#### Abstract

Suppose that is a finite group, such that , where is prime. We show that if is any generating set of , then there is a Hamiltonian cycle in the corresponding Cayley graph Cay.

#### 1. Introduction

Theorem 1.1. *If , where is prime, then every connected Cayley graph on has a Hamiltonian cycle.*

Combining this with results in [1–3] establishes that

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 -subgroup of is normal; Section 4 presents the proof under the assumption that the Sylow -subgroups of 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 [4, Sections 1.1 and 5.1]). *For any group , we use the following notation: (1) denotes the *commutator subgroup * of , (2) denotes the *center* of , (3) denotes the *Frattini subgroup* of . For , we use to denote the *conjugate *.

*Notation 2. *If * is any sequence, we use ** to denote the sequence ** that is obtained by deleting the last term. *

Theorem 2.1 (Marušič, Durnberger, Keating-Witte [5]). *If is a cyclic group of prime-power order, then every connected Cayley graph on has a hamiltonian cycle.*

Lemma 2.2 (see [3, Lemma 2.27]). *Let generate the finite group , and let . If *(i)*, *(ii)* has a hamiltonian cycle, and*(iii)*either (1) , or (2) is prime,*

*then has a hamiltonian cycle.*

Lemma 2.3 (see [1, Lemma 2.7]). *Let generate the finite group , and let . If *(i)*, *(ii)* is a divisor of , where and are distinct primes, *(iii)*, *(iv)* is divisible by , and*(v)* has a hamiltonian cycle,**then there is a hamiltonian cycle in . *

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

Lemma 2.4 (“Factor Group Lemma”). *Suppose that*(i)* is a generating set of , *(ii)* is a cyclic, normal subgroup of , *(iii)* is a hamiltonian cycle in , and*(iv)*the product generates .**Then is a hamiltonian cycle in .*

Corollary 2.5. *Suppose that*(i)* is a generating set of , *(ii)* is a normal subgroup of , such that is prime, *(iii)* for some with , and*(iv)*there is a hamiltonian cycle in that uses at least one edge labelled .**Then there is a hamiltonian cycle in .*

*Definition 2.6. *If is any subgroup of , then denotes the multigraph in which(i)the vertices are the right cosets of , and(ii)there is an edge joining and for each , such that .Thus, if there are two different elements and of , such that and are both in , then the vertices and are joined by a double edge.

Lemma 2.7 (see [3, Corollary 2.9]). *Suppose that *(i)* is a generating set of , *(ii)* is a subgroup of , such that is prime, *(iii)*the quotient multigraph has a hamiltonian cycle , and*(iv)* uses some double-edge of .**Then there is a hamiltonian cycle in .*

Theorem 2.8 (see [6, Corollary 3.3]). *Suppose that*(i)* is a generating set of , *(ii)* is a normal -subgroup of , and*(iii)*, for all .**Then has a hamiltonian cycle.*

*Remark 2.9. *In the proof of our main result, we may assume , for otherwise either(i) is of the form , where is prime, and so [3, Propostion 9.1] applies, or (ii) is a prime power, and so the main theorem of [7] applies.

#### 3. Assume the Sylow -Subgroup of Is Normal

*Notation 3. *Let(i) be a group of order , where is prime, and (see Remark 2.9), (ii) be a minimal generating set for , (iii) be a Sylow -subgroup of , (iv) be a generator of , and(v) be a Sylow 3-subgroup of .

*Assumption 3.1. *In this section, we assume that is a normal subgroup of *. *

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

*Notation 4. *Since is a 3-group and acts nontrivially on , we must have . Thus, one may choose , such that
Dividing by , we see that

##### 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 ** is a nonabelian group of order *
27,
* and **, it is easy to see that *

Lemma 3.2. *Assume that *(i)*, such that does not centralize , and*(ii)*.**Then we may assume that is either or or or .*

*Proof. *Since is a 2-generated group of prime-power order, there must be an element of , such that generates . We may write
Note the following. (i)By replacing with its inverse if necessary, we may assume . (ii)By applying an automorphism of that fixes and maps to , we may assume that is trivial (since ). (iii)By replacing with if , we may assume .Thus,
*Case 1 ( Assume ). *Then , and so . This yields the four listed generating sets.

*Case 2 (*Then , and there must be a third element of , with ; after replacing with an appropriate power, we may write with . We must have , for otherwise (which contradicts the minimality of ). Therefore We may assume the following.(i), for otherwise ; so Lemma 2.3 applies. (ii), by replacing with its inverse if necessary. (iii), for otherwise and provide a double edge in ; so Corollary 2.5 applies.Then generates .

*Assume*).Consider the hamiltonian cycles in . Letting , we see that their endpoints in are (resp.) The final two endpoints both have a nontrivial projection to (since , being a 3-element, cannot invert ), and at least one of these two endpoints also has a nontrivial projection to . Such an endpoint generates , and so the Factor Group Lemma 2.4 provides a hamiltonian cycle in .

##### 3.2. Sylow 3-Subgroup of Exponent 3

Lemma 3.3. *Assume that is of exponent 3; so
**
Then one may assume the following: *(1)*, but and centralize , and*(2)*either(a) , or (b).*

*Proof. *(1) Since acts nontrivially on , and is cyclic, but is not cyclic, there must be elements and of , such that centralizes , but does not. (And must centralize , because it is in .) By applying an automorphism of , we may assume and . Furthermore, we may assume by replacing with its inverse if necessary.

(2) must contain an element that does not centralize ; so we may assume . By applying Lemma 3.2 with and , we see that we may assume that is
But there is an automorphism of that fixes and and sends to ; so we need only consider two of these possibilities.

Proposition 3.4. *Assume, as usual, that , where is prime, and that has a normal Sylow -subgroup. If the Sylow 3-subgroup is of exponent 3, then has a hamiltonian cycle.*

*Proof. *We write for the natural homomorphism from to . From Lemma 3.3(2), we see that we need only consider two possibilities for .*Case 1 ( Assume ). *For and , we have the following hamiltonian cycle in :
Its endpoint in is
Since the walk is a hamiltonian cycle in , we know that this endpoint is in . So all terms except powers of must cancel. Thus, we need only calculate the contribution from each appearance of in this expression. To do this, note that if a term is followed by a net total of appearances of , then the term contributes a factor of to the product. So the endpoint in is
Since , this simplifies to
Since , this endpoint generates ; so the Factor Group Lemma 2.4 provides a hamiltonian cycle in .

*Case 2 (*For and , we have the hamiltonian cycle in . Its endpoint in is Since we are free to choose to be either of the two primitive cube roots of 1 in , and the equation has only one solution in , we may assume that has been selected to make the exponent nonzero. Then the Factor Group Lemma 2.4 provides a hamiltonian cycle in .

*Assume*).##### 3.3. Sylow 3-Subgroup of Exponent 9

Lemma 3.5. *Assume that is of exponent 9; so
**
There are two possibilities for , depending on whether contains an element of order 9 or not. *(1)*Assume that does not contain an element of order 9. Then we may assume that centralizes , but . Furthermore, we may assume that: (a) , or (b). *(2)

*Assume that contains an element of order 9. Then we may assume centralizes , but . Furthermore, we may assume that: (a)*

*,*(b)*,*(c)*, or*(d)*.**Proof. *(1) Since has order 9, we know that it does not centralize . But must centralize (since is in ). Therefore, we may assume (by replacing with its inverse if necessary). Also, since must be cyclic (because is cyclic), but does not contain an element of order 9, we see that contains every element of order 3; so must be in .

Since must contain an element that does not centralize , we may assume . By applying Lemma 3.2 with and , we see that we may assume that is:
The second generating set need not be considered, because ; so it is equivalent to the first. Also, the fourth generating set can be converted into the third, since there is an automorphism of that fixes , but takes to and to .(2)We may assume ; so .

We know that must contain an element that does not centralize , and there are two possibilities: either (I) has order 3, or (II) has order 9.We consider these two possibilities as separate cases.*Case I ( Assume that has order 3). *We may assume . Letting , we see from Lemma 3.2 that we may assume is either
The second and fourth generating sets need not be considered, because there is an automorphism of that fixes and , but takes to . Also, the third generating set may be replaced with , since there is an automorphism of that fixes and , but takes to .

*Case II (*We may assume . Letting , we see from Lemma 3.2 that we may assume that is either The second generating set is equivalent to , since the automorphism of that sends to , to , and to maps it to . The third generating set is mapped to by the automorphism that sends to and to . The fourth generating set need not be considered, because is an element of order 3 that does not centralize , which puts it in the previous case.

*Assume that**has order*9).Proposition 3.6. *Assume, as usual, that , where is prime, and that has a normal Sylow -subgroup. If the Sylow 3-subgroup is of exponent 9, then has a hamiltonian cycle.*

*Proof. *We will show that, for an appropriate choice of and in , the walk
provides a hamiltonian cycle in whose endpoint in generates (so the Factor Group Lemma 2.4 applies).

We begin by verifying two situations in which (3.23) is a hamiltonian cycle.(HC1)If , , and in , then we have the hamiltonian cycle:(HC2)If , , , and in , then we have the hamiltonian cycle:

To calculate the endpoint in , fix , with
and write
Note that if an occurrence of in the product is followed by a net total of appearances of and a net total of appearances of , then it contributes a factor of to the product. (A similar occurrence of contributes a factor of to the product.) Furthermore, since , there is no harm in reducing and modulo 3.

We will apply these considerations only in a few particular situations.(E1)Assume (so and ). Then the endpoint of the path in is
By the above considerations, this simplifies to , where
Note the following. (a)If and , then simplifies to , because in this case. (b)If and , then simplifies to , because in this case. (E2)Assume (so and ). Then the endpoint of the path in is
By the above considerations, this simplifies to , where
Note the following. (a)If and , then simplifies to , because in this case. (b)If and , then simplifies to , because in this case.

Now we provide a hamiltonian cycle for each of the generating sets listed in Lemma 3.5. (1a)If has exponent 3, and , we let and in (HC1). In this case, we have , , and ; so (E1(a)) tells us that the endpoint in is .(1b)If has exponent 3, and , we let and in (HC2). In this case, we have , , and ; so (E1(b)) tells us that the endpoint in is , where
(2a)If has exponent 9, and , we let and in (HC1). In this case, we have , , and ; so (E2(a)) tells us that the endpoint in is .(2b)If has exponent 9, and , we let and in (HC1). In this case, we have and ; so (E2(b)) tells us that the endpoint in is , where
(2c)If has exponent 9, and , we let and in (HC2). In this case, we have , , and ; so (E2(a)) tells us that the endpoint in is , where
(2d)If has exponent 9, and , we let and in (HC2). In this case, we have and ; so (E1(b)) tells us that the endpoint in is , where
In all cases, there is at most one nonzero value of (modulo ) for which the exponent of is 0. Since we are free to choose to be either of the two primitive cube roots of 1 in , we may assume that has been selected to make the exponent nonzero. Then the Factor Group Lemma 2.4 provides a hamiltonian cycle in .

#### 4. Assume the Sylow -Subgroups of Are Not Normal

Lemma 4.1. *Assume that *(i)*, where is an odd prime, and*(ii)*the Sylow -subgroups of are not normal.**Then , and , where a generator of acts on via multiplication on the right by the matrix
**
Furthermore, we may assume that
**
where , and
*

*Proof. *Let be a Sylow -subgroup of , and let be a Sylow 3-subgroup of . Since no odd prime divides or , and 13 is the only odd prime that divides , Sylow's Theorem [8, Theorem 15.7, page 230] implies that , and that ; so must have a normal -complement [4, Theorem 7.4.3]; that is, . Since must act nontrivially on (since is not normal), we know that it must act nontrivially on [4, Theorem 5.3.5, page 180]. However, cannot act nontrivially on an elementary abelian group of order 3 or 3^{2}, because is not a divisor of or . Therefore, we must have ; so must be elementary abelian (and the action of is irreducible).

Let be the matrix representing the action of on (with respect to some basis that will be specified later). In the polynomial ring , we have the factorization:
Since , the minimal polynomial of must be one of the factors on the right-hand side. By replacing with an appropriate power, we may assume that it is the first factor. Then, choosing any nonzero , the matrix representation of with respect to the basis is (the Rational Canonical Form).

Now, let be a primitive 13th root of unity in the finite field . Then any Galois automorphism of over must raise to a power. Since the subgroup of order 3 in is generated by the number 3, we conclude that the orbit of under the Galois group is . These must be the 3 roots of one of the irreducible factors on the right-hand side of (4.4). Thus, for any , the matrices , , and all have the same minimal polynomial; so they are conjugate under . That is,

There is an element of that generates . Then has order ; so, replacing it by a conjugate, we may assume , and so for some . From (4.5), we see that we may assume (perhaps after replacing by its inverse).

Now let be the second element of ; so we may assume for some . We may assume (by replacing with its inverse, if necessary). We may also assume , for otherwise , and so Theorem 2.8 applies.

If , then is either or , both of which appear in the list; henceforth, let us assume .*Case 1 ( Assume ). *Since , we must have .

Note that since is conjugate to under (since they are in the same row of (4.5)), we know that the pair is isomorphic to the pair . By replacing with its inverse, and then interchanging and , this is transformed to . So we may assume .

*Case 2 (*We may assume that is in the second or fourth row of the table (for otherwise we could interchange with to enter the previous case. So . Since , this implies . However, since is conjugate to (since they are in the same row of (4.5)), and we have and , we see that the pair is isomorphic to . So we may assume .

*Assume*).Proposition 4.2. *If , where is prime, and the Sylow -subgroups of are not normal, then has a hamiltonian cycle.*

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

To save space, we use to denote the vertex . Double edge: withand *b*: Double edge: and: Double edge: withand: Double edge: with and : Double edge: with and : Double edge: with and : Double edge:withand:

#### Acknowledgments

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