G i. h i f i G h f j. G j

Transcription

1 2. TWO BASIC CONSTRUCTIONS Free Products with amalgamation. Let {G i i I} be a family of gps, A a gp and α i : A G i a monomorphism, i I. A gp G is the free product of the G i with A amalgamated (via the α i ) if homomorphisms f i : G i G such that f i α i = f j α j i, j I, and if h i : G i H are homomorphisms with h i α i = h j α j for all i, j I, then there is a unique homomorphism h : G H such that h f i = h i for all i I. Call h an extension of the homoms. h i. A G i α i h i f i G h f j α j h j G j Note. Follows: G is generated by i I f i (G i ). Let G 0 be the subgroup of G generated by this set. Take H = G 0 and h i = f i, viewed as a mapping to G 0, in the defn. an extn to a homomorphism f : G G 0. Let e : G 0 G be the inclusion map. Then e f and id G : G G are both extns of the maps f i : G i G. Since the extn is unique, e f = id G, so e is onto, hence G = G 0. Uniqueness. Let f i : G i G be homomorphisms such that f i α i = f j α j for all i, j I, and if h i : G i H are homomorphisms with h i α i = h j α j for all i, j I, then there is a unique homomorphism h : G H such that h f i = h i for all i I. In the defn, take h i = f i, to obtain a homomorphism f : G G such that f f i = f i for all i. Interchanging G and G, get f : G G such that f f i = f i for all i. Take H = G, h i = f i in the defn; both f f and id G : G G are extensions of the f i : G i G. Since the extension is unique, f f = id G. Interchanging G and G, f f = id G, so f and f are inverse isomorphisms. Existence. let X i R i be a presn of G i, with X i X j = /0 for i j. Let Y be a set of generators for A and, for each y Y, i I let a i,y be a word in X ±1 i representing α i (y). Let S = { a i,y a 1 j,y y Y, i, j I, i j}. Let G be the group with presentation i I X i i I R i S. Then G is the free product with amalgamation. There is an obvious mapping X i G, 1 H

2 which (by defn of a presentation) induces a homomorphism f i : G i G. The existence and uniqueness of the mapping h in the definition follows by defn of a presentation. If G is the free product of {G i i I} with A amalgamated, write G = A G i. In case of two groups {G 1,G 2 }, write G = G 1 A G 2. Example. let G 1 = x, G 2 = y, A = a inf. cyclic. α 1 : a x 2, α 2 : a y 3. G = G 1 A G 2 has presentation x,y x 2 = y 3, after Tietze transformations. Structure of free prods with amalgamation. Let G = A G i via α i : A G i, i I. To represent elts of G by certain words, identify A with α i (A) and assume G i G j = A for i j. Define a word to be a finite sequence w = (g 1,...,g k ), where k 1 and g 1 G i1,...,g k G ik for some i 1,...,i k I. Elt. of G represented by w is f i1 (g 1 )... f ik (g k ). Since i I f i (G i ) generates G and the f i are homoms, every element of G is repd by some word. Definition. Let w = (g 1,...,g k ) be a word with g j G i j, 1 j k. Then w is reduced if (1) i j i j+1 for 1 j k 1; (2) g j A for 1 j k, unless k = 1. If (1) fails, can replace w by (g 1,...,g j g j+1,...,g k ), representing the same element of G. If (2) fails, can also replace w by a shorter word. Hence every elt of G is represented by a reduced word (take a word of shortest length representing it). For every i I, choose a transversal T i for the cosets {Ax x G i }, with 1 T i. Let g G be represented by the reduced word (g 1,...,g k ) with g j G i j. Can write: g k = a k r k (a k A, r k T ik ) g k 1 a k = a k 1 r k 1 (a k 1 A, r k 1 T ik 1 ). g 1 a 2 = a 1 r 1 (a 1 A, r 1 T i1 ).. 2

3 Then g 1 = a 1 r 1 a 1 2, g 2 = a 2 r 2 a 1 3,...,g k = a k r k and (a 1,r 1,r 2,...,r k ) is a word representing g. Definition. A normal word is a word (a,r 1,...,r k ) where a A, k 0, r j T i j \ {1} for some i j I (1 j k), and i j i j+1 for 1 j k 1. Thus any element of G is represented by a normal word. (a A is represented by the normal word (a).) 3 Theorem 2.1 (Normal Form Theorem). Any element of G is represented by a unique normal word. Proof. Need to show uniqueness. Let W = set of normal words. Will define an action of G on W, equivalently, a homom. h : G S(W) (symmetric group on W). By defn of G, suffices to define, for i I, a homom. h i : G i S(W) with h i α i = h j α j for all i, j I. So with the identifications made, need homomorphisms h i which agree on A. Let W i = {(1,r 1,...,r k ) W r 1 T i } (i I). Define θ i : G i W i W as follows. If g G i, write g = ar, where r T i, a A. Then θ i (g,(1,r 1,...,r k )) = θ i is bijective (exercise). { (a,r,r 1,...,r k ) if r 1 (i.e. g A) (a,r 1,...,r k ) if r = 1. Now G i acts on G i W i by left mult. on the first co-ordinate, giving a homomorphism λ i : G i S(G i W i ), where λ i (g)(x,w) = (gx,w). Set h i (g) = θ i λ i (g)θ 1 i. Since λ i is a homomorphism, so is h i. If a A, h i (a)(a,r 1,...,r k ) = (aa,r 1,...,r k ) and RHS is independent of i. Thus the h i define a homomorphism h : G S(W). Let g G be represented by the normal word w = (a,r 1,...,r k ). Then h(g)(1) = w (induction on k) so w is uniquely determined by g. Corollary 2.2. (1) The homomorphisms f i are injective; (2) no reduced word of length greater than 1 represents the identity element of G. Proof. (1) If f i (g) = 1, write g = ar with a A, r T i. Then the normal word (a,r) (or (a), if r = 1) represents 1, so a = r = 1, ie g = 1.

4 (2) If a reduced word has length k > 1, procedure above gives a normal word of length k + 1 representing same elt of G, which cannot be 1 by Theorem 2.1. By Cor. 2.2(1), the f i can be suppressed. Important special case: A = {1}. G is called the free product of the family {G i i I}, written G = i I G i (or G = G 1 G 2 for two groups). A reduced word is a word (g 1,...,g k ) such that g j G i j, g j 1 unless k = 1, and i j i j+1 for 1 j k 1. Normal Form Theorem simplifies: every element of G is represented by a unique reduced word. Also, the G i embed in G. Defining property simplifies to: given any collection of homomorphisms h i : G i H, there is a unique extension to a homomorphism h : G H. Presn used to show existence simplifies. Let X i R i be a presentation of G i (via some mapping which will be suppressed), with X i X j = /0 for i j. Then i I G i has presentation i I X i i I R i. [This is obtained from the presentation above by taking the empty presentation of the trivial group, with no generators and no relations.] Examples. (1) x,y x 2 = 1, y 2 = 1 is a presentation of C 2 C 2, the infinite dihedral group. Has D n as a homomorphic image. (2) C 2 C 3 has presentation x,y x 2 = 1, y 3 = 1. This group is called the modular group, and is isomorphic to PSL 2 (Z). (3) F free with basis {x i i I}; then F = i I x i, a free product of infinite cyclic groups. (4) Trefoil knot K; group of the knot, π 1 (R 3 K) has presn a,b aba = bab 4 Tietze transf. x,y x 2 = y 3 (previous example). HNN-extensions. Suppose B, C are subgroups of a group A and γ : B C is an isomorphism.

5 Let G have presentation {t} X R 1 R 2, where X = { x g g A }, t X { } R 1 = x g x h xgh 1 g,h A { R 2 = tx b t 1 x 1 γ(b) }. b B a homom. f : A G given by a x a ; will show f is injective. Definition. The group G is called an HNN-extension with base A, associated pair of subgroups B, C and stable letter t. Presn is often abbreviated to t,a rel(a),tbt 1 = γ(b) or just t,a rel(a),tbt 1 = C. More generally, Let Y S be a presentation of A, let { b j j J } be a set of words in Y ±1 representing a set of generators for B, and let c j be a word representing γ(b j ) [more accurately, γ(the generator represented by b j )]. Then {t} Y S { tb j t 1 = c j j J } is also a presentation of G. (Exercise.) Example. (Baumslag-Solitar) x,y xy 2 x 1 = y 3 is an HNN-extension, base an infinite cyclic group A = y, stable letter x, assoc.pair y 2, y 3. (Famous example of a non-hopfian group.) In general, G is generated by f (A) {t}, so every element of G is represented in an obvious way by a word (g 0,t e 1,g 1,t e 2,g 2,...,t e k,g k ) where k 0, e i = ±1 and g i A for 0 i k. Definition. Such a word is reduced if it has no part of the form t,b,t 1 with b B or t 1,c,t with c C. Any element of G is represented by a reduced word (take a word of minimal length representing it). Choose a transversal T B for the cosets {Bg g A} and a transversal T C for the cosets {Cg g A}, with 1 T B, T C. Definition. A normal word is a reduced word (g 0,t e 1,r 1,t e 2,r 2,...,t e k,r k ), where g 0 A, r i T B if e i = 1 and r i T C if e i = 1. Suppose (g 0,t e 1,g 1,t e 2,g 2,...,t e k,g k ) is a reduced word with k 1. If e k = 1, write g k = br with b B, r T B. Then (g 0,t e 1,g 1,t e 2,g 2,...,g k 1,te k,r) 5

6 where g k 1 = g k 1γ(b), represents the same element of G. If e k = 1, write g k = cr with c C and r T C. Then (g 0,t e 1,g 1,t e 2,g 2,...,g k 1,te k,r) where g k 1 = g k 1γ 1 (c), represents the same element of G. Repetition leads to a normal word representing the same element of G. Theorem 2.3. (Normal Form Theorem.) Every element of G is represented by a unique normal word. Proof. Omitted. Corollary 2.4. (1) The homomorphism f : A G is injective; (2) (Britton s Lemma) no reduced word (g 0,t e 1,g 1,t e 2,g 2,...,t e k,g k ) with k > 0 represents the identity element of G. More generally, given A and a family γ i : B i C i (i I) of isomorphisms, where B i, C i are subgroups of A, we can form the HNN-extension with presentation (in abbreviated form) t i (i I), G rel(g), t i B i ti 1 = γ i (B i ) (i I). There are generalisations of the Normal Form Theorem and its corollary. An application. Theorem 2.5. Any countable group can be embedded in a 2-generator group. Proof. Let G = {g 0,g 1,g 2,...}, where g 0 = 1. Let F be free with basis {a,b}. b n ab n (n 0) is a free subgp of F with the given elts as basis. Similarly a n ba n (n 0) is a free subgp. (See Exercise 2, Sheet 1.) Claim: in G F, g n a n ba n (n 0) is free with basis the given elts. For the homoms. G F, g 1 and id F : F F have an extn to a homom. f : G F F. If w is repd by a non-empty reduced word in {g n a n ba n (n 0)} ±1 of positive length, then f (w) is repd by a corresponding reduced word in {a n ba n (n 0)} ±1, so f (w) 1, hence w 1. Claim follows (see Exercise 1, Sheet 1). Can form HNN-extn H = t,g F rel(g F),t(b n ab n )t 1 = g n a n ba n (n 0). Then G G F H, g n t,a,b, and (n = 0) tat 1 = b, so H = t,a. 6