Discrete Mathematics

Transcription

1 Dscrete Mathematcs 32 (202) Contents lsts avalable at ScVerse ScenceDrect Dscrete Mathematcs journal homepage: On the symmetrc dgraphs from powers modulo n Guxn Deng a,, Pngzh Yuan b a School of Mathematcs, Guangx Teachers Educaton Unversty, Nannng 53000, PR Chna b School of Mathematcs, South Chna Normal Unversty, Guangzhou 5063, PR Chna a r t c l e n f o a b s t r a c t Artcle hstory: Receved 8 August 200 Receved n revsed form 24 July 20 Accepted 6 November 20 Avalable onlne 3 December 20 Keywords: Symmetrc Congruence Dgraph products Component Heght Cycle length For any postve ntegers n and k, let G(n, k) denote the dgraph whose set of vertces s H = {0,, 2,..., n } and there s a drected edge from a H to b H f a k b(modn). The dgraph G(n, k) s called symmetrc of order M f ts set of connected components can be parttoned nto subsets of sze M wth each subset contanng M somorphc components. In ths paper, we establsh a necessary and suffcent condton for G(n, k) to be symmetrc of order M, where M has an odd prme dvsor. 20 Elsever B.V. All rghts reserved.. Introducton Let n, k be two postve ntegers. Let G(n, k) denote the dgraph whose set of vertces s {0,, 2,..., n } and there s a drected edge from a vertex a to a vertex b f a k b (mod n). In [4], Somer and Křížek called a dgraph symmetrc f ts connected components can be parttoned nto somorphc pars. They gave the followng defnton n [6]. Defnton.. Let M 2 be an nteger. A dgraph s sad to be symmetrc of order M f ts set of components can be parttoned nto subsets of sze M, each contanng M somorphc components. In [7], Szalay showed that G(n, 2) s symmetrc of order 2 f n 2 (mod 4) or n 4 (mod 8). In [], Carlp and Mncheva proved that f p s a Fermat prme, then G(2 r p, 2) s not symmetrc of order 2 when r = 3 or r 5. The followng theorem s part of Theorem 5. n [6]. Theorem. ([6, Theorem 5.]). Let n = n n 2, where n >, n 2, and gcd(n, n 2 ) =. Then () Suppose that n = p α, where p s an odd prme and α. Suppose further that k (mod p ) and p α k. Then G(n, k) s symmetrc of order p. () Suppose that n = q q 2 q s, where the q s are dstnct prmes, and s 2. Suppose that k (mod λ(n )). Then G(n, k) s symmetrc of order n. Supported by the Guangdong Provncal Natural Scence Foundaton (No ) and NSF of Chna (No ). Correspondng author. E-mal addresses: oldlao@63.com (G. Deng), mcsypz@mal.sysu.edu.cn (P. Yuan) X/$ see front matter 20 Elsever B.V. All rghts reserved. do:0.06/j.dsc

2 G. Deng, P. Yuan / Dscrete Mathematcs 32 (202) () Suppose that n = p α q q 2 q s, where p s an odd prme, α 2, s, and the q s are dstnct prmes such that p q and p q for =, 2,..., s. Suppose further that k (mod λ(pq q 2 q s )) and p α k. Then G(n, k) s symmetrc of order pq q 2 q s. Here the functon λ s the Carmchael lambda-functon, whch wll be ntroduced n Secton 2. In [2], Kramer-Mller obtaned a necessary and suffcent condton for G(n, k) to be symmetrc of order p, where n s square free and p s an odd prme. Theorem.2 ([2, Theorem 3.5]). Let n = pq q 2 q m, where q and p are dstnct odd prmes. Suppose G(p, k) s not symmetrc of order p. Then G(n, k) s symmetrc of order p f and only f both of the followng condtons are satsfed. () gcd(p, k) =. () Let T = {q gcd(q, k) = }. Then T s not empty and for all x N, p A x (G( q T q, k)) or ord p k x. Here we use A t (G(n, k)) to denote the number of t-cycles contaned n G(n, k), and use A(G(n, k)) to denote the set of cycle lengths that appear n G(n, k). In ths paper, we generalze Theorem.2 to any postve ntegers n and M, where M has an odd prme dvsor. The outlne of ths paper s as follows. In Secton 3, we present some prelmnary results on the structure of G(n, k). In Secton 4, we treat the case n = p m. In Secton 5, we prove two lemmas for our man results. In Secton 6, we state and prove the man theorem of the present paper. 2. The Carmchael lambda-functon Before proceedng further, we need to revew some propertes of the Carmchael lambda-functon λ(n). Defnton 2.. Let n be a postve nteger. The Carmchael-lambda-functon λ(n) s defned as follows: λ() =, λ(2) =, λ(4) = 2, λ(2 k ) = 2 k 2 for k 3, λ(p k ) = (p )p k for any odd prme p and k, λ(p k pk 2 2 pk r r ) = lcm[λ(pk ), λ(pk 2 2 ),..., λ(pk r r )], where p, p 2,..., p r are dstnct prmes and k for all {, 2,..., r}. The followng theorem generalzes the well-known Euler s theorem whch says that a φ(n) (mod n) f and only f gcd(a, n) =. Theorem 2. (Camchael). Let a, n N. Then a λ(n) (mod n) f and only f gcd(a, n) =. Moreover, there exsts an nteger g such that ord n g = λ(n), where ord n g denotes the multplcatve order of g modulo n. For the proof, see [3, p. 2]. 3. Some prelmnary results on G(n, k) There are two partcular subdgraphs of G(n, k). Let G (n, k) be the nduced subdgraph of G(n, k) on the set of vertces whch are coprme to n, and G 2 (n, k) be the nduced subdgraph of G(n, k) on the set of vertces whch are not coprme wth n. We observe that G (n, k) and G 2 (n, k) are dsjont and that G(n, k) = G (n, k) G 2 (n, k), that s, no edge goes between G (n, k) and G 2 (n, k). It s clear that each component of G(n, k) contans a unque cycle. The followng lemma tells us that the structure of each component contaned n G (n, k) s determned by ts cycle length. Lemma 3. ([6, Corollary 6.4]). Let t be a fxed nteger. Then any two components n G (n, k) contanng t-cycle are somorphc. Now consder a dgraph G(n, k), and factor λ(n) as λ(n) = uv, where u s the largest dvsor of λ(n) relatvely prme to k. We need the followng results on the cycles of G(n, k).

3 722 G. Deng, P. Yuan / Dscrete Mathematcs 32 (202) Lemma 3.2 ([8]). There exsts a t-cycle n G (n, k) f and only f t = ord d k for some factor d of u. Lemma 3.3 ([6, Theorem 6.6]). Let n = r = pe be the prme factorzaton of n. If t A(G(n, k)), then A t (G(n, k)) = r (δ gcd(λ(p e ), k t ) + ) da d (G(n, k)), t = where δ = 2 f 2 k t and 8 p e, and δ = otherwse. d t,d t Let G be a dgraph and a be a vertex n G. The ndegree of a, denoted by ndeg(a) s the number of drected edges comng to a, and the outdegree of a s the number of edges leavng a. Partcularly, let ndeg k n (a) denote the ndegree of a vertex a G(n, k). It s clear that each vertex n G(n, k) has outdegree. In the rest of ths paper, all dgraphs are assumed to be fnte and have ths property. Defnton 3.. We defne a heght functon on the vertces and components of G(n, k). Let c be a vertex of G(n, k). Let h(c) to be the mnmal nonnegatve nteger such that c k s congruent modulo n to a cycle vertex n G(n, k). If C s a component of G(n, k), we set h(c) = sup c C h(c). The ndegree and heght functon play an mportant role n the structure of G(n, k). We need the followng results concernng the ndegrees and heghts. Lemma 3.4 ([8]). Let n = r = pe be the prme factorzaton of n, where e. Let a be a vertex of postve ndegree n G (n, k). Then r r ndeg k n (a) = = ndeg k p e (a) = = δ gcd(λ(p e ), k), where δ = 2 f 2 k and 8 p e, and δ = otherwse. Lemma 3.5 ([5, Theorem 3.2]). Let p be a prme. Let a be a vertex of postve ndegree n G 2 (p e, k), and assume that p l a and a 0. Then l = kt for some postve nteger t and ndeg k p e(a) = δp(k )t gcd(λ(p e l ), k), where δ = 2 f p = 2 and e l 3, and δ = otherwse. Lemma 3.6 ([6, Lemma 3.2]). Let p be a prme and e, k be two postve ntegers. Then ndeg k p e(0) = pe e k. Lemma 3.7. Let p be a prme and let e, k 2 be ntegers. Suppose that h be the unque postve nteger such that k h < e k h. Then h = h(g 2 (p e, k)). Proof. It s clear that p G 2 (p e, k) and h(p) = h(g 2 (p e, k)). And p k 0 (mod p e ) f and only f k e. The proof s completed. Lemma 3.8. Let p be a prme and e, k 2 be two postve ntegers. Let λ(p e ) = uv, where u s the maxmal dvsor of λ(p e ) relatvely prme to k. If C s the component of G(p e, k) contanng, then h(c) = mn{ : v k }. Proof. Let h = mn{ : v k }. Then there exsts a dvsor d of v such that d s not a dvsor of k h. By Theorem 2., there exsts a vertex g G(p e, k) such that ord p eg = uv. Let a g uv d (mod p e ). Then ord p ea = d, a kh (mod p e ) and a kh (mod p e ). Hence, h(c) h(a) = h by the defnton of the heght functon. Conversely f a C, then there exsts j such that a kj (mod p e ), so ord p ea k j. Snce ord p ea uv, we have ord p ea v, so a kh (mod p e ), that s, h(c) h. 4. The case n = p e Recall that a dgraph G s called semregular f there exsts a postve nteger d such that each vertex of G ether has ndegree d or 0.

4 G. Deng, P. Yuan / Dscrete Mathematcs 32 (202) Lemma 4.. Let p be an odd prme and let e, k 2 be two postve ntegers. Then G(p e, k) s semregular f and only f gcd((p )p e, k) = p e. Proof. The case e = s trval. Assume that e 2 and that G(p e, k) s semregular. Then ndeg k p e(0) = ndegk pe(). By Lemmas 3.4 and 3.6, ndeg k p e(0) = pe e k = gcd((p )p e, k) = ndeg k p e(). If e k 2, then ndegk p e(pk ) > 0. By Lemma 3.5 ndeg(p k ) = p k gcd((p )p e k, k) = p k +mn{e k,e e k }. Snce ndeg(p k ) = ndeg(0), t follows that e e = k +mn{e k, e e }, so e e = k +(e k ) = e 2. k k k We fnd that e = 2. Now we have k < e 2k and k pe 2 k. Therefore, p e 2 < e, we deduce e = 2 and k =, whch s a contradcton. Hence, e k = and gcd((p )pe, k) = p e. Conversely, f gcd((p )p e, k) = p e, then k e. The vertex 0 s the only vertex n G 2 (p e, k) wth postve ndegree and ndeg(0) = p e = ndeg(a) for any vertces a n G (p e, k) wth postve ndegree. So G(p e, k) s semregular. Ths proves Lemma 4.. Corollary 4.. Let p be an odd prme and e, k 2 be two postve ntegers. Suppose that G(p e, k) s semregular. Then A (G(p e, k)) = p. A(G(p e,k)) Theorem 4.. Let p be an odd prme. Then G(p e, k) s symmetrc of order p f and only f gcd((p )p e, k) = p e and k (mod p ). Proof. Assume that G(p e, k) s symmetrc of order p. Then there exst at least p dstnct components C, C 2,..., C p contaned n G (p e, k) such that C G 2 (p e, k). But G 2 (p e, k) = p e, therefore, C + C C p + G 2 (p e, k) = p e = G(p e, k). It follows that G(p e, k) s the unon of C, C 2,..., C p, and G 2 (p e, k). By Lemma 3.3, we have p = A (G(p e, k)) = gcd((p )p e, k ) +. (4.) We deduce p k from ths equaton. The converse mplcaton follows mmedately from () of Theorem.. 5. Propertes of dgraphs products Gven two dgraphs G and G 2. Let G G 2 denote the dgraph whose vertces are the ordered pars (a, a 2 ), where a G and there s a drected edge from (a, a 2 ) to (b, b 2 ) f there s a drected edge from a to b and a drected edge from a 2 to b 2. In [6], Somer and Křížek noted the followng fact. Let n = n n 2, where gcd(n, n 2 ) =. Then G(n, k) = G(n, k) G(n 2, k). And the canoncal somorphsm s gven by a (a, a 2 ), where a a (mod n ), =, 2. In general, G(n, k) = G(p e, k) G(pe 2 2, k) G(pe r r, k), f n = r = pe s the prme factorzaton of n. We need ths fact and the followng lemmas. Lemma 5. ([2, Lemma 3.]). Let n = n n 2, where gcd(n, n 2 ) =. Let C be a component of G(n, k) and C 2 be a component of G(n 2, k). And the cycle length of C s t. Then C C 2 s a subdgraph of G(n, k) consstng of gcd(t, t 2 ) components, each havng cycles of length lcm[t, t 2 ]. Corollary 5.. Let n = n n 2, where gcd(n, n 2 ) =. If G(n, k) s symmetrc of order M, then G(n, k) s also symmetrc of order M. Proof. It follows mmedately from Lemma 5. and the fact G(n, k) G(n, k) G(n 2, k). Lemma 5.2 ([2, Lemma 3.2]). Let n = r = pe be the prme factorzaton of n. Let a = (a, a 2,..., a r ) and b = (b, b 2,..., b r ) be two vertces n G(n, k) = G(p e, k) G(pe 2, k) 2 G(pe r r, k). If a and b are n the same cycle, then a and b are n the same cycle for each. Lemma 5.3. If G(n, k) s symmetrc of order M, then G(n, k r ) s also symmetrc of order M for any r.

5 724 G. Deng, P. Yuan / Dscrete Mathematcs 32 (202) Proof. Let C, C 2 be two components of G(n, k) and there exsts an somorphsm of dgraphs: ϕ : C C 2. We frst show that each component of G(n, k) splts nto one or several components of G(n, k r ). If two vertces x, y are n the same component of G(n, k r ), then there exsts a vertex z and two postve ntegers u, v such that x ku z (mod n) and y kv z (mod n). It follows that x, y are n the same component of G(n, k). And f D s a component of G(n, k r ), then we have D C, where C s a component of G(n, k). Now we can assume that C = s j= D j and C 2 = s 2 j= E j, where each D j or E j s a component of G(n, k r ). If x, y C and x kr y (mod n), then there exst y, y 2,..., y r = y such that x k y (mod n) and y k y + (mod n). Hence, ϕ(x) k ϕ(y ) (mod n) and ϕ(y ) k ϕ(y + ) (mod n). So ϕ(x) kr ϕ(y) (mod n), ϕ stll preserves arrows f we consder C and C 2 as subdgraphs of G(n, k r ). Snce ϕ maps a component D j nto a component E l, we have s = s 2. Thus, ϕ s stll an somorphsm f we consder C and C 2 as subdgraphs of G(n, k r ). It mples that G(n, k r ) s also symmetrc of order M. Ths proves Lemma 5.3. Let G be a dgragh. Let G or G denote the number of vertces n G. Let M(G) = max c G {ndeg(c)}, N(G) = mn c G,ndeg(c)>0 {ndeg(c)}, I(G) = {d > 0 : there exsts a vertex a n G such that ndeg(a) = d}. Note that G s semregular f and only f I(G) =. Lemma 5.4. Let G and H be two dgraphs, and a G and b H. Then ndeg((a, b)) = ndeg(a)ndeg(b), M(G H) = M(G)M(H), N(G H) = N(G)N(H) and G H = G H. Moreover, f I(G) 2, I(H) 2, then I(G H) max{i(g), I(H)} +. Proof. It follows mmedately from the defntons. Defnton 5.. For any postve ntegers t, m, we defne O m t to be the dgraph whch satsfes: () t has tm vertces and a t-cycle, () ndeg(a) = m f a s a cycle vertex and ndeg(a) = 0, otherwse. Lemma 5.5. O m t O m 2 t 2 gcd(t, t 2 )O m m 2 lcm[t,t 2 ]. Proof. By Lemma 5., there are exactly gcd(t, t 2 ) cycles contaned n the product dgraph, each havng a cycle of length lcm[t, t 2 ]. It s clear that (a, a 2 ) s a cycle vertex of O m t O m 2 t 2 f and only f a s a cycle vertex of O m t. Consequently, the ndegree of each cycle vertex n O m t O m 2 t 2 s m m 2, and the ndegree of other vertces s 0. Lemma 5.5 s proved. Lemma 5.6. Let k 2, e be ntegers and p a prme. If a, b are two cycle vertces n the same cycle of G(p e, k), then ndeg k p e(a) = ndegk p e(b). Proof. If gcd(a, p) =, then gcd(b, p) = snce a, b are two cycle vertces n the same cycle of G(p e, k), and thus ndeg k p e(a) = ndegk p e(b) = gcd(pe (p ), k). If gcd(p, a) >, then p a. Snce a, b are two cycle vertces n the same cycle of G(p e, k), so a = b = 0 and ndeg k p e(a) = ndegk pe(b). Lemma 5.6 s proved. Remark 5.. By Lemmas 5.2, 5.4 and 5.6, we see that f a and b are two cycle vertces n the same cycle of G(n, k), then ndeg k(a) = n ndegk n (b). So f C s a component of G(n, k) and h(c), then C Om t where t s the cycle length of C and m s the ndegree of a cycle vertex of C. The followng two lemmas are very useful n the proof of our man results. Lemma 5.7. O m G Om H f and only f G H for any dgraphs G and H. Proof. Assume that ϕ : O m G Om H s an somorphsm of dgraphs. Let G 0 = {x G ndeg(x) = 0}, G = {x G ndeg(x) > 0}, H 0 = {x H ndeg(x) = 0}, H = {x H ndeg(x) > 0}. Let a be the unque vertex of O m wth ndeg(a) > 0. If x G and ndeg((a, x)) = ndeg(a)ndeg(x) > 0, then ndeg(ϕ((a, x))) > 0 and ϕ((a, x)) = (a, x ), we have x H. Now we defne a map ϕ : G H by ϕ (x) = x, x G. Obvously, ϕ s njectve. If y H, then there exsts a vertex (a, y) of postve ndegree n O m G such that ϕ((a, y)) = (a, y ). Hence, ϕ (y) = y and ϕ s also surjectve. If x, y G and there s a drected edge from x to y. Let ϕ (x) = x and ϕ (y) = y. Then ϕ((a, x)) = (a, x ) and ϕ((a, y)) = (a, y ) by defnton of ϕ. There s a drected edge from (a, x) to (a, y), then there s also a drected edge from (a, x ) to (a, y ), snce ϕ preserves arrows. So there s a drected edge from x and y. ϕ preserves arrows. Next we defne a map ϕ 0 from G 0 to H 0. For any y G, let E 0 (y) = {x G 0 there s a drected edge from x to y}, E (y) = {x G there s a drected edge from x to y}.

6 G. Deng, P. Yuan / Dscrete Mathematcs 32 (202) Then G 0 = y G E 0 (y). And the unon s a dsjont unon snce each vertex has outdegree. If ϕ (y) = y, we have ndeg((a, y)) = m( E 0 (y) + E (y) ) = ndeg((a, y )) = m( E 0 (y ) + E (y ) ), and E (y) = E (y ) snce ϕ maps E (y) nto E (y ), and so E 0 (y) = E 0 (y ). Now we can take ϕ 0 such that ϕ 0 s bjectve and ϕ 0 (x) E 0 (ϕ (y)) for any x E 0 (y). Fnally we defne φ : G H, φ(x) = ϕ (x), f x G, for = 0,. It s clear that φ s bjectve. Now we prove that φ preserves arrows. Suppose x, y G and there s a drected edge from x to y. We only need to treat the case when x G 0 and y G. Let φ(y) = ϕ (y) = y. By the constructon of ϕ 0, we see that φ(x) = ϕ 0 (x) E 0 (y ), thus there s also an arrow from φ(x) to φ(y). It s easy to show that the number of drected edges of G s equal to the number of drected edges of H. Thus φ s an somorphsm. Ths proves Lemma 5.7. Remark 5.2. Let K be a dgraph, M a postve nteger, wrte K = n D n 2 D 2 n r D r, where each D s a component and D D j f and only f = j. Then, by the defnton of symmetrc, K s symmetrc of order M f and only f M n for each =, 2,..., r. In partcular, M K. Lemma 5.8. Let G = O m H be a dgraph, where H s a semregular subdgraph of G, and ndeg(a) = 0 or d for any vertex a H. Suppose that d m and K s a dgraph. Then G K s symmetrc of order M f and only f K s symmetrc of order M. Proof. Suppose that K s not symmetrc of order M, we wrte K = n D n 2 D 2 n r D r, where each D s a component and D D j f and only f = j. If d < m, wthout loss of generalty, we may assume that M(D ) M(D 2 ) M(D r ). Let j be the maxmal ndex such that M n j. Then, by Lemma 5. and Remark 5.2, G K s symmetrc of order M f and only f G (n D n 2 D 2 n j D j ) s symmetrc of order M. Let E = O m D j. By Lemma 5. agan, E s a component of G (n D n 2 D 2 n j D j ). Let F be a component of G (n D n 2 D 2 n j D j ), f F = O m D, where < j, then E s not somorphc to F accordng to Lemma 5.7. If F s a component of H (n D n 2 D 2 n j D j ), then M(F) M(H)M(D j ) < mm(d j ) = M(E), whch mples that E s not somorphc to F. If d > m, we assume that N(D ) N(D 2 ) N(D r ) n ths case. Agan let j be the maxmal ndex such that M n j. Let E = O m D j and F be a component of G (n D n 2 D 2 n j D j ). If F = O m D, where < j, then E s not somorphc to E accordng to Lemma 5.7. If F s a component of H (n D n 2 D 2 n j D j ), then N(E) = N(O m )N(D j) = mn(d j ) < dn(d j ) < N(F), whch agan mples that E s not somorphc to F. In both cases, we have showed that there are exactly n j components contaned n G (n D n 2 D 2 n j D j ) whch are somorphc to E. Thus, G K s not symmetrc of order M. The converse mplcaton s trval. Lemma 5.8 s proved. 6. The man result We begn wth a lemma before we prove our man theorem. Lemma 6.. Let p be an odd prme and n = p e q, where p q. Suppose that G(n, k) s symmetrc of order p. Then gcd((p )p e, k) = p e. Proof. Step : Assume that gcd(p, k) = u > or p k when e >. Let h = h(g(p e, k)). Then h(g(p e, k h )) = and G(p e, k h ) = G 2 (p e, k h ) G (p e, k h ) O pe a O m a 2O m 2 a to m t, (6.) where m = gcd((p )p e, k h ) and a s the number of -cycles contaned n G (p e, k h ). Obvously, m p e. By Lemma 5.8, G(p e q, k h ) G(p e, k h ) G(q, k h ) s not symmetrc of order p, whch contradcts Lemma 5.3.

7 726 G. Deng, P. Yuan / Dscrete Mathematcs 32 (202) Step 2: It remans to prove that p e k, the case e = s trval. Suppose that e 2 and p r k, r. Let C 0, C be the components of G(p e, k) contanng the vertex 0 and, respectvely. Let h 0 = h(c 0 ) and h = h(c ). Then h 0 and h. If h 0 = h, by Lemmas 3.7 and 3.8, k h 0 < e k h 0 and r(h ) < e rh, combnng wth p r k we have p r(h 0 ) e rh = rh 0. Therefore, h 0 =, and G(p e, k) = G 2 (p e, k) G (p e, k) O pe G (p e, k). Snce gcd((p )p e, k) = p r, t follows that ndeg(a) = p r or 0 for any a G (p e, k). From the proof of Step, we obtan m = p r = p e and r = e. If h 0 h, then p r(h 0 ) e rh. We must have h 0 < h. Then h(g 2 (p e, k h 0 )) =, G2 (p e, k h 0 ) O p e. In ths case G(p e, k h 0 ) = G 2 (p e, k h 0 ) G (p e, k h 0 ) O pe G (p e, k h 0 ). For any vertces a G (p e, k h 0 ) wth postve ndegree, ndeg kh 0 p e (a) = gcd((p )pe, k h 0 ) = p rh 0, snce rh 0 r(h ) < e. Agan we have M(G (p e, k h 0 )) < p e, whch mples that G(p e q, k h 0 ) = G(pe, k h 0 ) G(q, k h 0) s not symmetrc of order p by the same argument n Step. But ths contradcts Lemma 5.3. Lemma 6. s proved. Theorem 6.. Let n, k 2 be two postve ntegers. Let 2 M be a postve nteger such that M has an odd prme dvsor. If G(n, k) s symmetrc of order M, then 2 n. Moreover, n ths case we have 2 M and G( n 2, k) s symmetrc of order M 2. Proof. Assume that 2 r n. Let n = 2 r m and h = h(g(2 r, k)), where 2 m. Then G(2 r, k h ) = G 2 (2 r, k h ) G (2 r, k h ) O 2r G (2 r, k h ). G(n, k) s symmetrc of order M, where M has an odd prme dvsor p, so p n by Remark 5.2. By Lemma 6., we get gcd(p, k) =, so k s odd. Therefore, ndeg kh 2 r (a) = gcd(λ(2r ), k) = or 0 for any vertex a n G (2 r, k h ), by Lemma 5.8, G(n, k h ) G(2 r, k h ) G(m, k h ) s not symmetrc of order 2, snce G(m, k h ) s not symmetrc of order 2. Ths contradcts Lemma 5.3. Thus r = and G(2, k) 2O. Let G(m, k) n H n 2 H 2 n s H s, where H H j f and only f = j. We have G(n, k) G(2, k) G(m, k) 2n H 2n 2 H 2 2n s H s. Now, by Remark 5.2, M 2n for s, and m s odd. Thus, there exsts a j, j s such that 2 n j. Therefore, 2 M and G(m, k) s symmetrc of order M. Theorem 6. s proved. 2 Theorem.2 s a specal case of the followng lemma. We nclude the proof here for ts smplcty. Theorem 6.2. Let n = p e r = pe that gcd((p )p e, k) = p e and gcd((p )p e p A t (G( r = p, k)) or ord p k t for any t N., where p s an odd prme, and the p s are dstnct odd prmes such that p p. Suppose, k) = p e. Then G(n, k) s symmetrc of order p f and only f Proof. We know that G(p e, k) a O pe a 2 O pe 2 a l O pe l, where a = A (G(p e, k)) and a l 0. By Lemma 3.3, we see that A (G(p e, k)) = A (G(p, k)) for any N. It follows that G(p, k) a O a 2 O 2 a l O l. By Lemma 5.5, G(p e, k) O pe G(p, k). Smlarly, G(p e, k) O p e G(p, k) for r. Let m = p e r G(n, k) = G(p e, k) r = O pe G(p, k) O m G p G(p e, k) r = r p, k. = O pe G(p, k) = pe. We observe that

8 G. Deng, P. Yuan / Dscrete Mathematcs 32 (202) By Lemma 5.7, G(n, k) r s symmetrc of order p f and only f G(p = p, k) s symmetrc of order p. The rest of the proof follows from Theorem.2. Theorem 6.2 s proved. By Theorem 6. and Lemma 6., we only need to treat the case that M s an odd nteger and k 3 n the followng theorem. Theorem 6.3 (Man Theorem). Let M 3 be an odd nteger. Suppose that n = r = pe, where the p s are dstnct odd prmes. Let k 3 be an nteger and T = {p j gcd((p j )p e j j, k) = p e j j, j r}. Then G(n, k) s symmetrc of order M f and only f G( p j T p j, k) s symmetrc of order M. Proof. If X s a set of prme dvsors of n, we denote n X = p X p e. Suppose that G(n, k) s symmetrc of order M. Our method s to show that the subdgraph G(n T, k) ( p T G 2(p e, k)) of G(n, k) s also symmetrc of order M. If t s true, note that L = r =t+ G 2(p e, k) s a sngle component of cycle length and that G(n T, k) a O m a 2O m 2 a vo m v, where m = p T pe and a s the number of -cycles contaned n G(n T, k). Therefore, O m L s a sngle component. By comparng the cycle length, O m L O m j L f and only f = j. We get M a and G(n T, k) s symmetrc of order M. Step : We do some reductons n ths step. Let T = {p j : p j n, p j T}. By the proof of Theorem 6.2, G(n, k) = G(n T, k) G(n T, k) O m G p j, k G(n T,k) pj T O m G nt p j, k. p j T Here m = p j T pe j j. Hence, G(n, k) s symmetrc of order M f and only f G(n T p j T p j, k) s symmetrc of order M. In the followng, we can assume that e j = f p j T. If there exsts an ndex such that e k and G(p e, k) s not semregular, then p T. And G(p e, k) = G 2 (p e, k) G (p e, k) O p e G (p e, k). Let X = {p, p 2,..., p, p +,..., p r }. Then G(n, k) G(n X, k) G(p e, k). By Lemma 5.8, G(n X, k) s symmetrc of order M, snce G (p e, k) s always semregular. By nducton, we can assume that f e j k, then p j T. Step 2: Let S = {p p T, G 2 (p e, k) s semregular}, R = {p p T, G 2 (p e, k) s not semregular}. Then, by a permutaton of ndces f necessary, we may assume that T = {p, p 2,..., p t }, S = {p t+, p t+2,..., p s }, and R = {p s+, p s+2,..., p r }. We show that G(n T S, k) r =s+ G 2(p e, k) s symmetrc of order M n ths step. We wrte G(n, k) H H 2 H l, where H l and each H s a unon of components of G(n, k) such that I(D) = for each component D H. Then each H s symmetrc of order M. We clam that H l G(n T S, k) G 2 (p e s+, k) s+ G 2(p e s+2, k) s+2 G 2(p e r, r k). Let C be a component of G(n, k). By Lemma 5., there exst E, E 2,..., E r, such that E s a component of G(p e, k) and C r s a component of = E. Snce E s not semregular only f s + and E = G 2 (p e, k), t follows that r r r I(C) I E = I E I G 2 (p e, k). = =s+ =s+ r But we observe that =s+ G 2(p e, k) s a component by Lemma 5.. Hence, C H l f and only f E = G 2 (p e, k) for each s +. It mples that G(n T S, k) r =s+ G 2(p e, k) H l s symmetrc of order M. Step 3: We clam that f p S, then M(G 2 (p e, k)) > M(G (p e, k)). We have e > k, snce p T. Thus ndeg k p e (p k ) = pk gcd((p )p e k, k) = ndeg k p e (0) = p e k, (6.2) e

9 728 G. Deng, P. Yuan / Dscrete Mathematcs 32 (202) snce G 2 (p e, k) s semregular. Then k s odd and e 4. Assume that p r k, 0 r e 2. Then e k + mn{e k, r } = e. k If r e k, then k + r = e e. Thus M(G k (p e, k)) = gcd((p )p e If e k r, then e = 2 and M(G k (p e, k)) = p r k < e p e 2, snce e 4, thus r < e 2. Step 4: Let L = r =s+ G 2(p e, k). We wrte H l G(n T S, k) L G(n T, k) G(n S, k) L K K 2 K u,, k) = p r < p e e k = M(G 2 (p e, k)). p e 2, where the equalty holds f and only f r = e 2. But where K u and K s a unon of components of H l such that M(D) = for each component D K. Then each K s symmetrc of order M, snce H l s symmetrc of order M. Now f C s a component of H l, there exst F and E t+, E t+2,..., E s, where F s a component of G(n T, k) and E s a component of G(p e, k) such that C s a component of F s =t+ E L. But M(F) =, snce M(G(n T, k)) =. Consequently, by the result n Step 3 M(C) M(L) s =t+ M(E ) M(L) s =t+ M(G 2 (p e, k)), and the last equalty holds f and only f E = G 2 (p e, k) for t + s. But F L s =t+ G 2(p e, k) s a sngle component by Lemma 5.. We see that C K u f and only f E = G 2 (p e, k) for t + s. Hence, G(n T, k) r =t+ G 2(p e, k) K u s symmetrc of order M. Theorem 6.3 s proved. Corollary 6.. Let M 2 be an nteger such that M has an odd prme dvsor. Then there exst n and k 2 such that G(n, k) s symmetrc of order M f and only f M s square free. References [] W. Carlp, M. Mncheva, Symmetry of teraton dgraphs, Czechoslovak Math. J. 58 (2008) [2] J. Kramer-Mller, Structural propertes of power dgraphs modulo n, Manuscrpt. [3] M. Křížek, F. Lucas, L. Somer, 7 Lectures on the Fermat Numbers: From Number Theory to Geometry, Sprnger, New York, 200. [4] L. Somer, M. Křížek, On a connecton of number theory wth graph theory, Czechoslovak Math. J. 54 (2004) [5] L. Somer, M. Křížek, On semregular dgraphs of the congruence x k y(mod)n, Comment. Math. Unv. Caroln. 48 () (2007) [6] L. Somer, M. Křížek, On symmetrc dgrphs of the congruence x k y(mod)n, Dscrete Math. 309 (2009) [7] L. Szalay, A dscrete teraton n number theory, BDTF Tud. Közl. 8 (992) 7 9. [8] B. Wlson, Power dgraphs modulo n, Fbonacc Quart. 36 (998)