IJCTInternational Journal of Combinatorics1687-91711687-9163Hindawi Publishing Corporation20693010.1155/2011/206930206930Research ArticleCayley Graphs of Order 27p Are HamiltonianGhaderpourEbrahim1MorrisDave Witte1LiCai Heng1Department of Mathematics and Computer ScienceUniversity of LethbridgeLeth-BridgeABT1K 3M4Canadauleth.ca2011982011201122012011180420112011Copyright © 2011 Ebrahim Ghaderpour and Dave Witte Morris.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Suppose that G is a finite group, such that |G|=27p, where p is prime. We show that if S is any generating set of G, then there is a Hamiltonian cycle in the corresponding Cayley graph Cay (G;S).

1. IntroductionTheorem 1.1.

If |G|=27p, where p is prime, then every connected Cayley graph on G has a Hamiltonian cycle.

Combining this with results in  establishes that Every Cayley graph on  G  has a hamiltonian cycleif   |G|=kp,  where  p  is prime,  1k<32,  and  k24.

The remainder of the paper provides a proof of the theorem. Here is an outline. Section 2 recalls known results on hamiltonian cycles in Cayley graphs; Section 3 presents the proof under the assumption that the Sylow p-subgroup of G is normal; Section 4 presents the proof under the assumption that the Sylow p-subgroups of G are not normal.

2. Preliminaries: Known Results on Hamiltonian Cycles in Cayley Graphs

For convenience, we record some known results that provide hamiltonian cycles in various Cayley graphs, after fixing some notation.

Notation 1 (see [<xref ref-type="bibr" rid="B3">4</xref>, Sections  1.1 and 5.1]).

For any group G, we use the following notation:

G denotes the commutator subgroup [G,G] of G,

Z(G) denotes the center of G,

Φ(G) denotes the Frattini subgroup of G.

For a,bG, we use ab to denote the conjugate b-1ab.

Notation 2.

If (s1,s2,,sn) is any sequence, we use (s1,s2,,sn)# to denote the sequence (s1,s2,,sn-1) that is obtained by deleting the last term.

Theorem 2.1 (Marušič, Durnberger, Keating-Witte [<xref ref-type="bibr" rid="B5">5</xref>]).

If G is a cyclic group of prime-power order, then every connected Cayley graph on G has a hamiltonian cycle.

Lemma 2.2 (see [<xref ref-type="bibr" rid="B6">3</xref>, Lemma  2.27]).

Let S generate the finite group G, and let sS. If

sG,

Cay(G/s;S) has a hamiltonian cycle, and

either

sZ(G), or

|s| is prime,

then Cay(G;S) has a hamiltonian cycle.

Lemma 2.3 (see [<xref ref-type="bibr" rid="B1">1</xref>, Lemma  2.7]).

Let S generate the finite group G, and let sS. If

sG,

|s| is a divisor of pq, where p and q are distinct primes,

spZ(G),

|G/s| is divisible by q, and

Cay(G/s;S) has a hamiltonian cycle,

then there is a hamiltonian cycle in Cay(G;S).

The following results are well known (and easy to prove).

Lemma 2.4 (“Factor Group Lemma”).

Suppose that

S is a generating set of G,

N is a cyclic, normal subgroup of G,

(s1N,,snN) is a hamiltonian cycle in Cay(G/N;S), and

the product s1s2sn generates N.

Then (s1,,sn)|N| is a hamiltonian cycle in Cay(G;S).

Corollary 2.5.

Suppose that

S is a generating set of G,

N is a normal subgroup of G, such that |N| is prime,

st  (modN) for some s,tSS-1 with st, and

there is a hamiltonian cycle in Cay(G/N;S) that uses at least one edge labelled s.

Then there is a hamiltonian cycle in Cay(G;S).

Definition 2.6.

If H is any subgroup of G, then HCay(G;S) denotes the multigraph in which

the vertices are the right cosets of H, and

there is an edge joining Hg1 and Hg2 for each sSS-1, such that g1sHg2.

Thus, if there are two different elements s1 and s2 of SS-1, such that g1s1 and g1s2 are both in Hg2, then the vertices Hg1 and Hg2 are joined by a double edge.

Lemma 2.7 (see [<xref ref-type="bibr" rid="B6">3</xref>, Corollary  2.9]).

Suppose that

S is a generating set of G,

H is a subgroup of G, such that |H| is prime,

the quotient multigraph HCay(G;S) has a hamiltonian cycle C, and

C uses some double-edge of HCay(G;S).

Then there is a hamiltonian cycle in Cay(G;S).

Theorem 2.8 (see [<xref ref-type="bibr" rid="B7">6</xref>, Corollary  3.3]).

Suppose that

S is a generating set of G,

N is a normal p-subgroup of G, and

st-1N, for all s,tS.

Then Cay(G;S) has a hamiltonian cycle.

Remark 2.9.

In the proof of our main result, we may assume p5, for otherwise either

|G|=54 is of the form 18q, where q is prime, and so [3, Propostion  9.1] applies, or

|G|=34 is a prime power, and so the main theorem of  applies.

3. Assume the Sylow <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M112"><mml:mrow><mml:mi>p</mml:mi></mml:mrow></mml:math></inline-formula>-Subgroup of <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M113"><mml:mrow><mml:mi>G</mml:mi></mml:mrow></mml:math></inline-formula> Is NormalNotation 3.

Let

G be a group of order 27p, where p is prime, and p5  (see Remark 2.9),

S be a minimal generating set for G,

Pp be a Sylow p-subgroup of G,

w be a generator of P, and

Q be a Sylow 3-subgroup of G.

Assumption 3.1.

In this section, we assume that P is a normal subgroup of G.

Therefore G is a semidirect product: G=QP. We may assume that G is not cyclic of prime order (for otherwise Theorem 2.1 applies). This implies that Q is nonabelian and acts nontrivially on P; so G=Q×P  is  cyclic  of  order  3p.

Notation 4.

Since Q is a 3-group and acts nontrivially on Pp, we must have p1  (mod3). Thus, one may choose r, such that r31  (modp),  but  r1  (modp). Dividing r3-1 by r-1, we see that r2+r+10  (modp).

3.1. A Lemma That Applies to Both of the Possible Sylow 3-Subgroups

There are only 2 nonabelian groups of order 27, and we will consider them as separate cases, but, first, we cover some common ground.

Note

Since Q is a nonabelian group of order 27, and G=QPQp, it is easy to see that Q=Φ(Q)=Z(Q)=Z(G)=Φ(G).

Lemma 3.2.

Assume that

s(SS-1)Q, such that s does not centralize P, and

cCQ(P)Φ(Q).

Then we may assume that S is either {s,cw} or {s,c2w} or {s,scw} or {s,sc2w}.

Proof.

Since G/PQ is a 2-generated group of prime-power order, there must be an element a of S, such that {s,a} generates G/P. We may write a=sicjzwk,with  0i2,  1j2,  zZ(Q),  and  0k<p. Note the following.

By replacing a with its inverse if necessary, we may assume i{0,1}.

By applying an automorphism of G that fixes s and maps c to czj, we may assume that z is trivial (since (czj)j=cjzj2=cjz).

By replacing w with wk if k0, we may assume k{0,1}.

Thus, a=sicjwk  with  i,k{0,1},  and  j{1,2}.

Case 1 (<italic>Assume </italic><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M174"><mml:mi>k</mml:mi><mml:mo>=</mml:mo><mml:mn>1</mml:mn></mml:math></inline-formula>).

Then s,a=G, and so S={s,a}. This yields the four listed generating sets.

Case 2 (<italic>Assume </italic><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M177"><mml:mi>k</mml:mi><mml:mo>=</mml:mo><mml:mn>0</mml:mn></mml:math></inline-formula>).

Then s,a=Q, and there must be a third element b of S, with bQ; after replacing w with an appropriate power, we may write b=tw with tQ. We must have ts,Φ(Q), for otherwise s,b=G (which contradicts the minimality of S). Therefore t=sizwith  0i2,  and  zΦ(Q)=Z(G). We may assume the following.

i0, for otherwise b=zwS(Z(G)×P); so Lemma 2.3 applies.

i=1, by replacing b with its inverse if necessary.

ze, for otherwise s and b provide a double edge in Cay(G/P;S); so Corollary 2.5 applies.

Then s-1b=zw generates Z(G)×P.

Consider the hamiltonian cycles (a-1,s2)3,((a-1,s2)3#,b),((a-1,s2)3##,b2) in Cay(G/z,w;S). Letting z=(a-1s2)3z, we see that their endpoints in G are (resp.) z,z(s-1b)=zz  w,z(s-1b)s(s-1b)=z(z)2  wsw. The final two endpoints both have a nontrivial projection to P (since s, being a 3-element, cannot invert w), and at least one of these two endpoints also has a nontrivial projection to Z(G). Such an endpoint generates Z(G)×P=z,w, and so the Factor Group Lemma 2.4 provides a hamiltonian cycle in Cay(G;S).

3.2. Sylow 3-Subgroup of Exponent 3Lemma 3.3.

Assume that Q is of exponent 3; so Q=x,y,zx3=y3=z3=e,  [x,y]=z,  [x,z]=[y,z]=e. Then one may assume the following:

wx=wr, but y and z centralize P, and

either

S={x,yw}, or

S={x,xyw}.

Proof.

(1)  Since Q acts nontrivially on P, and Aut(P) is cyclic, but Q/Φ(Q) is not cyclic, there must be elements a and b of QΦ(Q), such that a centralizes P, but b does not. (And z must centralize P, because it is in Q.) By applying an automorphism of Q, we may assume a=y and b=x. Furthermore, we may assume wx=wr by replacing x with its inverse if necessary.

(2)  S must contain an element that does not centralize P; so we may assume xS. By applying Lemma 3.2 with s=x and c=y, we see that we may assume that S is {x,yw}  or  {x,y2w}  or  {x,xyw}  or  {x,xy2w}. But there is an automorphism of G that fixes x and w and sends y to y2; so we need only consider two of these possibilities.

Proposition 3.4.

Assume, as usual, that |G|=27p, where p is prime, and that G has a normal Sylow p-subgroup. If the Sylow 3-subgroup Q is of exponent 3, then Cay(G;S) has a hamiltonian cycle.

Proof.

We write G¯ for the natural homomorphism from G to G¯=G/P. From Lemma 3.3(2), we see that we need only consider two possibilities for S.

Case 1 (<italic>Assume </italic><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M258"><mml:mi>S</mml:mi><mml:mo>=</mml:mo><mml:mo stretchy="false">{</mml:mo><mml:mi>x</mml:mi><mml:mo>,</mml:mo><mml:mi>y</mml:mi><mml:mi>w</mml:mi><mml:mo stretchy="false">}</mml:mo></mml:math></inline-formula>).

For a=x and b=yw, we have the following hamiltonian cycle in Cay(G/P;S): e¯ax¯ax2¯bx2y¯a-1xyz¯a-1yz2¯by2z2¯bz2¯axz2¯ax2z2¯bx2yz2¯ayz¯axy¯bxy2¯ax2y2z¯bx2z¯bx2yz¯a-1xyz2¯bxy2z2¯ax2y2¯ay2z¯bz¯axz¯b-1xy2z¯ax2y2z2¯ay2¯b-1y¯b-1e¯. Its endpoint in G is a2ba-2b2a2ba2bab2a-1ba2bab-1a2b-2=x2ywx-2(yw)2x2ywx2ywx(yw)2x-1ywx2ywx(yw)-1x2(yw)-2=x2ywxy2w2x2ywx2ywxy2w2x2ywx2ywxy2w-1x2yw-2. Since the walk is a hamiltonian cycle in G/P, we know that this endpoint is in P=w. So all terms except powers of w must cancel. Thus, we need only calculate the contribution from each appearance of w in this expression. To do this, note that if a term wi is followed by a net total of j appearances of x, then the term contributes a factor of wirj to the product. So the endpoint in G is wr13w2r12wr10wr8w2r7wr5wr3w-r2w-2. Since r31  (modp), this simplifies to wrw2wrwr2w2rwr2ww-r2w-2=wr+2+r+r2+2r+r2+1-r2-2=wr2+4r+1=wr2+r+1w3r=w0w3r=w3r. Since p3r, this endpoint generates P; so the Factor Group Lemma 2.4 provides a hamiltonian cycle in Cay(G;S).

Case 2 (<italic>Assume </italic><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M280"><mml:mi>S</mml:mi><mml:mo>=</mml:mo><mml:mo stretchy="false">{</mml:mo><mml:mi>x</mml:mi><mml:mo>,</mml:mo><mml:mi>x</mml:mi><mml:mi>y</mml:mi><mml:mi>w</mml:mi><mml:mo stretchy="false">}</mml:mo></mml:math></inline-formula>).

For a=x and b=xyw, we have the hamiltonian cycle ((a,b2)3#,a)3 in Cay(G/P;S). Its endpoint in G is ((ab2)3  b-1a)3=((x(xyw)2)3  (xyw)-1x)3=((x(x2y2wr+1))3  (w-1y-1x-1)x)3=((y2wr+1)3  (w-1y-1))3=(w3(r+1)  (w-1y-1))3=(y-1w3r+2)3=w3(3r+2). Since we are free to choose r to be either of the two primitive cube roots of 1 in p, and the equation 3r+2=0 has only one solution in p, we may assume that r has been selected to make the exponent nonzero. Then the Factor Group Lemma 2.4 provides a hamiltonian cycle in Cay(G;S).

3.3. Sylow 3-Subgroup of Exponent 9Lemma 3.5.

Assume that Q is of exponent 9; so Q=x,yx9=y3=e,  [x,y]=x3. There are two possibilities for G, depending on whether CQ(P) contains an element of order 9 or not.

Assume that CQ(P) does not contain an element of order 9. Then we may assume that y centralizes P, but wx=wr. Furthermore, we may assume that:

S={x,yw}, or

S={x,xyw}.

Assume that CQ(P) contains an element of order 9. Then we may assume x centralizes P, but wy=wr. Furthermore, we may assume that:

S={xw,y},

S={xyw,y},

S={xy,xw}, or

S={xy,x2yw}.

Proof.

(1) Since x has order 9, we know that it does not centralize P. But x3 must centralize P (since x3 is in G). Therefore, we may assume wx=xr (by replacing x with its inverse if necessary). Also, since Q/CQ(P) must be cyclic (because Aut(P) is cyclic), but CG(P) does not contain an element of order 9, we see that CQ(P) contains every element of order 3; so y must be in CQ(P).

Since S must contain an element that does not centralize P, we may assume xS. By applying Lemma 3.2 with s=x and c=y, we see that we may assume that S is: {x,yw}  or  {x,y2w}  or  {x,xyw}  or  {x,xy2w}. The second generating set need not be considered, because (y2w)-1=yw-1=yw; so it is equivalent to the first. Also, the fourth generating set can be converted into the third, since there is an automorphism of G that fixes y, but takes x to xyw and w to w-1.

We may assume xCQ(P); so CQ(P)=x.

We know that S must contain an element s that does not centralize P, and there are two possibilities: either

s has order 3, or

s has order 9.

We consider these two possibilities as separate cases.

Case I (<italic>Assume that </italic><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M346"><mml:mrow><mml:mi>s</mml:mi></mml:mrow></mml:math></inline-formula><italic> has order </italic>3).

We may assume s=y. Letting c=x, we see from Lemma 3.2 that we may assume S is either {y,xw}  or  {y,x2w}  or  {y,yxw}  or  {y,yx2w}. The second and fourth generating sets need not be considered, because there is an automorphism of G that fixes y and w, but takes x to x2. Also, the third generating set may be replaced with {y,xyw}, since there is an automorphism of G that fixes y and w, but takes x to y-1xy.

Case II (<italic>Assume that </italic><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M362"><mml:mrow><mml:mi>s</mml:mi></mml:mrow></mml:math></inline-formula><italic> has order </italic>9).

We may assume s=xy. Letting c=x, we see from Lemma 3.2 that we may assume that S is either {xy,xw}  or  {xy,x2w}  or  {xy,xyxw}  or  {xy,xyx2w}. The second generating set is equivalent to {xy,xw}, since the automorphism of G that sends x to x4, y to x-3y, and w to w-1 maps it to {xy,(xw)-1}. The third generating set is mapped to {xy,x2yw} by the automorphism that sends x to x[x,y] and y to [x,y]-1y. The fourth generating set need not be considered, because xyx2w is an element of order 3 that does not centralize P, which puts it in the previous case.

Proposition 3.6.

Assume, as usual, that |G|=27p, where p is prime, and that G has a normal Sylow p-subgroup. If the Sylow 3-subgroup Q is of exponent 9, then Cay(G;S) has a hamiltonian cycle.

Proof.

We will show that, for an appropriate choice of a and b in SS-1, the walk (a3,b-1,a,b-1,a4,b2,a-2,b,a2,b,a3,b,a-1,b-1,a-1,b-2) provides a hamiltonian cycle in Cay(G/P;S) whose endpoint in G generates P (so the Factor Group Lemma 2.4 applies).

We begin by verifying two situations in which (3.23) is a hamiltonian cycle.

If |a¯|=9, |b¯|=3, and ab¯=a4¯ in G¯=G/P, then we have the hamiltonian cycle:

e¯aa¯aa2¯aa3¯b-1a3b2¯aa7b2¯b-1a7b¯aa5b¯aa3b¯aab¯aa8b¯ba8b2¯ba8¯a-1a7¯a-1a6¯ba6b¯aa4b¯aa2b¯ba2b2¯aa6b2¯aab2¯aa5b2¯ba5¯a-1a4¯b-1a4b2¯a-1b2¯b-1b¯b-1e¯.

If |a¯|=9, |b¯|=9, ab¯=a7¯, and b3¯=a6¯ in G¯=G/P, then we have the hamiltonian cycle:

e¯aa¯aa2¯aa3¯b-1a6b2¯aa4b2¯b-1a4b¯aa8b¯aa3b¯aa7b¯a  a2b¯ba2b2¯ba8¯a-1a7¯a-1a6¯ba6b¯aab¯aa5b¯ba5b2¯aa3b2¯aab2¯aa8b2¯ba5¯a-1a4¯b-1a7b2¯a-1b2¯b-1b¯b-1e¯.

To calculate the endpoint in G, fix r1,r2p, with wa=wr1,wb=wr2, and write a=a̲w1,b=b̲w2,  where  a̲,b̲Q,w1,w2P. Note that if an occurrence of wi in the product is followed by a net total of j1 appearances of a̲ and a net total of j2 appearances of b̲, then it contributes a factor of wir1j1r2j2 to the product. (A similar occurrence of wi-1 contributes a factor of wi-r1j1r2j2 to the product.) Furthermore, since r13r231  (modp), there is no harm in reducing j1 and j2 modulo 3.

We will apply these considerations only in a few particular situations.

Assume w1=e (so aQ and a̲=a). Then the endpoint of the path in G is a3b-1ab-1a4b2a-2ba2ba3ba-1b-1a-1b-2=a3(b̲w2)-1a(b̲w2)-1a4(b̲w2)2a-2(b̲w2)a2×(b̲w2)a3(b̲w2)a-1(b̲w2)-1a-1(b̲w2)-2=a3(w2-1b̲-1)a(w2-1b̲-1)a4(b̲w2b̲w2)a-2(b̲w2)a2×(b̲w2)a3(b̲w2)a-1(w2-1b̲-1)a-1(w2-1b̲-1w2-1b̲-1). By the above considerations, this simplifies to w2m, where m=-1-r12r2+r1r2+r1+r22+r1r2+r1-r12-r2-r22=-r12r2-r12+2r1r2+2r1-r2-1. Note the following.

If r11 and r2=1, then m simplifies to 6r1, because r12+r1+10  (modp) in this case.

If r11 and r21, then m simplifies to 3r1(r2+1), because r12+r1+1r22+r2+10  (modp) in this case.

Assume w2=e (so bQ and b̲=b). Then the endpoint of the path in G is a3b-1ab-1a4b2a-2ba2ba3ba-1b-1a-1b-2=(a̲w1)3b-1(a̲w1)b-1(a̲w1)4b2(a̲w1)-2b(a̲w1)2b(a̲w1)3b(a̲w1)-1b-1(a̲w1)-1b-2=(a̲w1a̲w1a̲w1)b-1(a̲w1)b-1(a̲w1a̲w1a̲w1a̲w1)b2(w1-1a̲-1w1-1a̲-1)×b(a̲w1a̲w1)b(a̲w1a̲w1a̲w1)b(w1-1a̲-1)b-1(w1-1a̲-1)b-2. By the above considerations, this simplifies to w1m, where m=r12+r1+1+r12r2+r1r22+r22+r12r22+r1r22-r1-r12+r12r22+r1r22+r2+r12r2+r1r2-r1-r12r2=2r12r22+3r1r22+r22+r12r2+r1r2+r2-r1+1. Note the following.

If r1=1 and r21, then m simplifies to -3(r2+2), because r22+r2+10  (modp) in this case.

If r11 and r21, then m simplifies to -r1r2-2r1+r2+2, because r12+r1+1r22+r2+10  (modp) in this case.

Now we provide a hamiltonian cycle for each of the generating sets listed in Lemma 3.5.

If CQ(P) has exponent 3, and S={x,yw}, we let a=x and b=yw in (HC1). In this case, we have w1=e, r1=r, and r2=1; so (E1(a)) tells us that the endpoint in G is w26r.

If CQ(P) has exponent 3, and S={x,xyw}, we let a=x and b=(xyw)-1 in (HC2). In this case, we have w1=e, r1=r, and r2=r-1=r2; so (E1(b)) tells us that the endpoint in G is w2m, where m=3r1(r2+1)=3r(r2+1)=3(r3+r)3(1+r)=3(r+1)(modp).

If CQ(P) has exponent 9, and S={xw,y}, we let a=xw and b=y in (HC1). In this case, we have w2=e, r1=1, and r2=r; so (E2(a)) tells us that the endpoint in G is w1-3(r+2).

If CQ(P) has exponent 9, and S={xyw,y}, we let a=xyw and b=y in (HC1). In this case, we have w2=e and r1=r2=r; so (E2(b)) tells us that the endpoint in G is w2m, where m=-r1r2-2r1+r2+2=-r2-2r+r+2=-(r2+r+1)+33(modp).

If CQ(P) has exponent 9, and S={xy,xw}, we let a=xw and b=(xy)-1 in (HC2). In this case, we have w2=e, r1=1, and r2=r-1=r2; so (E2(a)) tells us that the endpoint in G is w1m, where m=-3(r2+2)=-3(r2+2)-3(-(r+1)+2)=3(r-1)(modp).

If CQ(P) has exponent 9, and S={xy,x2yw}, we let a=xy and b=x2yw in (HC2). In this case, we have w1=e and r1=r2=r; so (E1(b)) tells us that the endpoint in G is w2m, where m=3r1(r2+1)=3r(r+1)=3(r2+r)3(-1)=-3(modp).

In all cases, there is at most one nonzero value of r (modulo p) for which the exponent of wi is 0. Since we are free to choose r to be either of the two primitive cube roots of 1 in p, we may assume that r has been selected to make the exponent nonzero. Then the Factor Group Lemma 2.4 provides a hamiltonian cycle in Cay(G;S).

4. Assume the Sylow <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M519"><mml:mrow><mml:mi>p</mml:mi></mml:mrow></mml:math></inline-formula>-Subgroups of <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M520"><mml:mrow><mml:mi>G</mml:mi></mml:mrow></mml:math></inline-formula> Are Not NormalLemma 4.1.

Assume that

|G|=27p, where p is an odd prime, and

the Sylow p-subgroups of G are not normal.

Then p=13, and G=13(3)3, where a generator w of 13 acts on (3)3 via multiplication on the right by the matrix W=. Furthermore, we may assume that S  is  of  the  form  {wi,wjv}, where v=(1,0,0)(3)3, and (i,j){(1,0),(2,0),(1,2),(1,3),(1,5),(1,6),(2,5)}.

Proof.

Let P be a Sylow p-subgroup of G, and let Q be a Sylow 3-subgroup of G. Since no odd prime divides 3-1 or 32-1, and 13 is the only odd prime that divides 33-1, Sylow's Theorem [8, Theorem  15.7, page  230] implies that p=13, and that NG(P)=P; so G must have a normal p-complement [4, Theorem  7.4.3]; that is, G=PQ. Since P must act nontrivially on Q (since P is not normal), we know that it must act nontrivially on Q/Φ(Q) [4, Theorem  5.3.5, page  180]. However, P cannot act nontrivially on an elementary abelian group of order 3 or 32, because |P|=13 is not a divisor of 3-1 or 32-1. Therefore, we must have |Q/Φ(Q)|=33; so Q must be elementary abelian (and the action of P is irreducible).

Let W be the matrix representing the action of w on (3)3 (with respect to some basis that will be specified later). In the polynomial ring 3[X], we have the factorization: X13-1X-1=(X3-X-1)(X3+X2-1)(X3+X2+X-1)(X3-X2-X-1). Since w13=e, the minimal polynomial of W must be one of the factors on the right-hand side. By replacing w with an appropriate power, we may assume that it is the first factor. Then, choosing any nonzero v(3)3, the matrix representation of w with respect to the basis {v,vw,vw2} is W (the Rational Canonical Form).

Now, let ζ be a primitive 13th root of unity in the finite field GF(27). Then any Galois automorphism of GF(27) over GF(3) must raise ζ to a power. Since the subgroup of order 3 in 13× is generated by the number 3, we conclude that the orbit of ζ under the Galois group is {ζ,ζ3,ζ9}. These must be the 3 roots of one of the irreducible factors on the right-hand side of (4.4). Thus, for any k13×, the matrices Wk, W3k, and W9k all have the same minimal polynomial; so they are conjugate under GL3(3). That is, powers  of  W  in  the  same  row  of  thefollowing  array  are  conjugate  under  GL3(3):  W,W3,W9W2,W5,W6W4,W12,W10W7,W8,W11.

There is an element a of S that generates G/QP. Then a has order p; so, replacing it by a conjugate, we may assume aP=w, and so a=wi for some i13×. From (4.5), we see that we may assume i{1,2} (perhaps after replacing a by its inverse).

Now let b be the second element of S; so we may assume b=wjv for some j. We may assume 0j6 (by replacing b with its inverse, if necessary). We may also assume ji, for otherwise SaQ, and so Theorem 2.8 applies.

If j=0, then (i,j) is either (1,0) or (2,0), both of which appear in the list; henceforth, let us assume j0.

Case 1 (<italic>Assume </italic><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M607"><mml:mi>i</mml:mi><mml:mo>=</mml:mo><mml:mn>1</mml:mn></mml:math></inline-formula>).

Since ji, we must have j{2,3,4,5,6}.

Note that since W3 is conjugate to W under GL3(3) (since they are in the same row of (4.5)), we know that the pair (w,w4) is isomorphic to the pair (w3,(w3)4)=(w3,w-1). By replacing b with its inverse, and then interchanging a and b, this is transformed to (w,w3). So we may assume j4.

Case 2 (<italic>Assume </italic><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M620"><mml:mi>i</mml:mi><mml:mo>=</mml:mo><mml:mn>2</mml:mn></mml:math></inline-formula>).

We may assume that Wj is in the second or fourth row of the table (for otherwise we could interchange a with b to enter the previous case. So j{2,5,6}. Since ji, this implies j{5,6}. However, since W5 is conjugate to W2 (since they are in the same row of (4.5)), and we have (w2)3=w6 and (w5)3=w2, we see that the pair (w2,w6) is isomorphic to (w2,w5). So we may assume j6.

Proposition 4.2.

If |G|=27p, where p is prime, and the Sylow p-subgroups of G are not normal, then Cay(G;S) has a hamiltonian cycle.

Proof.

From Lemma 4.1 (and Remark 2.9), we may assume G=13(3)3. For each of the generating sets listed in Lemma 4.1, we provide an explicit hamiltonian cycle in the quotient multigraph PCay(G;S) that uses at least one double edge. So Lemma 2.7 applies.

To save space, we use i1i2i3 to denote the vertex P(i1,i2,i3).

(i,j)=(1,0)  a=w,  a-1=w12,  b=(1,0,0),  and  b-1=(-1,0,0)  Double edge: 222022  with  a-1  and b:

000  b-1200  a020  a002  a220  b-1120  a012a221  a102  b202  a210  a021  a112  a201b-1101  a-1211  a-1212  a-1222  b022  b122  a-1121a-1111  b-1011  a-1110  a-1001  a-1010  a-1100  b-1000.000  b-1200  a002  a022  a212  b-1112  a-1210a-1122  a-1111  a-1110  b-1010  a-1201  b-1101  a012a102  a020  b-1220  a222  a211  a120  a221b021  a-1202  a-1121  a-1011  a-1001  a-1100  b-1000.000  b-1221  a-1012  a-1120  b-1102  b-1200  a020a002  a220  b022  a222  b011  a111  a121a122  a202  a210  a021  a112  b-1101  a-1211a-1212  b201  b110  a-1001  a-1010  a-1100  b-1000.000  b-1012  a-1120  b-1221  a102  a200  b020a002  a220  a022  a222  a212  a211  a101b-1201  a-1112  a-1021  a-1210  a-1202  a-1122  b121a-1111  a-1011  a-1110  a-1001  a-1010  a-1100  b-1000.000  b-1101  a120  a012  a221  b-1010  a001a110  a011  a111  b121  a122  b-1102  a200a020  a002  a220  b-1022  a222  a212  a211b202  a210  a021  a112  a201  a100  b-1000.000  b-1211  b-1201  a-1112  a-1021  b210  b101b120  a012  a221  a102  a200  a020  a002a220  a022  a222  a212  b202  a-1122  a-1121a-1111  a-1011  a-1110  a-1001  a-1010  a-1100  b-1000.  000    b-1  101    a  012    b  102    a  020    a  220    a  222    b  112    b  210    a-1  122    a-1  111    a-1  110    a-1  010    a-1  201    a-1  021    a-1  202    b-1  211    a  120    a  221    a  200    a  002    a  022    a  212    b-1  121    a-1  011    a-1  001    a-1  100    b-1  000.

Acknowledgments

This work was partially supported by research grants from the Natural Sciences and Engineering Research Council of Canada.

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