Research Article Cayley Graphs of Order 27p Are Hamiltonian

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1 Interntionl Journl of Comintorics Volume 2011, Article ID , 16 pges doi: /2011/ Reserch Article Cyley Grphs of Order 27p Are Hmiltonin Erhim Ghderpour nd Dve Witte Morris Deprtment of Mthemtics nd Computer Science, University of Lethridge, Leth-Bridge, AB, Cnd T1K 3M4 Correspondence should e ddressed to Dve Witte Morris, dve.morris@uleth.c Received 22 Jnury 2011; Accepted 18 April 2011 Acdemic Editor: Ci Heng Li Copyright q 2011 E. Ghderpour nd D. W. Morris. This is n open ccess rticle distriuted under the Cretive Commons Attriution License, which permits unrestricted use, distriution, nd reproduction in ny medium, provided the originl work is properly cited. Suppose tht G is finite group, such tht G 27p, wherep is prime. We show tht if S is ny generting set of G, then there is Hmiltonin cycle in the corresponding Cyley grph Cy G; S. 1. Introduction Theorem 1.1. If G 27p, wherep is prime, then every connected Cyley grph on G hs Hmiltonin cycle. Comining this with results in 1 3 estlishes tht Every Cyley grph on G hs hmiltonin cycle if G kp, where p is prime, 1 k<32, nd k / The reminder of the pper provides proof of the theorem. Here is n outline. Section 2 reclls known results on hmiltonin cycles in Cyley grphs; Section 3 presents the proof under the ssumption tht the Sylow p-sugroup of G is norml; Section 4 presents the proof under the ssumption tht the Sylow p-sugroups of G re not norml. 2. Preliminries: Known Results on Hmiltonin Cycles in Cyley Grphs For convenience, we record some known results tht provide hmiltonin cycles in vrious Cyley grphs, fter fixing some nottion.

2 2 Interntionl Journl of Comintorics Nottion 1 see 4, Sections 1.1 nd 5.1. For ny group G, we use the following nottion: 1 G denotes the commuttor sugroup G, G of G, 2 Z G denotes the center of G, 3 Φ G denotes the Frttini sugroup of G. For, G, weuse to denote the conjugte 1. Nottion 2. If s 1,s 2,...,s n is ny sequence, we use s 1,s 2,...,s n # to denote the sequence s 1,s 2,...,s n 1 tht is otined y deleting the lst term. Theorem 2.1 Mrušič, Durnerger, Keting-Witte 5. If G is cyclic group of prime-power order, then every connected Cyley grph on G hs hmiltonin cycle. Lemm 2.2 see 3, Lemm Let S generte the finite group G, nd let s S.If i s G, ii Cy G/ s ; S hs hmiltonin cycle, nd iii either 1 s Z G,or 2 s is prime, then Cy G; S hs hmiltonin cycle. Lemm 2.3 see 1, Lemm 2.7. Let S generte the finite group G, nd let s S. If i s G, ii s is divisor of pq,wherep nd q re distinct primes, iii s p Z G, iv G/ s is divisile y q, nd v Cy G/ s ; S hs hmiltonin cycle, then there is hmiltonin cycle in Cy G; S. The following results re well known nd esy to prove. Lemm 2.4 Fctor Group Lemm. Suppose tht i S is generting set of G, ii N is cyclic, norml sugroup of G, iii s 1 N,...,s n N is hmiltonin cycle in Cy G/N; S, nd iv the product s 1 s 2 s n genertes N. Then s 1,...,s n N is hmiltonin cycle in Cy G; S. Corollry 2.5. Suppose tht i S is generting set of G,

3 Interntionl Journl of Comintorics 3 ii N is norml sugroup of G, such tht N is prime, iii s t modn for some s, t S S 1 with s / t, nd iv there is hmiltonin cycle in Cy G/N; S tht uses t lest one edge lelled s. Then there is hmiltonin cycle in Cy G; S. Definition 2.6. If H is ny sugroup of G, then H \Cy G; S denotes the multigrph in which i the vertices re the right cosets of H, nd ii there is n edge joining Hg 1 nd Hg 2 for ech s S S 1, such tht g 1 s Hg 2. Thus, if there re two different elements s 1 nd s 2 of S S 1, such tht g 1 s 1 nd g 1 s 2 re oth in Hg 2, then the vertices Hg 1 nd Hg 2 re joined y doule edge. Lemm 2.7 see 3, Corollry 2.9. Suppose tht i S is generting set of G, ii H is sugroup of G, such tht H is prime, iii the quotient multigrph H \ Cy G; S hs hmiltonin cycle C, nd iv C uses some doule-edge of H \ Cy G; S. Then there is hmiltonin cycle in Cy G; S. Theorem 2.8 see 6, Corollry 3.3. Suppose tht i S is generting set of G, ii N is norml p-sugroup of G, nd iii st 1 N, for ll s, t S. Then Cy G; S hs hmiltonin cycle. Remrk 2.9. In the proof of our min result, we my ssume p 5, for otherwise either i G 54 is of the form 18q, where q is prime, nd so 3, Propostion 9.1 pplies, or ii G 3 4 is prime power, nd so the min theorem of 7 pplies. 3. Assume the Sylow p-sugroup of G Is Norml Nottion 3. Let i G e group of order 27p, where p is prime, nd p 5 see Remrk 2.9, ii S e miniml generting set for G, iii P Z p e Sylow p-sugroup of G, iv w e genertor of P,nd v Q e Sylow 3-sugroup of G.

4 4 Interntionl Journl of Comintorics Assumption 3.1. In this section, we ssume tht P is norml sugroup of G. Therefore G is semidirect product: G Q P. 3.1 We my ssume tht G is not cyclic of prime order for otherwise Theorem 2.1 pplies.this implies tht Q is nonelin nd cts nontrivilly on P;so G Q P is cyclic of order 3p. 3.2 Nottion 4. Since Q is 3-group nd cts nontrivilly on P Z p, we must hve p 1 mod 3. Thus, one my choose r Z, such tht r 3 1 ( mod p ), ut r / 1 ( mod p ). 3.3 Dividing r 3 1yr 1, we see tht r 2 r 1 0 ( mod p ) A Lemm Tht Applies to Both of the Possile Sylow 3-Sugroups There re only 2 nonelin groups of order 27, nd we will consider them s seprte cses, ut, first, we cover some common ground. Note Since Q is nonelin group of order 27, nd G Q P Q Z p, it is esy to see tht Q Φ Q Z Q Z G Φ G. 3.5 Lemm 3.2. Assume tht i s S S 1 Q, such tht s does not centrlize P, nd ii c C Q P \ Φ Q. Then we my ssume tht S is either {s, cw} or {s, c 2 w} or {s, scw} or {s, sc 2 w}. Proof. Since G/P Q is 2-generted group of prime-power order, there must e n element of S, such tht {s, } genertes G/P. We my write s i c j zw k, with 0 i 2, 1 j 2, z Z Q, nd 0 k<p. 3.6 Note the following. i By replcing with its inverse if necessry, we my ssume i {0, 1}.

5 Interntionl Journl of Comintorics 5 Thus, ii By pplying n utomorphism of G tht fixes s nd mps c to cz j, we my ssume tht z is trivil since cz j j c j z j2 c j z. iii By replcing w with w k if k / 0, we my ssume k {0, 1}. s i c j w k with i, k {0, 1}, nd j {1, 2}. 3.7 Cse 1 Assume generting sets. k 1. Then s, G, ndsos {s, }. This yields the four listed Cse 2 Assume k 0. Then s, Q, nd there must e third element of S,with/ Q; fter replcing w with n pproprite power, we my write tw with t Q. We must hve t s, Φ Q, for otherwise s, G which contrdicts the minimlity of S. Therefore t s i z with 0 i 2, nd z Φ Q Z G. 3.8 We my ssume the following. i i / 0, for otherwise z w S Z G P ;solemm 2.3 pplies. ii i 1, y replcing with its inverse if necessry. iii z / e, for otherwise s nd provide doule edge in Cy G/P; S ;socorollry 2.5 pplies. Then s 1 z w genertes Z G P. Consider the hmiltonin cycles ( 1,s 2) 3, ( ( 1,s 2) 3 #, ), ( ( 1,s 2) 3 ##, 2) 3.9 in Cy G/ z, w ; S. Letting z 1 s 2 3 z, we see tht their endpoints in G re resp. z, z ( ) s 1 z z w, z ( ) s ( ) s 1 s 1 z ( z ) 2 w s w The finl two endpoints oth hve nontrivil projection to P since s, eing 3-element, cnnot invert w, nd t lest one of these two endpoints lso hs nontrivil projection to Z G. Such n endpoint genertes Z G P z, w, nd so the Fctor Group Lemm 2.4 provides hmiltonin cycle in Cy G; S Sylow 3-Sugroup of Exponent 3 Lemm 3.3. Assume tht Q is of exponent 3; so Q x, y, z x 3 y 3 z 3 e, [ x, y ] z, x, z [ y, z ] e. 3.11

6 6 Interntionl Journl of Comintorics Then one my ssume the following: 1 w x w r,uty nd z centrlize P, nd 2 either S {x, yw}, or S {x, xyw}. Proof. 1 Since Q cts nontrivilly on P, nd Aut P is cyclic, ut Q/Φ Q is not cyclic, there must e elements nd of Q \ Φ Q, such tht centrlizes P,ut does not. And z must centrlize P, ecuse it is in Q. By pplying n utomorphism of Q, we my ssume y nd x. Furthermore, we my ssume w x w r y replcing x with its inverse if necessry. 2 S must contin n element tht does not centrlize P; so we my ssume x S.By pplying Lemm 3.2 with s x nd c y, we see tht we my ssume tht S is { } { } x, yw or x, y 2 w or { x, xyw } { } or x, xy 2 w But there is n utomorphism of G tht fixes x nd w nd sends y to y 2 ; so we need only consider two of these possiilities. Proposition 3.4. Assume, s usul, tht G 27p,wherep is prime, nd tht G hs norml Sylow p-sugroup. If the Sylow 3-sugroup Q is of exponent 3, then Cy G; S hs hmiltonin cycle. Proof. We write for the nturl homomorphism from G to G G/P. FromLemm 3.3 2, we see tht we need only consider two possiilities for S. Cse 1 Assume S {x, yw}. For x nd yw, we hve the following hmiltonin cycle in Cy G/P; S : e x x 2 x 2 y 1 y 2 z 2 z 2 xz 2 x 2 z 2 xyz x 2 yz 2 1 yz 2 yz xy xy 2 z 2 x 2 y 2 x 2 y 2 z 2 xy 2 x 2 y 2 z y 2 1 y 2 z y x 2 z z 1 e. x 2 yz 1 xyz 2 xz 1 xy 2 z 3.13 Its endpoint in G is x 2 ywx 2( yw ) 2 x 2 ywx 2 ywx ( yw ) 2 x 1 ywx 2 ywx ( yw ) 1 x 2 ( yw ) x 2 ywxy 2 w 2 x 2 ywx 2 ywxy 2 w 2 x 2 ywx 2 ywxy 2 w 1 x 2 yw 2.

7 Interntionl Journl of Comintorics 7 Since the wlk is hmiltonin cycle in G/P, we know tht this endpoint is in P w. So ll terms except powers of w must cncel. Thus, we need only clculte the contriution from ech ppernce of w in this expression. To do this, note tht if term w i is followed y net totl of j ppernces of x, then the term contriutes fctor of w irj to the product. So the endpoint in G is w r13 w 2r12 w r10 w r8 w 2r7 w r5 w r3 w r2 w Since r 3 1 modp, this simplifies to w r w 2 w r w r2 w 2r w r2 ww r2 w 2 w r 2 r r2 2r r 2 1 r 2 2 w r2 4r 1 w r2 r 1 w 3r w 0 w 3r w 3r Since p 3r, this endpoint genertes P; so the Fctor Group Lemm 2.4 provides hmiltonin cycle in Cy G; S. Cse 2 Assume S {x, xyw}. For x nd xyw, we hve the hmiltonin cycle ( (, 2) 3 #, ) in Cy G/P; S. Its endpoint in G is ( ( 2) ) 3 3 ( ( 1 x ( xyw ) ) ) 2 3 ( ) 3 ( ( 1x xyw x(x 2 y 2 w r 1)) 3 ( w 1 y 1 x 1) ) 3 x ( ( y 2 w r 1) 3 (w 1 y 1)) 3 (w ( 3 r 1 w 1 y 1)) 3 (y 1 w 3r 2) 3 w 3 3r Since we re free to choose r to e either of the two primitive cue roots of 1 in Z p,ndthe eqution 3r 2 0 hs only one solution in Z p, we my ssume tht r hs een selected to mke the exponent nonzero. Then the Fctor Group Lemm 2.4 provides hmiltonin cycle in Cy G; S Sylow 3-Sugroup of Exponent 9 Lemm 3.5. Assume tht Q is of exponent 9; so Q x, y x 9 y 3 e, [ x, y ] x

8 8 Interntionl Journl of Comintorics There re two possiilities for G, depending on whether C Q P contins n element of order 9 or not. 1 Assume tht C Q P does not contin n element of order 9. Then we my ssume tht y centrlizes P,utw x w r. Furthermore, we my ssume tht: S {x, yw}, or S {x, xyw}. 2 Assume tht C Q P contins n element of order 9. Then we my ssume x centrlizes P, ut w y w r. Furthermore, we my ssume tht: S {xw, y}, S {xyw, y}, c S {xy, xw}, or d S {xy, x 2 yw}. Proof. 1 Since x hs order 9, we know tht it does not centrlize P. Butx 3 must centrlize P since x 3 is in G. Therefore, we my ssume w x x r y replcing x with its inverse if necessry. Also,sinceQ/C Q P must e cyclic ecuse Aut P is cyclic, utc G P does not contin n element of order 9, we see tht C Q P contins every element of order 3; so y must e in C Q P. Since S must contin n element tht does not centrlize P, we my ssume x S. By pplying Lemm 3.2 with s x nd c y, we see tht we my ssume tht S is: { } { } x, yw or x, y 2 w or { x, xyw } { } or x, xy 2 w The second generting set need not e considered, ecuse y 2 w 1 yw 1 yw ;soitis equivlent to the first. Also, the fourth generting set cn e converted into the third, since there is n utomorphism of G tht fixes y, ut tkes x to xyw nd w to w 1. 2 We my ssume x C Q P ;soc Q P x. We know tht S must contin n element s tht does not centrlize P, nd there re two possiilities: either I s hs order 3, or II s hs order 9. We consider these two possiilities s seprte cses. Cse I Assume tht s hs order 3. We my ssume s y. Letting c x, weseefrom Lemm 3.2 tht we my ssume S is either { } { } y, xw or y, x 2 w or { y, yxw } { } or y, yx 2 w The second nd fourth generting sets need not e considered, ecuse there is n utomorphism of G tht fixes y nd w, ut tkes x to x 2. Also, the third generting set my e replced with {y, xyw}, since there is n utomorphism of G tht fixes y nd w, ut tkes x to y 1 xy.

9 Interntionl Journl of Comintorics 9 Cse II Assume tht s hs order 9. We my ssume s xy. Letting c x, weseefrom Lemm 3.2 tht we my ssume tht S is either { } { } xy, xw or xy, x 2 w or { xy, xyxw } { } or xy, xyx 2 w The second generting set is equivlent to {xy, xw}, since the utomorphism of G tht sends x to x 4, y to x 3 y,ndw to w 1 mps it to {xy, xw 1 }. The third generting set is mpped to {xy, x 2 yw} y the utomorphism tht sends x to x x, y nd y to x, y 1 y. The fourth generting set need not e considered, ecuse xyx 2 w is n element of order 3 tht does not centrlize P, which puts it in the previous cse. Proposition 3.6. Assume, s usul, tht G 27p,wherep is prime, nd tht G hs norml Sylow p-sugroup. If the Sylow 3-sugroup Q is of exponent 9, then Cy G; S hs hmiltonin cycle. Proof. We will show tht, for n pproprite choice of nd in S S 1, the wlk ( 3, 1,, 1, 4, 2, 2,, 2,, 3,, 1, 1, 1, 2) 3.23 provides hmiltonin cycle in Cy G/P; S whose endpoint in G genertes P so the Fctor Group Lemm 2.4 pplies. We egin y verifying two situtions in which 3.23 is hmiltonin cycle. HC1 If 9, 3, nd 4 in G G/P, then we hve the hmiltonin cycle: e e HC2 If 9, 9, 7,nd 3 6 in G G/P, then we hve the hmiltonin cycle: e e. 3.25

10 10 Interntionl Journl of Comintorics To clculte the endpoint in G,fixr 1,r 2 Z p,with w w r 1, w w r 2, 3.26 nd write w 1, w 2, where, Q, w 1,w 2 P Note tht if n occurrence of w i in the product is followed y net totl of j 1 ppernces of nd net totl of j 2 ppernces of, then it contriutes fctor of w rj 1 1 rj 2 2 i to the product. A similr occurrence of w 1 i contriutes fctor of w rj1 1 rj 2 2 i to the product. Furthermore, since r 3 1 r3 2 1 modp, there is no hrm in reducing j 1 nd j 2 modulo 3. We will pply these considertions only in few prticulr situtions. E1 Assume w 1 e so Q nd. Then the endpoint of the pth in G is ( w 2 ) 1 ( w2 ) 1 4 ( w 2 ) 2 2 ( w 2 ) 2 ( w 2 ) 3 ( w 2 ) 1 ( w 2 ) 1 1 ( w 2 ) 2 3( w 1 2 1) ( w 1 2 1) 4( w 2 w 2 ) 2 ( w 2 ) ( w 2 ) 3 ( w 2 ) 1 ( w 1 2 1) 1( w w 1 2 1). By the ove considertions, this simplifies to w m 2, where m 1 r 2 1 r 2 r 1 r 2 r 1 r 2 2 r 1r 2 r 1 r 2 1 r 2 r 2 2 r 2 1 r 2 r 2 1 2r 1r 2 2r 1 r Note the following. If r 1 / 1ndr 2 1, then m simplifies to 6r 1, ecuse r 2 1 r modp in this cse. If r 1 / 1ndr 2 / 1, then m simplifies to 3r 1 r 2 1, ecuse r 2 1 r 1 1 r 2 2 r modp in this cse.

11 Interntionl Journl of Comintorics 11 E2 Assume w 2 e so Q nd. Then the endpoint of the pth in G is ( w 1 ) 3 1 ( w 1 ) 1 ( w 1 ) 4 2 ( w 1 ) 2 ( w1 ) 2 ( w1 ) 3 ( w1 ) 1 1 ( w 1 ) 1 2 ( w 1 w 1 w 1 ) 1 ( w 1 ) 1 ( w 1 w 1 w 1 w 1 ) 2 ( w w 1 1 1) ( w 1 w 1 ) ( w1 w 1 w 1 ) ( w 1 1 1) 1( w 1 1 1) By the ove considertions, this simplifies to w m 1, where m r 2 1 r 1 1 r 2 1 r 2 r 1 r 2 2 r2 2 r2 1 r2 2 r 1r 2 2 r 1 r 2 1 r2 1 r2 2 r 1r 2 2 r 2 r 2 1 r 2 r 1 r 2 r 1 r 2 1 r r 2 1 r2 2 3r 1r 2 2 r2 2 r2 1 r 2 r 1 r 2 r 2 r 1 1. Note the following. If r 1 1ndr 2 / 1, then m simplifies to 3 r 2 2, ecuse r 2 2 r modp in this cse. If r 1 / 1ndr 2 / 1, then m simplifies to r 1 r 2 2r 1 r 2 2, ecuse r 2 1 r 1 1 r 2 2 r modp in this cse. Now we provide hmiltonin cycle for ech of the generting sets listed in Lemm If C Q P hs exponent 3, nd S {x, yw}, welet x nd yw in HC1. In this cse, we hve w 1 e, r 1 r, ndr 2 1; so E1 tells us tht the endpoint in G is w 6r 2. 1 If C Q P hs exponent 3, nd S {x, xyw},welet x nd xyw 1 in HC2. In this cse, we hve w 1 e, r 1 r, ndr 2 r 1 r 2 ;so E1 tells us tht the endpoint in G is w m 2, where ( ) ( ) m 3r 1 r 2 1 3r r r 3 r 3 1 r 3 r 1 ( mod p ) If C Q P hs exponent 9, nd S {xw, y}, welet xw nd y in HC1. In this cse, we hve w 2 e, r 1 1, nd r 2 r; so E2 tells us tht the endpoint in G is w 3 r If C Q P hs exponent 9, nd S {xyw, y}, welet xyw nd y in HC1. In this cse, we hve w 2 e nd r 1 r 2 r; so E2 tells us tht the endpoint in G is w m 2, where ( ) m r 1 r 2 2r 1 r 2 2 r 2 2r r 2 r 2 r ( mod p ). 3.33

12 12 Interntionl Journl of Comintorics 2c If C Q P hs exponent 9, nd S {xy, xw},welet xw nd xy 1 in HC2. In this cse, we hve w 2 e, r 1 1, nd r 2 r 1 r 2 ;so E2 tells us tht the endpoint in G is w m 1, where ( ) m 3 r r r r 1 ( mod p ) d If C Q P hs exponent 9, nd S {xy, x 2 yw},welet xy nd x 2 yw in HC2. In this cse, we hve w 1 e nd r 1 r 2 r; so E1 tells us tht the endpoint in G is w m 2, where ( ) m 3r 1 r 2 1 3r r 1 3 r 2 r ( mod p ) In ll cses, there is t most one nonzero vlue of r modulo p for which the exponent of w i is 0. Since we re free to choose r to e either of the two primitive cue roots of 1 in Z p,we my ssume tht r hs een selected to mke the exponent nonzero. Then the Fctor Group Lemm 2.4 provides hmiltonin cycle in Cy G; S. 4. Assume the Sylow p-sugroups of G Are Not Norml Lemm 4.1. Assume tht i G 27p, wherep is n odd prime, nd ii the Sylow p-sugroups of G re not norml. Then p 13, nd G Z 13 Z 3 3, where genertor w of Z 13 cts on Z 3 3 vi multipliction on the right y the mtrix W Furthermore, we my ssume tht S is of the form { } w i,w j v, 4.2 where v 1, 0, 0 Z 3 3, nd ( i, j ) { 1, 0, 2, 0, 1, 2, 1, 3, 1, 5, 1, 6, 2, 5 }. 4.3 Proof. Let P e Sylow p-sugroup of G,ndletQ e Sylow 3-sugroup of G. Since no odd prime divides 3 1or3 2 1, nd 13 is the only odd prime tht divides 3 3 1, Sylow s Theorem 8, Theorem 15.7, pge 230 implies tht p 13, nd tht N G P P;soG must hve norml

13 Interntionl Journl of Comintorics 13 p-complement 4, Theorem ; thtis,g P Q. Since P must ct nontrivilly on Q since P is not norml, we know tht it must ct nontrivilly on Q/Φ Q 4, Theorem 5.3.5, pge 180. However, P cnnot ct nontrivilly on n elementry elin group of order 3 or 3 2, ecuse P 13 is not divisor of 3 1or Therefore, we must hve Q/Φ Q 3 3 ; so Q must e elementry elin nd the ction of P is irreducile. Let W e the mtrix representing the ction of w on Z 3 3 with respect to some sis tht will e specified lter. In the polynomil ring Z 3 X, we hve the fctoriztion: X 13 1 ( ) ( ) ( ) ( ) X 1 X 3 X 1 X 3 X 2 1 X 3 X 2 X 1 X 3 X 2 X Since w 13 e, the miniml polynomil of W must e one of the fctors on the right-hnd side. By replcing w with n pproprite power, we my ssume tht it is the first fctor. Then, choosing ny nonzero v Z 3 3, the mtrix representtion of w with respect to the sis {v, v w,v w2 } is W the Rtionl Cnonicl Form. Now, let ζ e primitive 13th root of unity in the finite field GF 27. Then ny Glois utomorphism of GF 27 over GF 3 must rise ζ to power. Since the sugroup of order 3 in Z 13 is generted y the numer 3, we conclude tht the orit of ζ under the Glois group is {ζ, ζ 3,ζ 9 }. These must e the 3 roots of one of the irreducile fctors on the right-hnd side of 4.4. Thus, for ny k Z 13, the mtrices W k, W 3k,ndW 9k ll hve the sme miniml polynomil; so they re conjugte under GL 3 3. Thtis, W, W 3,W 9 powers of W in the sme row of the following rry re conjugte under GL 3 3 : W 2,W 5,W 6 W 4,W 12,W 10 W 7,W 8,W There is n element of S tht genertes G/Q P. Then hs order p; so, replcing it y conjugte, we my ssume P w, ndso w i for some i Z 13.From 4.5, we see tht we my ssume i {1, 2} perhps fter replcing y its inverse. Now let e the second element of S; so we my ssume w j v for some j. We my ssume 0 j 6 y replcing with its inverse, if necessry. We my lso ssume j / i, for otherwise S Q,ndsoTheorem 2.8 pplies. If j 0, then i, j is either 1, 0 or 2, 0, oth of which pper in the list; henceforth, let us ssume j / 0. Cse 1 Assume i 1. Since j / i, we must hve j {2, 3, 4, 5, 6}. Note tht since W 3 is conjugte to W under GL 3 3 since they re in the sme row of 4.5, we know tht the pir w, w 4 is isomorphic to the pir w 3, w 3 4 w 3,w 1.By replcing with its inverse, nd then interchnging nd, this is trnsformed to w, w 3. So we my ssume j / 4. Cse 2 Assume i 2. We my ssume tht W j is in the second or fourth row of the tle for otherwise we could interchnge with to enter the previous cse. So j {2, 5, 6}. Since

14 14 Interntionl Journl of Comintorics j / i, this implies j {5, 6}. However, since W 5 is conjugte to W 2 since they re in the sme row of 4.5, nd we hve w 2 3 w 6 nd w 5 3 w 2, we see tht the pir w 2,w 6 is isomorphic to w 2,w 5. So we my ssume j / 6. Proposition 4.2. If G 27p, wherep is prime, nd the Sylow p-sugroups of G re not norml, then Cy G; S hs hmiltonin cycle. Proof. From Lemm 4.1 nd Remrk 2.9, we my ssume G Z 13 Z 3 3. For ech of the generting sets listed in Lemm 4.1, we provide n explicit hmiltonin cycle in the quotient multigrph P \ Cy G; S tht uses t lest one doule edge. So Lemm 2.7 pplies. To sve spce, we use i 1 i 2 i 3 to denote the vertex P i 1,i 2,i 3. i, j 1, 0 w, 1 w 12, 1, 0, 0, nd 1 1, 0, 0 Doule edge: with 1 nd : i, j 2, 0 w 2, 1 w 11, 1, 0, 0, nd 1 1, 0, 0 Doule edge: with nd 1 : i, j 1, 2 w, 1 w 12, w 2 1, 0, 0, nd 1 w 11 1, 1, 1 Doule edge: with nd :

15 Interntionl Journl of Comintorics 15 i, j 1, 3 w, 1 w 12, w 3 1, 0, 0, nd 1 w 10 0, 1, 1 Doule edge: with nd : i, j 1, 5 w, 1 w 12, w 5 1, 0, 0, nd 1 w 8 1, 0, 1 Doule edge: with nd 1 : i, j 1, 6 w, 1 w 12, w 6 1, 0, 0, nd 1 w 7 1, 1, 1 Doule edge: with 1 nd : i, j 2, 5 w 2, 1 w 11, w 5 1, 0, 0, nd 1 w 8 1, 0, 1 Doule edge: with 1 nd :

16 16 Interntionl Journl of Comintorics Acknowledgments This work ws prtilly supported y reserch grnts from the Nturl Sciences nd Engineering Reserch Council of Cnd. References 1 S. J. Currn, D. W. Morris, nd J. Morris, Hmiltonin cycles in Cyley grphs of order 16p, Preprint. 2 E. Ghderpour nd D. W. Morris, Cyley grphs of order 30p re hmiltonin, Preprint. 3 K. Kutnr, D. Mrušič, J. Morris, D. W. Morris, nd P. Šprl, Hmiltonin cycles in Cyley grphs whose order hs few prime fctors, Ars Mthemtic Contemporne. In press. 4 D. Gorenstein, Finite Groups, Chelse, New York, NY, USA, K. Keting nd D. Witte, Hmilton cycles in Cyley grphs with cyclic commuttor sugroup, Annls of Discrete Mthemtics, vol. 27, pp , D. W. Morris, 2-generted Cyley digrphs on nilpotent groups hve hmiltonin pths, Preprint. 7 D. Witte, Cyley digrphs of prime-power order re hmiltonin, Journl of Comintoril Theory, Series B, vol. 40, no. 1, pp , T. W. Judson, Astrct Alger, Virgini Commonwelth University, Richmond, V, USA, 2009.

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