5-cycles and the Petersen graph

5-cycles and the P etersen graph M. DeV os ⊥ , V. V. Mkrtc h y an †‡ ∗ , S. S. P etrosy an † , ⊥ Departmen t of Mathematics, Simon F raser Univ ersit y , Canada † Departmen t of Informatics and Applied Mathematics, Y erev an State Univ ersit y , Y erev an, 0025, Armenia ‡ Institute for Informatics and Automation Problems, NAS RA, 0014, Armenia email: mdev os@sfu.ca v ahanmkrtc h y an2002@ { ysu.am, ipia.sci.am, yahoo.com } sam v elp etrosy an@gmail.com Octob er 29, 2018 Abstract W e sho w that i f G i s a connected bridgeless cubic graph whose every 2-factor is comprised of cycles of length five then G is the Petersen graph. ”The Petersen graph is an obstruc tio n to man y prop erties in g r aph theor y , and often is, or conjectured to b e, the only obstruction”. This phrase is tak en from one of the series of pa pers by Rob ertson, Sa nder s, Seymour and Thomas that is devoted to the pro of of prominent T utte co njectur e - a conjecture which states that if the Petersen graph is not a minor of a bridgeless cubic gr aph G then G is 3 -edge-color a ble, and whic h in its turn is a particula r case of a m uch more gener al conjecture of T utte stating that every bridgeless g r aph G has a nowhere zero 4- flo w unless the Petersen gr aph is not a minor of G . Another result that stres ses the exceptional role of the Petersen graph is prov ed by Alspach et al. in [1 ]. The following striking conjecture of J aeger states that everything related to the c olorings o f bridgeles s cubic graphs can b e reduced to that o f the Petersen graph, mor e specifica lly , Conjecture 1 Petersen c olori ng c onje ctur e of Jae ger [4]: the e dges of every bridgeless cubic gr aph G c an b e mapp e d into the e dges of t he Petersen gr aph in such a wa y that any thr e e mutual ly incident e dges of G ar e mapp e d t o thr e e mutual ly incident e dges of the Petersen gr aph . ∗ The author is supp orted by a gr an t of Armenian N ationa l Science and Education F und 1 The Petersen graph is so imp ortant in gr aph theory that even an entire bo ok is written ab out it [2]. It is known that for every bridgeless cubic graph G and its tw o edg es e and f w e can always find a 2-fac to r of G containing these tw o edges. Rece ntly , J ac k- son a nd Y o s himoto in [3] observed that in any bridgele ss cubic graph without m ultiple edges we can alw ays find a triangle-fre e 2- factor. An ea rlier result by Rosenfeld [7] also w or ths to be mentioned here, which states that there are in- finitely many 3-connected cubic gra phs whose every 2- fa ctor contains a cycle of length no more five. Sometimes it is convinient to hav e a 2-fa ctor F o f a bridge less cubic gra ph G such that not all cycles of F are 5-cycles (=cycles o f length fiv e) [6]. The main reason why we are in tere sted in 5-cycles is the following: it is known that it is the o dd cycles of a g raph (particular ly , a bridgeless cubic gra ph) that pr ev ent it to have a 3-edg e-coloring. F ortunately , the tr iangles can b e ov ercome easily . This is due to the op eration of the contraction of the triangles, which pres e r v es the cubicness a nd br idgelessness of a gra ph. Therefore, we need a tec hnique to co pe with o dd cycles of length at lea st five, and particular ly , the 5-cycles . The main result of this pap er s tates that unless G is the Petersen gra ph in every connected br idgeless cubic graph G we c an always find a 2-factor F that contains a cyc le which is not a 5-cycle. W e consider finite, undirected gra phs without lo ops. G r a phs may con tain m ultiple edges. W e follow [5, 9] for the terminology . Theorem 2 (T utte, [8]): A gr aph G c ontains a p erfe ct matching if and only if for every S ⊆ V ( G ) o ( G − S ) ≤ | S | , wher e o ( H ) denotes the numb er of o dd c omp onents of a gr aph H . Corollary 3 If G is a bridgeless cubic gr aph , then for every its e dge e ther e is a p erfe ct matching F with e ∈ F . This corollar y immediately implies Corollary 4 If G is a bridgeless cubic gr aph , then for every its e dge e ther e is a p erfe ct matching F with e / ∈ F . W e will a lso need the following prop erty of the Petersen graph: Prop osition 5 Th e Petersen gr aph is the u nique cubic gr aph of girth five on ten vertic es. W e are r eady to state the main result of the paper : Theorem 6 If G is a c onne cte d bridgeless cubic gr ap h whose every 2 -factor is c omprise d of 5 - cycle s t hen G is t he Petersen gr aph. Pro of. First of a ll note that G is not 3-edge- colorable, th us every 2-facto r of G contains a t least t wo o dd cy cles. 2 Claim 7 G do es not have a cycle of length two. Pro of. Supp ose that G contains a cycle C o f length tw o. Let u and v b e the vertices of C , and let u ′ , v ′ be the other ( 6 = v , 6 = u ) neig h b ours o f u a nd v , resp ectively . Note that since G is bridg eless, we hav e u ′ 6 = v ′ . Now let F be a p erfect matching of G co n taining the edge ( u, u ′ ) (corollar y 3). Cle a rly , ( v , v ′ ) ∈ F . Co ns ider the co mplemen tary 2-facto r of F . Note that C is a cyc le in this 2-factor c o n tradicting the co ndition of the theor em. Claim 8 G do es not have two t r iangles sharing an e dge. Pro of. Let u, v , w and u ′ , v , w b e tw o tr iangles of G which share the edge ( v , w ) . Clearly , ( u, u ′ ) / ∈ E ( G ), as G is not 3-e dge-colorable. Let u 1 and u ′ 1 be the other ( 6 = v, w ) neighbour s of u a nd u ′ , res pectively . Note that since G is bridgeless we hav e u 1 6 = u ′ 1 . Consider a p erfect matching F with ( v , w ) ∈ F (cor ollary 3). Clearly , ( u, u 1 ) , ( u 1 , u ′ 1 ) ∈ F . Note tha t the complementary 2 -factor of F contains the 4-cy c le o n vertices u, v , u ′ , w c on tradicting the condition of theor em. Claim 9 G do es not have a squar e and a t riangle sharing an e dge. Pro of. Suppo se, on the contrary , that G contains a square ( u, v ), ( v , x ), ( x, w ), ( w, u ) and a triang le ( v , y ) , ( y , x ) , ( x, v ) which shar e the edge ( x, v ). Due to claim 8, ( u, x ) / ∈ E ( G ), ( v , w ) / ∈ E ( G ). Now, let F b e a p erfect matching of G containing the edge ( u , w ) (corollar y 3 ). Clearly , ( v , x ) ∈ F and there is a vertex y ′ / ∈ { u, v , x, w , y } such that ( y , y ′ ) ∈ E ( G ). Now, consider a path u, ( u , v ) , v , ( v , y ) , y , ( y , x ) , x, ( x, w ) , w of length four. The path lies on a cycle C of the co mplemen tary 2-factor of F . Due to claim 7 there is only one edge connecting u and w . Thu s the length of C is at least six contradicting the condition o f theorem. Claim 10 G do es not have a triangle. Pro of. On the oppo site assumption, consider a tr iangle C o n vertices x, y , z o f G . Since G is bridgele s s we imply that there are vertices x ′ , y ′ , z ′ adjacent to x, y , z , res p ectively , that do not lie o n C . Now, consider a perfect matc hing F of G containing the edge ( x, y ) (coro llary 3). Clearly , ( z , z ′ ) ∈ F . Note that the path y ′ , ( y ′ , y ) , y , ( y , z ) , z , ( z , x ) , x, ( x, x ′ ) , x ′ of length four lies on some cycle C ′ of the complementary 2 -factor of F . Claim 8 implies that ( y ′ , x ′ ) / ∈ E ( G ) thus the leng th of C ′ is at least six c on tradicting the condition of theo rem. Claim 11 G do es not have a squar e, and girth of G is five. Pro of. Assume G to contain a square C = ( u, v ) , ( v , w ) , ( w , z ) , ( z , u ). Claim 10 implies that ( u, w ) / ∈ E ( G ), ( v , z ) / ∈ E ( G ). Let u 1 , v 1 , w 1 , z 1 be the vertices of G that are adjacent to u , v , w , z , r espectively and do not lie on C . Consider a p erfect matching F o f G containing the edge ( u, u 1 ) (corollar y 3) . Clea rly , 3 { ( v , v 1 ) , ( w , w 1 ) , ( z , z 1 ) } * F , as if it were true then the co mplementary 2- factor of F would hav e cont ained C as a cycle, which contradicts the condition of theore m. Thus | F ∩ { ( v , v 1 ) , ( w , w 1 ) , ( z , z 1 ) }| = 1 Without lo s s of generality , we may assume that ( v , v 1 ) ∈ F . Note that ( w , z ) ∈ F . Now, consider the cy c le C F in the complementary 2-factor o f F , which contains the path z 1 , ( z 1 , z ) , z , ( z , u ) , u, ( u, v ) , v , ( v , w ) , w , ( w, w 1 ) , w 1 . Due to claim 10 z 1 6 = w 1 th us the length o f C F is at least six c o n tradicting the co ndition of theore m. Thus, G canno t contain a square, too , therefore its girth is five. Claim 12 G is 3 - e dg e-c onne cte d. Pro of. Suppo se, for a contradiction, that G is only 2-edge-connected, a nd let ( u, v ) , ( u ′ , v ′ ) b e tw o edg e s which form a 2-e dg e cut so that u and u ′ are in the same co mponent o f G \{ ( u, v ) , ( u ′ , v ′ ) } . Now, there must exist a per fect matching not using ( u, v ) (cor ollary 4), so the complementary 2-factor must contain a 5-cycle whic h uses b oth the edges ( u, v ) and ( u ′ , v ′ ). It follows that either ( u, u ′ ) ∈ E ( G ) or ( v , v ′ ) ∈ E ( G ). Without loss o f g enerality , we may assume tha t ( v , v ′ ) ∈ E ( G ). Let w b e the neighbor of v other than u, v ′ , a nd let w ′ be the neighbor of v ′ other than u ′ , v . Now, there ex ists a p erfect matching containing the edge ( v , v ′ ) (corollar y 3), a nd the complemen tary 2- factor must contain a 5-cycle whic h uses all of the edges ( u, v ) , ( v , w ) , ( u ′ , v ′ ) , ( v ′ , w ′ ). It follows tha t either u = u ′ or v = v ′ , but either of these cont ra dicts the fact that G is bridgeless. This co n tradiction shows that G is 3-e dg e connected. Claim 13 Every 3 -e dge-cut of G c onsists of thr e e e dges incident to a c ommon vertex. Pro of. Let ( U, ¯ U ) = { ( u 1 , v 1 ) , ( u 2 , v 2 ) , ( u 3 , v 3 ) } b e a 3 - cut of G and supp ose that { u 1 , u 2 , u 3 } ⊆ U , { v 1 , v 2 , v 3 } ⊆ ¯ U . W e claim that either u 1 = u 2 = u 3 or v 1 = v 2 = v 3 . Before showing this let us show that there is no edge connecting u i and u j or v i and v j , 1 ≤ i < j ≤ 3. On the o pposite ass umption, s uppose that ( v 1 , v 2 ) ∈ E ( G ). Let v ′ 1 and v ′ 2 be the neighbours of v 1 and v 2 , resp ectively , that are different fr o m u 1 , v 2 and u 2 , v 1 . Clearly , v ′ 1 , v ′ 2 ∈ ¯ U and cla im 10 implies that v ′ 1 6 = v ′ 2 . Consider a p erfect matching F 1 , 2 of G containing the edge ( v 1 , v 2 ) (corollar y 3). Since | U | is odd (( U, ¯ U ) is an odd cut), w e ha ve ( u 3 , v 3 ) ∈ F 1 , 2 . Thu s, the complementary 2-factor of F 1 , 2 m ust co n tain a 5-c y cle containing the edges ( u 1 , v 1 ) , ( v 1 , v ′ 1 ) , ( u 2 , v 2 ) , ( v 2 , v ′ 2 ). Since v ′ 1 6 = v ′ 2 we have u 1 = u 2 . Note that u 1 , v 1 , v 2 forms a tr iangle contradicting claim 10. Thus ( v 1 , v 2 ) / ∈ E ( G ). Similarly , the absence of the other edg es can b e shown. Now, let us turn to the pro of o f c la im 13. Let F b e a perfect matching missing ( u 1 , v 1 ) (corollary 4). Since | U | is o dd, we imply that F contains one of ( u 2 , v 2 ) , ( u 3 , v 3 ) and misses the other one. Without loss of genera lit y , we may assume that ( u 3 , v 3 ) ∈ F , ( u 2 , v 2 ) / ∈ F . Note that there should b e a 5-cycle containing b oth the edg e s ( u 2 , v 2 ) a nd ( u 3 , v 3 ). As ( u 1 , u 2 ) / ∈ F , ( v 1 , v 2 ) / ∈ F , we imply tha t either u 1 = u 2 or v 1 = v 2 . Again, we ca n a ssume that u 1 = u 2 . Let us show that u 1 = u 3 , to o. 4 Suppo se that u 1 6 = u 3 . Let w be a vertex fr om U adjacen t to the vertex u 1 = u 2 , and let U ′ = U \ { u 1 } . Note that ( U ′ , ¯ U ′ ) = { ( u 1 , w ) , ( u 3 , v 3 ) } is a 2-e dg e-cut of G co n tradicting the choice of cla im 12. Thus u 1 = u 3 and we ar e done. Claim 14 If u, v , w , x ∈ V ( G ) and ( u, v ) , ( v , w ) , ( w , x ) ∈ E ( G ) then ther e is a p erfe ct matching of G c o nt aining b oth ( u, v ) and ( w, x ) . Pro of. Supp ose, for a contradiction, that the statement do es no t hold. Then G ′ = G \{ u, v , w , x } has no p erfect matching, so by theor em 2 there exists a subset of v er tice s Y ⊆ V ( G ′ ) so that G ′ \ Y has more than | Y | o dd comp onen ts. Let I be the set o f is olated vertices in G ′ \ Y , let O b e the set of o dd comp onents of G ′ \ Y with at least three vertices, and let E b e the set of even co mponents of G ′ \ Y . W e know that | Y | < | I | + | O | by assumption, but in fact | Y | + 2 ≤ | I | + | O | since | Y | − | I | − | O | must b e an even num b er (a s | V ( G ′ ) | is even). Now, let Y + = Y ∪ { u, v , w , x } and let C b e the edge cut which separ ates Y + from V ( G ) \ Y + . It follows from o ur co nstruction that | C | ≤ 3 | Y | + 6 since every vertex in Y ca n c o n tribute a t most three edges to C , and there are a t most six edges in C with o ne o f u, v , w, x a s endpoint. On the other hand, claim 13 implies that every comp onen t in O ∪ E must contribute at least four edges to C , and every vertex in I co ntributes exa ctly three edge s to C , so | C | ≥ 3 | I | + 4 | O | + 4 | E | ≥ 3( | Y | + 2) + | O | + 4 | E | It follows fro m this that O = E = ∅ , a nd that every vertex in Y must hav e all three incident edges in C . Thus G \{ ( u, v ) , ( v , w ) , ( w, x ) } is a bipartite graph. Now, there exists a perfect matching of G which contains the edge ( u, v ), and every o dd cycle in the complemen tary 2-fac to r must contain ( v , w ) and ( w , x ), so the complement ar y 2-factor cannot hav e t wo o dd cycles - giving us a cont r a- diction. Now w e a re ready to complete the pro of of the theorem. Claim 14 im- plies that every 3-edge pa th must b e contained in a c y cle of length five, a nd it follows from this that every 2-edge path of is contained in at least tw o 5- cycles. Let u b e a vertex of G , let v , w , x b e the neig h b ors of u , a nd assume that the neighbors of v , w, x are { u, v 1 , v 2 } , { u, w 1 , w 2 } , and { u, x 1 , x 2 } , resp ec- tively . It follows from the fact tha t G has girth five that all of these vertices we have named are distinct. Since the 2-edge path with edges ( v, u ) , ( u, w ) is contained in tw o cycles of length five, there must b e at least t wo edges b etw een { v 1 , v 2 } and { w 1 , w 2 } . Similarly , there ar e at lea st t wo edges b etw een { w 1 , w 2 } and { x 1 , x 2 } , and betw een { x 1 , x 2 } and { v 1 , v 2 } . As G is connected w e imply that V ( G ) = { u, v , w , x, v 1 , v 2 , w 1 , w 2 , x 1 , x 2 } , and there are exactly t wo edges betw een { v 1 , v 2 } and { w 1 , w 2 } , { w 1 , w 2 } and { x 1 , x 2 } , { x 1 , x 2 } and { v 1 , v 2 } . Prop osition 5 implies tha t G is isomo rphic to the Petersen graph. References [1] B. Alspach, L. Goddy n, C. Q . 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