Lecture2, Part2 Slides

Embeddings of free groups Free product Basic properties of subgroup graph Intersections of Subgroups

Free groups, Lecture 2, Part 2

Olga Kharlampovich

NYC, Sep 2

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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.

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Proof To prove the result it suffices to find a free subgroup of countable rank in a free group of rank 2. Let F2 be a free group with a basis {a, b}. Denote

n −n xn = b ab (n = 0, 1, 2,... ) and let S = {x0, x1, x2,... }. We claim that S freely generates the subgroup hSi in F2. Indeed, let

w = x1 x2 ... xn i1 i2 in be a reduced non-empty word in S±1. Then w can also be viewed as a word in {a, b}.

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Indeed, w = bi1 a1 b−i1 bi2 a2 b−i2 ... bin an b−in .

Since w is a reduced word in S, we have that either ij 6= ij+1, or ij = ij+1 and j + j+1 6= 0, for each j = 1, 2,..., n − 1. In either case, any reduction of w (as a word in {a, b}) does not affect aj and aj+1 in the subword

bij aj b−ij bij+1 aj+1 b−ij+1 i.e., the literals aj and aj+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 6= 1 in F2. Clearly, hSi is a free group of countable rank.

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Given a family of groups {Gi |i ∈ I } we may assume that the Gi are mutually disjoint sets. Let X = ∪i∈I Gi and let {1} be a one-element set disjoint from X . A word in X is a finite string (a1, a2,..., an, 1, 1,...) = a1a2 ... an, where ai ∈ X . A word is reduced provided 1) no ai is the udentity in its group Gj , 2) for all i, ai and ai+1 are not in the same Gj , 3) ak = 1 implies ai = 1 for i ≥ k. Q∗ Let i∈I Gi (or G1 ∗ G2 ∗ ... ∗ Gn) be the set of all reduced words Q∗ on X . i∈I Gi forms a group, called the free product of the family {Gi |i ∈ I } under the operation “Juxtaposition+Cancellation+Contraction”. For example, if −1 ai bi ∈ Gi then (a1a2a3)(a3 b2b1b3) = a1c2b1b3, where c2 = a2b2 ∈ G2.

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Homework problem 1 Prove that the reduced form of a given Q∗ element is unique. Prove that the map ik : Gk → i∈I Gi given by e → 1 and a → a = (a, 1, 1,...) is a monomorphism. We identify Q∗ Q∗ Gi with its image in i∈I Gi . Prove that i∈I Gi is a coproduct in the category of groups.

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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 0 of Γ with the following properties: (a) There is a unique Γ-geodesic path [v, v 0] from v to v 0 and every vertex of this geodesic, other than v and v 0, has degree two. (b) The vertex v 0 has degree at least three. Let Γ0 be the graph obtained by removing from Γ all the edges of [v, v 0] and all the vertices of [v, v 0] except for v 0. Then Γ0 is called the type of Γ and denoted Type(Γ). Finally, if Γ consists of a single vertex, we set Type(Γ) = Γ.

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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.

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If we consider several subgroups (H, K) we often will denote distinguished vertices of their Stallings graphs by 1H , 1K . 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 −1 there is a path with label z in Γ starting at 1H (such a path is unique since Γ is folded). Denote the end-vertex of this path by u. Let ∆0 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 u0 be the other end of this segment. Put ∆00 = Core(∆0, u0). 00 0 −1 Then (∆ , u ) = (Γ(K), 1K ) where K = gHg .

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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.

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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.

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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.

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Homework Problem 2 Show that a subgroup H of F (A) has finite index if and only if this natural morphism from ΓA(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 ΓA(H).

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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 φH,K from ΓA(H) to ΓA(K). If it exists, this morphism is unique.

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If H and K are finitely generated subgroups of F (A), then ΓA(H ∩ K) can be easily constructed from ΓA(H) and ΓA(K): one first considers the A-labeled graph whose vertices are pairs (u, v) consisting of a vertex u of ΓA(H) and a vertex v of ΓA(K), with an a-labeled edge from (u, v) to (u0, v 0) if and only if there are 0 0 a-labeled edges from u to u in ΓA(H) and from v to v in ΓA(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.

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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 0, u0) ∈ V (Γ × ∆) (so that v, v 0 ∈ V Γ and u, u0 ∈ V ∆ and a letter x ∈ X we introduce an edge labeled x with origin (v, u) and terminus (v 0, u0) provided there is an edge, labeled x, from v to v 0 in Γ and there is an edge, labeled x, from u to u0 in ∆. Thus Γ × ∆ is an X -digraph. In this situation we will sometimes denote a vertex (v, u) of Γ × ∆ by v × u.

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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.

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Corollary There exists an algorithm which, given finitely many freely reduced words h1,..., hs , k1,..., km ∈ F (X ), finds the rank and a Nielsen-reduced free basis of the subgroup hh1,..., hs i ∩ hk1,..., kmi of F (X ). In particular, this algorithm determines whether or not hh1,..., hs i ∩ hk1,..., kmi = 1.

Corollary (Howson Property) (Howson,54) The intersection of any two finitely generated subgroups of F (X ) is again finitely generated.

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