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Subgroups of Finite Index in Profinite Groups

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Subgroups of Finite Index in Profinite Groups Subgroups of Finite Index in Profinite Groups Sara Jensen May 14, 2013 1 Introduction In addition to having a group structure, profinite groups have a nontrivial topological structure. Many results pertaining to profinite groups exploit both structures, and there- fore both structures are important in understanding profinite groups. An amazing result due to Nikolov and Segal is the following theorem. Theorem 1.1. Suppose that G is a topologically finitely generated profinite group. Then every subgroup of G of finite index is open. One way to view Theorem 1.1 is as a statement that the algebraic structure of a finitely generated profinite group somehow also encodes the topological structure. That is, if one wishes to know the open subgroups of a profinite group G, a topological property, one must only consider the subgroups of G of finite index, an algebraic property. As profinite groups are compact topological spaces, an open subgroup of G necessarily has finite index. Thus it is also possible to begin with a subgroup of G having a particular topological property (the subgroup is open) and deduce that this subgroup must also have a particular algebraic property (the subgroup has finite index). The proof of Theorem 1.1 is quite extensive, and requires the classification of finite simple groups. However, if one restricts attention to a smaller class of groups, the result can be done in a fairly straightforward manner. Suppose that G is a finite group having a normal series 1 = Gl ⊆ Gl−1 ⊆ ::: ⊆ G1 ⊆ G0 = G such that Gi=Gi+1 is nilpotent for all 0 ≤ i < l. Then we say that G l belongs to the class N . Note that finite nilpotent groups all belong to N1, while finite supersolvable groups all belong to N2 since the commutator subgroup of a supersolvable group is nilpotent. The theorem we aim to prove in this document is the following. Theorem 1.2. Suppose that G is a topologically finitely generated profinite group such that there exists some fixed l with G=N 2 N l whenever N is an open normal subgroup of G. Then every subgroup of G of finite index is open. 1 2 Preliminary Facts We begin by establishing some notation. Let s Y w = w(x ; : : : ; x ) = xi 1 n ji i=1 where i 2 f1; −1g and ji 2 f1; : : : ; ng. That is, w is a finite expression of the variables x1; : : : ; xn and their inverses. Then if G is any group, we write w(G) = fw(g1; : : : ; gn) j g1; : : : ; gn 2 Gg and we write w(G) for the subgroup of G algebraically generated by w(G). We call the group w(G) the verbal subgroup of G. Now suppose Y is an arbitrary subset of G and t a positive integer. We write Prt(Y ) for the collection of all possible words of length t involving only elements of Y or their inverses. Example. Suppose that w = w(x; y) = [x; y] = y−1x−1yx. Then for an arbitrary group G, w(G) is the set of commutators in G, while w(G) is the commutator subgroup of G. If we let t = 4, then w(G) ⊆ Prt(G). If we pick g 2 [G; G], then we know that there exists some t so that g is the product of t commutators. For this t, we have that g 2 Prt(w(G)). An introductory fact about profinite groups that we will need is the following. Proposition 1. Suppose G is a profinite group and C1;C2;::: is a countably infinite set 1 [ of nonempty closed subsets of G having empty interior. Then G 6= Ci. i=1 Lemma 2.1. Let G be a profinite group and let X be a closed subset of G such that −1 X = X and 1 2 X. Then G = hXi if and only if G = Prm(X) for some positive integer m. Proof. If G = Prm(X) for some positive integer m, then it is clear that G = hXi. Now suppose that G = hXi. For every g 2 G, we know that g 2 Prn(X) for some positive 1 [ integer n, and thus G = Pri(X). Note that if i < j, then Pri(X) ⊆ Prj(X) since i=1 1 2 X. Now for each n, we know that Prn(X) is closed, and by Proposition 1, we know that G cannot be an infinite union of closed subsets with empty interior. This means that Prt(X) contains a non-empty open subset U for some positive integer t. Thus we [ may also write G = gU, and by compactness, we know there exists a finite collection g2G r [ g1; : : : ; gr 2 G for which G = giU. As G = hXi and 1 2 X, we know that there exists i=1 s for which g1; : : : ; gr 2 Prs(X). Finally, set m = s + t. Then if g 2 G, we can write g = giu for some gi 2 fg1; : : : ; grg and u 2 U. Since gi 2 Prs(X) and u 2 Prt(X), we see that we can write g as a combination of m members of X, as claimed. 2 A crucial lemma which will be used to prove Theorem 1.2, and which is used in Nikolov and Segal's proof of Theorem 1.1, is the following. Lemma 2.2. Let w be a group word. The following are equivalent. (a) For every finitely generated profinite group G, w(G) is closed. (b) There exists an integer valued function f such that for each natural number k and for each finite group H that can be generated by k elements, one has w(H) = Prf(k)(w(H)) ; that is, for each k-generated finite group H, the word w has width f(k) in H. Proof. The image of a compact topological space under a continuous map is compact, k Y and the map from G ! G that sends (g1; : : : ; gk) ! w(g1; : : : ; gk) or (g1; : : : ; gk) 7! i=1 t −1 Y w(g1; : : : ; gk) is continuous; hence w(G) is compact. Likewise, the map from w(G) ! i=1 t Y Prt(w(G)) that sends (w1; : : : ; wt) ! wi is continuous, and therefore Prt(w(G)) is also i=1 compact for each t. Write G = lim G with projection maps ' : G ! G . Now w(G) = lim w(G ), and − i i i − i i i therefore w(G) is also a profinite group. Of course, Prt(w(G)) ⊆ w(G) for every integer t. As we have just shown that Prt(w(G)) is compact for each t and w(G) is Hausdorff, it follows that Prt(w(G)) is closed for each t. It is easy to see that 'i(Prt(w(G))) = Prt(w(Gi)) for all t and for all i. Assume now that (b) holds, and let d be the number of topological generators for G. That is, each Gi is topologically generated by d elements, and so there exists an integer valued function f for which w(Gi) = Prf(d)(w(Gi)). Since Prf(d)(w(G)) is closed, we have Pr (w(G)) = lim Pr (w(G )) f(d) − f(d) i i = lim w(G ) − i i = w(G) However, we also clearly have that Prf(d)(w(G)) ⊆ w(G), and therefore w(G) = Prf(d)(w(G)) ⊆ w(G). This gives us that w(G) is closed, as claimed. We now assume that (a) holds. Let k be a natural number and F the free profi- nite group on k generators. By hypothesis, w(F ) = w(F ) and so w(F ) is a profinite 3 group which is algebraically generated by w(F ). It follows by Lemma 2.1 that w(F ) = Prm(w(F )) for some natural number m. Define f(k) = m. Then if H is an arbitrary profi- nite group generated by k elements, then we have a surjection ' : F ! H. As a result, we have that w(H) = '(w(F )) = '(Prm(w(F ))) = '(Prf(k)(w(F ))) = Prf(k)(w(H)). We will be particularly interested in the statement of Lemma 2.2 as it applies to the word w = x−1y−1xy. That is, we are interested in what Lemma 2.2 has to say about the commutator subgroup of a finitely generated profinite group. In any topological space, one is concerned with the structure of open subgroups of that space. Since profinite groups are topological structures, we wish to know about the open subgroups of a profinite group G. Lemma 2.3. Suppose that G is a profinite group and H is a subgroup of G. Then H is open in G if and only if H is closed of finite index. Proof. If H is open, then all cosets of H in G are also open, and the cosets of H in G form an open cover of G. Since G is compact, we know that finitely many cosets will suffice to cover G, and therefore H has finite index in G. To see that H is also closed, note that the complement of H in G is open, since the complement of H in G is the union of the cosets of H in G that are distinct from H, each of which is open. Now suppose that H ⊆ G is a closed subgroup of finite index. To see that H is open, we will show that its complement is closed. The complement of H in G is the union of the cosets of H in G distinct from H.
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