Free groups, Lecture 2, part 1

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1 Free groups, Lecture 2, prt 1 Olg Khrlmpovich NYC, Sep. 2 1 / 22

2 Theorem Every sugroup H F of free group F is free. Given finite numer of genertors of H we cn compute its sis. 2 / 22

3 Schreir s grph The grph of right cosets of H, denoted y Γ 0 (H), is clled the Scheir s grph of H. V (Γ 0 ) = G/H = {Hg g set of right coset representtives} Theorem Hg, s S, e E + e = (Hg, Hgs) λ(e) = s If F = F (S), H F, then Γ(H) is isomorphic to the miniml sugrph of Γ 0 (H), contining ll loops t v 0 = H [we cll tt sugrph Core(Γ 0 (H))]. 3 / 22

4 Exmple Γ(H) for H = 2, 1 2, 2, H H 1 H H 2 H H 2 4 / 22

5 Algorithm for finding sis for H Let H F (, ). Ex.: H = 2, 1 2, 2, Construct Γ(H). 2 Let T e mximl sutree in Γ(H) (not unique). 3 If v V (Γ), denote y g v the unique geodesic from v 0 to v in T. 4 Let e E + (Γ \ T ) e positively oriented edge tht is not on the tree T. Set S e = λ(g α(e) ) λ(e) λ(g 1 τ(e) ) S e H H H 1 H H 2 H H 2 5 / 22

6 For exmple, in H 0, S e = 1 }{{} λ(g α(e) ) λ(e) {}}{ 1 1 }{{} λ(g 1 τ(e) ) α(e) λ(e) S e τ(e) λ(g α(e) ) λ(g τ(e) ) v 0 6 / 22

7 Remrk If e E + (Γ \ T ), we define S e similrly, ut then S e = 1 (the empty word). Indeed, y uniqueness of geodesic, g 1 τ(e) = e 1 g 1 α(e) Clim The set {S e e E + (Γ \ T )} forms free sis of H. 7 / 22

8 Proof Let w H e ritrry. Since Γ(H) ccepts w. reduced loop t v 0, e 1 e 2 e n such tht But this loop cn e written s n i=1 ( ) g α(ei ) e i g 1 τ(e i ) λ(e 1 e 2 e n ) = w = g α(e1 ) e 1 g 1 = g α(e1 ) τ(e 1 ) g α(e 2 ) e 2 g 1 τ(e 2 ) g α(e n) e n g 1 τ(e = n) [ n ] n (e i ) g 1 τ(e = n) (e i ) i=1 ecuse g α(e1 ) = g τ(en) re trivil. Thus, n S ei = w. i=1 i=1 However, s S ei generting. = 1 if e i T, the set {S e e E + (Γ \ T )} is 8 / 22

9 On the other hnd, for the product S e1 S e2, where e 1 e 1 2, e 1, e 2 E + (Γ \ T ), the corresponding pth is P = g α(e1 ) e 1 g 1 τ(e 1 ) }{{} S e1 1 g α(e2) e 2 gτ(e 2 ) } {{ } S e2 Every edge in g α( ), g τ( ) elongs to the tree, while e 1 e2 1 do not elong to the tree. Thus, e 1 &e 2 cnnot e cnceled in P S e1 S e2 1 Hence, {S e e E + (Γ \ T )} is sis of H nd H = S e e E + (Γ \ T ) 9 / 22

10 Corollry Every sugroup of free group is free, rk(h) = E V / 22

11 Exmple Consider F = F (, ) H = 1 1, 2 1, 1 Construct Γ(H) nd fold it. v 0 e 2 e 1 The red edges indicte the mximl sutree chosen. Thus, E T = {e 1, e 2 } nd S e2 = () 1 1 Then, the rnk of H is 2. S e1 = () 1 11 / 22

12 Definition The grph Γ(H) is clled complete (or S-regulr) if v Γ(H) nd for s S (the lphet), there exists n edge lelled y s nd s 1 from v. Theorem Suppose F, H re finitely generted, then [F : H] < Γ(H) is complete in tht cse, [F : H] = Γ(H) = V (Γ). 12 / 22

13 Proof. Suppose Γ(H) is complete. Recll tht Γ(H) Γ 0 (H), the grph of right cosets of H. Γ(H) is complete Γ 0 (H) = Γ(H) Indeed, for ny coset Hg, g is reduced word in F (S). By completeness, we cn red g s pth (not necessrily loop!) in Γ(H). The lel of this pth is representtive of Hg, which mens tht Γ 0 (H) Γ(H) Γ 0 (H) = Γ(H). H finitely generted Γ(H) < Γ 0 (H) < [F : H] <. Conversely, if [F : H] <, Γ 0 (H) = Core(Γ 0 (H)) ecuse every reduced word in F is eginning of the lel of loop t H. Therefore Γ 0 (H) = (Γ(H)), nd Γ(H) is complete. 13 / 22

14 Exmple F = F (, ) H = 1, 2 2, 2, 1, 1 2, 2 3, 1 1 Construct Γ(H). 2 Find [F : H]. 3 The numer of elements in the sis for H is the rnk of H, rk (H)=E-V+1. 4 Is w = 3 1 H? 14 / 22

15 Proof[Solution] Γ(H) fter dding the first 3 genertors: v 0 As 2 3 H, we need to indentufy the yellow vertices, which implies the the green vertices need to e identified too. This produces the following (finl) result: 15 / 22

16 v 0 Thus, the rnk of H is the numer of fundmentl simple loops = / 22

17 To find the sis let us choose mximl sutree (lue edges): v 0 There re 6 non tree edges (green) nd the sis corresponding to them is {[ 1 ] 1, [] 1, [] 1 1, [] 1, 1 [], 1 []} There is vertex of degree 2 (yellow), so Γ(H) is not complete, i.e. 17 / 22

18 Spnning trees nd free ses We will use nottion G = L(Γ, v) mening tht the group G is lnguge ccepted y the utomton Γ(G). We sy tht Γ is folded if it hs OR property. When we sy tht Γ is connected we men tht it is connected s non-directed grph. Definition (Core grphs) Let Γ e n X -digrph nd let v e vertex of Γ. Then the core of Γ t v is defined s: Core(Γ, v) = {p where p is reduced pth in Γ from v to v } It is esy to see tht Core(Γ, v) is connected sugrph of Γ contining v. If Core(Γ, v) = Γ we sy tht Γ is core grph with respect to v. 18 / 22

19 Definition (Nielsen set) Let S e set of nontrivil elements of the free group F (X ) such tht S S 1 =. We sy tht S is Nielsen reduced with respect to the free sis X if the following conditions hold: (1) If u, v S S 1 nd u v 1 then u v X u X nd u v X v X. (2) If u, v, w S S 1 nd u w 1, v w 1 then u w v X > u X + v X w X. Condition (1) mens tht no more thn hlf of u nd no more thn hlf of v freely cncels in the product u v. Condition (2) mens tht t lest one letter of w survives fter ll free cncelltions in the product u w v. 19 / 22

20 Definition (Geodesic tree) Let Γ e connected grph with se-vertex v. A sutree tree T in Γ is sid to e geodesic reltive to v if v T nd for ny vertex u of T the pth [v, u] T is geodesic in Γ, tht is pth of the smllest possile length in Γ from v to u. It is esy to see tht geodesic spnning trees lwys exist: Lemm Let Γ e grph (whether finite or infinite) with se-vertex v. Then there exists geodesic reltive to v spnning tree T for Γ. 20 / 22

21 The first spnning tree is not geodesic H H 1 H H 2 H H 2 H H 1 H H 2 H 2 21 / 22

22 Proposition (Nielsen sis) Let Γ e folded X -digrph which is core grph with respect to vertex v of Γ. Let H = L(Γ, v) F (X ) nd let T e spnning tree in Γ which is geodesic with respect to v. Then the set Y T is Nielsen-reduced free sis of the sugroup H. 22 / 22