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 [1–3] establishes that Every Cayley graph onGhas a hamiltonian cycleif |G|=kp,wherepis prime,1≤k<32,andk≠24.

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,b∈G, 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.

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 s∈S. If

〈s〉⊲G,

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

either

s∈Z(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 s∈S. If

〈s〉⊲G,

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

sp∈Z(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 s1s2⋯sn 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,

s≡t(modN) for some s,t∈S∪S-1 with s≠t, 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 H∖Cay(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 s∈S∪S-1, such that g1s∈Hg2.

Thus, if there are two different elements s1 and s2 of S∪S-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 H∖Cay(G;S) has a hamiltonian cycle C, and

C uses some double-edge of H∖Cay(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-1∈N, for all s,t∈S.

Then Cay(G;S) has a hamiltonian cycle.Remark 2.9.

In the proof of our main result, we may assume p≥5, 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 [7] 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 p≥5(see Remark 2.9),

S be a minimal generating set for G,

P≅ℤp 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=Q⋉P.
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′×Piscyclicoforder3p.

Notation 4.

Since Q is a 3-group and acts nontrivially on P≅ℤp, we must have p≡1(mod3). Thus, one may choose r∈ℤ, such that
r3≡1(modp),butr≢1(modp).
Dividing r3-1 by r-1, we see that
r2+r+1≡0(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=Q⋉P≅Q⋉ℤp, it is easy to see that Q′=Φ(Q)=Z(Q)=Z(G)=Φ(G).

Lemma 3.2.

Assume that

s∈(S∪S-1)∩Q, such that s does not centralize P, and

c∈CQ(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/P≅Q 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,with0≤i≤2,1≤j≤2,z∈Z(Q),and0≤k<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 k≠0, we may assume k∈{0,1}.

Thus,
a=sicjwkwithi,k∈{0,1},andj∈{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 b∉Q; after replacing w with an appropriate power, we may write b=tw with t∈Q. We must have t∈〈s,Φ(Q)〉, for otherwise 〈s,b〉=G (which contradicts the minimality of S). Therefore
t=si′z′with0≤i′≤2,andz′∈Φ(Q)=Z(G).
We may assume the following.

i′≠0, for otherwise b=z′w∈S∩(Z(G)×P); so Lemma 2.3 applies.

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

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

Then s-1b=z′w 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)3∈〈z〉, we see that their endpoints in G are (resp.)
z′′,z′′(s-1b)=z′′z′w,z′′(s-1b)s(s-1b)=z′′(z′)2wsw.
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,z∣x3=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 Q∖Φ(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 x∈S. 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 r3≡1(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 p∤3r, 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)3b-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 ℤp, and the equation 3r+2=0 has only one solution in ℤp, 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,y∣x9=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 x∈S. 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 x∈CQ(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 S∪S-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:

To calculate the endpoint in G, fix r1,r2∈ℤp, with
wa=wr1,wb=wr2,
and write
a=a̲w1,b=b̲w2,wherea̲,b̲∈Q,w1,w2∈P.
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 r13≡r23≡1(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 a∈Q 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 r1≠1 and r2=1, then m simplifies to 6r1, because r12+r1+1≡0(modp) in this case.

If r1≠1 and r2≠1, then m simplifies to 3r1(r2+1), because r12+r1+1≡r22+r2+1≡0(modp) in this case.

Assume w2=e (so b∈Q 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 r2≠1, then m simplifies to -3(r2+2), because r22+r2+1≡0(modp) in this case.

If r1≠1 and r2≠1, then m simplifies to -r1r2-2r1+r2+2, because r12+r1+1≡r22+r2+1≡0(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)+3≡3(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 ℤp, 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=ℤ13⋉(ℤ3)3, where a generator w of ℤ13 acts on (ℤ3)3 via multiplication on the right by the matrix
W=[010001110].
Furthermore, we may assume that
Sisoftheform{wi,wjv},
where v=(1,0,0)∈(ℤ3)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=P⋉Q. 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 3^{2}, 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 (ℤ3)3 (with respect to some basis that will be specified later). In the polynomial ring ℤ3[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∈(ℤ3)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 ℤ13× 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 k∈ℤ13×, the matrices Wk, W3k, and W9k all have the same minimal polynomial; so they are conjugate under GL3(3). That is,
powersofWinthesamerowofthefollowingarrayareconjugateunderGL3(3):W,W3,W9W2,W5,W6W4,W12,W10W7,W8,W11.

There is an element a of S that generates G/Q≅P. Then a has order p; so, replacing it by a conjugate, we may assume a∈P=〈w〉, and so a=wi for some i∈ℤ13×. 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 0≤j≤6 (by replacing b with its inverse, if necessary). We may also assume j≠i, for otherwise S⊂aQ, 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 j≠0.

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 j≠i, 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 j≠4.

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 j≠i, 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 j≠6.

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=ℤ13⋉(ℤ3)3. For each of the generating sets listed in Lemma 4.1, we provide an explicit hamiltonian cycle in the quotient multigraph P∖Cay(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).

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

CurranS. J.MorrisD. W.MorrisJ.Hamiltonian cycles in Cayley graphs of order 16pPreprintGhaderpourE.MorrisD. W.Cayley graphs of order 30p are hamiltonianPreprintKutnarK.MarušičD.MorrisJ.MorrisD. W.ŠparlP.Hamiltonian cycles in Cayley graphs whose order has few prime factorsArs Mathematica Contemporanea. In pressGorensteinD.KeatingK.WitteD.Hamilton cycles in Cayley graphs with cyclic commutator subgroupMorrisD. W.2-generated Cayley digraphs on nilpotent groups have hamiltonian pathsPreprintWitteD.Cayley digraphs of prime-power order are hamiltonianJudsonT. W.