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The conjunction of Cayley digraphs
Discrete Mathematics 42 (1982) 209-219 North-Holland Publishing Company
209
THE CONJUNCTION OF CAYLEY DIGRAPHS Kevin K E A T I N G
Mathematics Department, Washington Univers~ity, St. Louis, MO 63130, USA Received 20 October 1980 Revised 14 December 1981 If G~ and G 2 a r e digraphs with vertex sets V~ a~td V2, then the conjunction of G~ and G2, de~oted G~ • G2, is a digraph with vertex set 'V~× V 2 in which there is an arc from (x~, x2) to (Y~, Y2) if there is art arc from x, to Yl in G 1 and an arc from x~, ",o Y2 in G2. I~ Cay(S) and CaI.'(T) are Cayley di.graphs, then C.zy(S). Cay(T)= Cay(S x "/3 as iong as the digraph on the left is connected. In this paper we find ~ulficie~tco~;ditionsfor the conjunction of two Cayley dig-~ohs to have a hamiltonian circuit. For certain classes of generating sets S and T. we characterize the cases for which Cay(S)-Cay(T) has a hami]toniaa circuit. We also present sortie basic results on the existence of i',~n,iltonian paths in the conjunction of two Cayley dig~,~ohs.
I. Inlrot~tion If the ~et S generates a finite group G, we can associate a directed graph with S in tl,,~ following manner. T h e ' v e r t i c e s of the graph are the elements of G, and there ~s an arc from x to y in the graph if y =: xs for sorne s ~ S. W e call this graph the Cay,::ev digraph of S in G, denoted Cay(S) or Cay(S : G). A n u m b e r of authors have corl,~,dered the problem of finding hamiltoniap paths and circuits in Cayley digraphs. In 1946, F',ankin [4] wrote a pioneering work on the case where Isl-- 2; he has, al~o published some more recent results [5]. Holsztyfiski and Strube [1] deft net~ c~rcular groups and sequential groups and found a large class of sequential grou~:. Witte [7] sho~ved that certain p-groups and dihedral groups ara circular and e~it:e:nded Holsztyfiski and Strube's work on sequential groups. St::ve::~t papers have been written on the problem of finding hamiltonian circuits in the Cartesian product of two Cayley digraphs. Trotter and Erdds [6] gave necessary and sufficient conditions for Zo x Zb to have a hamiltonian circuit. ~F,y Z , we mean C a y ( l : Z~).) Letzter [3] concentrated on finding hamiltonian circuits in Z , × Cay(S). Witte, Letzter, and Gallian [8] also considered that problem, and in addition found hamiltonian circuits in C a y ( S ) × Ca,y(T) for certain generating sets S and T. The con.iunction of two digraphs G~ and G2 with vertex sets V~ and V7 is defined as the digraph with vertex set V~ x V2 in which there is an arc from (x,, x2) to (y~, Y2) if there is all arc from x~ to y~ in G~ and an arc from x2 to Y2 in G2. This digraph is denoted G~. G2. In the case where Gt and G2 are Cayley 0012-365X/82/0000-0000/$02.75 O 1982 North-Holland
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K. Keafing
digraphs, Cay(S). Cay(T) = Cay(S ×: T) as long as the digraph on the leflt is connected, i.e., ( S ) × ( T ) = ( S × T ) . In this paper° we consider the: problem of f, ading hamiltonian circuits and hamiltonian paths in lbc conjunction "of two Cayley digraphs. We fiad sufficient conditions for Cay(S) • Cay(T) to have a hamiltonian circuit. In the case where (S) is nilpotent and circular we give necessary and sufficient conditions for Z~ • Cay(S) to have a hamiltonian circuit. We present a method ~hich might be useful in an inductive argument on the conjunction of Cayley digraphs. In the last two sections we investigate the conjunction of (undirected) Cayley graphs and the problem of finding hamiltonian paths in the conjunction of two Cayley digraphs.
2. Definitions and notation In all cases, S generates the finite group G with identity element e. We let G > H mean that H is v proper sffbgroup of G, while G ~> H naeans that H is a proper normal subgroup of G. The relationships G > ~ H and G ~ H have their usual meanings, If ae;(S), then a S = { a s l s e S } . Also, by S -~ we mean {s ~ [ s e S}. We denote :he order of the group element x by olx). We will specify a path in Cay(S) by listing its arcs. We write (x~, x2 . . . . . x,). The first vertex is taken to be the identity element e. Thus, the vertices encountered in this path are e, x~, x~x:, . . . . , x , x 2 . . ' x ~ , in that order. We have a hamiltonian circuit if n = !(S)[. the partial products x~, x~x2 . . . . , x ~ x , . . . x , are distinct, and x~x 2 • • • x, = e. For example, if we let D~ = (t, f l t3 -- f2 __ e, f t f = lI-') we can specify a hamitto,fian circuit in Cay({t, f}) as (t, t, f, t, t, [ ) We will let n * x denote a string of n x's, so that the circuit above could also be written (2* t, f, 2* t. f). Similarly, if A is a string of arcs, let n * A represent the concatenation of n copies of A , so that the circuit above can be further compressed to (2*(2* t, J0). Following Holszty6ski and Strube, we define a circular group as a group in which every Cayley digraph has a hamiltonian circuit and a sequential group as a group in which every Cayley digraph has a hamiltonian path.
3. Some limiting counterexamples At the outset, one might suppose that if Cay(S) and Cay(T) each have a hamiltonian circuit then C a r ( S ) . Cay(T) must also have a hamiltonian circuit. Our first example shows that this 2s not necessarily true, even if we assume that
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Cay(S) • Cay(T) is connected. We use here, and again in Section 6, a speeia! case of one of Rankin's theorems. Witte [7, Theorem 4.2] was the first to extract this useful theorem from Rankin's result. Theorem 3.1 (Rankin [4, Theorem 4]). Let (a, ~) be an abelian group, and let b = I(a, 13) : (a -/3~1. Pick an integer c so that b[3 = c(a - [3). Then Cay({a,/3}) has a hamiltonian circuit if and only if there is an integer d satisfying c ~ d ~ c + b and gcd(d, o ( a - ,8))= 1. E x a m # e 3.2. Let m t = 2 . 5 • 7 - I 1 • 17 and n~ = 2 . 3 • 13 • 19. Then Z,, x Z,,, has a hamiltonian circuiL but Z H " (Z,, x Z,~,) is connected and has no hamiltonian c~rcuit. PreoL Since gcd(nl, m t ) = 2 , we go,_ a hamiltonian circuit in Z , xZ,~, by Hols~,*,~,Sski and Strube [I, Theorem 5.2]. Since g c d ( l l ~ n t , m l ) = 1, Zt~ • (.Z, × Z,~,) is connected. To determine whether Z11" (Zn, x Z,~,) has a hamilton~an circuit, we apply Rankin's conditions to the general Cayley digraph Zk • (Z, × Z=). We have b = k • gcd(n, m). Assuming our graph is connected, we can ,:.hoc:~e c to satisfy b . (1, 0, 1) = c • (0, 1, - 1 ) ; since k divides b, we only need to h,~w, c-=0 (mod n) and b+c-=-O (rood m). By Theorem 3.1, Zk • (Z, xZm) has a h:¢mi'.'tonian circuit if and only if there is an integer d satisfying c <~d -<.c + b and gcd(d, o((1, 1, 0 ) - (I, 0, I))) = 1, i.e., gcd(d, lcm(n, m)) = 1. This last equation holds if and only if gcd(d, n) = gcd(d, m ) = 1. Since c-'-=0 (mod n), gcd(d, n)-- 1 if and o1@ if g o d ( d - c, n) = 1; since b + c m 0 (mod m), god(d, m) = 1 if and only if g c d ( b . ~ c - d , m ) = 1. Let u = d - c > ~ O and v = b + c - d > ~ O . We have shown that Zk " (Z~ x Zm) has a hamiltonian circuit if and only if we can find integers u and v with u v ~>0, u + v = b = k • gcd(n, m), and gcd(u, n) = god(v, m) = 1. In this example. we ~leed u and v such that u + v = 22 and gcd(u, n) = god(v, m) = 1. It is trivial to verify that no such u and o exist. On :he other hand, one might suppose that if Cay(S) • Cay(T) has a hamiltonian circuit, then either Cay(S) or Cay(T) must have a hamiltonian circuit. ~ a t supposition is also fal,;e, as our next example shows. Example 3.3. Z2 x Z3 has no hamiltonian circuit, but (Z2 x Z3)" (Z2 x Z3) does have a hamiltonian circuit. Proof. That Z2 x Za has no hamiltonian circuit follows from Hoisztyfiski and Strube [I, Proposition 5.6]. A hamiltonian circuit in (Z2 x Z3)" (Z2 x Z3) is 6*(((0, 1), (1, 0)), 5.((1, 0), (0, 1))).
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K. Kea6ng
4. The subgraph lemma Our first result gives a convenient method for finding a hamiltonian circuit in Cay(T). Cay(S) when Cay(T) has a hamiltonian circuit. This theorem was ;_nspired by an analogous result for the Cartesian product due to Witte, Letzter, and Gallian [8]. Theorem 4.1 ('Subgraph Lemma'). Let I(TX~l= m. I[ Cay(T) has a hamiltonian circui~ and Z , , ' C a y ( S ) has a hamiltonian circuit, then Cay(T) ~Cay(S) has a hamiltonian circuit.
Proof. Let the hamiltenian circuit in Cay(T) be (Xl, x 2 , . . . , x,,) and let the Mmiltoaian circuit in Z,, .Cay(S) be ((1, y l ) , ( l , Y2). . . . . (1, y,,.)). Then the circuit in Cay(T)-Cay(S) is ((x~, Y3, (x2, Y2). . . . . (:~,,, y,,), (x~, y,,+,), (x2, Y,.+2). . . . . (x,,, Y2,,). . . . . (x~, y. . . . . . . ~), (x2, y,~-,,+~) . . . . . (x~, y,,~)). As a consequence of this theorem, we can reasonably restrict our attention to the problem of determining when Z , • Cay(S) has a hamiltonian circuit. Our first result in ~:his direction is a corollary to Theorem 4.1 which relies on the fact that if gcd(m, n) = 1, then Z,~ • Z,, has a hamiltonian circuit. Corellary 4.2. If Cay(S) has a hamiltonian circuit and gcd(n,[(S)[)= 1, then Z~ • Cay(S) has a hamiltonian circuit. Let S be a minimal generating set for a p-group. In [7], Witte showed that in a large number of cases, Cay(S) must have a hamiltonian circuit. If Cay(S) does have a hamiltonian circuit, we ca~l characterize those integers n for which Z~ • Cay(S) has a hamiltonian circuit. We let qb(G) denote the Frattini :~ubgroup of G, the intersection of t~e maximal subgroups of G; we let G ' denote the commutator subgroup of G, the subgroup generated by elements of the form [ g , h ] = g ~h ~gh with g and h elements of G. Theorem 4.3. Ij" S is a minimal generating set ~br a p-group and Cay(S) has a hamiltonian circuit, then Z , . Cay(S) has a hamihonian circuit if and only if ~cd(n, p) ~- 1.
Proof. If gcd(n. !9)= 1, Corollary 4.2 gives us the desired result. If P l n, we claim that Z,, • Cay(S) is not even connected. It suffices to show that if sl, s2,. • , sk ~ S with s ~ s 2 " " s k =:e, then p [ k . To this end, choose s ~ S , Since S is a ,l~iaimal generating se~ for a p-group, and ls-'S\{e}l
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213
III, p. 272 and Theorem 3.2a, Ch. III, p. 268]; in fact, (s-~S, G')<1(S). Returning to the s~, s2. . . . , sk, notice that s ? ~ s ~ ( s - ~ S , G') for each i = 1, 2 . . . . . k. Therefore s ~ s~(s-tS, G') so that s ~ ~ sis2"" • s k ( s - t S , G ' ) = e ( s - l $ , O').
Since G/(s-~S, G') is a nowtrivial p-group and s(s-~S, G') generates G/(s-~S, G'), it follows that p l k.
5. The skewed generator conditions Next we introduce a very general method for finding hamiltonian circuits in Z , ' Cay(S) which does not require that Cay(S) have a circuit. DeSpite, on 5.1. We say that Z , • Cay(.q) satisfies the skewed generator conditions if ther*, is an element a of (S) satisfyir, g the following conditions: (1) (,~S) = (S). (2) Cay(aS) has a hamiltonian circuit. (3) There exist s~, s2. . . . . s,,_~ e S with slsz" • • s,_~ = a. Tiffs ~e,finition is justified by the following theorem. Theo:'~,m 5.2. I f Z , . Cay(S) satisfies the skewed generator conditions, Z~ • Cay(S) has a hamiltonian circuit.
then
Proof. l e t
s i s 2 . . , s,,_~=a and let the hamiltonian circuit in Cay(aS) be (axe, ¢~'x~. . . . . , axm) with each xi ~ S. "['hen the hamiltonian circuit in Z , . Cay(S)
is ((L o%),(1, s2). . . . . (i, s,-1), (1, x,), (I, sl), (1, s2). . . . . (1, s,_,), (1, x2) . . . . . (1, s,), (1, s2) . . . . . (1, s._,), (1, xm)). Cerol~:~zy 5.3. I f Cay(S) has a hamiltonian s~, s2 . . . . circuft.
circuit a n d there are elements s,__~ in S such that sis2 • • • ~,~-~ = e, then Z , • Cay(S) has a hamiltonian
Using Theorem 5.2 and its corollary we can determine whether Z , • Cay(S) has a hamiltonian circuit when S is a standard generating set for a dihedral or dieyclic group. Example 5.4. Let D m = (t, f[ t m = f2 = e, f t [ = t -~} be the dihedral group of order 2m. Then Z, • Cay({t, f}) has a hamiltonian circuit if and only if either n or m is odd.
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K. Keaang
Proof. W e h a v e several cases to consider. First, if n is odd, let s ~ = f for l<~i<~n - 1. T h e n s t s 2 " ' " s . - t = f " - ~ = e . Since Cay({t,/}) has a h a m i l t o n i a n circuit, by C o r o l l a r y 5.3, Z~ • Cay({t, [}) h a s a h a r a i l t o n i a n circuit. If n is e v e n a n d m is odd, let s , = f for l < ~ i < ~ n - 2 a n d let s,,_~=t. T h e n a = s a s 2 . . . s , _ t = f~-2t=t. Since ( f , t ) ' ) = ( t , f ) a n d Cay({t2, t/}) has a h a m i l t o n i a n circuit, Z,,. Cay(S) has a h a m i l t o n i a n circulit. Finally, if n a n d m are b o t h even, Z,, • Cay({t, f}) is not c o n n e c t e d ; in particular, there, is no Fath f r o m (0, e) to (1, e). For if ((1, xl), (1, x2) . . . . . (1, xk)) w e r e such a path, t h e n x l x 2 " " x k = e. This implies k is even (since m is even). H o w e v e r , k -~ 1 (rood n), a n d h e n c e k is odd, a contradiction. E x a m p l e 5.5,. Let D ~ = (f, g I f" = g2 = (fg),,, = e) be the d i h e d r a l g r o u p of o r d e r 2m.r this time g e n e r a t e d by two reflections. T h e n Z,,.Cay({f, g}) has a h a m i l t o n i a n circuit if and only if n is odd. 1FroM. If n is odd, let s , = f for l < ~ i < ~ n - 1 . T h e n s t s 2 " " s , - ~ = f " - ~ = e and Cay({f, g}) has a h a m i l t o n i a n circuit, so by C o r o l l a r y 5.3, Z , . Cay({f, g}) has a h a m i l t o n i a n circuit. If n is even, then Z , • C a y({f, g}) i~ not c o n n e c t e d . E x a m p l e 5.6. Let DC,,, = (x, y I x " = y~-, x 2" = y4 = e, y - l x y = x - t ) be the dicyclic g r o u p of o r d e r 4m. T h e n Z . • Cay({x, y}) has a h a m i l t o n i a n circuit if a a d only if either n or m is odd. Proof. If n is odd, let s~ = y for 1 ~< i ~-: n - 1, a n d let a := sis2 • • • s,,__t = y , - t = e or y 2 In e i t h e r case, a satisfies the conditions of Definition 5.1, so t h a t Z , • Cay({x° Yt) has a h a m i l t o n i a n circuit. If n is e v e n and m is odd, let s~ = y for l<<-i~n-2 and let s , ~ = x . T h e n set a = s l s 2 . . . s , _ t = y n - 2 x = y 2 x o r x. In either case, a satisfies the conditions of Definition 5.1, so t h a t again we find that Z,, • Cay({x, y}) has a h a m i l t o n i a n circuit. Finally, if n a n d m are b o t h even, t h e n Z,, • Cay(S) i~ not c o n n e c t e d .
6. Charac'lerizations It turns o u t that for certain g e n e r a t i n g sets S, the d i g r a p h Z n . Cay(S) has a h a m i l t o n i a n circuit if and only if Z n " Cay(S) satisfies the s k e w e d g e n e r a t o r conditions. In this section we establish the existence of two large classes of such g e n e r a t i n g sets. W e begin with a useful result on connectednesso T h e o r e m 6.1. I f (S) = G is nilpotent, then Z . • Cay(S) is connected if a n d only if for all sl, s2 . . . . . s~_t ~ S we have (sis2" • • s . _ l S ) = (S).
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215
l ~ o f . First assume that for all choices sl, sz . . . . , s , _ l ~ S , we have ( S ) = (s~'.~2" • • s,-:tS). To show that Z , • Cay(S) is connected, it suffices to find a path frown (0, e) to (0, x) in Z , • Cay(S), where x is an arbitrary element of (S). Cho~ise an)' s ~ S and let s~=s for l < ~ i ~ n - 1 . Then (S)=(s"-~S). Thus there exist xt, x 2 , . . . , xk e S such that s"-~x~s"-~x2 • •. S"-IXk = X. A path from (0, e) to (0, x) is ((n - 1)* (1, s), (1, xi), (n - 1)*(1, s), (1, xz) . . . . . (n - 1)* (1, s), (1, xk)). Conversely, suppose that Z~ • Cay(S) is connected. Choose s~, s 2 , . . . , s , ~ ~ S and let a = s l s 2 " " s,-i. Assume that ( a S ) ~ ( S ) . Then ( a S ) < ( S ) . Since G is nilpotent, we have G ' < ~ ( G ) , so that (aS, G ' ) < G . (See, for example, [2, Theorem 3.11, Ch. III, p. 2 7 L and Theorem 3.2a, Ch. II1, p. 268].) It follows that (aS, G'~.~ G so that G/(aS, G') "~ nontrivial and abelian; in fact, it is generated by a(aS, G'). Note that for any x, y e S, we have
y - i x = (y-l a-l)(ax) = (ay)-l(~x)E ( a S ) c (aS, G'). Hence, for any x, y e 5, we have x(aS, G') = y(aS, G'). Thus, in the quotient group,, ,,ll the elements of S are equivalent. $inc~. Z , . Cay(S) is connected, we can find elements x~, x2. . . . ,xk of S sa~sfyi~g x~x2"." xk = a and k-= 0 (mod n). Choose any s e S. Then as ~ (aS, G'), so ~s(aS G')= e(aS, G'). Thus,
e(aS, G')= (as)k/"(aS, G ' ) = (sis2 ' • • s,-ls)k/"(aS, G'). Uc.~r~g ~ e result of the previous paragraph, we find that
a(aS, G') = x l x : " " x~(aS, G') = (s~s2"" s._,s)~"(aS, G')= e(aS, G'). This i~ a contradiction, so (aS)= (S). Coa'~ary 6.2. If (S) is nilpotent and circular then the following are equivalent: (t~ ,;.-.f,,• Cay(S) is connected. t2) Z , . Cay(S) has a hamiltonian circuit. (3) Z , • Cay(S) satisfies the skewed generator conditions. In ~-articular, Witte [7] showed that any p-group with a cyclic commutator subgroup is circular. Therefore, if S generates such a group, Z , . Cay($) has a hamiltonian circuit if and only if Z , . Cay(S) satisfies the skewed generator conditions. Using Theorem 3.1 (Rankin) we can find another class of generating sets which have this property. Theorem 6.3. Let (a,/3) be an abelian group. I[ Z , • Cay({a, ~}) has a hamiltoniaa circuit, then ,Z, , Cay({a,/3}) satisfies the skewed generator conditions.
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K. J~eating
ProoL Since Z , . Cay({a,/3}) has a hamiltonian circuit, R a n k i n ' s conditions (Theorem 3.1) must be satisfied. Let m = I",~,/3):(a -/3)1
and B --- I((1, a ) , ( 1 , / 3 ) ) : ( ( 0 , a - / 3 ) ) [ = n . I ( a , / 3 ) : ( . - / 3 ) 1
= nm.
Choose C such that B(1,/3) = C(0, a - / 3 ) ; there must exist D such that C < ~ D < ~ C + B = C + n m and gcd(D, o(ct - / 3 ) ) = 1. Let D = C + (k - 1) m + j where 1 <~ k <~ 11 and 0 ~ j <~ m This may not determine k and j uniquely, but that doesn't matter. Let sl = t~ for ~ <~ i ~< k - 1 and s~ =/3 for k < < - i ~ n - 1. Then by T h e o r e m 6.1, we have ~a, ~ ) =
• • + s._t +/3)
= ( k a + (n - k)/3, (k -- 1)or + (n - k + 1)/3).
By Definition 5.1 and T h e o r e m 5.2, all that remains to be shown is that Cay({~:a + (n - k)/3, (k - 1)a + (n - k + 1)/3}) has a hamiltonian circuit. Applying Rankin's theorem, we let b = [ ( a / 3 } : ( a - / 3 } 1 = m and choose c to satisfy m ( ( k - l ) ~ + (n - k + ~)/3) = c ( a - / 3 ) ,
The/1 ( m ( k - 1 ) - c ) ( a - / 3 ) = - m n ~
=-C(tr-/3)
and hence
c ~-=-C + m ( k - 1) (rood o~a -/3)).
Let d = c + j .
Then
and
c~d<~c+b=c+m
d -~ c + j = - C + ( k
- 1)m + j ~ D (mod o(c~ -/3)),
so that gcd(d, o(c~ - / 3 ) ) = god(D, o ( a - / 3 ) ) = 1. Therefore, Cay({ka + ( n - k)/3, (k - l ) a + (n - k + 1)/3}) has a h amiltonian circuit. Thus, Z , • Cay({c~,/3}) satisfies the skewed generator conditions.
7. E x t e n d i n g circuits
We present here a theorem which allows us to use a hamiltonian circuit in Z , • Cay(S) to find a hamiltorfian circuit in Zr • Cay(S) for certain values of r. Theorem
7.1. I]" Z , , . C a y ( S )
s~, s: . . . . .
s,, ~ S w i t h s~s2 " • " so, = e, t h e n Z ..... • Coy(S) h a s a h a m i l t o n i a n circuit.
has
a
hamiltonian
circuit
and
there
exist
Proof. Let the hamittonian circuit in Z , • Cay(S) be ((1, xl), (1, x2) . . . . . (1, x,k)).
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217
Then the circuit in Z,+,, • Cay(S) is ((1, Sl) , (1, S2) . . . . .
(1, Sm) , (1, XI) , (1, X2) . . . . .
(1, x.), (1, sl) . . . . . (1, s.,), (1, x.+l) . . . . . (1, x2.) . . . . . (1, sO . . . . . (1, s,,), (1, x,k-,÷t) . . . . . (1, x,k)). Corollary 7.2. I f Z . . Cay(S) has a hamiltonian circuit l o t at least one n, it has a hamiltonian circuit [or an infinite n u m b e r o f integers n.
Suppose Z , • Cay(S) has a hamiltonian circuit by the skewed generator conditions, aad we apply Theorem 7.1 to find a hamiltonian circuit in Z.+., • Cay(S) for an appropriate positive integer at. Then we can get the same hamiltonian circuit in Z~ ..... • Cay(S) by showing that this conjunction digraph satisfies the skewed gener~t,~r conditions. We know that Z~. Cay(S) satisfies the skewed generator condi,~cons, so that there exists an a ~ S such that ( a S ) = ( S ) , Cay(S) has a hamiIIoaian circuit, and a = s i s 2 • " " Srt_1 for some sl, s2 . . . . . s , _ l e S . Since Th~:ore,m 7.1 holds, we have x~, x2. . . . . x~ e S with x l x 2 " • ' x= = e. Thus, S~S 2
" " " sn_lxlx2
" " " x m
=
a.
Sinc.~ (aS) = (S) and Cay(aS) has a hamiltonian circuit, Z~÷m " Cay(S) satisfies the ske'~ed generator conditions. Then Theorem 5.2 give~ us a hamiitonian circuit in Z ..... • Cay(S), the same circuit which Theorem 7.1 gives u,~. (However, Theorem 7.1 is u~;eful v-hen the circuit in Z . • Cay(S) is not derived using Theorem 5.2.)
8. Imd~ected Cayley graphs A C~yley graph can be thought of as a Cayley digraph of the form Cay(S O s--l), since this is just Cay(S) with the directed arcs replaced by undirected arcs. In tbi:~ secti~a we apply the results of Section 5 to the problem of finding hamiltonian ci,:'cuits in the conjunction of undirected Cayley graphs.
Theorem 8.1. I f Cay(S U S -~) has a hamiltonian circuii, then Z , . Cay(SU S -a) has a hamiltonian
circuit if either n is odd,
or n is even,
n~[(S)!, a n d
Z , • Cay(S U S -~) is connected (i.e,, Cay(S U 8 -1) is not bipartite). Proof. If n is odd, we use the skewed generator conditions as follows. Choose s ~ S , and let s ~ = s for i odd and st = s -1 for i even. Then sls2" " • s n _ t = e , so by Corollary 5.3, Zn • Cay(S O S -t) has a hamiltonian circuit. In the second case, with n even, n>~l(S)l, and C a y ( S O S -t) not bipartite, choose s~, s2. . . . . sk ~ S U S -~ such that sls2 • • • sk = e and k is odd and as small as possible subject to these requirements. Then I claim that n > k. If not, k > n, and hence k > [(S)[. Then by the pigeon
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K. Keafing
hole principle, there exist partial products of t the si's which are equal, say sv~.2.., s, = s ~ s 2 . " sv wfth u > v . Then Sz,,~ I S u . v 2
• " * Su =
SIS 2 ....
SUSu+ISu+
2 " ~ " S k "~" e ~
so that we have two sets of elements of S U S -t, each with product e, and each with I(;ss than k elements. One of these sets has u - v elements and the other has k-(u-v) e~ements; thus, one set :must have an odd number of elements, violating k's minimality. Therefore k < n. Now choose s ~ S, and for k < i < n let s, = s when i is odd and let si = s -t when i is even. Then SlS 2 " " " Sn__ t ~
StS 2 " " " S k " Sk+tSk+2
" " ° Sn_
1 ~"
e,
and the desiled result follows from Corollary 5.3. By applying Theorem 4.1 to Theorem 8.1, we get the following corollary. Corollary 8.2. Suppose [(S)[~[(T)[ a n d C a y ( S U S -t) a n d C a y ( T U T -t) each have ,~ hamiltonian circuit. The~z Cay(SU S-t) • C a y ( T L I T -t) has a hamiltonian circuit if either [(S) I is odd, [(T)! is odd, or C a y ( T U T -1) is not bipartite. The above corollary leads us to conjecture that if Cay(S O S -t) and C a y ( T U T -I) each I'~ave a hamiltonian circuit then Cay(S U S -t) • C a y ( T U T -t) has a hamiltonian circuil as long as thi~ conjunction graph is cotmected. The only case not covered above is the case with [(S)[ and [(T)I even and C a y ( T O T -~) bipartite. If the conjunction graph is to be connected, we must have C a y ( S U S -t) non-bipartite. In that case it is sufficient to find a bamiltonian circuit in Z2" C a y ( S U S - t ) , for once wc have that circuit we can use Theorem 7.1 and the Subgraph Lemma to solve the general case.
9. Hamiltonian paths In this last section we show that some of our theorems on hamiltonian circuits also hold for hamiltonian paths. We close by pointing out z large class of groups G such that if ( S ) = G then Z , . Cay(S) has a hamiltonian path if and only if Z,.Cay(S) satisfies the path version of the skewed generator conditions. The proo£, for the following two theorems are essentially identical with the proofs for their uircuit analogues, Theorems 5.2 and 7.1. Theorem 9.1. Suppose ( a S } = ( S ) , Cay(aS) has a hamiltonian path, a,:d a = s~ s2 • • • s, ~ for ,'ome st, s2 . . . . . s,. i ~ S. T h e n Z , • Cay(S) has a hamiltonian path. When the conditions of Theorem 9.1 are satisfied, we say that Z , . Cay(S) satisfies the path s k e w e d generator conditions.
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219
%heorem 9.2. If Z , . C a y ( S ) has a hamiltonian path and there exist st, s2. . . . . sm e S such that sis:,." • " s,. = e, then Z,+m • Cay(S) has a hamiltonian path. Our final result relies on a ,theorem of Holsztyfiski and Nathanson (see [1, Theorem 3.1]) which says that every abelian ot hamiltonian group is sequential. Theorem 9.3. If (S) is a group in which every subgroup is normal, the following are equivalent: (1) Z , . Cay(S) is connected. (2) Z~ • Cay(S) ht;s a hamiltonian Fath. (3~ Z , • Cay(S) satisfies the path skewed generator conditions. l~-e~l. That (2) implies (1) i,~ d e a l and (3) implies (2) follows from Theorem 9.1. To see that (1) implies (3), first not~c~ that since every subgroup of (S) is normal, (S) nt~st be nilpotent, since the Sylow p-subgroups of (S) are normal. Choose any s~.S and let s i = s for ~<-i<~n-1. Then by Theorem 6.1, (s*-1S)=($). By Holsz~vfiski and Nathanson's result, Ca y(s"-~S) has a hamiltonian path. Thus, Z~ • 'Cz,y(S) satisfies the path skewed generator conditions.
Aek~ewledgements TI~e author would like to tha~ak Steven Curran, Gall Letzter, and David Witte for rh.:ir help and sugge;~tions, and Professor Joseph A. Gallian for hit constant s~np[~ort and indispensib~e advice. This research was done at the University of Minr~esota, Duluth, in an Undergraduate Research Participation program funded by ~:ae NSF (Grant number NSF/SPI-7926564).
[1] W. Holsztyfiskiand R.F. Strube, Paths and circuits in finite groups, Discrete Math. 22 (1978) 263-272. [2] B. Huppert, Endliche Gruppc.n ! (Springer-Verlag,New York/Berlin, 1967). [3] G. Letzter, Hamiltonianc~rcuitsin cartesian products with a metacyclicfactor, prcprint. [4] R.A. Rankin, A campanologicalproblem in group theory, Proc. Camb. Phil. Soc. 44 (1948) 17-25. [5] R.A. Rankin, A campanologicalproblem in group theory II, Proc. Camb. Phil. Soc, 62 (1966) 11-18. [6] W.T. Trotter, Jr. and P. Erd6s, When the cartesian product of directed cycles is hamiltonian,J, Graph Theory2 (1978) 137-142. [7] D. Wine, On hamiltoaian ciircuitsin Cayley diagrams, Discrete Math. 38 (1982), 99-108. [8] D. Witty, G. Letzter, and J.A. Gallian, On hamiltoniancircuits in cartesian products of .Cayley digraph;i, Discrete Math. 43 (1983), to appear.
209
THE CONJUNCTION OF CAYLEY DIGRAPHS Kevin K E A T I N G
Mathematics Department, Washington Univers~ity, St. Louis, MO 63130, USA Received 20 October 1980 Revised 14 December 1981 If G~ and G 2 a r e digraphs with vertex sets V~ a~td V2, then the conjunction of G~ and G2, de~oted G~ • G2, is a digraph with vertex set 'V~× V 2 in which there is an arc from (x~, x2) to (Y~, Y2) if there is art arc from x, to Yl in G 1 and an arc from x~, ",o Y2 in G2. I~ Cay(S) and CaI.'(T) are Cayley di.graphs, then C.zy(S). Cay(T)= Cay(S x "/3 as iong as the digraph on the left is connected. In this paper we find ~ulficie~tco~;ditionsfor the conjunction of two Cayley dig-~ohs to have a hamiltonian circuit. For certain classes of generating sets S and T. we characterize the cases for which Cay(S)-Cay(T) has a hami]toniaa circuit. We also present sortie basic results on the existence of i',~n,iltonian paths in the conjunction of two Cayley dig~,~ohs.
I. Inlrot~tion If the ~et S generates a finite group G, we can associate a directed graph with S in tl,,~ following manner. T h e ' v e r t i c e s of the graph are the elements of G, and there ~s an arc from x to y in the graph if y =: xs for sorne s ~ S. W e call this graph the Cay,::ev digraph of S in G, denoted Cay(S) or Cay(S : G). A n u m b e r of authors have corl,~,dered the problem of finding hamiltoniap paths and circuits in Cayley digraphs. In 1946, F',ankin [4] wrote a pioneering work on the case where Isl-- 2; he has, al~o published some more recent results [5]. Holsztyfiski and Strube [1] deft net~ c~rcular groups and sequential groups and found a large class of sequential grou~:. Witte [7] sho~ved that certain p-groups and dihedral groups ara circular and e~it:e:nded Holsztyfiski and Strube's work on sequential groups. St::ve::~t papers have been written on the problem of finding hamiltonian circuits in the Cartesian product of two Cayley digraphs. Trotter and Erdds [6] gave necessary and sufficient conditions for Zo x Zb to have a hamiltonian circuit. ~F,y Z , we mean C a y ( l : Z~).) Letzter [3] concentrated on finding hamiltonian circuits in Z , × Cay(S). Witte, Letzter, and Gallian [8] also considered that problem, and in addition found hamiltonian circuits in C a y ( S ) × Ca,y(T) for certain generating sets S and T. The con.iunction of two digraphs G~ and G2 with vertex sets V~ and V7 is defined as the digraph with vertex set V~ x V2 in which there is an arc from (x,, x2) to (y~, Y2) if there is all arc from x~ to y~ in G~ and an arc from x2 to Y2 in G2. This digraph is denoted G~. G2. In the case where Gt and G2 are Cayley 0012-365X/82/0000-0000/$02.75 O 1982 North-Holland
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K. Keafing
digraphs, Cay(S). Cay(T) = Cay(S ×: T) as long as the digraph on the leflt is connected, i.e., ( S ) × ( T ) = ( S × T ) . In this paper° we consider the: problem of f, ading hamiltonian circuits and hamiltonian paths in lbc conjunction "of two Cayley digraphs. We fiad sufficient conditions for Cay(S) • Cay(T) to have a hamiltonian circuit. In the case where (S) is nilpotent and circular we give necessary and sufficient conditions for Z~ • Cay(S) to have a hamiltonian circuit. We present a method ~hich might be useful in an inductive argument on the conjunction of Cayley digraphs. In the last two sections we investigate the conjunction of (undirected) Cayley graphs and the problem of finding hamiltonian paths in the conjunction of two Cayley digraphs.
2. Definitions and notation In all cases, S generates the finite group G with identity element e. We let G > H mean that H is v proper sffbgroup of G, while G ~> H naeans that H is a proper normal subgroup of G. The relationships G > ~ H and G ~ H have their usual meanings, If ae;(S), then a S = { a s l s e S } . Also, by S -~ we mean {s ~ [ s e S}. We denote :he order of the group element x by olx). We will specify a path in Cay(S) by listing its arcs. We write (x~, x2 . . . . . x,). The first vertex is taken to be the identity element e. Thus, the vertices encountered in this path are e, x~, x~x:, . . . . , x , x 2 . . ' x ~ , in that order. We have a hamiltonian circuit if n = !(S)[. the partial products x~, x~x2 . . . . , x ~ x , . . . x , are distinct, and x~x 2 • • • x, = e. For example, if we let D~ = (t, f l t3 -- f2 __ e, f t f = lI-') we can specify a hamitto,fian circuit in Cay({t, f}) as (t, t, f, t, t, [ ) We will let n * x denote a string of n x's, so that the circuit above could also be written (2* t, f, 2* t. f). Similarly, if A is a string of arcs, let n * A represent the concatenation of n copies of A , so that the circuit above can be further compressed to (2*(2* t, J0). Following Holszty6ski and Strube, we define a circular group as a group in which every Cayley digraph has a hamiltonian circuit and a sequential group as a group in which every Cayley digraph has a hamiltonian path.
3. Some limiting counterexamples At the outset, one might suppose that if Cay(S) and Cay(T) each have a hamiltonian circuit then C a r ( S ) . Cay(T) must also have a hamiltonian circuit. Our first example shows that this 2s not necessarily true, even if we assume that
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211
Cay(S) • Cay(T) is connected. We use here, and again in Section 6, a speeia! case of one of Rankin's theorems. Witte [7, Theorem 4.2] was the first to extract this useful theorem from Rankin's result. Theorem 3.1 (Rankin [4, Theorem 4]). Let (a, ~) be an abelian group, and let b = I(a, 13) : (a -/3~1. Pick an integer c so that b[3 = c(a - [3). Then Cay({a,/3}) has a hamiltonian circuit if and only if there is an integer d satisfying c ~ d ~ c + b and gcd(d, o ( a - ,8))= 1. E x a m # e 3.2. Let m t = 2 . 5 • 7 - I 1 • 17 and n~ = 2 . 3 • 13 • 19. Then Z,, x Z,,, has a hamiltonian circuiL but Z H " (Z,, x Z,~,) is connected and has no hamiltonian c~rcuit. PreoL Since gcd(nl, m t ) = 2 , we go,_ a hamiltonian circuit in Z , xZ,~, by Hols~,*,~,Sski and Strube [I, Theorem 5.2]. Since g c d ( l l ~ n t , m l ) = 1, Zt~ • (.Z, × Z,~,) is connected. To determine whether Z11" (Zn, x Z,~,) has a hamilton~an circuit, we apply Rankin's conditions to the general Cayley digraph Zk • (Z, × Z=). We have b = k • gcd(n, m). Assuming our graph is connected, we can ,:.hoc:~e c to satisfy b . (1, 0, 1) = c • (0, 1, - 1 ) ; since k divides b, we only need to h,~w, c-=0 (mod n) and b+c-=-O (rood m). By Theorem 3.1, Zk • (Z, xZm) has a h:¢mi'.'tonian circuit if and only if there is an integer d satisfying c <~d -<.c + b and gcd(d, o((1, 1, 0 ) - (I, 0, I))) = 1, i.e., gcd(d, lcm(n, m)) = 1. This last equation holds if and only if gcd(d, n) = gcd(d, m ) = 1. Since c-'-=0 (mod n), gcd(d, n)-- 1 if and o1@ if g o d ( d - c, n) = 1; since b + c m 0 (mod m), god(d, m) = 1 if and only if g c d ( b . ~ c - d , m ) = 1. Let u = d - c > ~ O and v = b + c - d > ~ O . We have shown that Zk " (Z~ x Zm) has a hamiltonian circuit if and only if we can find integers u and v with u v ~>0, u + v = b = k • gcd(n, m), and gcd(u, n) = god(v, m) = 1. In this example. we ~leed u and v such that u + v = 22 and gcd(u, n) = god(v, m) = 1. It is trivial to verify that no such u and o exist. On :he other hand, one might suppose that if Cay(S) • Cay(T) has a hamiltonian circuit, then either Cay(S) or Cay(T) must have a hamiltonian circuit. ~ a t supposition is also fal,;e, as our next example shows. Example 3.3. Z2 x Z3 has no hamiltonian circuit, but (Z2 x Z3)" (Z2 x Z3) does have a hamiltonian circuit. Proof. That Z2 x Za has no hamiltonian circuit follows from Hoisztyfiski and Strube [I, Proposition 5.6]. A hamiltonian circuit in (Z2 x Z3)" (Z2 x Z3) is 6*(((0, 1), (1, 0)), 5.((1, 0), (0, 1))).
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K. Kea6ng
4. The subgraph lemma Our first result gives a convenient method for finding a hamiltonian circuit in Cay(T). Cay(S) when Cay(T) has a hamiltonian circuit. This theorem was ;_nspired by an analogous result for the Cartesian product due to Witte, Letzter, and Gallian [8]. Theorem 4.1 ('Subgraph Lemma'). Let I(TX~l= m. I[ Cay(T) has a hamiltonian circui~ and Z , , ' C a y ( S ) has a hamiltonian circuit, then Cay(T) ~Cay(S) has a hamiltonian circuit.
Proof. Let the hamiltenian circuit in Cay(T) be (Xl, x 2 , . . . , x,,) and let the Mmiltoaian circuit in Z,, .Cay(S) be ((1, y l ) , ( l , Y2). . . . . (1, y,,.)). Then the circuit in Cay(T)-Cay(S) is ((x~, Y3, (x2, Y2). . . . . (:~,,, y,,), (x~, y,,+,), (x2, Y,.+2). . . . . (x,,, Y2,,). . . . . (x~, y. . . . . . . ~), (x2, y,~-,,+~) . . . . . (x~, y,,~)). As a consequence of this theorem, we can reasonably restrict our attention to the problem of determining when Z , • Cay(S) has a hamiltonian circuit. Our first result in ~:his direction is a corollary to Theorem 4.1 which relies on the fact that if gcd(m, n) = 1, then Z,~ • Z,, has a hamiltonian circuit. Corellary 4.2. If Cay(S) has a hamiltonian circuit and gcd(n,[(S)[)= 1, then Z~ • Cay(S) has a hamiltonian circuit. Let S be a minimal generating set for a p-group. In [7], Witte showed that in a large number of cases, Cay(S) must have a hamiltonian circuit. If Cay(S) does have a hamiltonian circuit, we ca~l characterize those integers n for which Z~ • Cay(S) has a hamiltonian circuit. We let qb(G) denote the Frattini :~ubgroup of G, the intersection of t~e maximal subgroups of G; we let G ' denote the commutator subgroup of G, the subgroup generated by elements of the form [ g , h ] = g ~h ~gh with g and h elements of G. Theorem 4.3. Ij" S is a minimal generating set ~br a p-group and Cay(S) has a hamiltonian circuit, then Z , . Cay(S) has a hamihonian circuit if and only if ~cd(n, p) ~- 1.
Proof. If gcd(n. !9)= 1, Corollary 4.2 gives us the desired result. If P l n, we claim that Z,, • Cay(S) is not even connected. It suffices to show that if sl, s2,. • , sk ~ S with s ~ s 2 " " s k =:e, then p [ k . To this end, choose s ~ S , Since S is a ,l~iaimal generating se~ for a p-group, and ls-'S\{e}l
The con]unction of Cayley digraphs
213
III, p. 272 and Theorem 3.2a, Ch. III, p. 268]; in fact, (s-~S, G')<1(S). Returning to the s~, s2. . . . , sk, notice that s ? ~ s ~ ( s - ~ S , G') for each i = 1, 2 . . . . . k. Therefore s ~ s~(s-tS, G') so that s ~ ~ sis2"" • s k ( s - t S , G ' ) = e ( s - l $ , O').
Since G/(s-~S, G') is a nowtrivial p-group and s(s-~S, G') generates G/(s-~S, G'), it follows that p l k.
5. The skewed generator conditions Next we introduce a very general method for finding hamiltonian circuits in Z , ' Cay(S) which does not require that Cay(S) have a circuit. DeSpite, on 5.1. We say that Z , • Cay(.q) satisfies the skewed generator conditions if ther*, is an element a of (S) satisfyir, g the following conditions: (1) (,~S) = (S). (2) Cay(aS) has a hamiltonian circuit. (3) There exist s~, s2. . . . . s,,_~ e S with slsz" • • s,_~ = a. Tiffs ~e,finition is justified by the following theorem. Theo:'~,m 5.2. I f Z , . Cay(S) satisfies the skewed generator conditions, Z~ • Cay(S) has a hamiltonian circuit.
then
Proof. l e t
s i s 2 . . , s,,_~=a and let the hamiltonian circuit in Cay(aS) be (axe, ¢~'x~. . . . . , axm) with each xi ~ S. "['hen the hamiltonian circuit in Z , . Cay(S)
is ((L o%),(1, s2). . . . . (i, s,-1), (1, x,), (I, sl), (1, s2). . . . . (1, s,_,), (1, x2) . . . . . (1, s,), (1, s2) . . . . . (1, s._,), (1, xm)). Cerol~:~zy 5.3. I f Cay(S) has a hamiltonian s~, s2 . . . . circuft.
circuit a n d there are elements s,__~ in S such that sis2 • • • ~,~-~ = e, then Z , • Cay(S) has a hamiltonian
Using Theorem 5.2 and its corollary we can determine whether Z , • Cay(S) has a hamiltonian circuit when S is a standard generating set for a dihedral or dieyclic group. Example 5.4. Let D m = (t, f[ t m = f2 = e, f t [ = t -~} be the dihedral group of order 2m. Then Z, • Cay({t, f}) has a hamiltonian circuit if and only if either n or m is odd.
214
K. Keaang
Proof. W e h a v e several cases to consider. First, if n is odd, let s ~ = f for l<~i<~n - 1. T h e n s t s 2 " ' " s . - t = f " - ~ = e . Since Cay({t,/}) has a h a m i l t o n i a n circuit, by C o r o l l a r y 5.3, Z~ • Cay({t, [}) h a s a h a r a i l t o n i a n circuit. If n is e v e n a n d m is odd, let s , = f for l < ~ i < ~ n - 2 a n d let s,,_~=t. T h e n a = s a s 2 . . . s , _ t = f~-2t=t. Since ( f , t ) ' ) = ( t , f ) a n d Cay({t2, t/}) has a h a m i l t o n i a n circuit, Z,,. Cay(S) has a h a m i l t o n i a n circulit. Finally, if n a n d m are b o t h even, Z,, • Cay({t, f}) is not c o n n e c t e d ; in particular, there, is no Fath f r o m (0, e) to (1, e). For if ((1, xl), (1, x2) . . . . . (1, xk)) w e r e such a path, t h e n x l x 2 " " x k = e. This implies k is even (since m is even). H o w e v e r , k -~ 1 (rood n), a n d h e n c e k is odd, a contradiction. E x a m p l e 5.5,. Let D ~ = (f, g I f" = g2 = (fg),,, = e) be the d i h e d r a l g r o u p of o r d e r 2m.r this time g e n e r a t e d by two reflections. T h e n Z,,.Cay({f, g}) has a h a m i l t o n i a n circuit if and only if n is odd. 1FroM. If n is odd, let s , = f for l < ~ i < ~ n - 1 . T h e n s t s 2 " " s , - ~ = f " - ~ = e and Cay({f, g}) has a h a m i l t o n i a n circuit, so by C o r o l l a r y 5.3, Z , . Cay({f, g}) has a h a m i l t o n i a n circuit. If n is even, then Z , • C a y({f, g}) i~ not c o n n e c t e d . E x a m p l e 5.6. Let DC,,, = (x, y I x " = y~-, x 2" = y4 = e, y - l x y = x - t ) be the dicyclic g r o u p of o r d e r 4m. T h e n Z . • Cay({x, y}) has a h a m i l t o n i a n circuit if a a d only if either n or m is odd. Proof. If n is odd, let s~ = y for 1 ~< i ~-: n - 1, a n d let a := sis2 • • • s,,__t = y , - t = e or y 2 In e i t h e r case, a satisfies the conditions of Definition 5.1, so t h a t Z , • Cay({x° Yt) has a h a m i l t o n i a n circuit. If n is e v e n and m is odd, let s~ = y for l<<-i~n-2 and let s , ~ = x . T h e n set a = s l s 2 . . . s , _ t = y n - 2 x = y 2 x o r x. In either case, a satisfies the conditions of Definition 5.1, so t h a t again we find that Z,, • Cay({x, y}) has a h a m i l t o n i a n circuit. Finally, if n a n d m are b o t h even, t h e n Z,, • Cay(S) i~ not c o n n e c t e d .
6. Charac'lerizations It turns o u t that for certain g e n e r a t i n g sets S, the d i g r a p h Z n . Cay(S) has a h a m i l t o n i a n circuit if and only if Z n " Cay(S) satisfies the s k e w e d g e n e r a t o r conditions. In this section we establish the existence of two large classes of such g e n e r a t i n g sets. W e begin with a useful result on connectednesso T h e o r e m 6.1. I f (S) = G is nilpotent, then Z . • Cay(S) is connected if a n d only if for all sl, s2 . . . . . s~_t ~ S we have (sis2" • • s . _ l S ) = (S).
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215
l ~ o f . First assume that for all choices sl, sz . . . . , s , _ l ~ S , we have ( S ) = (s~'.~2" • • s,-:tS). To show that Z , • Cay(S) is connected, it suffices to find a path frown (0, e) to (0, x) in Z , • Cay(S), where x is an arbitrary element of (S). Cho~ise an)' s ~ S and let s~=s for l < ~ i ~ n - 1 . Then (S)=(s"-~S). Thus there exist xt, x 2 , . . . , xk e S such that s"-~x~s"-~x2 • •. S"-IXk = X. A path from (0, e) to (0, x) is ((n - 1)* (1, s), (1, xi), (n - 1)*(1, s), (1, xz) . . . . . (n - 1)* (1, s), (1, xk)). Conversely, suppose that Z~ • Cay(S) is connected. Choose s~, s 2 , . . . , s , ~ ~ S and let a = s l s 2 " " s,-i. Assume that ( a S ) ~ ( S ) . Then ( a S ) < ( S ) . Since G is nilpotent, we have G ' < ~ ( G ) , so that (aS, G ' ) < G . (See, for example, [2, Theorem 3.11, Ch. III, p. 2 7 L and Theorem 3.2a, Ch. II1, p. 268].) It follows that (aS, G'~.~ G so that G/(aS, G') "~ nontrivial and abelian; in fact, it is generated by a(aS, G'). Note that for any x, y e S, we have
y - i x = (y-l a-l)(ax) = (ay)-l(~x)E ( a S ) c (aS, G'). Hence, for any x, y e 5, we have x(aS, G') = y(aS, G'). Thus, in the quotient group,, ,,ll the elements of S are equivalent. $inc~. Z , . Cay(S) is connected, we can find elements x~, x2. . . . ,xk of S sa~sfyi~g x~x2"." xk = a and k-= 0 (mod n). Choose any s e S. Then as ~ (aS, G'), so ~s(aS G')= e(aS, G'). Thus,
e(aS, G')= (as)k/"(aS, G ' ) = (sis2 ' • • s,-ls)k/"(aS, G'). Uc.~r~g ~ e result of the previous paragraph, we find that
a(aS, G') = x l x : " " x~(aS, G') = (s~s2"" s._,s)~"(aS, G')= e(aS, G'). This i~ a contradiction, so (aS)= (S). Coa'~ary 6.2. If (S) is nilpotent and circular then the following are equivalent: (t~ ,;.-.f,,• Cay(S) is connected. t2) Z , . Cay(S) has a hamiltonian circuit. (3) Z , • Cay(S) satisfies the skewed generator conditions. In ~-articular, Witte [7] showed that any p-group with a cyclic commutator subgroup is circular. Therefore, if S generates such a group, Z , . Cay($) has a hamiltonian circuit if and only if Z , . Cay(S) satisfies the skewed generator conditions. Using Theorem 3.1 (Rankin) we can find another class of generating sets which have this property. Theorem 6.3. Let (a,/3) be an abelian group. I[ Z , • Cay({a, ~}) has a hamiltoniaa circuit, then ,Z, , Cay({a,/3}) satisfies the skewed generator conditions.
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K. J~eating
ProoL Since Z , . Cay({a,/3}) has a hamiltonian circuit, R a n k i n ' s conditions (Theorem 3.1) must be satisfied. Let m = I",~,/3):(a -/3)1
and B --- I((1, a ) , ( 1 , / 3 ) ) : ( ( 0 , a - / 3 ) ) [ = n . I ( a , / 3 ) : ( . - / 3 ) 1
= nm.
Choose C such that B(1,/3) = C(0, a - / 3 ) ; there must exist D such that C < ~ D < ~ C + B = C + n m and gcd(D, o(ct - / 3 ) ) = 1. Let D = C + (k - 1) m + j where 1 <~ k <~ 11 and 0 ~ j <~ m This may not determine k and j uniquely, but that doesn't matter. Let sl = t~ for ~ <~ i ~< k - 1 and s~ =/3 for k < < - i ~ n - 1. Then by T h e o r e m 6.1, we have ~a, ~ ) =
• • + s._t +/3)
= ( k a + (n - k)/3, (k -- 1)or + (n - k + 1)/3).
By Definition 5.1 and T h e o r e m 5.2, all that remains to be shown is that Cay({~:a + (n - k)/3, (k - 1)a + (n - k + 1)/3}) has a hamiltonian circuit. Applying Rankin's theorem, we let b = [ ( a / 3 } : ( a - / 3 } 1 = m and choose c to satisfy m ( ( k - l ) ~ + (n - k + ~)/3) = c ( a - / 3 ) ,
The/1 ( m ( k - 1 ) - c ) ( a - / 3 ) = - m n ~
=-C(tr-/3)
and hence
c ~-=-C + m ( k - 1) (rood o~a -/3)).
Let d = c + j .
Then
and
c~d<~c+b=c+m
d -~ c + j = - C + ( k
- 1)m + j ~ D (mod o(c~ -/3)),
so that gcd(d, o(c~ - / 3 ) ) = god(D, o ( a - / 3 ) ) = 1. Therefore, Cay({ka + ( n - k)/3, (k - l ) a + (n - k + 1)/3}) has a h amiltonian circuit. Thus, Z , • Cay({c~,/3}) satisfies the skewed generator conditions.
7. E x t e n d i n g circuits
We present here a theorem which allows us to use a hamiltonian circuit in Z , • Cay(S) to find a hamiltorfian circuit in Zr • Cay(S) for certain values of r. Theorem
7.1. I]" Z , , . C a y ( S )
s~, s: . . . . .
s,, ~ S w i t h s~s2 " • " so, = e, t h e n Z ..... • Coy(S) h a s a h a m i l t o n i a n circuit.
has
a
hamiltonian
circuit
and
there
exist
Proof. Let the hamittonian circuit in Z , • Cay(S) be ((1, xl), (1, x2) . . . . . (1, x,k)).
The con]unction of Cayley digraphs
217
Then the circuit in Z,+,, • Cay(S) is ((1, Sl) , (1, S2) . . . . .
(1, Sm) , (1, XI) , (1, X2) . . . . .
(1, x.), (1, sl) . . . . . (1, s.,), (1, x.+l) . . . . . (1, x2.) . . . . . (1, sO . . . . . (1, s,,), (1, x,k-,÷t) . . . . . (1, x,k)). Corollary 7.2. I f Z . . Cay(S) has a hamiltonian circuit l o t at least one n, it has a hamiltonian circuit [or an infinite n u m b e r o f integers n.
Suppose Z , • Cay(S) has a hamiltonian circuit by the skewed generator conditions, aad we apply Theorem 7.1 to find a hamiltonian circuit in Z.+., • Cay(S) for an appropriate positive integer at. Then we can get the same hamiltonian circuit in Z~ ..... • Cay(S) by showing that this conjunction digraph satisfies the skewed gener~t,~r conditions. We know that Z~. Cay(S) satisfies the skewed generator condi,~cons, so that there exists an a ~ S such that ( a S ) = ( S ) , Cay(S) has a hamiIIoaian circuit, and a = s i s 2 • " " Srt_1 for some sl, s2 . . . . . s , _ l e S . Since Th~:ore,m 7.1 holds, we have x~, x2. . . . . x~ e S with x l x 2 " • ' x= = e. Thus, S~S 2
" " " sn_lxlx2
" " " x m
=
a.
Sinc.~ (aS) = (S) and Cay(aS) has a hamiltonian circuit, Z~÷m " Cay(S) satisfies the ske'~ed generator conditions. Then Theorem 5.2 give~ us a hamiitonian circuit in Z ..... • Cay(S), the same circuit which Theorem 7.1 gives u,~. (However, Theorem 7.1 is u~;eful v-hen the circuit in Z . • Cay(S) is not derived using Theorem 5.2.)
8. Imd~ected Cayley graphs A C~yley graph can be thought of as a Cayley digraph of the form Cay(S O s--l), since this is just Cay(S) with the directed arcs replaced by undirected arcs. In tbi:~ secti~a we apply the results of Section 5 to the problem of finding hamiltonian ci,:'cuits in the conjunction of undirected Cayley graphs.
Theorem 8.1. I f Cay(S U S -~) has a hamiltonian circuii, then Z , . Cay(SU S -a) has a hamiltonian
circuit if either n is odd,
or n is even,
n~[(S)!, a n d
Z , • Cay(S U S -~) is connected (i.e,, Cay(S U 8 -1) is not bipartite). Proof. If n is odd, we use the skewed generator conditions as follows. Choose s ~ S , and let s ~ = s for i odd and st = s -1 for i even. Then sls2" " • s n _ t = e , so by Corollary 5.3, Zn • Cay(S O S -t) has a hamiltonian circuit. In the second case, with n even, n>~l(S)l, and C a y ( S O S -t) not bipartite, choose s~, s2. . . . . sk ~ S U S -~ such that sls2 • • • sk = e and k is odd and as small as possible subject to these requirements. Then I claim that n > k. If not, k > n, and hence k > [(S)[. Then by the pigeon
218
K. Keafing
hole principle, there exist partial products of t the si's which are equal, say sv~.2.., s, = s ~ s 2 . " sv wfth u > v . Then Sz,,~ I S u . v 2
• " * Su =
SIS 2 ....
SUSu+ISu+
2 " ~ " S k "~" e ~
so that we have two sets of elements of S U S -t, each with product e, and each with I(;ss than k elements. One of these sets has u - v elements and the other has k-(u-v) e~ements; thus, one set :must have an odd number of elements, violating k's minimality. Therefore k < n. Now choose s ~ S, and for k < i < n let s, = s when i is odd and let si = s -t when i is even. Then SlS 2 " " " Sn__ t ~
StS 2 " " " S k " Sk+tSk+2
" " ° Sn_
1 ~"
e,
and the desiled result follows from Corollary 5.3. By applying Theorem 4.1 to Theorem 8.1, we get the following corollary. Corollary 8.2. Suppose [(S)[~[(T)[ a n d C a y ( S U S -t) a n d C a y ( T U T -t) each have ,~ hamiltonian circuit. The~z Cay(SU S-t) • C a y ( T L I T -t) has a hamiltonian circuit if either [(S) I is odd, [(T)! is odd, or C a y ( T U T -1) is not bipartite. The above corollary leads us to conjecture that if Cay(S O S -t) and C a y ( T U T -I) each I'~ave a hamiltonian circuit then Cay(S U S -t) • C a y ( T U T -t) has a hamiltonian circuil as long as thi~ conjunction graph is cotmected. The only case not covered above is the case with [(S)[ and [(T)I even and C a y ( T O T -~) bipartite. If the conjunction graph is to be connected, we must have C a y ( S U S -t) non-bipartite. In that case it is sufficient to find a bamiltonian circuit in Z2" C a y ( S U S - t ) , for once wc have that circuit we can use Theorem 7.1 and the Subgraph Lemma to solve the general case.
9. Hamiltonian paths In this last section we show that some of our theorems on hamiltonian circuits also hold for hamiltonian paths. We close by pointing out z large class of groups G such that if ( S ) = G then Z , . Cay(S) has a hamiltonian path if and only if Z,.Cay(S) satisfies the path version of the skewed generator conditions. The proo£, for the following two theorems are essentially identical with the proofs for their uircuit analogues, Theorems 5.2 and 7.1. Theorem 9.1. Suppose ( a S } = ( S ) , Cay(aS) has a hamiltonian path, a,:d a = s~ s2 • • • s, ~ for ,'ome st, s2 . . . . . s,. i ~ S. T h e n Z , • Cay(S) has a hamiltonian path. When the conditions of Theorem 9.1 are satisfied, we say that Z , . Cay(S) satisfies the path s k e w e d generator conditions.
The eonfunctionof Cayley digraphs
219
%heorem 9.2. If Z , . C a y ( S ) has a hamiltonian path and there exist st, s2. . . . . sm e S such that sis:,." • " s,. = e, then Z,+m • Cay(S) has a hamiltonian path. Our final result relies on a ,theorem of Holsztyfiski and Nathanson (see [1, Theorem 3.1]) which says that every abelian ot hamiltonian group is sequential. Theorem 9.3. If (S) is a group in which every subgroup is normal, the following are equivalent: (1) Z , . Cay(S) is connected. (2) Z~ • Cay(S) ht;s a hamiltonian Fath. (3~ Z , • Cay(S) satisfies the path skewed generator conditions. l~-e~l. That (2) implies (1) i,~ d e a l and (3) implies (2) follows from Theorem 9.1. To see that (1) implies (3), first not~c~ that since every subgroup of (S) is normal, (S) nt~st be nilpotent, since the Sylow p-subgroups of (S) are normal. Choose any s~.S and let s i = s for ~<-i<~n-1. Then by Theorem 6.1, (s*-1S)=($). By Holsz~vfiski and Nathanson's result, Ca y(s"-~S) has a hamiltonian path. Thus, Z~ • 'Cz,y(S) satisfies the path skewed generator conditions.
Aek~ewledgements TI~e author would like to tha~ak Steven Curran, Gall Letzter, and David Witte for rh.:ir help and sugge;~tions, and Professor Joseph A. Gallian for hit constant s~np[~ort and indispensib~e advice. This research was done at the University of Minr~esota, Duluth, in an Undergraduate Research Participation program funded by ~:ae NSF (Grant number NSF/SPI-7926564).
[1] W. Holsztyfiskiand R.F. Strube, Paths and circuits in finite groups, Discrete Math. 22 (1978) 263-272. [2] B. Huppert, Endliche Gruppc.n ! (Springer-Verlag,New York/Berlin, 1967). [3] G. Letzter, Hamiltonianc~rcuitsin cartesian products with a metacyclicfactor, prcprint. [4] R.A. Rankin, A campanologicalproblem in group theory, Proc. Camb. Phil. Soc. 44 (1948) 17-25. [5] R.A. Rankin, A campanologicalproblem in group theory II, Proc. Camb. Phil. Soc, 62 (1966) 11-18. [6] W.T. Trotter, Jr. and P. Erd6s, When the cartesian product of directed cycles is hamiltonian,J, Graph Theory2 (1978) 137-142. [7] D. Wine, On hamiltoaian ciircuitsin Cayley diagrams, Discrete Math. 38 (1982), 99-108. [8] D. Witty, G. Letzter, and J.A. Gallian, On hamiltoniancircuits in cartesian products of .Cayley digraph;i, Discrete Math. 43 (1983), to appear.