Free groups, Lecture 2, Part 2

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1 Free groups, Lecture 2, Part 2 Olga Kharlampovich NYC, Sep 2 1 / 19

2 We say that a group G embeds into a group H, if there is a monomorphism φ: G H. If φ(g) H, then we say that G properly embeds into H and that φ is a proper embedding. Proposition Any countable free group G can be embedded into a free group of rank 2. 2 / 19

3 Proof To prove the result it suffices to find a free subgroup of countable rank in a free group of rank 2. Let F 2 be a free group with a basis {a, b}. Denote x n = b n ab n (n = 0, 1, 2,... ) and let S = {x 0, x 1, x 2,... }. We claim that S freely generates the subgroup S in F 2. Indeed, let w = x ɛ 1 i 1 x ɛ 2 i 2... x ɛn i n be a reduced non-empty word in S ±1. Then w can also be viewed as a word in {a, b}. 3 / 19

4 Indeed, w = b i 1 a ɛ 1 b i 1 b i 2 a ɛ 2 b i 2... b in a ɛn b in. Since w is a reduced word in S, we have that either i j i j+1, or i j = i j+1 and ɛ j + ɛ j+1 0, for each j = 1, 2,..., n 1. In either case, any reduction of w (as a word in {a, b}) does not affect a ɛ j and a ɛ j+1 in the subword b i j a ɛ j b i j b i j+1 a ɛ j+1 b i j+1 i.e., the literals a ɛ j and a ɛ j+1 are present in the reduced form of w as a word in {a, b} ±1. Hence the reduced form of w is non-empty, so w 1 in F 2. Clearly, S is a free group of countable rank. 4 / 19

5 Given a family of groups {G i i I } we may assume that the G i are mutually disjoint sets. Let X = i I G i and let {1} be a one-element set disjoint from X. A word in X is a finite string (a 1, a 2,..., a n, 1, 1,...) = a 1 a 2... a n, where a i X. A word is reduced provided 1) no a i is the udentity in its group G j, 2) for all i, a i and a i+1 are not in the same G j, 3) a k = 1 implies a i = 1 for i k. Let i I G i (or G 1 G 2... G n ) be the set of all reduced words on X. i I G i forms a group, called the free product of the family {G i i I } under the operation Juxtaposition+Cancellation+Contraction. For example, if a i b i G i then (a 1 a 2 a 3 )(a 1 3 b 2b 1 b 3 ) = a 1 c 2 b 1 b 3, where c 2 = a 2 b 2 G 2. 5 / 19

6 Homework problem 1 Prove that the reduced form of a given element is unique. Prove that the map i k : G k i I G i given by e 1 and a a = (a, 1, 1,...) is a monomorphism. We identify G i with its image in i I G i. Prove that i I G i is a coproduct in the category of groups. 6 / 19

7 Definition (Type of a core graph) Let Γ be a folded X -digraph which is a core graph with respect to some vertex. Suppose that Γ has at least one edge. If every vertex of Γ has degree at least two, we set the type of Γ, denoted Type(Γ) to be equal to Γ. Suppose now that Γ has a vertex v of degree one (such a vertex is unique since Γ is a core graph). Then there exists a unique vertex v of Γ with the following properties: (a) There is a unique Γ-geodesic path [v, v ] from v to v and every vertex of this geodesic, other than v and v, has degree two. (b) The vertex v has degree at least three. Let Γ be the graph obtained by removing from Γ all the edges of [v, v ] and all the vertices of [v, v ] except for v. Then Γ is called the type of Γ and denoted Type(Γ). Finally, if Γ consists of a single vertex, we set Type(Γ) = Γ. 7 / 19

8 Lemma Let Γ be a folded core graph (with respect to one of its vertices). Let v and u be two vertices of Γ and let q be a reduced path in Γ from v to u with label g F (X ). Let H = L(Γ, v) and K = L(Γ, u). Then H = gkg 1. 8 / 19

9 If we consider several subgroups (H, K) we often will denote distinguished vertices of their Stallings graphs by 1 H, 1 K. Lemma Let H F (X ) and let Γ = Γ(H). Let g F (X ) be a nontrivial freely reduced word in X. Let g = yz where z is the maximal terminal segment of the word g such that there is a path with label z 1 in Γ starting at 1 H (such a path is unique since Γ is folded). Denote the end-vertex of this path by u. Let be the graph obtained from Γ as follows. We attach to Γ at u the segment consisting of y edges with label y 1, as read from u. Let u be the other end of this segment. Put = Core(, u ). Then (, u ) = (Γ(K), 1 K ) where K = ghg 1. 9 / 19

10 Proposition (Conjugate subgroups) Let H and K be subgroups of F (X ). Then H is conjugate to K in F (X ) if and only if the graphs Type(Γ(H)) and Type(Γ(K)) are isomorphic as X -digraphs. Proof. Suppose that Type(Γ(H)) = Type(Γ(K)) = Γ. Let v be a vertex of Γ. The subgroup L(Γ, v) is conjugate to both H and K, so that H is conjugate to K. Suppose now that K is conjugate to H, that is K = ghg 1 for some g F (X ). Then Type(Γ(H)) = Type(Γ(K)), as required. 10 / 19

11 Morphism of Labelled Graphs Let Γ and be reduced A-labeled graphs as above. A mapping φ from the vertex set of Γ to the vertex set of (we write φ: Γ ) is a morphism of reduced (A-)labeled graphs if it maps the designated vertex of Γ to the designated vertex of and if, for each a A, whenever Γ has an a-labeled edge e from vertex u to vertex v, then has an a-labeled edge f from vertex φ(u) to vertex φ(v). The edge f is uniquely defined since is reduced. We then extend the domain and range of φ to the edge sets of the two graphs, by letting φ(e) = f. Note that such a morphism of reduced A-labeled graphs is necessarily locally injective (an immersion), in the following sense: for each vertex v of Γ, distinct edges starting (resp. ending) at v have distinct images. 11 / 19

12 Morphism of Labelled Graphs Further we say that the morphism φ: Γ is a cover if it is locally bijective, that is, if the following holds: for each vertex v of Γ, each edge of starting (resp. ending) at φ(v) is the image under φ of an edge of Γ starting (resp. ending) at v. 12 / 19

13 Homework Problem 2 Show that a subgroup H of F (A) has finite index if and only if this natural morphism from Γ A (H) to the bouquet of A circles is a cover, and in that case, the index of H in F (A) is the number of vertices of Γ A (H). 13 / 19

14 Homework Problem 3 Prove Lemma Let H, K be subgroups of a free group F with basis A. Then H K if and only if there exists a morphism of labeled graphs φ H,K from Γ A (H) to Γ A (K). If it exists, this morphism is unique. 14 / 19

15 If H and K are finitely generated subgroups of F (A), then Γ A (H K) can be easily constructed from Γ A (H) and Γ A (K): one first considers the A-labeled graph whose vertices are pairs (u, v) consisting of a vertex u of Γ A (H) and a vertex v of Γ A (K), with an a-labeled edge from (u, v) to (u, v ) if and only if there are a-labeled edges from u to u in Γ A (H) and from v to v in Γ A (K). Finally, one considers the connected component of vertex (1, 1) in this product, and we repeatedly remove the vertices of valence 1, other than the distinguished vertex (1, 1) itself, to make it a reduced A-labeled graph. 15 / 19

16 16 / 19

17 Definition (Product-graph) Let Γ and be X -digraphs. We define the product-graph Γ as follows. The vertex set of Γ is the set V Γ V. For a pair of vertices (v, u), (v, u ) V (Γ ) (so that v, v V Γ and u, u V and a letter x X we introduce an edge labeled x with origin (v, u) and terminus (v, u ) provided there is an edge, labeled x, from v to v in Γ and there is an edge, labeled x, from u to u in. Thus Γ is an X -digraph. In this situation we will sometimes denote a vertex (v, u) of Γ by v u. 17 / 19

18 Lemma Let Γ and be folded X -digraphs. Let H = L(Γ, v) and K = L(, u) for some vertices v V Γ and u V. Let y = (v, u) V (Γ ). Then Γ is folded and L(Γ, y) = H K. 18 / 19

19 Corollary There exists an algorithm which, given finitely many freely reduced words h 1,..., h s, k 1,..., k m F (X ), finds the rank and a Nielsen-reduced free basis of the subgroup h 1,..., h s k 1,..., k m of F (X ). In particular, this algorithm determines whether or not h 1,..., h s k 1,..., k m = 1. Corollary (Howson Property) (Howson,54) The intersection of any two finitely generated subgroups of F (X ) is again finitely generated. 19 / 19

Lecture 11 COVERING SPACES

Lecture 11 COVERING SPACES A covering space (or covering) is not a space, but a mapping of spaces (usually manifolds) which, locally, is a homeomorphism, but globally may be quite complicated. The simplest