds for every divisible abelian group (for example Q). 5. Example of a residually finite group that is not profinite. The additive group Z is residually finite but it is not profinite, because it is discrete and infinite (so it is not compact). 6. Completion does not preserve injectivity (it is not left exact). For example the injective homomorphism (inclusion) Z → Q induces Zb → Qb = {0} which is (clearly) not injective. Qb = {0} because the only subgroup of Q of finite index is Q (see the proof above for R).

60 7. Proabelian implies abelian, prosolvable does not imply solvable, pronilpotent does not imply nilpotent. A pro-C group is an inverse limit of finite C -groups, in particular it is a subgroup of a product of C -groups. This explains why proabelian implies abelian. Now for all integer i ≥ 1 let Gi be a finite solvable (resp. nilpo- Q∞ tent) group of derived length (resp. nilpotency class) i, then G = i=1 Gi is prosolvable (resp. pronilpotent) but it is not solvable (resp. nilpotent). This is because the derived series (resp. the central series) cannot termi- nate at any finite n-th step because (the derived series can be computed componentwise and) the n-th term of Gn+1 will not be trivial. 8. Prove that if G and H have the pro-C topology (where C is a class closed under taking subgroups and finite direct products) then the product topology on G × H coincides with the pro-C topology.

• If U is a pro-C open normal subgroup of G×H then it is open in the product topology. This is because U contains L = (U ∩ G) × (U ∩ H) so it is a union of cosets of L, and to prove that L is open (in the product topology) it is enough to prove that U ∩ G is open in G and U ∩H is open in H. This is because G/U ∩G =∼ GU/U is a subgroup of (G × H)/U ∈ C (and C is closed under taking subgroups), and the same for U ∩ H. • If A is open normal in G and B is open normal in H then A×B is pro- C open in G×H. This is because (G×H)/(A×B) =∼ G/A×H/B and G/A, H/B ∈ C and C is closed under taking ﬁnite direct products.

9. Prove that G\× H =∼ Gb × Hb. We will do this for the pro-C completion where C is a class closed under taking subgroups and doing finite direct products (for example C can be the class of finite groups, or finite p-groups, or finite nilpotent groups, or finite solvable groups, etc.). First we can construct the canonical projections G\× H → Gb, G\× H → Hb by considering the canonical maps G\× H → G × H/N × M → G/N and G\× H → G × H/N × M → H/M (where G/N, H/M ∈ C - observe that here we use that C is closed under taking finite direct products) and deducing the canonical projections from the universal property of the inverse limit. To prove the result we use the universal property of the product. Let X → Gb, X → Hb be two maps (continuous homomorphisms), we want to show that there is a unique compatible map X → G\× H. For this it is enough to produce canonical maps X → G/N, X → H/M (compatible) where G/N ∈ C and H/M ∈ C , because by cofinality (C is closed under taking subgroups) G\× H is the inverse limit of the G × H/N × M. But such compatible maps are simply obtained by composing the canonical maps Gb × Hb → G/N, Gb × Hb → H/M with X → Gb × Hb.

61 10. Wilson Chapter 2 Exercise 1. Show that the order of a profinite group G is the product of the orders of its Sylow subgroups. Q n(p) Let G be a profinite group and write |G| = |G : {1}| = p p where each n(p) is a natural number (possibly zero) or infinity. Let P be a Sylow p-subgroup of G. We want to show that |P | = pn(p). By definition P is a pro-p closed subgroup of G with p not dividing |G : P |. Write |P | = pm(p) Q n(q) and |G : P | = k, h = q6=p q , we have

h · pn(p) = |G| = |G : {1}| = |G : P | · |P : {1}| = k|P | = k · pm(p)

hence n(p) = m(p) being h and k coprime to p.

11. Wilson Chapter 2 Exercise 6. Let G be a proﬁnite group.

(a) Show that if K1,K2 are closed pronilpotent normal subgroups of G then K1K2 is closed, normal and pronilpotent. Clearly K1K2 is a closed normal subgroup of G hence we may assume that G = K1K2. We show that G is pronilpotent. If U is an open normal subgroup of G and π : G → G/U is the canonical projection then G/U = π(G) = π(K1K2) = π(K1)π(K2) hence G/U is nilpotent because π(K1)π(K2) are normal nilpotent subgroups of the ﬁnite group G/U whose product is G/U (Fitting theorem). (b) Show that G has a normal subgroup which is pronilpotent and which contains all pronilpotent normal closed subgroups of G. Let N be the family of closed normal pronilpotent subgroups of G and let F = hN i, clearly F is a normal closed subgroup of G, we must show that F is pronilpotent. If U is an open normal subgroup of F then F/U is ﬁnite generated by {NU/U : N ∈ N } hence there are N1,...,Nt ∈ N with F/U = N1U/U ··· NtU/U, so F/U is nilpotent.

18 Exercises - list 3

1. Torsion of U(Zp) and Aut (details for product of inverse systems).

Let now G := U(Zp) be the group of units of the local domain Zp, where p is an odd prime. It is a group. We want to ﬁnd the elements of G of ﬁnite order, that is, t(G), the torsion of G. Clearly at least −1, 1 ∈ t(G). Recall that we already know that G = Cp−1 × Zp, we want now to construct elements in the torsion.

62 • First step. Take a ∈ Z such that (a, p) = 1, and consider

p 2 pn n+1 xa := (a + pZ, a + p Z, ..., a + p Z, ...)

n n−1 We have ap ≡ ap mod(pn) for any n ∈ N, because a ∈ U(Z/pnZ) and |U(Z/pnZ)| = ϕ(pn) = pn − pn−1, so

n pn−pn−1 n (a + Z/p Z) = 1 + Z/p Z Therefore,

p pn n pn−1 n xa = (a + p Z)n∈N = (a + p Z)n∈N = xa

p−1 Since xa is invertible, being (a, p) = 1, this means xa = 1. The choices for a such that (a, p) = 1 modulo p are exactly p − 1, so these a are exactly the roots of xp−1 − 1 (pass to the ﬁeld of fractions). • Second step. Assume that m > 0 is an integer, coprime with p, m p−1 such that x = 1 for some x ∈ Zp. We claim that x = 1. Say

n x = (xn + p Z)n∈N

m n Then xn ≡ 1 mod(p ) for any n ∈ N, and

pn−pn−1 (m,pn−pn−1) n xn ≡ 1 ≡ xn mod(p )

for any n ∈ N. But (m, p) = 1, so (m, pn − pn−1) = (m, p − 1). So p−1 n xn ≡ 1 mod(p ) for any n ∈ N, because (m, p − 1) divides p − 1. p • Third step. We claim that if x = 1 then x = 1 for any x ∈ Zp. By P i p contradiction assume that x = i aip 6= 1 and x = 1. It follows that a0 = 1, and there exists i > 1 such that ai 6= 0. Let n be the smallest positive integer bigger than 1 with this property, so that

n n+1 n+2 n n+2 x = 1 + anp + an+1p + p z = 1 + αp + p z

for a suitable z ∈ Zp (here α = an + pan+1). So for some suitable ∗ n p Pp p n j z , z,˜ z¯ ∈ Zp we get, since (1 + αp ) = j=0 j (αp ) ,

p n p n+2 ∗ n+1 n+2 n+1 n+2 1 = x = (1+αp ) +p z = 1+αp +p z˜ = 1+anp +p z¯

p p 2 2n because p is odd, so 2 is divisible by p (so p shows up in 2 α p with multiplicity > n + 1). This is a contradiction because an is between 0 and p − 1, and it is diﬀerent from 0.

∗ It follows that t(G) is cyclic of order p−1, generated by xa where Zp = (a). We just remark that if p = 2, t(G) = {1, −1}.

63 More in general we will prove that lim X × Y = lim X × lim Y . Let ←−i,j i j ←−i i ←−j j X = lim X and Y = lim Y . Let L = lim X × Y . Here the ←−i∈I i ←−j∈J j ←−i,j i j family of the Xi × Yj has the structure of inverse system where the order is given componentwise and this gives canonical morphisms Xr × Ys → Xi × Yj whenever (r, s) ≥ (i, j). To prove that L = X × Y we need to ﬁx Z → Xi × Yj compatible and prove that there exists a unique compatible map Z → X × Y . This last map is given by the universal property of the inverse limits X and Y applying it to the family of compatible maps obtained composing Z → Xi × Yj with the projections Xi × Yj → Xi and Xi × Yj → Yj.

From this we can deduce that G\× H = Gb × Hb more directly (we are talking about the pro-C completion for some class C ).

G×H = lim G/N×lim G/M = lim G×H/N×M = lim G×H/U = G\× H. b b ←− ←− ←− ←− N M N,M U

For the second equality to make sense it is necessary that the class is closed under finite direct products and the fourth equality is true because the normal subgroups N × M form a cofinal family, and for this we need that the class is closed under taking subgroups (indeed if G × H/U ∈ C then setting N = (G × {1}) ∩ U and M = ({1} × H) ∩ U we have G/N =∼ GU/U ≤ G × H/U and G/M =∼ HU/U ≤ G × H/U so G/N × H/M ∈ C being C closed under taking subgroups and finite direct products. 2. Let G be a compact topological group and let H be an open subgroup of G. Show that if H is profinite then G is profinite. G is compact by assumption, it is Hausdorff because 1 ∈ H and H is S Hausdorff hence H − {1} is open, so G − {1} = H − {1} ∪ g6∈H Hg is open. We are left to show that G is totally disconnected. Let A be a connected nonempty subset of G. Then writing G as union of the cosets of H we see that A is contained in a coset Hg of H. Therefore Ag−1 is a connected subset of H, so |Ag−1| = 1, say Ag−1 = {x}, then A = {xg}. 3. Wilson Chapter 2 Exercise 2. Q n(p) For n = p p a supernatural number and K a procyclic (i.e. 1-generated) group of order n show that K =∼ Zb/H where

Y n(p) H = p Zp. n(p)6=∞

Write K = hxi. As we have seen the canonical group homomorphism Z → hxi → K (where the second arrow is simply inclusion) is continuous when Z is given the profinite topology because the pre-image of a finite index subgroup of K is a finite index subgroup of Z (all the non-trivial subgroups

64 of Z have ﬁnite index) therefore there is a unique compatible map θ : Zb → K. It is surjective because its image is dense (it contains hxi and it is closed being the image of Zb, which is compact). Let H = ker(θ). By the isomorphism theorem we have a homeomorphic isomorphism Zb/H =∼ K. We are left to show that H is as in the statement. Recall that Zb = Q −1 p Zp. Also, H is a closed subgroup of Zb being the preimage θ (1). Since Zb is pronilpotent, H is also pronilpotent hence we may write H = Q m(p) p∈X p Zp where X is the set of prime divisors of |H| and each m(p) is a non-negative integer (ﬁnite). Clearly

∼ Y Y m(p) K = Zb/H = Zp × Zp/p Zp p6∈X p∈X

Q ∞ Q m(p) The order of K is then p6∈X p · p∈X p . Hence if p ∈ X then m(p) = n(p).

19 Inﬁnite Galois Theory

We want to give a motivation as to why we study profinite groups. We will show that a profinite group is precisely the Galois group of an algebraic field extension.

Let F/K be an algebraic extension. A K-automorphism of F is a ring isomorphism f : F → F with the property that f(x) = x whenever x ∈ K. Define G := G(F/K) to be the group of the K-automorphisms of F . Definition 14. Say that F/K is a Galois extension if it satisfies the following conditions: • F/K is normal: any irreducible polynomial of K[X] with a zero in F factorizes completely in F [X]. In other words F is generated over K by the roots of a family of polynomials of K[X]. • F/K is separable: any irreducible polynomial of K[X] with a root in F is separable, meaning that it has no multiple roots.

Observe that if g ∈ G and K ≤ L ≤ F with L/K normal then g|L has values in L. This is because L is generated over K by the roots of a family of polynomials in K[X] and if P is such a polynomial and α is a root of P then P (g(α)) = g(P (α)) = g(0) = 0 so g(α) is also a root of P . This means that the restriction induces a canonical homomorphism G → G(L/K).

Let now

F := {L : K ≤ L ≤ F ; L/K is finite and Galois}

ϕN : G(F/K) → G(N/K), α 7→ α|N ∀N ∈ F.

65 This is well defined because N is normal. We have ker(ϕN ) = G(F/N). It is a normal subgroup of G of finite index. The following family of normal subgroups of finite index is a filtering family:

{G(F/N) | N ∈ F}

To see this just observe that G(F/N1) ∩ G(F/N2) = G(F/(N1 ∪ N2)). Here (N1 ∪N2) is the ﬁeld generated by N1 and N2. This is enough to put a topology on G, called the Krull topology: the cosets of the subgroups in this ﬁltering family are basis open subsets. In this way G becomes a topological group. Indeed setting ν : G × G → G, ν(x, y) = xy−1 and N = G(F/N) where N ∈ F we have [ ν−1(gN ) = xN × g−1xN x∈G for every g ∈ G.

Theorem 21. {G(L/K)L∈F , ϕL2L1 } is an inverse system with the restriction maps ϕL2L1 : G(L2/K) → G(L1/K) for L1 ≤ L2. We have a canonical topological isomorphism G(F/K) ∼ lim G(L/K). = ←− L∈F In particular, G is a proﬁnite group and the Krull topology on G coincides with the inverse limit topology. Proof. For any L ∈ F the restriction

αL : G(F/K) → G(L/K)

−1 is continuous: αL ({1}) = G(F/L). We get then the product map Y α : G(F/K) → G(L/K) L∈F

We have to check that α(G(F/K)) ≤ Y = lim G(L/K). This is clear because ←−L∈F if σ ∈ G(F/K) and L2 ≥ L1 are in F then

ϕL2L1 (α(σ)L2 ) = ϕL2L1 (σ|L2 ) = (σ|L2 )|L1 = σ|L1 = α(σ)L1

By construction, α is continuous. Moreover α is open: if N ∈ F then     Y Y α(G(F/N)) = Y ∩  G(M/K) ×  {1} F3M6⊆N F3M⊆N

The inclusion ⊆ is clear, to prove the converse observe that if (gM )M belongs to the second set then we may deﬁne g ∈ G(F/N) by setting g(m) := gM (m) whenever m ∈ M ∈ F. This deﬁnition makes sense (in other words it does not depend on the choice of M) because of the compatibility (gM )M ∈ Y (recall that

66 M contains the splitting field of the minimal polynomial of m). The fact that α is open now follows from the fact that there are only finitely many M ∈ F contained in N (because N/K is a finite extension).

We are left to show that α is bijective. We will construct the inverse

β : Y → G(F/K)

We want to send (gL)L∈F ∈ Y to a K-automorphism of F .

Take f ∈ F . Then K(f) is a finite extension of K, hence it admits the normal closure Lf ∈ F (the field generated by K and all the roots of the minimal polynomial of f over K). Then given (gL)L∈F ∈ Y , it makes sense to consider the element gLf (f) ∈ F . We can construct the map γ : F → F which sends f to gLf (f). It is well defined because if L1 and L2 are two normal closures of K(f) in F then

gL1 (f) = gL1 |L1∩L2 (f) = gL1∩L2 (f) = gL2 (f)|L1∩L2 = gL2 (f)

(recall that we are working in the inverse limit). Then we can send (gL)L∈F to γ and check that it is what we want. Before the examples we mention that the direct limit of a direct system is defined simply taking the definition (in particular, the universal property) of the inverse limit of an inverse system and inverting all the arrows. Denoting by F the algebraic closure of the field F given a direct system of subfields (with the order given by inclusion) the direct limit is just the field generated by them.

Example: Take a prime p, and let K := Fp. Using the above notations, F = {Fpn | n ∈ N} and it turns out that the direct limit of the members of F is F = Fp (the algebraic closure of Fp). So G(F/K) is the inverse limit on F of ∼ the G(Fpn /Fp) = Z/nZ. It follows that

G(F/K) =∼ Zˆ

Example: Consider the direct limit of the Fpqr , where q is a fixed prime. In other words F is the field generated by the qr-th roots of unity (for all r). ∼ r In this case G(F/K) is the inverse limit of the G(Fpqr /Fp) = Z/q Z. It follows that ∼ G(F/K) = Zq S Lemma 21. Let F/K be a Galois extension, and assume F = i∈I Fi with Fi an intermediate field such that |Fi : K| is finite for any i. Then {G(F/Fi) | i ∈ I} is a basis of open neighbourhoods of 1 for G = G(F/K).

Proof. G(F/Fi) is open: if L is the normal closure of Fi over K then G(F/L) ⊆ G(F/Fi), hence G(F/Fi) is a union of cosets of G(F/L), which is open, so

67 G(F/Fi) is open. Conversely, for any M ∈ F there exists Fi such that G(F/Fi) ⊆ G(F/M), in fact we have M = K(α) for some α (by the primitive element theorem), so α ∈ Fi for a suitable Fi. It follows that the given family is a basis for the topology of G(F/K).

Proposition 19. Let F/K be a Galois extension, and let K ≤ M1,M2 ≤ F . Let γ : M1 → M2 be a K-isomorphism. Then there exists γ ∈ G(F/K) such that γ|M1 = γ. Proof. We know that the statement is true in the finite case. Let Y A := G(L/K) L∈F with the discrete topology on the factors. A is clearly compact (Tychonoff). For any N ∈ F define

BN := {g = (gL)L∈F : gN |N∩M1 = γ|N∩M1 , gL = gN |L ∀K ≤ L ≤ N} ⊆ A. T By construction, if g ∈ N∈F BN then g belongs to the inverse limit of the G(F/L) on F, that is exactly G(F/K). Again by construction its restriction to M1 is γ, so that we can choose γ := g. We are then left to check that the intersection of the BN ’s is non empty. Let {N1, ..., Nt} ⊆ F, and let M be the normal extension generated by N1, ..., Nt. Clearly BM ⊆ BNi for any Tt i = 1, ..., t, so BM ⊆ i=1 BNi . If we show that for any N ∈ F, BN is closed and non empty, we can conclude that any ﬁnite intersection of the BN ’s is non empty (recall that M ∈ F), so since we are in the compact group A, the whole intersection is non empty.

So let us show that the BN are non empty and closed. Let N ∈ F. We have γ|N∩M1 : N ∩ M1 → N ∩ M2, because N is normal. By the finite case we know that γ|N∩M1 can be extended to γN ∈ G(N/K). Let h be the element of A which has γN in the position N, the restriction γN |L in the finitely many positions L such that L ≤ N and random otherwise (observe that BN is defined by finitely many conditions). Then h ∈ BN so BN 6= ∅.

BN is an intersection of preimages of projections, so it is closed. More precisely, B is the intersection of the π−1(γ|j )), where the γ|j are N Li Li∩M Li∩M the ﬁnite number of extensions of γ|Li∩M to N, and where the Li are the intermediate ﬁelds of N/K.

20 The fundamental theorem of Galois Theory

Theorem 22 (Fundamental theorem). Let F/K be a Galois extension; let A be the set of the ﬁelds M such that K ≤ M ≤ F , and let B be the set of the closed subgroups of G(F/K). Consider the maps ϕ : A → B,M 7→ G(F/M), ψ : B → A,H 7→ F ix(H),

68 where F ix(H) is the set of elements of F fixed by every element of H. Then ϕ and ψ are bijections, one the inverse of the other. S Proof. Let us show that ϕ is well defined. If M ∈ A then M = Mλ with T λ |Mλ : K| finite for any λ. Moreover G(F/M) = λ G(F/Mλ). But |Mλ : K| < ∞, so G(F/Mλ) is open, hence closed, for any λ. But then the intersection is closed. Now we must show that ϕ ◦ ψ = idB and ψ ◦ ϕ = idA.

First we show that ψ ◦ ϕ = idA. Let M ∈ A. We show that ψ(ϕ(M)) = M. Equivalently, if H := ϕ(M), we show that M = ψ(H) = F ix(H). Clearly M ⊆ F ix(H); now take α ∈ F − M. It is enough to show that α 6∈ F ix(H). Let f(x) ∈ M[X] be the minimum polynomial of α over M. Clearly the degree of f is strictly bigger than 1, so that there exists β ∈ F , β 6= α, such that f(β) = 0. Therefore there exists a unique M-isomorphism γ : M(α) → M(β) which sends α to β. By the previous proposition γ extends to some γ ∈ G(F/K) (that is, γ|M(α) = γ). Clearly γ ∈ G(F/M) = H. Since γ(α) = β 6= α we get α 6∈ F ix(H).

Now we show that ϕ ◦ ψ = idB. We claim that if H ∈ B there exists M ∈ A such that H = ϕ(M) = G(F/M). If this is the case then by the above discussion ϕ(ψ(H)) = ϕ(ψ(ϕ(M))) = ϕ(M) = H, so we are done. Since H is closed in G(F/K), a profinite group, H is given by the intersection of all the open subgroups of G(F/K) containing H. Hence H is an intersection of open T subgroups, say H = Hλ. If Hλ = G(F/Mλ) for any λ (i.e. the thesis holds λ S for the open subgroups Hλ), then H = G(F/( λ Mλ)) (i.e. the thesis holds for H). This says that we can suppose H to be open. Take a basis open subset ∗ G(F/L), contained in H, for some L ∈ F. Let H := {h|L | h ∈ H} ≤ G(L/K). Now H∗ is a subgroup of G(L/K) (recall that L is normal in F/K), and L/K is a finite Galois extension, so when we consider the field K ≤ M ≤ L associated to H∗, i.e. M = {x ∈ L | k(x) = x ∀k ∈ H∗}, we know that H∗ = G(L/M) (by the finite degree Galois theory). We claim that H = G(F/M). Clearly H ⊆ G(F/M), because if h ∈ H then h|M = (h|L)|M = idM (recall that h|L ∈ H∗). We are left to show that G(F/M) ⊆ H. Let then g ∈ G(F/M). Clearly ∗ −1 g|L ∈ G(L/M) = H , say g|L = h|L for some h ∈ H. Then gh ∈ G(F/L) ≤ H, so gh−1 ∈ H, i.e. g ∈ H.

Example: Take the prime numbers in increasing order: p , p , p , .... For √ √ 1 2 3 any i ∈ N consider Ki := Q( p1, ..., pi). Then Ki/Q is a Galois extension and |K : | = 2i. Let G := G(K / ) =∼ /2 ×...× /2 (i times): to see this take, i Q i i Q Z Z√ Z √Z √ √ for any 1 ≤ j ≤ i, αj ∈ Gi such that αj( pj) = − pj, and αj( pi) = pi for ∼ Qi S every i 6= j. It follows that Gi = hαji. Let K := ( Ki). Then K/ j=1 i∈N Q is Galois because (Ki)i∈N is a chain of Galois extensions. Let G := G(K/Q), and Hi := G(K/Ki). {Hi}i is a ﬁltering basis of open normal subgroups of G, so G is isomorphic and homeomorphic to the inverse limit of the G(Ki/Q), i.e. the inverse limit of the (Z/2Z)i, that is (Z/2Z)N (cartesian product). Now we can see in a diﬀerent way a fact that we already know: there are only countably

69 many open subgroups of G of index 2 because there are only countably many irreducible polynomials of degree 2 in Q[X].

Now our aim is the following: Let K be a field, G a profinite group. Then we want to find K ≤ L ≤ F such that F/L is a Galois extension whose Galois group is G. Recall Artin’s lemma (which basically is precisely what we want to prove in the finite degree case): Theorem 23 (Artin’s Lemma). Let F be a field and let G be a finite subgroup of Aut(F ). Let L = F G = {f ∈ F : g(f) = f ∀g ∈ G}. Then F/L is a finite Galois extension and G(F/L) = G. Proposition 20. Let G be a profinite group, F/K an algebraic extension,

θ : G → G(F/K) a homomorphism (not necessarily continuous). For x ∈ F, g ∈ G let g(x) := θ(g)(x). Suppose that for any x ∈ F the stabilizer Gx := {g ∈ G | g(x) = x} is open in G, and that F ix(G) := {x ∈ F | g(x) = x ∀g ∈ G} coincides with K. Then F/K is a Galois extension and θ is continuous and surjective.

T −1 Proof. For every x ∈ F let Rx := g∈G g Gxg. Since Gx is open, it has finite index in G, so Rx is an open normal subgroup of G. For any finite subset Y of F consider LY := K({g(y): y ∈ Y, g ∈ G}). Since F/K is algebraic, LY /K is a finite extension: Y is finite and for every y ∈ Y the set {g(y): g ∈ G} is finite, contained in the set of roots of the minimal polynomial of y over K. We have K ≤ LY ≤ F ; G acts on LY in the natural way, and the kernel of the T action is RY := y∈Y Ry. Indeed h ∈ G belongs to the kernel of the action if −1 and only if h(g(y)) = g(y) for every g ∈ G, y ∈ Y , equivalently h ∈ gGyg for every g ∈ G, y ∈ Y , i.e. h ∈ Ry. Since Y is finite, RY is an open and normal subgroup of G. We have that G/RY ≤ G(LY /K) and F ix(G/RY ) = K, by the second assumption.

By Artin’s lemma, G/RY = G(LY /K) and LY /K is a ﬁnite Galois extension. Then F/K is Galois because F is generated by all the LY over K. The canonical homomorphism G → G(LY /K) is continuous because the kernel is RY , which S is open. It is also surjective, by Artin’s lemma. Moreover since F = Y LY , we have that G(F/K) can be realized as the inverse limit of the G(LY /K). We have the following commutative diagram:

θ G / G(F/K)

πY # G(LY /K)

Which implies that θ is continuous: such is the composition with every compatible map. By the density criterion we see that θ(G) is dense, so it is equal to

70 G(F/K) because it is compact, hence closed (G is profinite and θ is continuous). So θ is surjective. Theorem 24. Let K be a field, G a profinite group. Then there exist K ≤ L ≤ F such that F/L is Galois and G =∼ G(F/L).

Proof. Let S be the set of all cosets Ng where N E G is open and g ∈ G. For any s ∈ S consider an indeterminate Xs. Let F := K({Xs | s ∈ S}). g G acts on F : if s = Ny ∈ S, g ∈ G then (Xs) := Xt where t = Nyg. When extended by ﬁeld operation compatibility, this action is well deﬁned. Let

K ≤ L := F ix(G) ≤ F , and take f ∈ F . Then f ∈ K(Xs1 , ..., Xst ) for some Tt s1, ..., st ∈ S, say si = Niyi for i = 1, ..., t. Consider N := i=1 Ni. N is open, g and N ≤ Gf := {g ∈ G | f = f}, so Gf is open. Let θ : G → G(F/L) be the homomorphism determined by the action of G on F . We are in the hypothesis of the previous proposition, so that we can apply it: θ is continuous and surjective, g and F/L is a Galois extension. Let now g ∈ ker(θ). Then (Xs) = Xs for every s ∈ S, i.e. Ng = N, i.e. g ∈ N, for every N E G open. But G is proﬁnite, so that the intersection of its open normal subgroups is {1}. This says that g = 1, hence G =∼ G(F/L).

21 Free proﬁnite groups

Let X be a set and let G be a profinite group. A map θ : X → G is called 1-convergent if the set {x ∈ X : θ(x) 6∈ N} is finite for every open normal subgroup N of G (observe that if X is finite this is an empty condition). If G is finite θ is 1-convergent if and only if θ−1(1) contains all but finitely many elements of X.

Let C be a class of groups closed under subgroups, quotients and ﬁnite direct products. A free pro-C group on a set X is a pro-C group F together with a 1-convergent map j : X → F with the following universal property: whenever ξ : X → G is a 1-convergent map to a pro-C group G there is a unique continuous homomorphism ξ : F → G such that ξ = ξ ◦ j.

Example. If X = {x} then the free proﬁnite group on X is Zb.

We want to discuss existence and uniqueness of the free pro-C group.

Uniqueness. The free pro-C group (F, j) on X, if it exists, is unique up to isomorphism and it is generated by j(X). Indeed let (F, j) and (F1, j1) be two free pro-C groups on X. By the universal property of F we have α : F → F1 with j1 = αj. By the universal property of F1 we have β : F → F1 with j = βj1. Then j1 = αβj1 and j = βαj therefore by uniqueness of the compatible map applied to j and j1 we deduce that α and β are isomorphisms one the inverse of the other. Now let E be the subgroup of F generated by j(X). The inclusion

71 map from X to E extends to a homomorphism µ : F → E compatible, so that idF is the composition between µ and the inclusion from E to F , so that E = F .

We now prove that the free pro-C group on X, at least when X is ﬁnite, is precisely the pro-C completion of the abstract free group on X. Theorem 25. Let E be the abstract free group on X and deﬁne

I = {U E E : E/U ∈ C,X − U finite}. Then the completion F = lim E/U with the map j(x) = (Ux) is the free ←−U∈I U pro-C group on X.

Proof. The kernels of the maps pU : F → E/U form a base of open neighbourhoods of 1 in F and j(x) ∈ ker(pU ) if and only if x ∈ U. It follows that j is 1-convergent. We have j = ει where ι : X → E is the inclusion map and ε is the canonical map from E to its completion F . Now let ξ : X → H be a 1-convergent map to a C-group H. By the universal property of the free abstract group there is a unique homomorphism µ : E → H with ξ = µι. Since all but ﬁnitely many elements of X map to 1 in H we have ker(µ) ∈ I so µ is continuous with respect to the topology on E having I as a base of open neighbourhoods of 1. Therefore the universal property of the completion F gives a map ξ : F → H completing the commutative diagram

ι ε X / E / F ξ µ ~ ξ H and so satisfying ξj = ξ. We need to prove that such ξ is unique. If ξ1 : F → H is a homomorphism satisfying ξ1j = ξ then we have (ξ1ε)ι = ξ. It follows from the universal property of E that ξ1ε = ξε therefore (composing with ι) ξ1j = ξj, hence from the universal property of F that ξ1 = ξ. We can use a standard argument to extend the universal property to the case when H is a pro-C group. By our discussion of Hopﬁan groups it is clear that: Proposition 21. Let n be a positive integer and let G be a pro-C group such that • Every C-group discrete image of G can be generated by n elements. • Every n-generator C-group is a discrete image of G. Then G is the free pro-C group on {1, . . . , n}. We now concentrate on pro-p groups. Let p be a prime and let G be a proﬁnite group. The p-Frattini subgroup Φp(G) is the closure of the abstract group generated by {x−1y−1xy : x, y ∈ G} ∪ {xp : x ∈ G}.

72 Proposition 22. Φp(G) ≥ Φ(G) with equality if and only if G is a pro-p group.

Proof. If N is an open normal subgroup of G containing Φp(G) then G/N is an abelian group whose non-trivial elements have order p, so G/N is a direct product of groups of order p. So N is an intersection of open subgroups of index T p, so Φ(G) ≤ N and since Φp(G) = {N Eo G :Φp(G) ≤ N} we deduce that Φ(G) ≤ Φp(G). If G is a pro-p group then G/Φ(G) is abelian and every open maximal subgroup of G/Φ(G) has index p and so contains the image in G/Φ(G) of p {x : x ∈ G}. It follows that Φp(G) ≤ Φ(G) and hence we have equality. Conversely assume G is not a pro-p group. Let M be a maximal open subgroup containing a Sylow p-subgroup of G. Since |G : M| is prime to p and |G :Φp(G)| is a power of p we cannot have Φp(G) ≤ M. However Φ(G) ≤ M, therefore Φ(G) 6= Φp(G). Theorem 26. Let X be a set.

X 1. Fp is the free elementary abelian pro-p group on the set X.

2. If Fp ∈ C and F is the free pro-C group on X then F/Φp(F ) is the free X elementary abelian pro-p group on X and so is isomorphic to Fp . 3. For each class C the canonical map j from X to the free pro-C group F on X is injective.

X X Proof. 1. Define j : X → Fp the following way: for x ∈ X let j(x) ∈ Fp have x-component 1 and all other components 0. Then j is 0-convergent, since every X1 open normal subgroup N contains the kernel of the projection to Fp for some finite set X1 by definition of the product topology, so that {x ∈ X : j(x) 6∈ N} ⊆ X1. Suppose that ξ : X → A is 0-convergent, where A is a finite additive elementary abelian p-group, and let X1 = X − ker(ξ), so that X1 is finite. We X X1 X−X1 X have Fp = Fp ⊕ Fp and any map ξ : Fp → A with ξ ◦ j = ξ must vanish on the second summand. Since j(X1) is a basis for the first summand, regarded as a vector space over Fp (here we use that X1 is finite! j(X1) would not be a base if X1 was infinite), the map ξ exists and is unique (it sends the generators j(x) with x ∈ X1 to ξ(j(x)) = ξ(x)). 2. Let K be the intersection of the open normal subgroups of F of index p. It is easy to see that F/K is the free elementary abelian pro-p group on X, with the canonical map X → F → F/K, and since K = Φ(F ) and F is a pro-p group we deduce K = Φp(F ). The result follows from the previous item. 3. Since C contains non-trivial groups we have Fp ∈ C for some prime p. Since ∼ X the map X → F → F/Φp(F ) = Fp is injective, j : X → F is injective.

22 Projective proﬁnite groups

A pro-C group G is called C-projective (or a projective pro-C group) if whenever E,W are pro-C groups and ϕ : G → W , β : E → W are continuous homomorphisms with β surjective there exists a continuous homomorphism θ : G → E

73 with ϕ = β ◦ θ. G θ ϕ ~ β E / W Proposition 23. Let G be a pro-C group. The following are equivalent. 1. G is a projective pro-C group.

2. For every pro-C group E and for every surjective continuous homomorphism β : E → G there exists a continuous homomorphism θ : G → E such that β ◦ θ = idG. 3. For every pro-C group E and any closed normal subgroup K of E with E/K =∼ G there is a complement C of K in E (that is, a closed subgroup C of E such that C ∩ K = {1} and CK = E). Proof. (2) implies (3). Suppose (2) holds. Let E be a pro-C group and let K be a closed normal subgroup of E with E/K =∼ G, let β : E → E/K =∼ G be the canonical projection. Since E/K =∼ G, by (2) there exists a continuous homomorphism θ : G → E such that β ◦ θ = idG, set C := θ(G). We need to prove that C ∩ K = {1} and that CK = E. For the first claim let g ∈ G be such that θ(g) ∈ K, then 1 = β(θ(g)) = idG(g) = g, and for the second claim let e ∈ E, and c := θ(β(e)), it is enough to prove that c−1e ∈ K, but this is clear because c−1e = β(θ(β(e))−1e) = β(θ(β(e))−1)β(e) = β(e)−1β(e) = 1. (3) implies (2). Suppose (3) holds and let E be a pro-C group, and β : E → G a surjective homomorphism. Set K := ker(β) so that G =∼ E/K. By (3) there is a complement C of K in E, so that E/K =∼ C canonically, and composing G =∼ E/K with E/K =∼ C and then with the inclusion C → E we obtain a continuous homomorphism θ : G → E and by construction β ◦ θ is the identity of G. Now we prove the equivalence between (1) and (2). Clearly (1) implies (2): apply the definition of projective group to β and ϕ = idG : G → G. Now we prove that (2) implies (1). Consider the diagram defining projective groups. We want to construct θ. Let

E1 = {(e, g): β(e) = ϕ(g)} ≤ E × G.

Since C is closed for subgroups and direct products, E1 is a pro-C group. Deﬁne β1 : E1 → G, β1(e, g) = g. Since β is surjective, β1 is surjective, so there is a homomorphism θ1 : G → E1 satisfying β1 ◦ θ1 = idG by (2). Let π : E1 → E 0 0 be the map (e, g) 7→ e and deﬁne θ = π ◦ θ1. If g ∈ G, say θ1(g) = (e , g ), then 0 0 0 0 0 we have g = β1(e , g ) = β1θ1(g) = g so that (e , g) ∈ E1 hence ϕ(g) = β(e ) = 0 βπ(e , g) = βπθ1(g) = βθ(g) thus βθ = ϕ. Proposition 24. Every free pro-C group is C-projective.

74 Proof. Let (G, j) be the free pro-C group on a set X and consider the diagram in the definition of projective group. Let T be a transversal to ker(β) in E with 1 ∈ T (a closed subset of E containing precisely one element in each coset of ker(β); it is possible to show that such a set always exists: see page 20 of Wilson’s book). The function β|T : T → W is a continuous bijection, so both β|T and its inverse τ : W → T are homeomorphisms (T is closed, hence compact, and W is Hausdorff). The map τ ◦ ϕ ◦ j : X → T ⊆ E is 1-convergent. Indeed −1 if M Eo E then (τϕ) (M) is open in G and contains 1 (it contains 1 because 1 ∈ T ), so {x ∈ X : τϕj(x) 6∈ M} = {x ∈ X : j(x) 6∈ (τϕ)−1(M)} is finite: it is contained in {x ∈ G : j(x) 6∈ N} where N is an open normal subgroup of G contained in (τϕ)−1(M) and j is 1-convergent. Therefore the universal property of G first yields a homomorphism θ : G → E with θj = τϕj, and then, since βθj = βτϕj = ϕj, it yields that βθ = ϕ by the universal property of G: ϕ is the unique map G → W extending ϕj : X → W and βθ also extends it. We now want to prove that projective pro-p groups are free. Proposition 25. Suppose that either (i) C is contained in the class of finite p-groups for some prime p or (ii) C is closed for quotients. Let G be a pro-C group and let P be a C-projective group. 1. Every surjective homomorphism α : P/Φ(P ) → G/Φ(G) lifts to a surjective homomorphism αb : P → G. 2. If also G is C-projective and α is an isomorphism then αb is an isomorphism.

Proof. By the hypotheses on C and the fact that Φ(G) = Φp(G) if G is a pro-p group we have that P/Φ(P ) and G/Φ(G) are pro-C groups. Since P is C-projective there exists a map αb such that the following diagram commutes. P

αb P/Φ(P )

α Ô G / G/Φ(G)

Since (αb(P )Φ(G))/Φ(G) = α(P/Φ(P )) = G/Φ(G) we deduce αb(P )Φ(G) = G therefore αb(P ) = G. Now assume that also G is C-projective and α is an isomorphism. Since αb : P → G is surjective, ker(αb) has a complement C in P . Since α is injective ker(αb) ≤ Φ(P ), so P = C ker(αb) = CΦ(P ) hence P = C therefore ker(αb) = {1}. This proves that αb is an isomorphism. Proposition 26. Suppose that C is contained in the class of ﬁnite p-groups, for some prime p. Then

75 1. Every pro-C group is an image of a free pro-C group. 2. Every C-projective group is a free pro-C group.

∼ X Proof. Let G be a pro-C group, then G/Φ(G) = Fp for some set X. Let F be the free pro-C group on X. Since F/Φ(F ) is the free elementary abelian pro-p group there is an isomorphism F/Φ(F ) =∼ G/Φ(G), now apply the above result in both cases (recall that F is free hence projective). Proposition 27. We have 1. Every proﬁnite group G has a 1-convergent generating set (meaning that its inclusion in G is 1-convergent).

2. Every pro-C group is an image of a free pro-C group. 3. Every C-projective group is isomorphic to a subgroup of a free pro-C group. Proof. 1. If G is a pro-p group for some prime p then the result follows from the previous result since if θ : F → G is a surjective homomorphism from the free pro-p group on X then the image of X under θ is a 1-convergent generating set (because X is a 1-convergent generating set of F ). In the general case for each prime p let Gp be a Sylow p-subgroup of G and let Xp be a 1-convergent generating set for Gp. Let X be the union of all the Xp’s. Clearly X generates r1 rn G (it generates a subgroup of index 1), now let N Eo G, say |G/N| = p1 ··· pn , then Xp ⊆ Gp ⊆ N for p 6= p1, . . . , pn and N ∩ Gpi Eo Gpi so that Xpi − N is Sn finite for i = 1, . . . , n, and X − N = i=1(Xpi − (Gpi ∩ N)) is finite. 2. Let G be a pro-C group generated by a 1-convergent generating set X (such X exists by the previous item), and let (F, j) be the free pro-C group on X. The inclusion map ι : X → G is 1-convergent, so it yields a homomorphism ξ : F → G with ξj = ι, and since the image of ξ contains the generating set X for G, the map ξ is surjective. 3. Let P be C-projective, and let π : F → P be a surjective homomorphism from a free pro-C group F . There is a homomorphism β : P → F such that π ◦ β = idP , therefore β is injective. Another example of a projective group that is not free. For all Q prime p let Gp be the free pro-p group. The group G = p Gp is pronilpotent, in particular it is not a free profinite group. However it is a projective profinite group. More in general a profinite group G is a projective profinite group if and only if the Sylow p-subgroups are free pro-p groups, for every prime p (see Serre, Galois cohomology, page 58).

23 Positively Finitely Generated groups

Let G be a proﬁnite group. If A is a family of subsets of G, with G ∈ A, the σ-algebra generated by A is the smallest collection A0 of subsets of A which contains A and is closed under taking complements and countable unions.

76 The Borel ﬁeld of G is the σ-algebra B = B(G) generated by the family of all closed subsets of G.A normalized Haar measure of G is a function µ : B → R such that:

1. 0 ≤ µ(B) ≤ 1 for every B ∈ B. 2. µ(∅) = 0 and µ(G) = 1. 3. σ-additivity. S∞ P∞ If B1,B2,B3,... ∈ B are pairwise disjoint, µ( i=1 Bi) = i=1 µ(Bi). 4. Translation invariance. If B ∈ B and g ∈ G then µ(gB) = µ(B) = µ(Bg). 5. Regularity. For every B ∈ B and every ε > 0 there are an open set U and a closed set C such that C ⊆ B ⊆ U and µ(U − C) < ε.

Some consequences.

1. If A ⊆ B are Borel sets then µ(B − A) = µ(B) − µ(A). Indeed µ(B) = µ(A ∪ (B − A)) = µ(A) + µ(B − A). S∞ P∞ 2. If B1,B2,B3,... are Borel sets then µ( i=1 Bi) ≤ i=1 µ(Bi). In par- S∞ ticular if µ(Bi) = 0 for all i then µ( i=1 Bi) = 0. This follows from the σ-additivity. T∞ 3. If B1,B2,B3,... are Borel sets with µ(Bi) = 1 ∀i then µ( i=1 Bi) = 1. This follows from the above items by taking complements. S∞ 4. If A1,A2,A3,... is an increasing sequence of Borel sets then µ( i=1 Ai) = limi→∞ µ(Ai) = supi µ(Ai). To prove this apply the σ-additivity to the family Ai+1 − Ai. T∞ 5. If A1,A2,A3,... is a decreasing sequence of Borel sets then µ( i=1 Ai) = T∞ S∞ limi→∞ µ(Ai) = infi µ(Ai). To prove this write i=1 Ai = G − i=1(G − T∞ Ai) and deduce by the above item that µ( i=1 Ai) = 1 − limi→∞(1 − µ(Ai)) = 1 − (1 − limi→∞ µ(Ai)) = infi µ(Ai).

6. If H ≤o G with |G : H| = n then µ(H) = 1/n. This is because writing G as union of cosets of H and using the translation invariance gives nµ(H) = µ(G) = 1. 7. If H is a closed non-open subgroup of G then µ(H) = 0. Indeed H is contained in a strictly decreasing sequence of open subgroups. ˜ ˜ 8. Let N Eo G, A a subset of G/N and A the preimage of A via G → G/N. Then µ(A) = |A˜|/|G : N|. Indeed A is a union of |A˜| cosets of N and each coset of N has measure µ(N) hence µ(A) = |A˜|µ(N).

77 9. If U is a non-empty open subset of G then µ(U) > 0. Indeed U contains some coset of an open normal subgroup.

It is possible to prove that every proﬁnite group has a unique normalized Haar measure, that we shall call µ. Since µ(G) = 1, the measure µ can be interpreted as a probability and G can be seen as a probability space. The members of the σ-algebra B are called measurable sets. For example the closed subsets of G are measurable.

A subset A of G is called a zero-set if there exists B ∈ B(G) with A ⊆ B and µ(B) = 0. Let Bb be the σ-algebra generated by B and the zero-sets. For each Bb ∈ B it is possible to find B ∈ B and a zero-set A such that Bb = B ∪A (this is proved showing that such unions B ∪ A form a σ-algebra); define µb(Bb) := µ(B). This defines a complete measure on G (a measure is said to be complete if its zero-sets are measurable).

Observe that if G is a finite discrete group (i.e. a finite profinite group) then every subset is measurable (being closed) and if A ⊆ G then µ(A) = |A|/|G|. This is because points are cosets of {1} so they have measure 1/|G|.

An important example for us is the following: for G a proﬁnite group and k a positive integer consider the set

k A = {(g1, . . . , gk) ∈ G : hg1, . . . , gki = G}.

This is a subset of Gk, which is a proﬁnite group with its own unique normalized Haar measure. The set A is closed in Gk, hence measurable and it makes sense to write µ(A). Indeed the complement of A is the union of the sets Kk where K is a proper open subgroup of G. Observe that since µ(G) = 1 the group G can be considered as a probability space. The probability that k random elements generate G is deﬁned to be

k P (G, k) := µ({(g1, . . . , gk) ∈ G : hg1, . . . , gki = G}).

For example if G is a ﬁnite group then

|{(g , . . . , g ) ∈ Gk : hg , . . . , g i = G}| P (G, k) = 1 k 1 k |G|k

k k is a quotient of two ﬁnite numbers, for example P (Cp, k) = (p − 1)/p if p is a prime, because a k-tuple (x1, . . . , xk) of elements of Cp generate Cp if and only if at least one xi is not 1.

P. Hall proved the following for a ﬁnite group G:

X µG(H) P (G, k) = |G : H|k H≤G

78 where µG (the Möbiusfunction of G) is the function defined on the subgroup P lattice of G by the rules µG(G) = 1 and H≤K≤G µG(K) = 0 for fixed H < G.

h(12)i h(13)i h(23)i h(123)i

{1}

For example here µS3 (S3) = 1, µS3 ({1}) = 3 and the other subgroups have M¨obiusfunction equal to −1. So 3 1 3 P (S , k) = 1 − − + . 3 3k 2k 6k

For example P (S3, 1) = 0 (obviously), and more interestingly P (S3, 2) = 1/2, this can also be seen “by hand” listing the pairs generating S3 (they are the 18 pairs of the form (x, y) where x 6= y have order 2 or have orders 2 and 3, or 3 and 2) Applying the idea to Zb we see that since in Zb there is precisely one subgroup of any given index, P ( , k) = P µ(n) which is the multiplicative inverse of Zb n∈N nk ζ(k) (where ζ is the Riemann zeta function). Here µ(n) is the classical M¨obius function. This explains why “the probability that two integers are coprime” (i.e. generate Z) is P (Zb, 2) = ζ(2)−1 = 6/π2.

The following is taken from the beautiful paper by Avinoam Mann “Posi- tively finitely generated groups” (1996). Let G be a finitely generated profinite group, and denote by d(G) the least integer d such that G can be generated by d elements. Definition 15 (PFG groups). The profinite group G is said to be Positively Finitely Generated (PFG) if there exists an integer k such that P (G, k) > 0. The least value of k for which this happens is denoted dP (G). Clearly PFG groups are finitely generated, but the converse is not true, as we shall see. Observe that dP (G) ≥ d(G) because if k < d there are no k-tuples generating G. In general for a PFG group d(G) and dP (G) are different, as we shall establish later.

Proposition 28. Let G be a profinite group and let {Ni : i ∈ I} be a filtering basis of open normal subgroups of G intersecting in {1} (here “filtering” means that for every i, j there is k with Nk ⊆ Ni ∩ Nj). Then

P (G, k) = inf P (G/Ni, k). i∈I Proof. Let N be an open normal subgroup of G and let π : Gk → Gk/N k be the canonical projection. Let A be the set of k-tuples in Gk generating G and

79 let B be the set of k-tuples in Gk generating G modulo N, then π(B) is the set of k-tuples in (G/N)k generating G/N. Since A ⊆ B we have

P (G, k) = µ(A) ≤ µ(B) = |π(B)|/|G : N| = P (G/N, k).

This implies that if infi P (G/Ni, k) = 0 then P (G, k) = 0. Now suppose infi P (G/Ni, k) > 0, then each factor group G/Ni can be generated by k elements hence G can be generated by k elements (because each open normal subgroup contains some Ni). So G has only finitely many open subgroups of a given index, so it has only countably many open normal subgroups, hence we may assume I is countable and taking subsequent intersections of the Ni’s we obtain a descending chain {Mi}i∈N of open normal subgroups (on some cofinal subposet isomorphic to N) intersecting in {1} (note that countability is important, cf. the definition of σ-algebra). Let S be the set of k-tuples in Gk k generating G and for each i let Si be the set of k-tuples in G generating G T modulo Mi, then S = i Si and Si ⊇ Si+1 for every i, so

P (G, k) = µ(S) = inf µ(Si) = lim µ(Si) = lim P (G/Mi, k). i i→∞ i→∞ This concludes the proof. It follows that if H is a closed normal subgroup of G then P (G, k) ≤ P (G/H, k), indeed

P (G, k) = inf P (G/N, k) ≤ inf P (G/N, k) = P (G/H, k). NEoG H≤NEoG In particular G/H is PFG if G is PFG.

When H = Φ(G) is the Frattini subgroup of G we have

P (G, k) = P (G/Φ(G), k).

To see this observe that if H = Φ(G) and NEoG then P (G/N, k) = P (G/HN, k) as HN/N ≤ Φ(G/N), and when applying the proposition we can take as filtering family of open normal subgroups of G/H the family of such subgroups HN/H of G/H. In particular if G is a finitely generated pro-p group then G is ∼ d positively finitely generated and dP (G) = d(G) (this is because G/Φ(G) = Cp where d = d(G)).

24 PMSG groups

Deﬁnition 16 (PMSG groups). Let G be a proﬁnite group. We say that G has Polynomial Maximal Subgroup Growth if there exists a positive real number s such that for all n ∈ N the number mn(G) of maximal open subgroups of index n is at most ns. It was proved by Mann and Shalev that PFG is equivalent to PMSG in [6]. The implication (⇐) is easy and we will prove it now.

80 Proposition 29. A proﬁnite PMSG group is PFG. Moreover for such a group P d if n>1 mn(G)/n < 1 then dP (G) ≤ d. P d Proof. Let d be an integer large enough so that n>1 mn(G)/n < 1 (for example d = [s + 3]). If a d-tuple (g1, . . . , gd) satisﬁes hg1, . . . , gdi= 6 G then there exists a maximal open subgroup M containing all the elements g1, . . . , gd. The probability of this happening is µ(M d) = 1/|G : M|d, hence

X 1 X mn(G) P (G, d) ≥ 1 − = 1 − > 0. |G : M|d nd M n>1 This implies that G is PFG. Recall that two events A, B of a probability space X (with the probability function denoted by P ) are said to be independent if P (A ∩ B) = P (A)P (B).

Theorem 27 (Borel-Cantelli Lemma). Let {Ai}i be a sequence of events in a probability space X, with probabilities pi. P 1. If i pi converges then the probability of an element of X belonging to infinitely many of the events Ai’s is zero. P 2. If i pi diverges and the events Ai are pairwise independent then the probability of an element of X belonging to infinitely many of the events Ai’s is 1. For example let A, B be two open subgroups of a profinite group G with its unique normalized Haar measure (which is a probability). The probability of an element belonging to an open subgroup H is 1/|G : H| therefore the events g ∈ A and g ∈ B are independent if and only if |G : A ∩ B| = |G : A||G : B| and this happens if and only if AB = G (to see this work in the finite quotient G/(A ∩ B)G for example). In other words the events g ∈ A and g ∈ B are independent if and only if AB = G.

We will show now that there exist pronilpotent PFG groups G with dP (G) 6= d(G).

Theorem 28. Let G be a ﬁnitely generated pronilpotent group with d(G) = d and let X be the set of prime divisors p of |G| with the property that d(P ) = d where P is the Sylow p-subgroup of G. Then G is PFG and P d if p∈X 1/p converges, dP (G) = P d + 1 if p∈X 1/p diverges. Proof. All maximal open subgroups of G are normal of prime index. For a prime p the maximal subgroups of index p correspond to the maximal subgroups of the Sylow p-subgroup P of G, and since d(P ) = dp ≤ d the number of maximal dp d d subgroups of index p is mp(G) = (p − 1)/(p − 1) ≤ (p − 1)/(p − 1) < p , in

81 particular G is PMSG, hence it is PFG by the above proposition. What about dP (G)? Observe that

d X mn(G) X mp(G) X p − 1 X 1 X 1 = ≤ < < = 1. nd+1 pd+1 (p − 1)pd+1 p(p − 1) n(n − 1) n p p p n

We deduce that dP (G) ≤ d + 1. However dP (G) ≥ d so there are only two possible values for dP (G): d and d + 1.

P d Suppose that the sum p∈X 1/p diverges. Since mp(G) = (p −1)/(p−1) > d−1 d 1 d p , using p − 1 > 2 p , the sum X X pdp − 1 X pdp − 1 X pd − 1 1 X 1 m (G)/nd = ≥ = > n (p − 1)pd (p − 1)pd (p − 1)pd 2 p p p∈X p∈X p∈X diverges. Since all maximal subgroups are normal, the product of any two of them is G, so they deﬁne independent events. By the Borel-Cantelli lemma with probability 1 a d-tuple from G lies in inﬁnitely many maximal subgroups, so it cannot generate G. This implies that dP (G) 6= d, therefore dP (G) = d + 1. P Suppose that the sum p∈X 1/p converges. Then

X d X X mn(G)/n ≤ 1/p(p − 1) + 1/(p − 1) < ∞. n p p∈X By the Borel-Cantelli lemma with probability 1 a d-tuple generates a subgroup which is contained in only finitely many maximal subgroups. So 1 equals the sum over all subgroups contained in only finitely many maximal subgroups of the probability that a d-tuple generates that subgroup, so at least one summand in this sum is nonzero, hence there exists a subgroup H which is generated by d elements with positive probability and is contained in only finitely many maximal subgroups. This means that H contains all but finitely many Sylow subgroups of G. Then we can write G = K × L where L is the product of the Sylow subgroups of G which are contained in H, and K is the product of the other Sylow subgroups (observe that K is a pronilpotent group with finitely many Sylow subgroups). Then H = (H ∩ K) × L, so H projects onto L, hence L is also generated by d elements with positive probability. Since the same holds for K (because it is a pronilpotent group with finitely many Sylow subgroups, all d-generated), it holds also for G. In other words the fact that G = K × L, dP (K) ≤ d, dP (L) ≤ d implies that dP (G) ≤ d hence dP (G) = d (being dP (G) ∈ {d, d + 1}). We want to give an example of finitely generated profinite group that is not PFG. The material is taken from [4].

Let Dk(G) denote the set of the ordered k-tuples generating the group G. If G is a nonabelian simple group and dk(G) denotes the largest integer m

82 m such that G has k generators then dk(G) = |Dk(G)|/|Aut(G)|. For example 2 d2(A5) = 19. By a result of Dixon P (An, 2) = |D2(An)|/|An| tends to 1 as 2 n → ∞ so when n is large enough (say n ≥ N) |D2(An)| ≥ |An| /2. We obtain

2 2 |D2(An)| |An| /2 1 (n!/2) n! d2(An) = ≥ ≥ = |Aut(An)| |Sn| 2 n! 8 for n ≥ N (we are choosing N so that N > 6 so that Aut(An) = Sn whenever n!/8 n ≥ N). This means that An can be generated by 2 elements whenever n ≥ N, therefore Y n!/8 G := An n≥N

n!/8 can be generated (topologically) by 2 elements (recall that the An do not have common composition factors so it is enough to check that they are all 2-generated).

We will prove now that G is not PFG. It is possible to show that if S is a ﬁnite non-abelian simple group√ then for all positive integer m√ we have P (Sm, k) ≤ P (S, k)m. Set k = b nc be the lower integer part of n. Since An−1 is a subgroup of An,

n!/8 k ! n!/8 n!/8 |An−1 | P (An , k) ≤ P (An, k) ≤ 1 − k |An |

√ n√!  n  8n n 1 n!/8 1 n ≤ 1 − √ =  1 − √  → 0 n n n n

√ n n 1 n! as n → ∞. This is because 1 − √ tends to 1/e when n → ∞ and √ n n 8n n goes to inﬁnity with n (by Stirling’s formula, for example).

n!/8 In particular for ﬁxed k the probability P (An , k) tends to zero as n → ∞, n!/8 n!/8 so since P (G, k) ≤ P (An , k) for all n, being An a discrete quotient of G, we deduce that P (G, k) = 0 for all ﬁxed k hence G is not PFG.

References

[1] John S. Wilson, Proﬁnite Groups. Oxford University Press, USA 1999. [2] Luis Ribes, Pavel Zalesski, Proﬁnite Groups (2nd ed). Springer-Verlag Berlin Heidelberg 2010.

[3] J. D. Dixon, M. P. F. Du Sautoy, A. Mann, D. Segal; Analytic pro-p groups. Cambridge University Press 2003.

83 [4] William M. Kantor; Alexander Lubotzky; The probability of generating a ﬁnite classical group; Springer-Verlag; Geometriae Dedicata; Year:1990; Volume:36; Issue:1; First page:67; Last page:87. [5] Mann, Avinoam; Positively ﬁnitely generated group. Walter de Gruyter; Forum Mathematicum. Year:1996. Volume:8, Issue:8.

[6] Avinoam Mann; Aner Shalev; Simple groups, maximal subgroups, and probabilistic aspects of proﬁnite groups. Springer-Verlag; Israel Journal of Mathematics. Year:1996, Month:06, Volume:96, Issue:2. First page:449, Last page:468.

[7] John D. Dixon; The probability of generating the symmetric group; Springer-Verlag; Mathematische Zeitschrift; Year:1969; Volume:110; Is- sue:3; First page:199; Last page:205.