Honors Thesis: Investigting the Algebri Properties of Cyley Digrphs Alexis Byers, Wittenberg University Mthemtis Deprtment April 30, 2014 This pper utilizes Grph Theory to gin insight into the lgebri struture of group using Cyley digrph tht depits the group. Using the properties of Cyley digrphs, we investigte how to tell if given digrph is Cyley digrph, nd we ttempt to build Cyley digrphs. We then use the Cyley digrph to find informtion bout the struture of the orresponding group. Finlly, we exmine the results of removing genertors from Cyley digrph nd wht it mens if the digrph remins onneted or is disonneted by the proess. 1
Contents 1 Introdution 3 1.1 Grph Theory................................. 3 1.2 Group Theory................................. 5 2 Properties of Cyley Digrphs 9 3 Informtion given by the Struture of Cyley Digrph 14 3.1 Abelin.................................... 14 3.2 Cyli..................................... 18 3.3 Cyli Subgroups............................... 22 4 Disonneting Cyley Digrphs 26 4.1 Conneted................................... 26 4.2 Disonneted................................. 28 4.3 Norml Subgroups.............................. 33 4.4 Conjetures.................................. 35 5 Further Reserh 40 6 Conlusion 41 7 Aknowledgments 41 2
1 Introdution 1.1 Grph Theory Leonhrd Euler is lrgely referred to s the fther of Grph Theory for his solution to fmous problem involving the seven bridges of Königsberg. In his 1736 publition entitled, Solutio Problemtis d Geometrim Situs Pertinentis (Trnslted, The solution to problem relting to the geometry of position), Euler onsidered the seven bridges tht onneted the ity of Königsberg, in Prussi, ross the Pregel River.[1] Figure 1: Euler s mp of the seven bridges of Königsberg inluded in his publition.[2] Euler fed the question: Is it possible to ross every bridge in Königsberg extly one, nd then return to the strting point? Arguing tht this ws impossible, Euler used grph for the first time in history.[4] His grph, set of verties, together with set of edges onneting pirs of verties, looked similr to the following grph, G: C A D B Figure 2: A grph G, modeling the Königsberg bridges. Eh vertex in the grph represents lnd mss in Königsberg, nd eh edge represents bridge onneting two lnd msses. Notie tht this grph is onneted, mening 3
tht there exists sequene of edges between ny two verties in the grph. Thus, if we re t ny lnd mss, we n get to ny other lnd mss by rossing one or more bridges. Euler observed tht every time you step onto lnd mss from bridge, you must lso leve the lnd mss by bridge. Then, it follows tht, to ross eh bridge extly one, you must hve n even number of bridges so tht hving entered lnd mss, one n lso exit it. Sine eh lnd mss in Königsberg hs n odd number of bridges onneting it to other lnd msses, t some point, one will enter lnd mss but not be ble to exit the lnd mss by n untrversed bridge. In Grph Theory, this prllels the ide tht, for onneted grphs, whenever you enter vertex from one edge, you must then leve tht vertex by nother edge. Consequently, in order to trverse eh edge extly one, the grph needs to hve n even number of edges djent to eh vertex; in other words, the degree of every vertex needs to be even. This kind of trversl problem, where one must trverse every edge extly one nd return to strt, is desribed in modern Grph Theory s finding n Eulerin Tour, due to Euler s ontribution to this problem.[6] Although we briefly onsider Eulerin Tours in this pper, we find more onnetions in our reserh with wlks in grphs, whih re similr to Eulerin Tours exept tht we llow edges nd verties to be repeted. In prtiulr, we utilize losed wlks, whih re simply wlks in whih the endpoints re the sme.[6] We ould lso look t subgrph, of the grph of Königsberg, suh s the grph H given below. We define subgrph H of grph G s grph suh tht the vertex set of H is subset of the vertex set of G, nd the edge set of H is subset of the edge set of G.[6] C A D Figure 3: Subgrph H of the grph G of the Königsberg bridges. A onept tht we will use extensively in this pper is tht of direted grph, or digrph, whih is grph in whih the edges hve ssoited diretions from one vertex to the other.[6] A digrph nlog of the Königsberg grph my look like this: 4
C A D B Figure 4: Digrph nlog of the Königsberg bridge grph. Note tht in digrph, I my hve undireted edges; we n think of eh undireted edge s two direted edges, one direted towrd one end vertex, nd one direted towrd the other end vertex. 1.2 Group Theory The beuty of Grph Theory is tht it n be onsidered in isoltion, but it lso hs mny pplitions to other fields. In prtiulr, this pper onsiders the wys tht Grph Theory n id in the study of the mthemtil field of Group Theory. To begin, we will give the bsi definitions tht one without ny knowledge of Group Theory should know before reding this pper. First, binry opertion * on set S is funtion mpping S x S into S.[3] A binry opertion n be ddition, multiplition, modulr rithmiti, et. In this pper, without loss of generlity, we resort to using multiplitive nottion for the binry opertion of rbitrry groups unless otherwise stted. The most importnt definition for this field is obviously tht of group: set G, together with binry opertion * is group < G, >, if G is losed 1 under *, if * is ssoitive 2, if there exists n identity element 3, nd if there exists n inverse 4 for every element of G.[3] 1 A set S is losed under * if for ll elements, b in S, b is lso in S.[3] 2 A binry opertion * is ssoitive in set S if for ll elements, b, in S, ( b) = (b ).[3] 3 The identity element of set S is n element e of S suh tht e x = x e = x for ll x in S.[3] 4 The inverse of n element in set S is n element suh tht = = e.[3] 5
Exmple 1.1. If you look t the set Z of ll integers, then you might notie tht Z is group under ddition, but not under multiplition sine not every element of Z hs n inverse under multiplition. For exmple, 3 is n element of Z, but 3 does not hve n inverse in < Z, >; there is no integer tht you n multiply by 3 nd get 1, the identity in < Z, >. Exmple 1.2. A type of group tht is used often in this pper is the group of integers modulo n, Z n, where Z n = {0, 1, 2, 3,..., n 1}, nd the opertion used is modulr rithmeti, whih n be desribed in the following wy: for n integer z, nd nturl numbers n nd r, then (z mod n) = r if r is the reminder when z is divided by n. For elements, b in Z n, we define + b = ( + b) mod n. We desribe the order of group G s the number of elements in G.[3] Exmple 1.3. The group of integers, Z, with ddition, hs infinite order, while Z 6, with modulr rithmeti, hs order 6 (in generl, Z n hs order n). We ll group G belin if its binry opertion is ommuttive 5.[3] Exmple 1.4. Both the groups < Z, + > nd < Z n, + > re belin. If H, subset of group G, is losed under the binry opertion of G, nd if H is group under the binry opertion of G, then H is subgroup of G. A proper subgroup of group G is ny subgroup of order stritly less thn the order of G.[3] And the index of subgroup H of G is the order of G divided by the order of H. Exmple 1.5. The group < 2Z, + >, with elements of only the even integers, is subgroup of < Z, + >. But < Z +, + >, with elements of only the positive integers, is not subgroup of < Z, + >. Even though Z + is subset of Z tht is losed under the opertion, < Z +, + > is not group under the binry opertion (sine there re no inverses). A group G is generted by set of elements S if S is subset of G nd every element of G n be written s ombintion of the elements in S. For group G, let x G. Then the set {x n n Z}= < x >, 6 subgroup of G, is the yli subgroup of G generted by x. We sy tht the order of n element x is the order of the yli subgroup generted by x.[3] We denote the order of n element x by o(x). And group is yli if there exists n element g in G suh tht G =< g > (i.e there exists n element tht genertes ll of G). We ll g genertor of G.[3] 5 An opertion * is ommuttive in set S if for ll elements g nd g in S, g g = g g.[3] 6 The symbol mens n element of, nd we be used often to denote n element s membership in set. 6
Exmple 1.6. The esiest exmple of yli group is < Z n, + >. Consider < Z 6, + > nd 1 Z 6. The yli subgroup generted by 1 would be < 1 > = {1, 1 2, 1 3, 1 4, 1 5, 1 6 } (1) = {1, 2, 3, 4, 5, 0} (2) whih is found by repetedly ddin to itself. Sine < 1 >= {1, 2, 3, 4, 5, 0} = Z 6, then we would sy tht 1 genertes Z 6, or tht 1 is genertor of Z 6. And sine there exists single element tht genertes the whole group, Z 6 is yli. For subgroup H of group < G, >, the left osets of H in G re sets gh suh tht g is n element of G nd gh = {g h h H}. Similrly, the right osets of H in G re sets Hg suh tht for n element g of G, Hg = {h g h H}. And subgroup H is norml subgroup in G if the left nd right osets gree, i.e. gh = Hg for ll elements g in G.[3] Exmple 1.7. Consider the group G = Z 2 Z 5 7, whih is diret produt 8 of groups Z 2 nd Z 5. Let s look t the osets of subgroup H = {(0, 1), (0, 2), (0, 3), (0, 4), (0, 0)}. Suppose we hoose (1, 1) Z 2 Z 5. Then, (1, 1)H = {(1, 1) (0, 1), (1, 1) (0, 2), (1, 1) (0, 3), (1, 1) (0, 4), (1, 1) (0, 0)} (3) = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 1)} (4) And (1, 1)H is left oset of H in G. Similrly, H(1, 1) = {(0, 1) (1, 1), (0, 2) (1, 1), (0, 3) (1, 1), (0, 4) (1, 1), (0, 0) (1, 1)} (5) = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 1)} (6) So H(1, 1) is right oset of H in G. Notie tht (1, 1)H = H(1, 1). If we show gh = Hg is true for ll elements g in G, not just (1, 1), then H is norml in G. Next is the most importnt definition for the understnding of this pper. Definition 1.8. A Cyley digrph is visul representtion of group. It is direted grph in whih eh element of group G is represented by vertex, nd eh edge represents multiplition on the right 9 by n element of generting set of G. [3] 7 Notie tht Z 2 Z 5 is tully Z 10. 8 A diret produt between two groups < G, > nd < G, > is the group < G G, > where we define (g, g ) s n element of G G if g G nd g G. Then, for elements (, g 1) nd (, g 2) of G G, we define (, g 1) (, g 2) = (, g 1 g 2).[3] 9 By onvention, we lwys multiply on the right. We will brek this onvention briefly lter in the pper, but it is ssumed tht we re using right multiplition in ll Cyley digrphs unless expliitly stted otherwise. 7
For exmple, if we sy tht the genertor g is represented in Cyley Digrph of group G by solid rrow, then we would interpret the following s x g = y. x y We wnt to point out two onventions tht we use when deling with Cyley digrphs. First, the mnner in whih we denote inverse of genertor: In the bove figure, the solid edge is represented by right multiplition by genertor g. If we trvel in the opposite diretion of this rrow, we n think of tht s multiplition on the right by the inverse of g, or g 1. So the figure tells us tht x g = y, but lso tht yg 1 = x. Seond, note tht in Cyley digrph, we n still hve undireted edges, s in Exmple 1.10, tht we would interpret gin just s two direted edges, one direted towrd one end vertex, nd one direted towrd the other end vertex. And genertor represented by ny undireted edge will hve order two in the orresponding group. Exmple 1.9. The Cyley digrph of < Z 6, + >, where we represent eh solid rrow s multiplition on the right by the element 1, is shown in Figure 5: 0 5 1 4 2 3 Figure 5: Cyley digrph of < Z 6, + > with generting set {1}. Exmple 1.10. Note tht, by using different sets of genertors, we n onstrut different Cyley digrph of < Z 6, + >. Suppose insted, we onsider the generting set {2, 3}. Eh element does not generte < Z 6, + > by itself, but together they do. Let solid rrow represent multiplition on the right by 2 nd dshed rrow represent multiplition on the right by 3. Then the Cyley digrph of < Z 6, + > would look like: 8
0 5 1 4 2 3 Figure 6: Cyley digrph of < Z 6, + > with generting set {2, 3}. By our onvention, the nme of eh vertex in Cyley digrph is not unique. We n nme eh vertex independently, or we n nme eh vertex by the sequene of edges between the identity nd tht vertex. For exmple, we n refer the vertex represented by 3 in Figure 5 equivlently s {1, 1, 1} sine we n trvel from 0 to 3 by sequene of three edges represented by multiplition on the right by 1. In Figure 6, we n refer to the vertex represented by 3 equivlently s the sequene {2, 2, 3, 2} sine if you begin t 0 nd follow two 2 edges, 3 edge, nd then nother 2 edge, you will be t vertex 3. Furthermore, the vertex 3 ould be represented by {2, 3, 2, 2}, or {2, 3, 2, 3, 2, 3}, or mny other different sequenes. Eh of these representtions re eptble in this pper. 2 Properties of Cyley Digrphs Sine this pper fouses on Cyley digrphs, let s disuss their onstrution. In order to preserve the lgebri properties of the group, we onstrut Cyley digrphs in speifi wy. We re bout four properties in prtiulr. Theorem 2.1. There re four properties tht every digrph, G, must stisfy in order to be Cyley digrph 10 : 1. G must be onneted. 2. Given verties x nd y, there exists t most one edge going from x to y. 10 These properties will be referred to lter in the pper by the numbers s listed here. 9
3. Every vertex x in G hs extly one edge of eh type strting t x nd one of eh type ending t x. 4. If two different sequenes of edges strting t some vertex x go to the sme vertex y, then whenever those sequenes begin t the sme vertex in G, they should lwys led to the sme vertex. [3] Proof. The first property pplies sine for elements g, x, h in group G, every eqution gx = h hs solution. In group, we know tht every element hs in inverse; thus, if gx = h, then we know x = g 1 h. So g is onneted to h by pth in G. Sine the solution to gx = h is unique in group, the seond property pplies. The third property pplies beuse, for ny element x G nd for eh genertor g, we n ompute xg nd (xg 1 )g = x. We know we n ompute xg sine the G is losed, nd we know tht (xg 1 )g = x beuse we hve ssoitivity of our opertion nd n identity in G: (xg 1 )g = x(g 1 g) = x(e) = x. And the fourth property pplies sine, for elements x, h, q, r, u in G, if xq = h nd xr = h, then uq = ux 1 h = ur. Thus, we hve tht the opertion is well defined. [3] On the other hnd, we n lso show tht, given digrph tht stisfies the four properties, it must be Cyley digrph for group. Theorem 2.2. Every digrph tht stisfies the bove four properties is Cyley digrph for group. [3] Proof. Let G be direted grph with n different types of edges, nd suppose tht G stisfies the four properties in Theorem 2.1. Choose ny vertex from G nd lbel it e, s we will im to prove tht this is the identity element in the orresponding group G. Sine G is vertex-symmetri, by the third property, then our hoie of e does not mtter. Now lbel eh different type of edge,,..., g n. For eh g i edge leding wy from e, lbel the vertex t the end of tht edge g i. Let ny edge g i strting t x V (G) nd ending t y V (G) represent the right multiplition of x by g i equl to y (i.e., xg i = y). By property 1 we know we n denote ny other vertex, y in G by sequene of edges from e to y, or, equivlently, s produt of,,..., g n. Thus, we n define ny vertex x G s sequene of edges, S 1. Suppose there is nother sequene of edges, S 2, tht n define x. So, strting t vertex e, two different sequenes of edge types led to the sme vertex x. By property 4, we know then tht those sme sequenes strting from ny vertex u will led to the sme v. Thus, we 10
hve tht the opertion we hve defined on the verties of G is well defined: given ny sequene of edges from e to x, tht sequene n be used to represent multiplition on the right by x. And s onsequene of property 1 nd the onstrution of G, we know tht the opertion we hve defined is losed. Now, let x, y, z V (G). Consider (xy)z. We would represent the vertex rrived t if we begn t x nd followed sequene of edges tht defines y s (xy). We know (xy) exists in G by the third property. Then (xy)z would be the vertex rrived t from vertex (xy) fter following the sequene of edges defined by z. Now x(yz) would represent the vertex rrived t fter strting t x nd following the sequene of edges defined by yz. Sine we ontente the sequenes of edges defined by two elements of V (G) when we multiply them, (xy)z nd x(yz) will be sequene of edges tht go from the sme vertex x to the sme vertex xyz. Thus, (xy)z = x(yz), nd the opertion of G is ssoitive. By how we hve defined e, we hve tht e will be the identity sine e(g i ) = (g i )e = g i for ny vertex lbeled,,..., g n in the wy desribed erlier. And sine eh element of G n be defined s the produt of g i s, then using ssoitivity, we know e times ny element of V (G) will be tht element beuse for ny vertex v V (G), v = (g )(g b )... (g α ), nd e(v) = e(g )(g b )... (g α ) (7) = (e(g ))(g b )... (g α ) (8) = (g )(g b )... (g α ) (9) = v. (10) We n define the inverse of ny edge g s going in the opposite diretion of how g is direted in G. 11 Thus, we would interpret (gg 1 ) s strting t e, trveling to g, nd then trveling in the opposite diretion on the edge g, bk to e. Therefore, (gg 1 ) = e s desired. So, we would define the inverse of ny element x of V (G), s the sequene of edges to get from x to e. This sequene n esily be found by trveling bkwrds through the sequene of edges used to desribe x in the first ple. (i.e., if x = ( )( )(g 3 ), then x 1 = (g3 1 )(g2 1 )(g3 1 )). Then xx 1 will be interpreted s strting t e, trveling to x, nd then trveling bk to e. Thus, xx 1 = e. Hene, ll elements of V (G) hve inverses. Therefore, we hve defined set of elements, V (G), nd n opertion suh tht V (G) is losed, hs n identity, hs inverses for ll elements, nd the opertion is ssoitive. Thus, < V (G), > is group. 11 If g hs order two, nd therefore is represented by n undireted edge, then g = g 1, nd tht undireted edge will represent both g nd g 1. 11
From this theorem, we know tht if we n drw ny grph tht hs these four properties, then the grph will be Cyley digrph for group. A question, however, is how hrd is it to tully drw grph, without preoneived notion of the group you intend it to represent, tht stisfies the four properties. In prtiulr, we found the lst property is espeilly hrd to predit. In his text, A First Course in Abstrt Algebr, Frleigh lims tht the four properties tht hrterize every Cyley digrph hve been used in disovering groups.[3] Thus, we ttempt to disover group from digrph tht stisfies the four properties. Our proess is s follows: we hoose n rbitrry number of verties, hoose one to two types of edges tht re intended to be genertors, nd then ttempt to drw digrph tht enompsses ll four properties. Construting digrph the stisfies the first three is esy enough, but it tkes us severl ttempts to find digrph tht stisfies the lst ondition. The following is one of the first ttempts tht proves to fil the fourth property. The genertors nd re represented by solid rrows nd dshed rrows, respetively. e f d b Figure 7: Filed ttempt t drwing digrph tht stisfied ll four properties. As you n see, in Figure 7, the sequene of edges S 1 = { } is the sme s sequene S 2 = {,,, } if you strt t e. If the fourth property is stisfied, then no mtter wht vertex t whih we strt, we should end up t the sme vertex by following S 1 or S 2. However, if we strt t vertex nd follow S 1, we re t vertex d; nd if we strt t vertex nd follow S 2, we re t vertex. Sine we do not end t the sme vertex, 12
the fourth property is not stisfied, nd this digrph is not Cyley digrph of group. Using the sme proess, we finlly mnged to disover digrph tht turned out to be the Cyley digrph for the Dihedrl group on 10 elements, DiH 5, whih is the group of symmetries of regulr pentgon. In the following digrph, let the solid edges represent multiplition on the right by the genertor nd the dshed edges represent multiplition on the right by the genertor. 12 e j f h b d Figure 8: Cyley Digrph of DiH 5 e b d f h j e e b d f h j b d h f e j j e f h d b e j f h d b b b d h f j e b e j h f d d d f h j e b f f e j b d h h h d b j e f j j h f d b e Tble 1: Cyley Tble for DiH 5 We n onfirm tht the digrph we disovered is Cyley digrph for group by om- 12 Note tht the dshed edges hve no rrow tips; this is beuse eh undireted edge represent two direted edges, one in eh diretion, between the two verties. 13
pleting the Cyley Tble 13 for the set of verties in our digrph, s seen bove, nd then mthing this tble with the Cyley tble for DiH 5.[5] However, even though we suessfully disovered Cyley digrph, the question remins s to whether or not, given digrph, we n visully detet if the fourth property is stisfied. We ome to the onlusion tht there is no simple wy to hek the stisftion of the fourth property without heking every sequene of edges in the grph, whih is often more time onsuming thn just nming the elements nd investigting the multiplition tble. This question sprked n interest in finding properties of group tht ould be visully determined by its Cyley digrph, s the next setion revels. 3 Informtion given by the Struture of Cyley Digrph 3.1 Abelin Wondering wht properties of group we n determine from the struture of Cyley digrph, we begin with the following question: Cn you tell from Cyley digrph whether the orresponding group is belin? Frliegh lims tht the nswer is yes.[3] Thus, we ome up with method for nlyzing Cyley digrph to sertin if group is belin. Proposition 3.1. Given Cyley digrph G of some group G with generting set {,,..., g n }, represented by n different edge types in G, one n determine if G is belin by the following method: 1. Choose ny two types of edges G, sy nd. 2. Selet vertex in G, sy x. 3. Strting t x, follow edge, then edge. Note the vertex t whih you finish. 4. Now, strting t x, follow edge, then edge. Note the vertex t whih you finish. 13 A Cyley Tble is tble tht lists the elements of the group on the top row nd left-most olumn, nd eh box in ith row nd jth olumn is the element tht represents the ith element times the jth element.[3] 14
5. If the vertex t whih you finish fter the sequene of edges {, } is the sme vertex t whih you finished when you followed the sequene of edges {, }, then repet this proess for every pir of genertors. If you finish t the sme vertex eh time, then G is belin. 6. If you finish t different verties for ny of the pirs of genertors, then G is not belin. If we follow the bove proess, it is obvious tht eh pir of genertors will ommute. But why does this men tht G is belin? Proof. Suppose we hve Cyley digrph G of some group G with generting set {,,..., g n }. If we find tht there exist genertors g i nd g j suh tht g i g j g j g i by the bove proess, then we know tht G nnot be belin, sine every element of G must ommute. However, if we find tht eh pir of genertors {,,..., g n } in G ommute with eh other by the bove proess, then for ny two elements x, y G, we n write x nd y s some produt of genertors, x = g g b... g k nd y = g g b... g k. Then, we know tht: xy = (g g b... g k )(g g b... g k ) (11) = g g b... (g g k )g b... g k (12) = g g b... g (g b g k )... g k (13). (14) = (g g b... g k )(g g b... g k ) (15) = yx (16) Thus, we know tht for ny two elements x, y G, we hve xy = yx. definition, G must be belin. Hene, by Exmple 3.2. To illustrte this method, onsider the following Cyley digrph G, representing some group G. 15
e j d h f b Figure 9: Given Cyley Digrph G. Sine we only hve two genertors, we ll them nd. Now, let s selet vertex e. If we begin t e nd follow the sequene of edges {, } we will end t vertex d. Similrly, if we strt t vertex e nd follow the sequene of edges {, }, we will end t vertex d gin. Thus, e = = d nd e = = d so = = d. And, by Proposition 3.1, G is belin. After lose inspetion, you my notie tht G is tully Cyley digrph of Z 2 Z 5, with genertors (0, 1) nd (1, 0) s nd respetively. 16
(0, 0) (1, 0) (0, 4) (1, 4) (1, 1) (0, 1) (1, 3) (1, 2) (0, 3) (0, 2) Figure 10: Cyley Digrph of Z 2 Z 5 Sine we know tht Z 2 Z 5 is tully Z 10, whih we know to be belin, we get the result tht we would expet from Proposition 3.1. Exmple 3.3. On the other hnd, suppose you re given the following digrph G of some group G. e j f h b d Figure 11: Given Cyley Digrph G. Agin, we only hve two types of edges in our Cyley digrph, so we must use genertors nd gin. Let s hoose the vertex e gin. Then, beginning t vertex e, if we follow 17
the sequene of edges {, }, we will finish t vertex. And, strting t vertex e, if we follow the sequene of edges {, }, we will finish t vertex j. Sine = = j, then Proposition 3.1 tells us tht G is not belin. You my notie tht Figure 11 is isomorphi to Cyley digrph of DiH 5, whih we know to be nonbelin. 3.2 Cyli A similr question we onsider is: Cn you tell from Cyley digrph whether the orresponding group is yli? We hve no immedite nswer to this question from ny referene we ould find, s we did with the previous question, nd we find no elegnt nswer s we hd hoped, but we do mnge to desribe method for investigting if group is yli from Cyley digrph. Proposition 3.4. Given Cyley digrph G of some group G, it n be determined if G is yli if you n onstrut losed wlk in G, strting t the identity, onsisting of sequene of one single pth nd tht losed wlk inludes ll of G. Exmple 3.5. Suppose we onsider the following Cyley digrph of G of some group G. 0 b Figure 12: Given Cyley digrph G of group G. 18
By Proposition 3.4, if we n find losed wlk in G by repeting sequene of one single pth tht onsists of every vertex nd edge of G, then G is yli. Consider the following pth: {, }. We lim tht if you repet this sequene, then you will hve losed wlk of G tht is ll of G, nd thus prove tht G is yli. Let s hek: Strt t e. If we follow this sequene one, we hve trversed the following red edges, nd we inluded the following red verties in our wlk: 0 b Figure 13: Given Cyley digrph G of group G fter we hve trversed the edges in the sequene {, }. If we repet this sequene of edges over nd over gin, we get the following grphs, with the trversed edges in red, nd inluded verties in red: 19
0 b Figure 14: Given Cyley digrph G of group G fter we hve trversed the edges in the sequene {, } twie. 0 b Figure 15: Given Cyley digrph G of group G fter we hve trversed the edges in the sequene {, } three times. 20
0 b Figure 16: Given Cyley digrph G of group G fter we hve trversed the edges in the sequene {, } four times. 0 b Figure 17: Given Cyley digrph G of group G fter we hve trversed the edges in the sequene {, } five times. 21
0 b Figure 18: Given Cyley digrph G of group G fter we hve trversed the edges in the sequene {, } six times. Thus, by repeting tht sequene of edges {, } six times, we hve inluded every vertex nd every edge of G in our wlk nd returned to where we begn, t vertex e. By Proposition 3.4, we hve tht G is yli. In ft, G is isomorphi to the Cyley digrph of < Z 6, + > with generting set {2, 3}, where genertor orresponds to 2 nd orresponds to 3. And we know ll groups < Z n, + > to be yli. 3.3 Cyli Subgroups Now tht we know how to determine if group is yli bsed its representtion in Cyley digrph, we onsider how to determine, from Cyley digrph, the yli subgroups of the group being represented. Proposition 3.6. Given Cyley digrph G of some group G, you n find ll of the yli subgroups of G by the following method: 1. Choose vertex of G. Sy x. 2. Sine eh vertex of G n be represented by sequene of edges of G, use this representtion for x. 3. Let S be the set of verties rehed by strting t the identity nd repetedly pplying the sequene of edges tht represent x until you rrive bk t the identity. 22
4. Then the set S will ontin the elements of G in the yli subgroup of G generted by x. Note: To find ll yli subgroups of G, you n repet this proess for every element of G. Proof. Let G be Cyley digrph for some group G nd let y be n element of G. Suppose you used the bove proedure of repetedly pplying the sequene of edges tht represent y in G to get the set S of verties rehed by this proess. Then, we n think of eh vertex in S s power of y sine you hve rrived t tht vertex from the identity just by following y. And S = {y, y 2,..., y o(y) = e} is the yli subgroup generted by y. Exmple 3.7. Let s find yli subgroup of the following Cyley digrph group Ĝ. Ĝ of some e b d f Figure 19: Given Cyley digrph Ĝ of some group Ĝ. 14 We n hoose ny vertex in Ĝ, so let s hoose vertex. Our first step is to write s sequene of nd edges, strting t the identity. Thus, =. We will let S represent the set of verties rehed by repetedly pplying the sequene of edges {, }. Now, we will begin t the identity nd follow {, } to our vertex. Our pth will be highlighted in red, s well s the elements being dded to S. 14 = 1 = 2, b = g 3 1, =, d = g 3 2, f = 23
e b d f Figure 20: Given Cyley digrph {, }. Ĝ of some group Ĝ, following the sequene of edges At this point, S ontins. We will now repetedly follow the sequene of edges {, } until we rrive bk t e, keeping trk of our set S long the wy. e b d f Figure 21: Given Cyley digrph {, }. Ĝ of some group Ĝ, following the sequene of edges Now, S ontins nd. 24
e b d f Figure 22: Given Cyley digrph {, }. Ĝ of some group Ĝ, following the sequene of edges Now, S ontins,, nd f. e b d f Figure 23: Given Cyley digrph {, }. Ĝ of some group Ĝ, following the sequene of edges Thus, we re bk t e, nd S = {,, f, e}. Therefore, by Proposition 3.6, the set S represents the yli subgroup generted by, i.e., < > = {,, f, e} (17) = { =, = ( ) 2, f = ( ) 3, e = ( ) 4 }. (18) 25
We lso onsider whether, given Cyley digrph, if we n get informtion bout the norml subgroups of group. We found prtil nswer to this question fter we disovered the proess presented in the next setion, where we remove edge types from the digrph nd possibly disonnet the Cyley digrph. We will disuss how to look t Cyley digrph to find ertin norml subgroups of group lter in this next setion. 4 Disonneting Cyley Digrphs We find tht removing one type of edge from given Cyley digrph my give us vluble informtion bout the group. Does the grph sty onneted? Does it beome disonneted? If so, wht do the onneted omponents look like? The nswers to these questions led to interesting results bout the underlying group struture. 4.1 Conneted First, let s investigte wht hppens when we remove ll edges of one type from Cyley digrph, but the digrph remins onneted. Exmple 4.1. Consider the following Cyley digrph of < Z 6, + >, with generting set {1, 2}, where the solid edges represent 1 nd the dshed edges represent 2: 0 5 1 4 2 3 Figure 24: Cyley digrph of < Z 6, + > with generting set {1, 2}. If we tke wy the dshed edges, then we hve the following onneted digrph: 26
0 5 1 4 2 3 Figure 25: Resulting digrph when genertor 2 is removed. This is still Cyley digrph of < Z 6, + >, s you n see. Exmples similr to the one bove led us to the following theorem. Theorem 4.2. Let G be Cyley digrph of group G. Suppose ll edges of one type re removed from G, nd the resulting grph G remins onneted. Then G remins Cyley digrph of G. Proof. Let G represent Cyley digrph of group G. Then G stisfies the four properties. If we remove one type of edge from G, nd the resulting grph G is onneted, then we know tht G stisfies the first ondition. Sine G G, then it is esy to see how properties 2, 3, nd 4 hold in G. Sine the only hnge mde between grphs G nd G is the removl of edges, nd G stisfied properties 2, 3, nd 4, then the following is true: G will not hve more thn one edge from some vertex x to some vertex y, else G would hve filed property two. Every vertex in G will still hve extly one edge of eh type strting nd ending t tht vertex sine ll edges of just one type hve been removed. And, finlly, we know tht there re no sequenes in G tht fil the fourth property else those sequenes would hve filed the fourth property in G s well. Thus, G stisfies ll four properties nd must be Cyley digrph of G by Theorem 2.1. Theorem 4.2 mkes intuitive sense if we think bout the interprettion of the Cyley digrph in lgebri terms. If one genertor is removed from the originl generting set, but tht set still genertes the entire group, then we still hve generting set for the 27
group. Thus, there should be Cyley digrph to represent it reltive to the new generting set. In the previous exmple, it is obvious tht the seond digrph is still Cyley digrph of < Z 6, + > beuse the new generting set, {1}, still genertes < Z 6, + >. 4.2 Disonneted More interesting results ome from studying wht hppens when Cyley digrph beomes disonneted when one edge type is removed. Theorem 4.3. Let G be the Cyley digrph of group G. Suppose ll edges of one type re removed from G nd the resulting grph G is disonneted. Then: 1. The onneted omponent ontining the identity of G is the Cyley digrph of proper subgroup H of G. 2. The other onneted omponents orrespond to the distint left osets of H in G. Proof. (1) Eh onneted omponent in G is proper subgrph of G nd represents proper subset of G. Consider the onneted omponent H ontining the identity. Even though one type of edge hs been removed, the properties 2, 3, nd 4 were stisfied in G, nd these properties will still hold if we remove one edge type, s seen in the proof of Theorem 4.2. Thus, the onneted omponent ontining the identity stisfies ll four properties nd hs the identity element of G. And, by Theorem 2.2, this omponent represents Cyley digrph of proper subgroup, H, of G. (2) We will prove the seond prt of this theorem with double inlusion; we will show tht for some element x of G, the left oset xh is subset of the onneted omponent ontining x in G, nd then we will show tht the onneted omponent ontining x is subset of xh. Consider onneted omponent in G, nd let x be n element of tht omponent. Let h 1 H. Then there exists pth from the identity to h 1 in H. 28
Figure 26: There is some pth in H between e nd h 1. Sine this pth exists in G, nd G is Cyley digrph, then, by the third property, we know we n onstrut the sme pth from x to xh 1 in the onneted omponent ontining x in G. We n do this in the following wy: we know tht the first edge in the pth from e to h 1 will be djent to x beuse there existed extly one edge of eh type strting t x in G. Figure 27: We know tht this first edge is djent to x beuse the third property of Cyley digrphs is stisfied in G. Similrly, we know tht the next edge in the pth from e to h 1 will be djent to the vertex tht is djent to x by the first edge for the sme reson. 29
Figure 28: We know tht this seond edge is inident to the first edge beuse of the sme reson. In this wy, we n onstrut the sme pth from x to the vertex xh 1 tht existed between e nd h 1. We n ll this vertex xh 1 by the onvention of our Cyley digrph nd right multiplition. Figure 29: We n onstrut the pth represented by h 1, nd we know hve xh 1 in the onneted omponent ontining x. Thus, for ny element h H, we know xh is in the onneted omponent ontining x. Sine the set {xh h H} is the left oset xh, we hve xh is subset of the onneted omponent ontining x. We hve proven the forwrd inlusion. Now, onsider n element y of the onneted omponent ontining x. We know x nd y must be onneted by sequene of edges in G. Let s ll this sequene S. 30
Figure 30: Sine x nd y re in the sme onneted omponent, we know tht they re onneted by some sequene of edges S in G. Using the sme logi s before, we n onstrut tht sme sequene of edges S in H, strting t the identity. The vertex t whih the sequene S, strting from the identity, ends is n element of H, sy h. Figure 31: This sequene of edges S exists in H. By property four of Cyley digrphs, we know tht wherever this sequene S exists, it is equivlent to multiplition on the right by h. Thus, in the onneted omponent ontining x, we n interpret y s xh, sine if we follow the sequene S (equivlent to h ) strting t x, we get to y. Thus, every element of the onneted omponent ontining x n be written s xh where h H. Therefore the onneted omponent ontining x is subset of the left oset xh. And, we hve the reverse inlusion. So we hve the onneted omponent ontining x s subset of xh, nd xh s subset of the onneted omponent ontining x. Hene, the onneted omponent ontining x is the sme s the left oset xh. 31
In generl, we hve tht eh of the onneted omponents of G represent left oset of H in G. By this result, it follows tht: Corollry 4.4. The number of onneted omponents in G will be equl to the index of H in G. To illustrte the onept of Theorem 4.3, let s reonsider the Cyley digrph tht represents the group DiH 5, with generting set {, }, where eh solid edge is represented by multiplition on the right by nd eh dshed edge is represented by multiplition on the right by : e j f h b d Figure 32: Cyley Digrph of DiH 5 with generting set {, } nd right multiplition. If we tke wy genertor, represented by the solid, direted edges, then the orresponding digrph looks like the following: 32
e j f h b d Figure 33: Resulting digrph when edges re removed. Thus, by our theorem, H = {e, } is subgroup of DiH 5 nd the left osets re: H = {, = }, (19) ( ) 2 H = {g1 2 = b, g1 2 = }, (20) ( ) 3 H = {g1 3 = d, g1g 3 2 = h}, (21) ( ) 4 H = {g1 4 = f, g1g 4 2 = j}. (22) 4.3 Norml Subgroups In our proof, we show how one n get the left osets of ertin subgroup using given Cyley digrph, but it is lso possible to see the right osets of the subgroup s well. In order to produe the right osets using this proess, we need to hnge our onvention of multiplition on the right, to multiplition on the left. Thus, if we redefine our opertion nd let eh solid edge represent multiplition on the left by genertor nd eh dshed edge represent multiplition on the left by genertor, then we n pply the sme proess bove nd be ble to see the right osets of DiH 5. As you n see in the following figure, the underlying digrph struture is the sme, but the lbeling of the verties hnges slightly: 33
e j f h b d Figure 34: Cyley Digrph of DiH 5 with generting set {, } nd left multiplition. Now, let s remove the solid edges representing genertor gin: e j f h b d Figure 35: Resulting digrph when edges re removed. Now, n rgument nlogous to the proof of Theorem 4.3 tells us tht H = {e, } is subgroup of DiH 5, nd the orresponding right osets re: H = {, = j}, (23) H( ) 2 = { 1 = b, 1 = h}, (24) H( ) 3 = {g 3 1 = d, g 3 1 = }, (25) H( ) 4 = {g 4 1 = f, g 4 1 = }. (26) 34
Not only does this help us visulize the left nd right osets for group, but we n lso use this s method to test whether or not group hs norml subgroups. Corollry 4.5. Given Cyley digrph G of group G, you n pply Theorem 4.3 to obtin disonneted grph G tht represents the subgroup H of G nd the left osets of H. Then, you n find the right osets of H by using left multiplition insted of right multiplition in the pplition of Theorem 4.3. If the right nd left osets re the sme, then H is norml subgroup of G. In our exmple of the DiH 5 group, we n see tht H is not norml sine H = {, = } H = {, = j}. 4.4 Conjetures In studying the implitions of Theorem 4.3, we investigted ouple onjetures involving the struture of the onneted omponents of the disonneted digrph G by following our intuition on few smll exmples. Consider Cyley digrph of DiH 5 gin. e j g 4 1 h 1 g 3 1 Figure 36: Originl Cyley Digrph of DiH 5 with generting set {, } And let s remove ll edges from our Cyley digrph. 35
e j g 4 1 h 1 g 3 1 Figure 37: Resulting digrph when edges re removed. After we remove ll edges, there is extly one power of in eh of the onneted omponents in Figure 9, nd the number of onneted omponents is five, the order of in G. This exmple, long with ouple others, leds us to the following onjeture. Conjeture 4.6. Given Cyley digrph G of group G, when we remove ll edges belonging to genertor g, nd the resulting grph G is disonneted, then eh onneted omponent ontins extly one power g. Then, the number of onneted omponents in G orrespond to the order of g. But, we find tht this is not lwys true. Counterexmple 4.7. Consider the Quternion Group, Q 8 : 36
e g 3 1 d f 1 Figure 38: Cyley digrph of Q 8. If we remove the genertor, then we get: e g 3 1 d f 1 Figure 39: Resulting digrph when genertor is removed. As you n see, we only hve two onneted omponents, even though the order of is four, nd there re two powers of in eh omponent. Thus, this fils our onjeture. After we find Conjeture 4.6 to be wrong, we try to reson tht there hs to be t lest one power of the genertor removed in eh onneted omponent of G. 37
Conjeture 4.8. Orphn Problem: Is it possible, when genertor g is removed nd the grph beomes disonneted, for there to be n orphn onneted omponent tht hs no power of g in it? Severl exmples, inluding Q 8, give the impression tht no orphn omponent n exist. However, in trying to prove tht this is true, we found nother ounterexmple. Counterexmple 4.9. Here is group where we get n orphn onneted omponent (tully severl) when we remove one genertor. This is the Cyley digrph of the group A 4, the lternting group of even permuttions on four elements, with genertors r 1, represented by the solid blk edges, nd u 1, represented by the dshed edges. r 8 r 6 u 1 e r 2 r 1 r 5 r 3 r 4 r 7 u 2 u 3 Figure 40: Cyley digrph of A 4. If we remove the genertor u 1 from the Cyley digrph, we get the following disonneted grph: 38
r 8 r 6 u 1 e r 2 r 1 r 5 r 3 r 4 r 7 u 2 u 3 Figure 41: Resulting digrph when ll u 1 edges re removed from the Cyley digrph of A 4. As you n see, only two of the four onneted omponents ontin power of u 1, the omponent ontining u 1 nd the omponent ontining e = (u 1 ) 2. Thus, the bottom two omponents re orphns. If one n begin with given Cyley digrph of some group G nd then determine subgroup H of the group nd the orresponding osets of H, then n we begin with given group G nd proper subgroup H nd find representtion of tht group suh tht H nd orresponding osets re visible s in Theorem 4.3? This is the question whih sprked the following onjeture: Conjeture 4.10. Given group G, suppose we hve proper subgroup H of G. Then, we n onstrut the Cyley digrph of G in the following wy: 1. Choose generting set of H. Sy H =<,..., g k >. 2. Construt Cyley digrph of H. 3. Mke n 1 opies of H, where n is the index of H in G. 4. Choose n element x G \ H. 15 15 The symbol \ here mens minus, i.e., x G \ H mens tht x is n element of G, but not n element of H. 39
5. Then G =<,..., g k, x >. But, this onjeture turns out to be untrue. Counterexmple 4.11. Suppose we re given the group Z 2 Z 5 Z 7 Z 11. We hoose H to be Z 2 Z 5 {e} {e}. We n follow numbers one through four of the Conjeture 4.10 nd hoose generting set, sy < (1, 0, 0, 0), (0, 1, 0, 0) >, onstrut the orresponding digrph of H nd 76 other onneted omponents identil to H, but the issue omes with the onlusion in step five. Sy we hd hosen x to be (0, 0, 1, 0), whih is in Z 2 Z 5 Z 7 Z 11 but not Z 2 Z 5 {e} {e}. Then G < (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) >. For exmple, (1, 1, 1, 1) is in G, but not in < (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) >. However, we believe only slight ltertion to this onjeture is needed. Although it is left unproven, we believe the following to be true: Conjeture 4.12. Given group G, suppose we hve proper subgroup H of G. Then, we n onstrut the Cyley digrph of subgroup H + of G for whih H is lso proper subgroup in the following wy: 1. Choose generting set of H. Sy H =<,..., g k >. 2. Construt Cyley digrph of H. 3. Mke n 1 opies of H, where n is the index of H in G. 4. Choose n element x G \ H. 5. Then H + =<,..., g k, x >, where H + is subgroup of G. Thus, we my not lwys be ble to onstrut the Cyley digrph of G with this method (lthough we will in some instnes), we will lwys be ble to onstrut the Cyley digrph of subgroup of G on whih we n pply Theorem 4.3 with this method. 5 Further Reserh A thorough investigtion of the informtion you n disover bout the lgebri properties of group from Cyley digrph would tke more time nd resoures thn two semesters nd n undergrdute edution would llow. Therefore, there is muh more reserh to be done on this topi. Sine we were never ble to ome up with foolproof wy to generte digrph tht 40
stisfied the four properties, one might further investigte wys either to determine if the fourth property is stisfied in given digrph, hving lredy generted digrph stisfying the first three properties, or to ensure tht the fourth property is stisfied while in the proess of generting digrph. It would be interesting to investigte if there ws wy to esily hnge Cyley digrph from right multiplition to left. If this were possible, it would be muh esier to implement Corollry 4.5 in order to ompre left nd right osets of subgroup of the orresponding group. One might lso onsider other importnt properties of group tht my be visully determined by Cyley digrph besides being belin, being yli, nd investigting yli subgroups nd norml subgroups. There re mny lgebri properties tht mthemtiins vlue in group tht would be interesting to look for in Cyley digrph. Furthermore, noting tht some of our onstrutions for visully determining lgebri properties in Cyley digrph were ineffiient to implement for lrger groups, one ould ttempt to find more prtil wy of testing Cyley digrphs for yli subgroups, for exmple. Our method works, but uses n ineffiient brute fore pproh tht ould possibly be improved upon. 6 Conlusion The results of this pper primrily funtion to present different perspetive in two different fields of mthemtis. Grph Theory introdues unique wy of viewing group; we rely not on multiplition tbles or textbooks to tell us if Z 2 Z 5 is belin, or yli, we n onfirm these fts for ourselves using grphs nd Grph Theory nlysis. In ddition, we n look t digrphs in different wy; every digrph hs the potentil to be Cyley digrph if it fulfills our properties. A reful drwing of dots nd rrows n turn into representtion of group. The ombintion of these two fields introdues new ides nd onnetions for both fields to onsider. We mnged to pture the bstrtness of group nd put it into grph so we ould see eh element, represent ertin osets, nd investigte yli subgroups. It retes new wy to experiene, nd understnd, group theory. 7 Aknowledgments I owe muh grtitude to my thesis dvisor, Dr. Willim Higgins, for his mny ides nd suggestions throughout ll stges of this thesis. His guidne nd support ws invluble in this proess. Thnks to Dr. Adm Prker, who provided ssistne nd knowledge whenever I sked. 41
I lso thnk the Mthemtis Deprtment t Wittenberg University who provided me with the knowledge nd resoures neessry for my reserh. Referenes [1] G. L. Alexnderson: EULER AND KONIGSBERGS BRIDGES: A HISTORICAL VIEW Bulletin of the Amerin Mthemtil Soiety (2006). [2] L. Euler: Solutio Problemtis d Geometrim Situs Pertinentis (1736). [3] J. B. Frleigh: A First Course In Abstrt Algebr (Seventh Edition) Addison Wesley (2003). [4] F. Hrry: The bridges of Königsberg/Kliningrd: A Tle of One City Mth Horizons (2001). [5] J. Wvrik: Multiplition tbles for groups of smll order. http://www.mth.niu. edu/~behy/ol/grouptbles1.html. [6] D.B. West: Introdution to Grph Theory (Seond Edition) Prentie Hll (2001). 42