Typing with Weird Keyboards Notes

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1 Typing with Weird Keyords Notes Ykov Berchenko-Kogn August 25, 2012 Astrct Consider lnguge with n lphet consisting of just four letters,,,, nd. There is spelling rule tht sys tht whenever you see n next to n, you cross those two letters out. Similrly, if you see next to, you cross those two letters out s well. For exmple, if you see, you d cross out to get, nd then youd cross out to get. For some inexplicle reson, memers of the Mfi use weird keyords when typing in this lnguge. Insted of hving keys leled,,, nd, they hve keys leled,,, nd. A memer of the Mfi cn, for exmple, type the word y pressing the key followed y the key to get, nd then crossing out the. One dy, Ruthi gets ccused of eing in the Mfi. However, she recently wrote newspper rticle contining the word. Ruthi clims tht noody in the Mfi cn type, nd thus she must e innocent. How cn we verify Ruthi s clim? We ll see how we cn esily figure out wht words given keyord cn produce using directed grphs (dots connected y rrows). We ll see how to use these grphs to nswer other questions. For exmple, we ll see how to tell when one keyord cn type every word nother keyord cn, nd we ll see how, given two keyords, we cn mke keyord tht cn type only the words tht oth of the given keyords cn type. We ll put these concepts into their lrger group theoretic context, so it will e helpful (ut not required) to e fmilir with the sic group theory definitions. 1 Free Groups: Typing with norml keyords We strt with n lphet, for exmple {,, c}. We write words using the letters in the lphet either right side up or upside-down. Exmple 1.1. In this lphet, we cn write the word c. There is spelling rule tht whenever we see letter right side up next to the sme letter upside-down, we cn erse oth of them. c Exmple 1.2. When we pply the spelling rule, c c ecomes c, which then ecomes Hnnh wnts to write song in this lnguge, nd she hs keyord tht looks like this: c. Figure 1: Hnnh s keyord. c c 1

2 Exercise 1.3. Hnnh ccidentlly typed. She wnts to erse it, ut there s no ckspce utton on her keyord! Wht should she type next to erse it? Exercise 1.4. Hnnh ccidentlly typed cc when she relly wnted to type c. Wht should she type next to fix her mistke? Alfonso s keyord looks just like Hnnh s, ut it hs n extr smiley fce key: c Figure 2: Alfonso s keyord. c c When Alfonso holds down the smiley fce key, pushing nother key cuses tht letter to pper t the end of the word nd t the eginning of the word flipped over. Exmple 1.5. Alfonso cn type y pressing nd then holding down the smiley key nd pressing Exercise 1.6. If Alfonso types nd then types the first few letters of the word while holding the smiley key, wht hppens? Rememer to pply the spelling rule. Definition 1.7. The set of ll finite-length words you cn get fter pplying the spelling rule is clled the free group over the lphet X. We will let F denote the free group, nd sometimes y F (X) if we wnt to mke it cler wht the lphet is. In our exmples ove, X = {,, c}. The spelling rule is usully clled free cncelltion, nd most people write ā or 1 insted of, ecuse writing upside-down letters is hrd. If you know group theory, you might e wondering wht the group opertion is. The group opertion here is conctention, tht is, writing one word fter the other (nd then pplying free cncelltion if necessry). If you did Exercise 1.3, you should e le to figure out wht the inverses re. 2 Finitely Generted Sugroups: Typing with silly keyords Susn doesn t like typing with norml keyords like Hnnh s. Insted, she hs keyords tht look like something like this:. Figure 3: Susn s keyord. The first row of the keyord is collection of words, nd the second row is the words in the first row upside-down. 2

3 Wht words cn Susn type? There re infinitely mny, ut she proly cn t type every word, so we d like n esy wy to test whether or not given word, such s, cn e produced with this keyord. One wy to do this would e to systemticlly mke list of the words tht cn e typed y pressing one key, then two keys, then three keys, nd so forth. If we find on this list, then we re done. Eventully, the words on the list will ecome very long, much longer thn. If we don t see y tht point, we cn e resonly sure tht it cn t e typed on this keyord. The prolem is tht, ecuse of free cncelltion, even though we might press lot of keys, the resulting word might still e very short. Exercise 2.1. In fct, it is possile to type with this keyord. Find wy to do it. There s much etter wy of testing whether or not word cn e typed on the keyord. We first mke leled directed grph out of loops leled with the words in the first row, like this: Figure 4: Mking the grph corresponding to Susn s keyord, step 1: loops. The left loop corresponds to, nd the right loop corresponds to. The vertex where ll of the loops strt nd end is drwn lrger. Notice tht we represent n upside-down letter y reversing the direction of the edge. Pressing key on the keyord corresponds to going round one of these loops, strting nd ending t the ig vertex. (Pressing the keys in the ottom row corresponds to going round the loop in the other direction.) Thus, word cn e typed on this keyord if nd only if you cn red it off when going round the grph strting nd ending t the ig vertex. However, if we try to use this grph to figure out whether or not given word cn e typed on this keyord, we will run into some prolems. Exmple 2.2. There re two edges leled going from the ig vertex. If the word we re testing strts with, we don t know which edge to tke. We d hve to try oth possiilities. If the word is very long, we might end up with too mny possiilities to go through. Exmple 2.3. You cn type using this keyord ecuse it s wht you get fter pplying free cncelltion to, which you cn red off y going round the grph. However, if we were sked to test whether or not could e typed on this keyord, we wouldn t hve ny wy of knowing to insert in the middle, except y guessing. We cn fix these prolems y folding the grph, like this: 3

4 Figure 5: Mking the grph corresponding to Susn s keyord, step 2: folding. Whenever you see two edges with the sme lel either oth going from the sme vertex or oth going to the sme vertex, you fold them together. Once you cn t fold the grph ny further, the grph is clled folded. Now, we cn t run into the prolems we hd efore, ecuse given vertex nd letter there s t most one wy to go, nd there re no pths where letter is followed y its inverse. Sometimes, there will e more thn one wy to fold grph, ut, t the end, the resulting folded grph will lwys e the sme. Exercise 2.4. For ech of the following words, check if Susn s keyord cn type them using the corresponding folded grph Exercise 2.5. Come up with few other words tht this keyord cn nd cn t type. Exmple 2.6. Here s Kevin s keyord: Figure 6: Kevin s keyord. We construct the corresponding folded grph elow. Note tht we do couple folding steps t time. The edges mrked in red re the ones out to e folded, nd the resulting edges re thick. 4

5 Figure 7: Folding the grph corresponding to Kevin s keyord. Definition 2.7. Let H e the set of words tht cn e typed on keyord with set S of words in the first row nd the corresponding upside-down words in the second row. We cll H the sugroup of F (X) generted y S, nd we write H = S. We cll S set of genertors of H. (Note tht sugroup hs mny possile generting sets.) If S is finite, then H is finitely generted. (In this clss we ll lmost entirely e deling with finitely generting sugroups.) Exmple 2.8. Kevin s keyord corresponds to the sugroup ā, ā, āā. Definition 2.9. Given set of letters X, n X-digrph is directed grph whose edges re leled with elements of X. Edges from vertex ck to itself re llowed, s re multiple edges etween two vertices. The endpoints of n edge re its origin nd terminus. Mny of our X-digrphs will hve specil se vertex which is drwn lrger. (Note tht digrph stnds for directed grph.) Exmple All of the grphs in Figures 4, 5, nd 7 re X-digrphs with X = {, }. 5

6 Definition Given n X-digrph Γ with se vertex v, the lnguge of the X-digrph with respect to the se vertex v, denoted L(Γ, v), is the set of lels of the reduced pths from v to v. Another wy to descrie our lgorithm ove is to sy tht it tkes finitely generted sugroup H nd outputs folded grph whose lnguge is precisely H. Definition Given sugroup H of free group F (X), the corresponding folded grph is clled the Stllings sugroup X-digrph of H nd is denoted Γ(H). Stllings sugroup grphs nd the folding process now known s Stllings folding were developed y geometric group theorist nd topologist John Stllings ( ) in 1983 pper. Stllings grphs hd pplictions fr eyond mking much fster lgorithm for testing whether or not word is in sugroup, nd they drmticlly chnged the wy in which sugroups of free groups re studied. Exmple If H = ā, ā, āā, then Γ(H) is the lst grph ove in Figure 7. Exercise Construct the Stllings sugroup digrph for the following sugroups. 1., 2. ā, ā 3., 4.,, ā 5. c, c, cc Exercise Go ck to the question in the lur. Is it possile for Ruthi to e in the Mfi? Exercise Pick your fvorite words over your fvorite lphet, nd construct the corresponding Stllings digrph. Exercise If the Stllings digrph of sugroup hs vertex tht hs just one edge (like in the Stllings grph of Kevin s keyord), wht cn you sy out the genertors of the sugroup? The strred exercises elow re preview of upcoming topics, ut re not necessrily hrder. Exercise 2.18 (*). There re lots of X-digrphs, ut only some of them re the Stllings X-digrph of some sugroup. For instnce, Stllings X-digrphs re lwys folded. First find folded X-digrph tht is not Stllings X-digrph of ny sugroup, nd then try to come up with property tht determines whether or not n X-digrph is Stllings X-digrph of some sugroup. Exercise 2.19 (*). If someone gives you Stllings X-digrph, how cn you find keyord tht it corresponds to? In other words, given folded digrph, how cn you find the genertors of the corresponding sugroup? (Note tht there re lots of possile keyords/generting sets for given sugroup, nd this question sks you how to find just one of them). Exercise 2.20 (*). Just y looking t the corresponding grphs, given two keyords how cn you tell if one keyord cn type everything the other cn? Exercise 2.21 (*). Given two keyords, how cn you find grph corresponding to the set of words tht cn e typed y oth of them? In other words, given two sugroups, how cn you find the grph corresponding to their intersection? (Hint: Use the grphs corresponding to the two sugroups.) Exercise 2.22 (*). Roxn hs keyord. One dy, Tom tkes ech key nd dds to the end of the word on the key nd to the eginning. Wht sorts of things cn hppen to the corresponding grph? Exercise 2.23 (*). Let X = {, }, nd let H e the set of words in F (X) where nd ā pper the sme numer of times, nd nd pper the sme numer of times. Cn you come up with corresponding X-digrph for this sugroup? Tht is, find folded X-digrph Γ such tht the lel of every pth from the se vertex to the se vertex is in H, nd such tht every word in H is the lel of some pth in Γ from the se vertex to the se vertex. (Hint: The sugroup H is not finitely generted, so Γ is not going to e finite grph.) 6

7 3 Finding Genertors: Mnufcturing silly keyords We first nswer Exercise 2.18: Wht sorts of X-digrphs re Stllings X-digrphs of some sugroup? Definition 3.1. A reduced pth in digrph is pth tht never cktrcks. Tht is, you never go long n edge nd the immeditely go long tht edge in the other direction. (You cn, however, go ck long tht edge lter in the pth.) Notice tht every vertex of Stllings X-digrph is on some reduced pth from the se vertex ck to the se vertex. Tht is, you never hve sitution like this: Figure 8: A grph tht cnnot e Stllings digrph ecuse it is not core. Definition 3.2. A digrph where every vertex is on some reduced pth from the se vertex ck to itself is clled core. (Note tht whether or not grph is core depends on which vertex is the se vertex, so sometimes we sy core with respect to v, where v is the se vertex.) The properties folded nd core re enough to chrcterize Stllings X-digrphs: If n X-digrph is oth folded nd core, then it is the Stllings X-digrph of the sugroup contining the lels of ll of the reduced pths from the se vertex ck to itself. We cn sy this fct s theorem, whose proof we will leve out. Theorem 3.3. The Stllings grph of sugroup H of free group F (X) is the unique folded core X-digrph with lnguge H. Now we ddress Exercise Given n X-digrph tht is folded nd core, how cn we find keyord tht corresponds to tht grph? The ide is to find pths from the se vertex to the se vertex tht go once round every hole in the grph, nd then these pths cn generte every other pth. However, this vgue rule will ecome hrd to keep trck of in more complicted grphs, so we need something more precise. Definition 3.4. A tree is connected grph with no cycles. Tht is, there re no reduced pths from vertex ck to itself. Exercise 3.5. Show tht, in tree, there is unique pth etween ny two vertices. Given n X-digrph, we cn construct sugrph tht contins ll of the vertices nd is tree. This sugrph is clled mximl sutree of the grph. Exmple 3.6. We will construct sutree of the Stllings grph of Kevin s keyord. We strt with the se vertex, nd then ttch one vertex t time with one edge. 7

8 Figure 9: Growing mximl sutree of the Stllings grph of Kevin s keyord. Of course, there re lots of other wys we could hve grown our mximl sutree. Here re some other possile trees: Figure 10: Other mximl sutrees of Kevin s keyord. For ech edge e not in the tree, there is pth from the se vertex to the se vertex tht contins e once nd whose other edges re ll in the tree. The lel of this pth is key in the top row on the keyord we mke, nd we mke one key in the top row for every edge not on the tree. Exercise 3.7. Show tht this pth is unique (up to reversing the direction of the pth). Exmple 3.8. In the rightmost grph in Figure 9, there re three edges not on the tree: one leled, loop leled, nd nother edge leled. The key corresponding to the edge leled is, the key corresponding to the loop leled is, nd the key corresponding to the other edge leled is. Thus, the corresponding keyord looks like this: 8

9 Figure 11: The keyord corresponding to the mximl sutree in Figure 9. Notice tht this is different from Kevin s keyord tht we used to mke this grph. How then do we know tht this new keyord cn type the sme words s Kevin s keyord? Since the keys of the new keyord re lels of pths from the se vertex to the se vertex, we know tht Kevin s keyord cn type them. We still need to show the other direction, nmely, given word tht Kevin s keyord cn type, we need to find wy to type them with our new keyord. Given word tht Kevin s keyord cn type, we hve pth with tht lel from the se vertex to the se vertex in the Stllings grph. To find out which keys to press on the new keyord, we use our tree gin. We look t the edges of the pth not on the tree, in order, nd we hit the keys tht correspond to those edges. Exmple 3.9. Consider the word āā ā, which Kevin s keyord cn type, s we cn check y looking t the corresponding pth on the Stllings grph. If we look t the mximl tree in Figure 9, we cn pick out the edges tht re not on the tree in the pth corresponding to our word. We write the corresponding lels in old: āā ā. Ech of these edges, long with the direction we go long it, corresponds to key on the new keyord. We cn thus convert the five non-tree edges of the pth into keypresses: ā ā āāā āā ā. You cn check tht this word reduces to our originl word. Exercise For ech of the four mximl sutrees in Figure 10, find the corresponding keyord. Compre the keyords to ech other nd to Kevin s originl keyord. Exercise For ech of the grphs you mde in Exercise 2.14, use this method to find generting set for the sugroup tht is different from the one given in Exercise You might need to try couple different trees. Exercise For which grphs tht you mde in Exercise 2.14 is it possile to find tree tht gives you ck the generting set in Exercise 2.14? Exercise 3.13 (*). For the grphs you mde in Exercise 2.14, mke nother vertex the se vertex nd find generting set for the corresponding sugroup. Compre your nswer with your nswer in Exercise (It will e esier to see wht s going on if you use the sme tree.) Exercise 3.14 (*). In Exmple 3.9, we took word nd used mximl tree to find wy type tht word using the keyord corresponding to tht tree. Show tht our method works in generl. Tht is, consider n ritrry sugroup nd mximl tree of its Stllings grph, which gives us keyord whose keys correspond to edges not on the tree. Given n ritrry reduced pth from the se vertex to the se vertex, we construct word y hitting the keys corresponding to the edges of the pth not on the tree. Show tht when we reduce this word, we get ck the lel of our pth. You my find it helpful to use the uniqueness of reduced pths etween two vertices of tree. 4 Intersections of Sugroups: Keeping in touch After Mthcmp, Kevin nd Nic wnt to keep in touch. Unfortuntely, they hve different keyords! This is prolem if they wnt to reply out something the other person hs typed. Perhps, however, they hve 9

10 some words they cn oth type, so they cn tlk out those. Are there ny such words, nd how cn they find them? To nswer this question, we define the product grph Definition 4.1. If Γ nd re two X-digrphs, we construct the product grph Γ s follows: For every pir contining vertex of Γ nd vertex of, we drw vertex of Γ. For every pir contining n edge of Γ nd n edge of with the sme lel, we drw n edge of Γ. If e is n edge of Γ with origin v nd f is n edge of with origin w, then the origin of the edge (e, f) in Γ is the vertex (v, w). The terminus of (e, f) is defined similrly. The lel of (e, f) is the lel of e, which is the sme s the lel of f. (This is why we required the edges of Γ to e pirs of edges with the sme lel.) Exmple 4.2. Here is the product of two grphs. Notice tht the product grph hs 4 2 = 8 vertices, 3 1 = 3 edges leled, nd 3 2 = 6 edges leled. Figure 12: Constructing the product grph. Exercise 4.3. Using Exmple 4.2 to guide you, convince yourself tht pth in the product grph Γ corresponds to pth in Γ nd pth in with the sme lels. Convince yourself of the other direction: given pth in Γ nd pth in with the sme lel, there is pth with the sme lel in Γ. The two fctors correspond to Kevin s sugroup ā, ā, āā nd Nic s sugroup,. The words they cn oth type correspond to pths in the two grphs with the sme lel. By the previous exercise, these correspond to pth in the product grph with the sme lel. Thus the set of lels of pths from the se vertex to the se vertex in the product grph is the set of words tht oth Kevin nd Nic cn type. This would men tht the product grph is the Stllings digrph of the sugroup of words tht they oth cn type, ut there is prolem: The product grph is not core. 10

11 Definition 4.4. If Γ is n X-digrph with se vertex v, then we construct new grph Core(Γ, v) clled the core of Γ t v y removing everything tht is not on reduced pth from v to v. Exmple 4.5. We highlight the core of the product grph in Figure 12 t the se vertex. Figure 13: The core of the product grph in Figure 12 is the Stllings grph of the intersection of the sugroups. Once we tke the core of the product grph, we find the grph we were looking for: the grph tht corresponds to the words tht cn e typed y oth keyords. Definition 4.6. Given two sugroups H nd K, the set of words in oth of them is clled the intersection of H nd K, nd is denoted H K. Thus we cn restte our result ove like this: Γ(H K) = Core(Γ(H) Γ(K), (v, w)), where v is the se vertex of Γ(H) nd w is the se vertex of Γ(K). Exmple 4.7. In Exmples 4.2 nd 4.5, we found the grph of ā, ā, āā,. By inspecting the grph or y using Section 3, we see tht ā, ā, āā, = ā. Thus Kevin nd Nic cn only type copies of the word or its inverse to ech other. They might wnt to find nother wy to communicte. Exercise 4.8. Find the intersection of the following sugroups y computing the product grph, finding its core t the se vertex, nd then finding set of genertors. 1.,, , c ā, c. 4., c ā, cā. 5.,,,,. (Leve lots of spce to drw this one.) Notice tht in the lst exmple, the numer of genertors of the intersection is more thn the numer of genertors in ech of the fctors. Prdoxiclly, in order to type smller set of words, you might need more keys. In fct, there is sttement clled the Hnn Neumnn conjecture tht sttes tht if H hs m genertors, K hs n genertors, nd H K hs s genertors, then s 1 (m 1)(n 1). Hnn Neumnn proved in 1957 tht s 1 2(m 1)(n 1), ut the conjecture remined unresolved until proof y Igor Mineyev

12 Exercise 4.9. If H is sugroup of K, wht is Core(Γ(H) Γ(K)) (with respect to the usul se vertex)? Exercise Using Exercise 4.9, show tht if H is sugroup of K then there is function from the vertices nd edges of Γ(H) to the vertices nd edges of Γ(K) tht preserves lels of edges, sends the origin nd terminus of n edge to the origin nd terminus of its imge, respectively, nd sends the se vertex of one grph to the se vertex of the other. Definition If φ is function from the vertices nd edges of one X-digrph to nother tht preserves lels of edges nd sends the origin nd terminus of n edge to the origin nd terminus of its imge, then φ is clled morphism of X-digrphs. Exercise Let the lphet e X = {,, c}, nd let K e the entire free group,, c. Drw Γ(K) nd show how to construct morphism from ny other X-digrph to Γ(K). Exercise Prove the converse of Exercise Tht is, if H nd K re two sugroups nd there exists morphism from Γ(H) to Γ(K) tht sends the se vertex of Γ(H) to the se vertex of Γ(K), then H is sugroup of K. 5 Conjugte Sugroups: Modifying keyords We now come ck to Exercise Wht hppens when we move the se vertex? Let s strt with n exmple: Exmple 5.1. Susn s keyord nd the corresponding grph look like this: Figure 14: Susn s keyord nd its Stllings grph. Let s move the se vertex up. We get grph tht looks like this: Figure 15: Moving the se vertex in the Stllings grph of Susn s keyord. Using Section 3 or just y looking t the grph, we see tht {, } is generting set for the sugroup. Compre it with the originl generting set {, }. The first letter moved to the end! Another wy of thinking out it is tht we wrote ā t the eginning of ech genertor nd t the end. (Notice tht ā = nd ā =.) Why the letter nd not some other letter? The edge from the originl se vertex to the new se vertex is leled. 12

13 Exercise 5.2. In Exmple 5.1, move the se vertex to the top nd find generting set. Definition 5.3. Writing ā t the eginning of word w nd t the end is clled conjugting w y. Be wre tht some other people cll this conjugting w y ā. We cn lso conjugte y words y doing it one letter t time. For exmple, conjugting w y gives āw. (Or, you cn think of it s writing t the end nd then flipping over to write t the eginning.) An importnt property of conjugtion is tht it is n utomorphism. Tht is, if v nd w re two words, then we cn conjugte oth of them y nd then conctente them or conctente them nd then conjugte the result y, with the sme result, ecuse (āv)(āw) = ā(vw). Definition 5.4. If H is sugroup, then we cn conjugte the entire sugroup y conjugting ll of its elements. We let 1 H denote H conjugted y. Two sugroups H nd K re clled conjugte to ech other if there is word w such tht H = w 1 Kw. Exmple 5.5. In the Exmple 5.1, we conjugted the genertors y. As result, the sugroup they generte ws lso conjugted y. Exercise 5.6. In the grph of Susn s keyord, there re three nturl wys of moving the se vertex to the top: long, long, nd long. As ws suggested erlier, this corresponds to conjugting the keys on Susn s keyord y,, nd, respectively. Try it. You get three different pirs of genertors. Do they ll generte the sugroup corresponding to the grph in Figure 14 with the se vertex on top? Exercise 5.7. Wht hppens if we conjugte the genertors of Susn s keyord y ā? How out? Compre the resulting Stllings grphs. Exercise 5.8. Think of simple wy to tke pth from one vertex ck to itself nd turn it into pth from nother vertex ck to itself. Wht hppens to the lels of the pth? Exercise 5.9. Given two finitely generted sugroups H nd K, how could you test if they re conjugte to ech other? Definition Given n X-digrph Γ tht is core with respect to some vertex, the type of Γ, denoted Type(Γ) is the grph we get y removing ny vertices with just one edge, s mny times s necessry until every vertex of the grph hs t lest two edges. Equivlently, we tke the intersection of Core(Γ, v) over ll vertices v of Γ. The type of grph does not hve designted se vertex. Exmple Here is the grph of Kevin s keyord nd the type of tht grph: Figure 16: The Stllings grph of Kevin s keyord nd the type of the Stllings grph of Kevin s keyord. 13

14 Exmple Here is the grph of Susn s keyord nd the type of tht grph. Notice tht nothing chnges, except tht there is no longer specil se vertex. Figure 17: The Stllings grph of Susn s keyord nd the type of the Stllings grph of Susn s keyord. As you might hve relized from the ove exercises, two sugroups H nd K re conjugte to ech other if nd only if the grphs Type(Γ(H)) nd Type(Γ(K)) re the sme. The pth from the se vertex of Γ(H) to the se vertex of Γ(K) is the word tht is conjugted y. 6 Grphs of Infinitely Generted Sugroups nd Applictions So fr, we ve used Stllings grphs to crete lgorithms to test memership, compute intersections, nd test for conjugcy. We cn lso use Stllings grphs to prove theorems out sugroups of free groups y trnslting properties of sugroups into properties of the corresponding Stllings grphs. Before we do tht, though, we need to define the Stllings grphs of ritrry sugroups of F (X), even ones tht re not finitely generted. When testing if word is in sugroup, we moved round the Stllings grph, nd sometimes we got stuck when vertex did not hve n edge with the pproprite lel nd direction. We cn dd in the missing edges. Tht is, whenever we hve vertex tht does not hve n edge with certin lel nd direction, we ttch tht missing edge nd crete new vertex t the other end of the edge. Of course, the new vertex is missing some edges, so we hve to repet this process indefinitely. At the end, wht we hve is Stllings grph with some infinte trees ttched. Exmple 6.1. Here is n exmple of the result of this process with the Stllings grph of Kevin s keyord. 14

15 Figure 18: The Stllings grph of Kevin s sugroup ā, ā, āā, in lck, with infinite gry trees dded so tht every vertex hs edges of every lel nd direction. The grph together with the trees is the covering grph of this sugroup. We cll this new grph the covering grph of the sugroup. It hs the property tht ny word is the lel of some pth strting from the se vertex. Tht is, when reding word y going long the grph, you will never get stuck. Another wy of sying this fct is tht the grph is X-regulr, which mens tht every vertex hs exctly one edge with ech lel nd direction. Also, the covering grph hs the property tht Stllings grphs hve tht word is in the sugroup if nd only if it is the lel of reduced pth from the se vertex to the se vertex. Indeed, if we tke the core of the covering grph with respect to the se vertex, the trees ll get removed, nd we get the Stllings grph ck. There is second wy of defining covering grph tht does not use Stllings grphs: Definition 6.2. A right coset of sugroup H of group F is the set {hw h H}, where w is some word in F. The coset is written Hw. Definition 6.3. The covering grph of sugroup H of F is n X-digrph with vertex for ech right coset of H, nd n edge leled x from the coset Hv to the coset Hvx for ll cosets Hv nd ll lels x in X. It is not immeditely ovious tht this definition of covering grphs grees with our erlier notion of ttching trees to Stllings grphs, though we cn see how it might e true in our exmple: Exmple 6.4. Here we hve the covering grph of Kevin s sugroup s efore, with the vertices corresponding to cosets. Note tht the wy tht we write coset is not unique. For exmple, for this sugroup, the coset H 3 is equl to H, nd the coset H 2 is equl to H 2. 15

16 Figure 19: The covering grph of Kevin s sugroup ā, ā, āā, with the vertices leled with the cosets they represent. H 4 H 3 H 2 H H 1 H H H 1 We cn see tht, for exmple, there is n edge from H to H 3 leled. We cn check tht, indeed, H 3 = H, ecuse H, nd so H 3 () 1 = H. To show tht this notion of covering grph coincides with our erlier notion, we first prove the following: Proposition 6.5. Given sugroup H of free group F, the lnguge of the covering grph of H (defined with cosets) is H itself. Proof. Given word w H, let w = w 1 w 2 w k, where the w i re letters or inverse letters. Reding the word long the grph strting t the se vertex, we will move long the vertices H, Hw 1, Hw 1 w 2, ll the wy to Hw 1 w k = Hw. But w H, so Hw = H. Hence, once we re done reding the word, we re ck t the se vertex, giving us reduced pth from the se vertex to the se vertex with lel w. Conversely, given reduced pth from the se vertex to the se vertex with lel w = w 1 w k, the sequence of vertices is, s efore, H, Hw 1, Hw 1 w 2, ll the wy to Hw 1 w k = Hw. But this pth ends ck t the se vertex, so Hw = H, nd so w H. Now, we cn prove the next theorem: Theorem 6.6. Given finitely generted sugroup H of free group F, the core of the covering grph of H is the Stllings grph of H. Proof. The covering grph is folded with lnguge H. Thus, the core of the covering grph is folded core grph with lnguge H. By Theorem 3.3, this grph must, in fct, e the Stllings grph of H. 16

17 Corollry 6.7. Our two constructions of the covering grph y ttching trees to the Stllings grph nd y using right cosets re the sme. Proof. The theorem tells us tht the Stllings grph is the core of the covering grph. The only wy to extend the Stllings grph without dding ny new reduced pths is y ttching trees, nd one cn check tht there is only one wy to ttch trees so tht every vertex hs exctly one edge with every lel nd direction. Thus, the Stllings grph is the core of exctly one X-regulr grph. Both of our constructions of the covering grph re X-regulr nd their cores re the Stllings grph, so we conclude tht our two constructions of the covering grph re the sme. Note tht in order to construct the covering grph vi cosets nd tke its core, we never use the fct tht H is finitely generted. In fct, we cn drop this ssumption, nd use this procedure s the definition of Stllings grphs for infinitely generted sugroups, nd our theorem sys tht this definition would gree with our erlier definition of the Stllings grph for finitely generted sugroups. Definition 6.8. Given sugroup H of the free group F (X), not necessrily finitely generted, the Stllings grph of H is the core of the covering grph of H with respect to the se vertex corresponding to the identity coset H. Exmple 6.9. An importnt exmple of the Stllings grph of n infinitely generted sugroup H is in the free group generted y two genertors nd when H is the set of words where the numer of times ppers is the sme s the numer of times 1 ppers, nd the numer of times ppers is the sme s the numer of times 1 ppers. One cn check tht there is coset of H for every pir of integers, where the first integer represents how mny more times ppers thn 1, nd the second integer represents how mny more times ppers thn 1. One cn verify tht H is closed under concntention nd tking inverses. The covering grph then looks like this: 17

18 Figure 20: The covering grph of the sugroup H conting ll words in which ppers the sme numer of times s 1 nd ppers the sme numer of times s 1. H 2 2 H 1 2 H 2 H 1 2 H 2 2 H 2 1 H 1 1 H 1 H 1 1 H 2 1 H 2 H 1 H H 1 H 2 H 2 1 H 1 1 H 1 H 1 1 H 2 1 H 2 2 H 1 2 H 2 H 1 2 H 2 2 This grph is core, so the Stllings grph of H is the sme grph. Now tht we ve defined the Stllings grph for ritrry sugroups, it might e useful to chrcterise when sugroup ctully is finitely generted, in terms of its Stllings grph. The nswer turns out to e simple. Theorem Let H e sugroup of free group F (X). Then H is finitely generted if nd only if the Stllings grph Γ(H) is finite. Proof. If H is finitely generted, then we cn use our initil method of constructing the Stllings grph vi drwing cycles for the gennertors nd folding. The grph we get this wy is finite. Coversely, if we hve finite folded grph, then, in prticulr, there re finitely mny edges tht re not on some mximl sutree. As discussed in Section 3, we conclude tht there re finitely mny genertors for the sugroup. OK, so we ve chrcterised the finitely generted sugroups in terms of their Stllings grphs. Wht other properties of sugroups cn we chrcterise in terms of the Stllings grphs? Here s one: Definition The index of sugroup H in free group F is the numer of cosets of H, denote [F : H]. In generl, we cn t expect the index of sugroup to e finite. The concept of finite index sugroup is firly importnt in group theory: Mny theorems hve the phrse finite index somewhere in either the ssumptions or the conclusions of the theorem (or oth!). Cn we determine whether sugroup of free group hs finite index in terms of its Stllings grph? Indeed, we cn: 18

19 Theorem Let H e sugroup of free group F (X) where the lphet X is finite. Then H hs finite index in F if nd only if the Stllings grph Γ(H) is finite nd X-regulr. Proof. The sugroup H hving finite index is the sme thing s sying tht the covering grph of H hs finitely mny vertices, since, y definition, the covering grph hs vertex for every coset of H. If the covering grph is finite, then I clim tht it is lredy core. Indeed, consider ny edge leled x from vertex v 0 to vertex v 1. We need to find reduced pth tht contins this edge. But since covering grphs re X-regulr, there is n edge leled x from v 1 to some vertex v 2. Likewise, there is n edge leled x from v 2 to some vertex v 3, nd so forth. Since the grph is finite, we will eventully repet vertex. One cn check tht the folded condition implies tht the first time we repet vertex we must ctully e t the initil vertex v 0. So now we hve reduced pth p from v 0 to v 0. We cn crete pth from the se vertex to the se vertex like we did in Section 5 y going long pth q from the se vertex to v 0, long our cycle p, nd then ck vi the inverse q 1. One cn check tht when we reduce this pth qpq 1, we still go long ll of the edges of p. Hence, our ritrry edge is on reduced pth from the se vertex to the se vertex, nd hence is in the core of the grph. We conclude tht the covering grph is core. Since the covering grph is core, it is equl to the Stllings grph. In prticulr, the Stllings grph is finite nd X-regulr. Conversely, sy the Stllings grph is finite nd X-regulr. Since the grph is X-regulr, when we ttch the missing edges to mke the covering grph, we hve no edges to ttch. Therefore, once gin, the Stllings grph is equl to the covering grph. In prticulr, the covering grph is finite, s desired. Our chrcteristions of finite index sugroups nd finitely generted sugroups yield n immedite corollry. Corollry Let H e sugroup of free group F (X) where the lphet X is finite. If H hs finite index in F, then H is finitely generted. Proof. If H hs finite index, then Γ(H) is finite y Theorem 6.12, nd so H is finitely generted y Theorem This result seems counterintuitive. Finite index sugroups re lmost s ig s the entire group F. Indeed, if H hs index d in F, then tht mens tht d copies of H cover ll of F. On the other hnd, finitely generted sugroups re uilt from finite list of words, so they re intuitively smll. Our corollry then sys tht if sugroup is ig, then it must e smll. Insted, perhps etter intuition is tht finitely generted groups re tidy wheres infinitely generted groups re messy. A finite index sugroup is close to the entire group F, which mkes it pretty tidy. Our result clims tht if H hs finite index, then it is finitely generted. But how mny genertors does H hve? Cn we nswer this question in terms of the index of H nd the numer of letters in the lphet X? Exercise Let F e free group over n lphet X with n letters, nd let H e sugroup of F (X) with index d. Prove tht the numer of genertors of H is exctly (n 1)d + 1. This result is known s Schreier s formul nmed fter the erly 20th century Austrin mthemticin Otto Schreier. Hint: Count the numer of vertices nd edges of the Stllings grph. We cn lso use Stllings grphs to sy something out the intersection of two finite index sugroups. Theorem Let H nd K e two sugroups of free group F. If H nd K hve finite indices c nd d, respectively, in F, then the index of H K is lso finite nd t most the product cd. Proof. As discussed in the proof of Theorem 6.12, for finite index sugroups, the covering grphs nd Stllings grphs re the sme. Thus, Γ(H) hs c vertices nd Γ(K) hs d vertices, so the product grph Γ(H) Γ(K) hs cd vertices. One cn check tht since Γ(H) nd Γ(K) re X-regulr, the product Γ(H) Γ(K) is lso X-regulr. 19

20 From Section 4, we know tht Γ(H K) is the core of Γ(H) Γ(K). As we showed in the proof of Theorem 6.12, connected finite X-regulr grph is utomticlly core, so tking the core of Γ(H) Γ(K) mounts to tking the connected component contining the se vertex. Therefore, the core of Γ(H) Γ(K) is still X-regulr. We conclude tht Γ(H K) is X-regulr nd finite with t most cd vertices, so H K hs finite index t most cd in F. As side note, this theorem is true even if F is replced y n ritrry group tht is not necessrily free. Exercise Use this result out the index of n intersection long with Schreier s formul to help you find exmples tht show tht the Hnn Neumnn ound is s strong s possile. Tht is, given ritrry positive integers r nd s, construct sugroup H with r genertors nd sugroup K with s genertors such tht the numer of genertors of H K is exctly (r 1)(s 1) + 1. We now return to the question of conjugting sugroups. As we discussed erlier, conjugting sugroup corresponds to moving the se vertex in the Stllings grph. More precisely, conjugting y x moved the se vertex long the edge leled x. Of course, there ws cvet: Sometimes, the se vertex didn t hve n edge leled x. Then we needed to move the se vertex off the Stllings grph to new vertex ttched to the old grph y single edge. Now tht you know out covering grphs, you see the ig picture. Wht s ctully going on is tht, s we conjugte the sugroup, the se vertex moves round the covering grph. When we go ck to the Stllings grph y tking the core, most of the covering grph disppers, nd we re left with the type of the grph possily with spur ttching it to the se vertex. Here we wnt to use the more generl definition of the type of grph, so tht we cn use it oth for Stllings grphs nd for covering grphs: Definition Given n X-digrph Γ, the type of Γ, denoted Type(Γ), is the intersection of Core(Γ, v) over ll vertices v of Γ. The type of grph does not hve designted se vertex. Roughly speking, the type of grph contins the interesting prt of grph. Nmely, it contins ll of the cycles nd cuts off the spurs nd trees. Note tht the type of the covering grph of nontrivil sugroup is the sme s the type of its Stllings grph. In group theory, very importnt concept is sugroup tht doesn t chnge when it is conjugted: Definition A sugroup N of free group F is norml if g 1 Ng = N for ll g F. Equivlently, we hve gn = Ng for ll g. If you conjugte n element of norml sugroup N, it will in generl ecome different element, ut it will sty inside N. It is the set N tht remins fixed y conjugtion. Exmple The sugroup H =,, is norml in the free group over the lphet {, }. Here is its Stllings grph: Figure 21: The Stllings grph of the sugroup H =,,. 20

21 Norml sugroups re very importnt in group theory ecuse we cn construct nother group clled the quotient group y setting everything in N to the identity. If the sugroup is not norml, we quickly run into prolems. After ll, if we conjugte the identity, we should get the identity ck. Like we did for finitely generted nd finite index sugroups, we cn chrcterise norml sugroups of free group F using Stllings grphs. Theorem Let H e nontrivil sugroup of free group F (X). Then H is norml if nd only if the Stllings grph Γ(H) is X-regulr nd looks the sme no mtter which vertex is selected s the se vertex. Proof. As discussed erlier, conjugting sugroup corresponds to moving the se vertex of the covering grph. Assume tht H is norml. Since conjugting H gives us H gin, we know tht the covering grph looks the sme no mtter which vertex is picked s the se vertex. In prticulr, the type of the covering grph must e either the entire covering grph or empty, ecuse otherwise the grph would look different sed t vertex in the type or sed t vertex not in the type. We ssumed tht H is nontrivil, so the type of the covering grph cnnot e empty. Therefore, the covering grph is equl to its type, nd hence the covering grph is core with respect to every vertex. Since the Stllings grph is the core of the covering grph with respect to the se vertex, we conclude tht the Stllings grph of H is equl to the covering grph of H. In prticulr, the Stllings grph of H is X-regulr nd looks the sme no mtter which vertex is selected s the se vertex. Conversely, ssume tht Γ(H) is X-regulr nd looks the sme from ech vertex. As discussed erlier, if the Stllings grph is X-regulr, then there re no missing edges to ttch, nd so the covering grph is equl to the Stllings grph. As result, the covering grph looks the sme no mtter which vertex is selected s the se vertex. Therefore, if we conjugte H, we move the se vertex in the covering grph, nd re left with the sme grph. Since H nd its conjugte hve the sme covering grph (nd hence the sme Stllings grph), we conclude tht H is equl to its conjugte, s desired. Therefore, H is norml. It is importnt to note tht the notion of norml sugroup depends hevily on the free group it is in. For exmple, the sugroup,, in Exmple 6.19 is norml in the free group over the lphet {, }, ut it is not norml in the free group over the lphet {,, c}. Indeed, its Stllings grph is {, }-regulr, ut not {,, c}-regulr. Erlier, we used our chrcteristions of finitely generted nd finite index sugroups to show tht finite index sugroup is finitely generted. We cn now hve prtil converse. Corollry Let N e nontrivil norml sugroup of free group F. Then N is finitely generted if nd only if it hs finite index in F. Proof. If N hs finite index, then we know tht it is finitely generted y Corollry Conversely, if N is finitely generted, then Γ(N) is finite y Theorem Since N is nontrivil nd norml, the Stllings grph Γ(N) is X-regulr y Theorem Therefore, N hs finite index y Theorem As side remrk, if N is norml sugroup in free group F, then the Stllings grph of Γ(N) is more commonly known s the Cyley grph of the quotient group F/N. The Cyley grph of group is very importnt concept in geometric group theory. In Exmple 6.9, one cn check y exmining the grph tht the sugroup is norml, nd the Stllings grph pictured is the Cyley grph of the quotient group, which is isomorphic to Z Z. 21