DIGRAPHS FROM POWERS MODULO p

DIGRAPHS FROM POWERS MODULO p Caroline Luceta Box 111 GCC, 100 Campus Drive, Grove City PA 1617 USA Eli Miller PO Box 410, Sumneytown, PA 18084 USA Clifford Reiter Department of Matematics, Lafayette College, Easton, PA 1804 USA 1. Introduction Given any function f defined modulo m, we can consider te digrap tat as te residues modulo m as vertices and a directed edge (a, b) if and only if f(a) b (mod m). Te digrap associated wit squaring modulo p, a prime, as been studied in [1]. In tat paper te cycle lengts and te number of cycles appearing were caracterized. Te structure of te trees attaced to cycle elements was also completely described. Our paper will generalize tose results to te digrap associated wit te function x modulo a prime p wit any positive integer. Since zero is in an isolated cycle for all p and, we will consider te digrap generated by te non-zero reduced residues. We will let G p denote te digrap on te non-zero residues modulo p wit edges given by x (mod p). Hence, te vertex set of G p is equal to Z * p. For example, G 3 53 is sown in Figure 1 and G 4 41 is sown in Figure. Note tat wen p =, G p consists of te vertex 1 in a loop. Tus, we need only consider G p wen p is an odd prime. We will use p to denote an odd prime trougout tis paper. Elementary results about tese digraps are described in Section. In particular, we see eac component contains a single cycle and we can determine wen tere are non-cycle vertices. Section 3 caracterizes te cycle lengts tat appear. Section 4 explores te relationsip of geometric subsets of te digrap to subgroups of te group of units mod p. Section 5 considers some special cases were long cycles occur. Section 6 returns to te basic structure of te digrap and sows tat all te forests appearing must be isomorpic and caracterizes teir eigts. Section 7 explores te simplifications of te structures tat appear wen is prime. We begin by enumerating six well nown elementary teorems wic will be used. Proofs can be found in many standard texts. Teorem 1: If a _ 0, tere are 0 or gcd(, p - 1) solutions to x a (mod p). Proof: See [,47]. Teorem : If d is a positive integer suc tat d p - 1, ten tere are exactly φ(d) incongruent residues of order d modulo p. Proof: See [,48]. Teorem 3: If n is a positive integer, ten Proof: See [4,83]. d_ n,d>0 φ(d) = n. Teorem 4: If a is an integer suc tat gcd(a, m) = 1 and i is a positive integer, ten

i ordm a ord a = gcd(i, ord Proof: See [4,13]. m. m a) Teorem 5: A primitive root modulo m exists if and only if m is of te form, 4, p n, or p n were p is an odd prime. Proof: See [,49]. Teorem 6: If a and b are elements of Z * p suc tat α = ord p a, β = ord p b, and gcd(α, β) = 1, ten ord p ab = αβ. Proof: See [3,46].. Basic Properties Te following lemmas are easy to prove but fundamental to te understanding of te digrap structure of G p. We will see tat for all G p, eac grap component contains a unique cycle wic may ave forest structures attaced to it. Lemma 7: Te outdegree of any vertex in G p is one. Proof: Te function x (mod p) maps te vertex a to a and only a. Lemma 8: Given any element in G p, repeated iteration of x (mod p) will eventually lead to a cycle. Proof: Because tere are p - 1 vertices in G p, iterating x (mod p) must eventually produce a repeated value. Lemma 9: Every component of G p contains exactly one cycle. Proof: Suppose a component as more tan one cycle; ten somewere along te undirected pat connecting any two cycles tere exists a vertex wit outdegree at least, wic is impossible. Lemma 10: Te set of non-cycle vertices leading to a fixed cycle vertex forms a forest. Proof: Since eac component contains exactly one cycle, te vertices leading to a cycle vertex cannot contain a cycle; tus tey are a forest. Lemma 11: Te indegree of any vertex in G p is 0 or gcd(, p - 1). Proof: Tis result is an immediate application of Teorem 1. Lemma 1: Every component of G p is cyclical if and only if gcd(, p - 1) = 1. Proof: ( ) If all digrap components are cyclical bot te indegree and outdegree are one, wic implies from Lemma 11 tat gcd(, p - 1) = 1. ( ) Conversely, if gcd(, p - 1) = 1 te indegree of every vertex is 0 or 1. If some component were not cyclical tere would exist a cycle vertex wit indegree, a contradiction.

For example, eac component of G 3 53 is cyclical since gcd(3, 5) = 1; tis is apparent in Figure 1. Liewise, G 4 41 as vertices outside te cycles because gcd(4, 40) = 4 (see Figure ). We will refer to a cild of a vertex a as a vertex v tat satisfies te equation v a (mod p). Tese are te predecessors of a in G p. Note tat our cild vertices are cildren in te sense of te forest structure but not in te standard sense of direction. Predecessors tat are not in a cycle will be called non-cycle cildren. For example, in G 4 41, 37 as te tree non-cycle cildren 8, 31 and 33. Lemma 13: Any cycle vertex, c, as gcd(, p - 1) - 1 non-cycle cildren. Proof: From Lemma 11, te indegree of c is 0 or gcd(, p - 1). Since c is a cycle vertex, te indegree is not zero but gcd(, p - 1). Te number of non-cycle cildren is gcd(, p - 1) - 1. Above we saw some of te basic results about G p. Notice tat if we fix and vary p te number of vertices canges and infinitely many different digraps result. However, te next teorem sows tat if p is fixed, only finitely many distinct digraps result as varies. Teorem 14: 1 (mod p - 1) if and only if G p 1 = G p. Proof: ( ) Suppose 1 (mod p - 1) and witout loss of generality 1. If a is any vertex in 1-1 te reduced residue set, ord p a (p - 1) ( 1 - ), so a 1 implies a a ( mod p). Hence, G p 1 = G p. ( ) Suppose G p 1 = G p 1, and assume 1. Ten a a ( mod p) implies ord p a ( 1 - ) for all a in G p. So (p - 1) ( 1 - ), and te conclusion follows. Te 1 different digraps for G13 are sown in Figure 3. Notice tat some ave only cycles and some ave forest structures. Also notice tat tis teorem gives a condition for equality of digraps, but does not settle te question of wen two digraps can be isomorpic for different 8 8 values of. Note tat G11 ne G11 but G11 G11. 3 3. Caracterizing Cycles Wen considering te cycle structure of G p it is convenient to factor p - 1 as wt were t is te largest factor of p - 1 relatively prime to. So gcd(, t) = 1 and gcd(w, t) = 1. For example, if p = 41 and = 6, ten p - 1 = 3 5, so w = 8 and t = 5. Similarly if p = 47 and = 4 ten w = and t = 3; also if p = 19 and = 6 ten w = 18 and t = 1. In all te teorems below we will be considering te digrap G p wit p - 1 = wt as described. Teorem 15: Te vertex c is a cycle vertex if and only if ord p c t. Proof: ( ) Since c is in a cycle tere exists some x 1 suc tat c x c and tus c x-1 1( mod p). Hence, ord p c x - 1, wic implies tat gcd(ord p c, ) = 1, so gcd(ord p c, w) = 1 also. We now ord p c p - 1 = wt, and so ord p c must divide t. ( ) Suppose c G p and ord p c t; terefore, gcd(ord p c, ) = 1. On repeated iteration, c

must eventually end up in a cycle. If y is te number of steps to reac te cycle and x is te cycle y( x -1) lengt, ten c 1( mod p). Terefore, ord p c y ( x - 1), but since gcd(ord p c, ) = 1, ord p c x - 1. Hence, c ( mod p) wic implies tat c is a cycle vertex. c x Corollary 16: Tere are t vertices in cycles. Proof: From Teorem 15, te total number of cycle vertices is of elements of order d (mod p). Teorems and 3 imply tat tis is t. N(d) were N(d) is te number Teorem 17: Vertices in te same cycle ave te same order (mod p). Proof: Assume a and b are in te same cycle. Hence, tere exists an e suc tat a e b (mod p). Let α = ord p a and β = ord p b. It follows tat b α a eα 1 (mod p) and tus β α. Similarly, α β and ence te orders are equal. Teorem 17 sows tat te order (mod p) of vertices in te same cycle are equal. Hence, te notion of te order of a cycle is well-defined. We now loo at te relationsip between te order of a cycle and its lengt. Teorem 18: Let x be te lengt of a cycle of order d (mod p), ten x - 1 is te smallest number of te form n - 1 divisible by d. Proof: Let c be a vertex in te cycle. Since te cycle is of lengt x, c x-1 1 ( mod p). It follows tat d = ord p c divides x - 1. If ord p c s - 1 for some s < x, ten c s c, a contradiction. Teorem 18 sows tat te lengt of a cycle depends entirely on its order. If we let l(d) denote te lengt of cycles wit order d, we get te following teorem. Teorem 19: Let a, b and d be orders of cycles. Ten : (i) l(d) = ord d. (ii) Tere are φ(d) / l(d) cycles of order d. (iii) l(lcm(a, b)) = lcm(l(a), l(b)). (iv) Te longest cycle lengt in G p is l(t) = ord t. Proof: (i) By Teorem 18, l(d) = min{n : d n - 1}, ence l(d) = ord d. (ii) By Teorem, tere are φ(d) elements of order d, and tere are l(d) in eac cycle by (i), ence te result. (iii.a) By (i), l(a) 1 (mod a) and l(b) 1 (mod b), tus lcm(l(a), l(b)) lcm(l(a), l(b)) 1 (mod a) and 1 (mod b). It follows tat lcm(l(a), l(b)) 1 (mod lcm(a, b)). So, l(lcm(a, b)) lcm(l(a), l(b)). (iii.b) We now l(lcm(a, b)) 1 (mod lcm(a, b)). So l(lcm(a, b)) 1 (mod a) and l(lcm(a, b)) 1 (mod b). Tus, l(a) l(lcm(a, b)) and l(b) l(lcm(a, b)) wic implies tat lcm(l(a), l(b)) l(lcm(a, b)). Putting (iii.a) and (iii.b) togeter gives l(lcm(a, b)) = lcm(l(a), l(b)). (iv) All orders of cycles divide t and if d t ten l(t) = l(lcm(t, d)) = lcm(l(t),l(d)) wic implies l(d) l(t). Tus, l(t) is te maximal cycle lengt. We are now in a position to identify te number of cycles of every lengt appearing in te digrap of G p. For example, consider G 3 53 (Figure 1); in tis case p - 1 = 5 so w = 1 and t = 5. d t 4

Te possible orders of te cycle elements are te divisors of t: 1,, 4, 13, 6, and 5. Tere are φ(5) = 4 elements of order 5, and tese 4 elements are in cycles of lengt l(5) = ord 5 3 = 6, contributing four cycles of lengt 6. Similarly, te elements of order 6 appear in 4 cycles of lengt 3; and tose of order 13 are in 4 cycles of lengt 3. Tere are elements of order 4 in one cycle and two cycles of lengt one wit orders 1 and. Table 1 gives some details about cycles in G p for selected p and. 5 4. Subgroups of Z * p in G p We now consider orders of elements trougout G p. We will be able to associate elements of various orders wit subgroups of Z * p, wic allows for te identification of certain 3 4 5 6 p d l(d) # d l(d) # d l(d) # d l(d) # d l(d) # 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 5 4 1 1 1 5 1 1 5 1 4 4 1 4 1 41 5 4 1 8 8 10 4 1 0 4 40 4 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 1 1 1 3 1 1 1 7 3 7 3 7 6 1 7 3 3 1 43 1 6 14 6 1 1 3 4 6 1 7 6 1 14 6 1 1 6 4 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 47 3 11 1 1 3 11 1 1 3 11 3 11 3 1 46 11 46 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 13 1 1 1 1 13 6 1 1 13 1 1 53 4 1 4 1 13 3 4 13 4 3 6 3 4 6 4 3 5 6 4 5 4 6

6 Table 1: Cycle lengts in G p. subgroups of Z * p wit geometric subsets of te digrap. In later sections we will return to caracterizing te cycle and forest structure of tese digraps. Lemma 0: If H d is te set of residues wit orders dividing d, wit d 1, ten H d is a cyclic subgroup of Z * p. Proof: Since Z * p is a finite cyclic group, we need only sow tat H d is non-empty and closed under multiplication. Clearly H d is not empty because it as te identity. To sow closure suppose tat a, b H d and let α = ord p a, and β = ord p b. Since (ab) lcm(α, β) 1 (mod p), ord p ab divides lcm(α, β) wic in turn divides d. Terefore, H d is a subgroup of Z * p. For instance, if we consider te group Z * 41, te elements of order 1 and form te subgroup H wile H 10 contains tose elements of order 1,, 5 and 10. We will now introduce a notation for te forest originating from any given cycle vertex. Let Fc n represent te set of vertices in te n t level of te forest originating from te cycle vertex c. Of course, Fc n depends on G p. For example, in G 4 41 (Figure ), F 1 16 = {, 3, 39} and F16 = {7, 19,, 34}. Similarly, F n refers to te vertices in te n t level of all forests and F c refers to all forest vertices associated wit te cycle vertex c at all levels n 1. Note tat te cycle vertices are not a part of te forests but for convenience we will denote te set of cycle elements by F 0 and also F 0 c = {c} but c Fc. Te next teorem and corollary explain te subgroup structures present in te digraps. First it will be sown tat te order of an element is constrained by its eigt in te forest structure. Teorem 1: Let a F c and ord p c = d t. Ten ord p a d if and only if a Fc were x. Proof: ( ) Suppose ord p a d. Ten ( a d ) 1( mod p). Now using Teorem 4, 1 = ord p 1 = ord p (a ) d = ordp a gcd(d, ord p a ) Because gcd(d, ordp a ) = ordp a, ordp a divides d wic divides t. From Teorem 15, a must be a cycle element, and ence a Fc were x. ( ) If a Fc and x, ten a x is a cycle element of order d. Furtermore, x a ordp d = ordp a =, ence d gcd( x, ord x p a) = ord p a and tus ord p a d x d. gcd(, ordp a) From te Teorem 1 it can be ascertained tat various sets of digrap structures form subgroups of Z * p, as stated in te next corollary.

7 U x Fc Corollary : For all d dividing t and all, 0 x is a subgroup of Z * p, namely H d ordp c d * Proof: Te union over all c suc tat ord p c d and over all x, contains all a Z p suc tat ord p a d by Teorem 1. Tis is te subgroup H of d Z * p by Lemma 0. Tus, te union of cycle vertices wit orders dividing a fixed d and vertices in teir associated forest structures up to a fixed eigt form a subgroup of Z * p. In particular, if d = 1, all te vertices in F 1 up to any fixed level (along wit 1) form a subgroup. On te oter extreme, if d = t, all te vertices in all te components up to a fixed eigt form subgroups. Examining te digrap of G 4 41 (Figure ) one finds te following subgroups : d = 1, = 0 : F1 0 = {1} = H 1. 0 1 d = 1, = 1 : F1 union F1 = {1, 9, 3, 40} = H 4. 0 1 d = 1, = : F1 union F1 union F1 = {1, 3, 9, 14, 7, 3, 38, 40} = H 8. d = 5, = 0 : F 0 = {1, 10, 16, 18, 37} = H 5. 0 1 d = 5, = 1 : F union F = {1,, 4, 5, 8, 9, 10, 16, 18, 0, 1, 3, 5, 31, 3, 33, 36, 37, 39, 40} = H 0. 0 1 d = 5, = : F union F union F = {1,..., 40} = Z * 41 = H 40. Lemma 3: Te product of a forest element and a cycle element is a forest element. Proof: Te cycle elements of G p form a closed multiplicative subgroup of Z * p ; ence, te product of a forest element and a cycle element must be a forest element. Tese algebraic properties imply tat te igest level of te digrap, wic will be referred to as te canopy, must contain at least alf of te vertices. For example, in G 4 41 (Figure ) te canopy is F. p - 1 Corollary 4: If 0 is te maximal eigt attained by te forest elements in G p ten F 0. p - 1 Proof: (i) If 0 = 0 ten all vertices are in cycles, so F 0 = p - 1. n (ii) If 0 1, ten from Corollary, H 0 t = F is a proper subgroup of Z * -1 p. Te union 0 n 0-1 * number of elements in H -1 0 t must be a proper divisor of Z p = p - 1. Because te largest proper divisor of p - 1 is (p - 1) / tere are at least (p - 1) / vertices remaining in te canopy. 5. Occurrence of Long Cycles Control over te lengts of cycles is igly desirable. Tis is essential for applications to pseudo-random number generation and data encryption. Te first teorem below provides an upper bound for te cycle lengts appearing in G p. Special cases were long cycles can be guaranteed are.

8 ten considered. Teorem 5: Let p > 5 be prime. Ten te lengt of te longest cycle in G p is less tan or equal to (p - 3) /. Proof: Consider two cases depending on gcd(, p - 1). (i) Suppose gcd(, p - 1) 1. By Lemma 1, G p is not entirely cyclical and by Teorem 4, it as a forest structure wit at least (p - 1) / vertices in te canopy. So, tere are at most (p - 1) / vertices in cycles. Since p > 5, we now we are not interested in longest cycles of lengt 1, and since 1 is in a loop, it is not part of any longest cycle of interest, ence te maximal lengt cannot p - 1 p - 3 exceed - 1 =. (ii) Suppose gcd(, p - 1) = 1; tus G p consists entirely of cycles. From Teorem 19, te longest cycle lengt is associated wit te elements of order t = p - 1 wic can be factored as s τ, were s 1 and τ is odd. (a) If τ 1, ten te number of elements of order p - 1 is φ(p - 1) = φ( s τ) = s-1 φ(τ) < s-1 τ = p - 1 p - 3. Now since φ(p - 1) is an integer, φ(p - 1). Hence, even if all elements of order p - 1 were togeter in one cycle, te lengt could not exceed (p - 3) /. (b) If τ = 1, ten p = s + 1 is a Fermat prime larger tan 5, so s > 1. By Teorem 19 te lengt of te longest cycle is l(t) = l( s ) = ord s. However, Z * s does not ave a primitive root for p-1 s > 1 (Teorem 5). Tus, ord < φ s ( ) = s and ence l(t) p - 3 in tis case as well. Wile tis teorem gives an upper bound for te cycle lengts in G p, it does not specify weter tis bound is ever attained. Te next teorem sows tat for Sopie Germain primes tese maximal cycle lengts can be attained. Teorem 6: Let p = q + 1 were q is an odd Sopie Germain prime. If is a primitive root mod q, ten G p contains a cycle of lengt (p - 3) /. Proof: Because gcd(, q) = 1 and q t, te elements wit order q are in cycles of lengt ord q = q - 1 = (p - 3) /. Corollary 7: Let p = q + 1 were q is an odd Sopie Germain prime. If is an odd primitive root mod q, ten G p contains two cycles of lengt (p - 3) /. Proof: Since is odd, t = q and te grap is entirely cyclical. Te elements of order q are in a cycle of lengt ord q = q - 1 = (p - 3) /. Te elements of order t = q are, by Teorem 19, in te longest cycles of te digraps. So te elements of order q are also in a cycle of lengt (p - 3) /. For example, consider G 3 (Figure 4); p = 3 = (11) + 1, and is a primitive root mod 11. As expected, G 3 contains one cycle of lengt 10. Corollary 7 is illustrated by G 5 47 were p = 47 = (3) + 1, 5 is a primitive root mod 3, and te digrap as two cycles of lengt.

Given a prime of te form q + 1, were is a primitive root mod q, it is simple to iterate troug a cycle of lengt (p - 3) /. Any residue between and p - is eiter in a long cycle or is one step away. Beginning wit any suc residue we can iterate x (mod p) to produce (p - 3) / incongruent values. For example, consider te prime 9887 = (4943) + 1 were 4943 is also prime. Since 7 is a primitive root of 4943, iteration of x 7 (mod 9887) beginning wit any x from to 9885 will yield 494 incongruent values. 6. Caracterizing Forests Having completely caracterized te cycles for te digrap generated by x (mod p) we turn our attention to caracterizing te noncyclical elements of G p. In eac of our examples, we notice tat te forests in any particular digrap are isomorpic. Tis turns out to be true in general and will be proved by constructing a one-to-one correspondence between F 1 and F c. Te next lemma gives te essence of ow te correspondence will be constructed. Lemma 8: If a F1 and c is a cycle vertex, ten ac F. c Proof: Using Lemma 3, it follows immediately tat ac F 0. Furtermore, (ac ) ac c ( mod p) is a cycle element but (ac ) - 1 is a forest element, wic implies tat. F c Teorem 9: Let c be a cycle element, ten F 1 F c. Proof: (i) First we sow tere exists a 1-1 correspondence between te vertices of F1 and Fc for all eigts, and ence between F 1 and F c. Let be fixed and let c denote te unique cycle element suc tat c c ( mod p). Define f : F1 Fc by f (a) a c (mod p). First we cec tat f is one-to-one and onto. Let b. Ten -1-1 - 1-1 c ) b ( c ) c c 1 and (b c ) F 0 because F 0 and (b -1 F1 0 c -1 F 0-1. It follows tat b c F1-1 -1. Furtermore, f (b c ) b c c b ( mod p) so f is onto Fc. Suppose f (a 1 ) f (a ) (mod p) for a1, a F1. Ten a 1 c a c implies a 1 a (mod p). Tus, f is one-to-one. (ii) It remains to be sown tat tere exists a 1-1 correspondence between te edges of F 1 and F c. We want to define g : E(F 1 ) E(F c ) by g(a, a ) = (f (a), f -1 (a )) were is te eigt of a in F 1. If (f (a), f -1 (a )) is in fact in E(F c ), ten g will inerit te 1-1 and onto properties from f and f -1. We ave an edge (f (a), f -1 (a )) if and only if (f (a)) f -1 (a ) (mod p). Now f (a) a c (mod p) were c and f -1 (a) a c -1 (mod p) were c ( mod p) ; tus c -1 - c 1-1 c c implies ( c ) c ( mod p). By te uniqueness of c, c-1 c and f -1(a) a c ( mod p). Now (f (a)) (a c ) a c f -1 (a ) (mod p). Hence, (f (a), f -1 (a )) E(F c ) and by te argument above, te edges and vertices are in 1-1 correspondence, so F 1 F c. Tere is anoter property of tis mapping tat can be addressed. Considering a F 1 and c F 0, te order of te element ac (mod p) will be (ord p a)(ord p c). Tat is, te isomorpism "preserves" orders between F 1 and F c in tat te orders of corresponding elements in F c are multiplied by te F c b -1 9

10 order of c. Teorem 30: If a F 1 and b F c wit c te cycle element suc tat b a c (mod p), ten ord p b = (ord p a)(ord p c). Proof: By Teorem 1, ord p a and tus gcd(ord p a, t) = 1. By Teorem 15, ord p c = ord p c t, ence gcd(ord p a, ord p c) = 1 and applying Teorem 6 gives te desired result. Some examples of te isomorpism described in Teorem 9 and Teorem 30 can be seen in Table. a c ac ord p a ord p c ord p ac 1 10 10 1 5 5 9 37 5 4 5 0 3 10 30 8 5 40 Table : Some orders and products in G 4 41. Finally, we prove a result tat determines te eigt of te forests: Teorem 31: If 0 is te minimal suc tat p - 1 t, ten 0 is te eigt of te forests in G p. Proof: (i) If gcd(, p - 1) = 1, ten all te vertices are in cycles and ence 0 = 0. (ii) If gcd(, p - 1) 1, ten 0 1. Let a be a vertex of order p - 1. From Teorem 1 a 0 0-1 must be at eigt 0 because ordp a t but ordp a _ t. Tere are no vertices at a greater eigt since for any vertex b in G 0 p, ordp b p - 1 t and tus b is at level 0 or lower in G p. For example, in G 4 41 we see tat t = 5 and 41-1 = 40 4 5 wic implies tat te eigt of te forest is. Tis value is apparent in Figure. 7. Prime Exponents In te special case were te exponents are prime, many of our results simplify. In particular, wile we were able to prove tat te forest structures were isomorpic wit a general exponent, we can completely caracterize tat structure if te exponent is prime. We will consider te digrap associated wit a prime exponent q, letting p - 1 = q s t were t is relatively prime to q. Corollary 3: Te indegree of a vertex in G q p is 0 or q if p 1(mod 1 oterwise Proof: Tis follows from Lemma 11 and Lemma 1 wit = q. q). Tus all te digrap components are cyclical if and only if p _ 1 (mod p). Corollary 33: Any cycle vertex as q - 1 non-cycle cildren. Proof: Tis follows from Lemma 13.

11 Teorem 34: If p 1 (mod q), te q - 1 non-cycle cildren of eac cycle element are roots of complete q-nary trees. Proof: Since te t forests in G q p are isomorpic and tere are q s t elements in te digrap, F 1 = q s - 1. Furtermore, Teorem 31 states tat te eigt of F 1 is s. If F 1 is not composed of q - 1 complete q-nary trees, tere exists a vertex tat as indegree 0 but is not at eigt s. Tis would imply tat F 1 < q s - 1, a contradiction. Since all te forests are isomorpic to F 1, and F 1 consists of complete q-nary trees, all te forests in G q p are complete q-nary trees as well. As an example, consider F 1 in te digrap G109 3, as sown in Figure 5. Since 109-1 = 3 3 (4), and 45 3 63 3 1 (mod 109), 45 and 63 are roots of complete ternary trees wit eigt 3-1 =. Wen te exponent is prime we can also say more about te orders of elements in te digrap G q p. Teorem 35: A vertex a F1 if and only if ord p a = q. Proof: Consider Teorem 1 wit = q and c = 1, and ence d = 1. Ten ord p a q if and only if a F1 x for x. Having elements of order q in a level x less tan would imply tat q q x were x <, a contradiction. Returning to G109 3 (Figure 5), one can cec tat te orders of 3, 9, and 7, correspond to F 1 1, F1, and F1 3. Corollary 36: If a Fc ten ord p a = q ord p c. Proof: Teorem 35 and te multiplying principle of Teorem 30 give te desired result. Conclusions We ave seen tat many of te features of te digrap G p can be determined in terms of properties of p and. In particular, we ave seen tat te digraps brea up into components wit exactly one cycle per component and tat te forest structures associated wit eac cycle vertex trougout te digrap are isomorpic. Te cycle lengts depend on te orders of te elements; we can also determine te eigts of te forests. In special cases long cycles can be found and complete q-nary trees can be guaranteed. Wile we ave found a very ric structure for te digraps associated wit x mod p, it is natural to as wat oter digraps arising from functions suc as tese ave a ric structure. Te function x mod m, were m is not prime will ave a muc different digrap since 0 will not necessarily be in a trivial cycle and primitive roots may not exist. On te oter and, looing at x in a finite field were 0 must be trivial and primitive roots always exist ougt to lead to a teory lie tat seen in tis paper. Digraps from functions suc as x + 1 mod p will be difficult to andle because we can not lean on te teory of orders of elements as in tis paper. It would be interesting to now wat ind of control on te digraps can obtained in suc cases.

1 Acnowledgement Tis wor was supported in part by NSF-REU grant DMS-9300555. References [1] Earle L. Blanton, Jr., Spencer P. Hurd, and Judson S. McCranie, "On a Digrap Defined by Squaring Modulo n", Te Fibonacci Quarterly 30.4 (199) 3-334. [] Hua Loo Keng, Introduction to Number Teory, translated by Peter Siu, Berlin: Springer- Verlag, 198. [3] Natan Jacobson, Basic Algebra I, San Francisco: W.H. Freeman and Co., 1974. [4] James Strayer, Elementary Number Teory, Boston: PWS Publising Co., 1994. AMS Classification Numbers: 05C0, 11B50. *****