# The Structure of Isomorphic Digraph from Powers Modulope

Authors: Jinxing Zhao; Guixin Deng; Tiejun Liu
Journal: Journal of Mathematics
 (2018)

---

## Abstract

For positive integersn and k, let G(n,k) denote the digraph whose set of vertices is {0,1,2,…,n-1} and there is a directed edge from a to b if ak≡b(modn). It is great value to consider conditions of G(p,k1)≃G(p,k2). In this paper, we obtain necessary and sufficient conditions for G(n,k1)≃G(n,k2) when n=pe.


---

## Body

## 1. Introduction

LetX be a nonempty set and f be a map from X to itself. Let G(X,f) denote the digraph whose set of vertices is X and there is a directed edge from a∈X to b∈X if f(a)=b. In the case that X is a commutative ring or a finite abelian group (written multiplicative) and f maps x into its k-th power in X, we denote G(X,f) by G(X,k). In particular, let G(n,k) denote the digraph whose set of vertices is Zn={0,1,2,…,n-1} and there is a directed edge from a to b if ak=b, or equivalently ak≡b(modn). The investigation of graphs of this sort is important, in particular, because they have valuable applications, as explained, for example, in the book [1] and survey [2].There are two particular subdigraphs ofG(n,k). Let G1(n,k) be the induced subdigraph of G(n,k) on the set of vertices which are coprime to n and G2(n,k) be the induced subdigraph of G(n,k) on the set of vertices which are not coprime to n. We observe that G1(n,k) and G2(n,k) are disjoint and that G(n,k)=G1(n,k)∪G2(n,k). It is clear that G1(n,k)≃G(Zn∗,k), where Zn∗ is the unit group of Zn.Many interesting properties ofG(n,k) are obtained by people. The distribution of indegree and structure of cycles of G(n,k) are obtained in [3–9]. Somer and Křížek [6, 10] determined when G(n,k) is semiregular. Carlip and Mincheva [11] called a digraph G symmetric of order M if the component of G can be partitioned into subsets of size M, each containing M isomorphic components. They also proved that G(n,2) is symmetric of order 2 for n≡2(mod4) and n≡4(mod8). Somer and Křížek obtained all G(n,k) which are symmetric of order 2. Deng and Yuan [12] determined all G(n,k) which are symmetric of order M, where M has an odd prime divisor. Meemark and Wiroonsri [13, 14] investigated the structure of G(R,k), where R is the quotient ring of polynomials over finite fields.In this paper, we are mainly interested in determining whenG(n,k1)≃G(n,k2). Lucheta, Miller, and Reiter [4] showed that G(p,k1)=G(p,k2) if and only if k1≡k2(modp-1). They also found that G(11,2)≃G(11,8) and thus two digraphs G(p,k1) and G(p,k2) can be isomorphic without the condition. Wilson [9] also obtained that G1(n,k1)=G1(n,k2) if and only if k1≡k2(modλ(n)), where λ is the Carmichael lambda-function. Deng and Yuan [15] gave a necessary and sufficient condition for G(p,k1)≃G(p,k2). They also determined that G(H,k1)≃G(H,k2) for an abelian group H by using graph theory and group theory. In this paper, we obtain a necessary and sufficient condition for G(n,k1)≃G(n,k2).This paper is organized as follows. In Section2, we review some basic properties of G(n,k) including indegrees, cycles, and structure of components. We point out that G(Zn∗,k) is an induced subdigraph of G(n,k). In Section 3, we review the properties of product of digraphs. In Section 4, we investigate G(n,k) when n is a power of a prime p and obtain a necessary and sufficient condition for that G(pe,k1)≃G(pe,k2).


## 2. Some Preliminary Results onG(H,k) and G(n,k)

In this section we review some basic properties ofG(n,k) and G(H,k), where H is a finite abelian group. The structure of G(H,k) and hence G1(n,k) is well understood in [16]. It has highly symmetric structure.Before proceeding further, we need to review some properties of the Carmichael lambda-functionλ(n) and the exponent of an abelian group.Definition 1.
Letn be a positive integer. Then the Carmichael-lambda-function λ(n) is defined as follows: (1)λ1=1,λ2=1,λ4=2,λ2k=2k-2fork≥3;λpk=p-1pk-1foranyoddprimepandk≥1;λ(p1k1…prkr)=lcm[λ(p1k1),…,λ(prkr)], where p1,…,pr are distinct primes and ki≥1 for all i∈{1,2,…,r}.The following theorem generalizes the well-known Euler’s theorem which says thataϕ(n)≡1(modn) if and only if gcd(a,n)=1.Theorem 2 (Carmichael).
Leta,n∈N. Then aλ(n)≡1(modn) if and only if gcd(a,n)=1. Moreover, there exists an integer g such that ordng=λ(n), where ordng denotes the multiplicative order of g modulo n.For the proof see [17, p. 21].LetH be a finite group. The exponent of G, denoted by e(G), is the minimal positive integer m such that gm=1 for any g∈H. It is well known that if n=∏i=1rpiei is the prime factorization of n, then Zn≃Zp1e1×Zp2e2×⋯×Zprer and hence Zn∗≃Zp1e1∗×Zp2e2∗×⋯×Zprer∗. Thus, λ(n) is exactly the exponent of the unit group of Zn.We have the following result aboutG(H,k1) and G(H,k2).Theorem 3 ([16] Theorem 3.2-3.4).
LetH be an abelian group and let e(H) be the exponent of H. Then G(H,k1)≃G(H,k2) if and only if the following conditions hold:
(i)gcd(e(H),k1)=gcd(e(H),k2),
(ii) we have a factorization ofe(H)=uv, where u is the maximal divisor of e(H) relatively prime to k1 and is also the maximal divisor of e(H) relatively prime to k2; for any divisor d of u, orddk1=orddk2.
Moreover, letEi be the component of G(H,ki) containing the identity element 1 of H, i=1,2. Condition (i) is equivalent to E1≃E2.Definition 4.
We define a height function on the vertices and components ofG(n,k). Let c be a vertex of G(n,k). Then h(c) is defined to be the minimal nonnegative integer i such that cki is congruent modulo n to a cycle vertex in G(n,k). If C is a component of G(n,k), then we define h(C)=supc∈Ch(c).The indegree and height function play an important role in the structure ofG(n,k). We need the following results concerning the indegrees and heights.Lemma 5 ([10] Theorem 3.1).
Letn=∏i=1r‍piei be the prime factorization of n. Let a be a vertex of positive indegree in G1(n,k). Then (2)indegnka=∏i=1rindegpieika=∏i=1rδigcdλpiei,k,where δi=2 if 2|k and 8|piei, and δi=1 otherwise.Lemma 6 ([10] Theorem 3.2).
Letp be a prime. Let a be a vertex with positive indegree in G2(pe,k). Suppose that a≠0 and pl∥a. Then l=kt for some positive integer t and (3)indegpeka=δpk-1tgcdλpe-l,k,where δ=2 if p=2, 2|k and e-l≥3, and δ=1 otherwise.Lemma 7 ([18] Lemma 3.2).
Letp be a prime and e, k be two positive integers. Then (4)indegpek0=pe-e/k.Lemma 8.
Letp be a prime and e≥2, k≥2 be two positive integers. Let h be the unique positive integer such that kh-1<e≤kh. Then h=h(G2(pe,k)).Proof.
It is clear thatp∈G2(pe,k) and h(p)=h(G2(pe,k)). Moreover, pki≡0(modpe) if and only if ki≥e. The proof is complete.Lemma 9.
Letp be a prime and e, k≥2 be two positive integers. Let λ(pe)=uv, where u is the maximal divisor of λ(pe) relatively prime to k. If C is the component of G(pe,k) containing 1, then (5)hC=min⁡i:v∣ki.Proof.
Leth=min⁡{i:v∣ki}. Then there exists a divisor d of v such that d is not a divisor of kh-1. By Theorem 2, there exists a vertex g∈G(pe,k) such that ordpeg=uv. Let a≡guv/d(modpe). Then ordpea=d, akh-1≢1(modpe), and akh≡1(modpe). Hence, h(C)≥h(a)=h by the definition of the height function.
Conversely, ifa∈C, then there exists j≥1 such that akj≡1(modpe), so ordpea∣kj. Since ordpea∣uv, we have ordpea∣v, so akh≡1(modpe). That is, h(C)≤h.


## 3. Properties of Digraphs Products

Given two digraphsG1 and G2. Let G1×G2 denote the digraph whose vertices are the ordered pairs (a1,a2), where ai∈Gi and there is a directed edge from (a1,a2) to (b1,b2) if there is a directed edge from a1 to b1 and a directed edge from a2 to b2. Somer and Křížek [18] noted the following fact: if n=n1n2 and gcd(n1,n2)=1, then G(n,k)≅G(n1,k)×G(n2,k). The canonical isomorphism is given by a↦(a1,a2) where a≡ai(modni), i=1,2. In general,(6)Gn,k≅Gp1e1,k×Gp2e2,k×⋯×Gprer,k,if n=∏i=1r‍piei is the prime factorization of n. We use the following notation:(7)∏i=1rGi≔G1×G2×⋯×Gr,where each Gi is a digraph. We need this fact and the following lemmas.Lemma 10 ([19] Lemma 3.1).
Letn=n1n2 where gcd(n1,n2)=1. Let Ei be a component of G(ni,k) with cycle length ti. Then E1×E2 is a subdigraph of G(n,k) consisting of gcd(t1,t2) components, each having cycles of length lcm[t1,t2].Corollary 11.
LetG1, G2, and G3 be digraphs. Suppose that At(G1×G3)=At(G2×G3). Then At(G1)=At(G2).Proof.
Letai be the number of i-cycles contained in G3, bi be the number of i-cycles contained in G1, and ci be the number of i-cycles contained in G2. Using Lemma 10 to compare the number of l-cycles of Gi×G3, we have (8)∑lcmi,j=laibjgcdi,j=∑lcmi,j=laicjgcdi,j.Let l=1; we see that b1=c1. Now assume that bi=ci for any i<l. We have (9)∑i∣laiblgcdi,l=∑i∣laiclgcdi,l.Thus bl=cl and we finish the proof by induction.Lemma 12 ([19] Lemma 3.12).
Letn=∏i=1r‍piei be the prime factorization of n. Let a=(a1,a2,…,ar) and b=(b1,b2,…,br) be two vertices in G(n,k)≅G(p1e1,k)×G(p2e2,k)×⋯×G(prer,k). If a and b are in the same cycle, then ai and bi are in the same cycle for each i.LetG denote the number of vertices of a digraph G. Let M(G)=maxc∈G{indeg(c)}, N(G)=minc∈G,indeg(c)>0{indeg(c)}, and I(G)=indega:a∈G∖0. The following lemma is immediately from the properties of product of digraphs.Lemma 13.
LetG1 and G2 be two digraphs, and a1∈G1 and a2∈G2. Then α=(a1,a2)∈G×H. Moreover, indeg(α)=indeg(a1)indeg(a2), M(G1×G2)=M(G1)M(G2), N(G1×G2) and G1×G2=G1G2. If I(G1)≥2 and I(G2)≥2, then I(G1×G2)>max{I(G1),I(G2)}.Definition 14.
For any positive integerst and m, we define Otm to be the digraph which satisfies the following: (i) it has tm vertices and a t-cycle, (ii) indeg(a)=m if a is a cycle vertex and indeg(a)=0 otherwise. In particular, let Ot=Ot1 be the t-cycle.Remark 15.
LetC be a component of G(n,k) with h(C)≤1. By Lemma 12, each cycle vertex of C has the same indegree. Thus, C≃Otm, where t is the cycle length of C and m is the indegree of a cycle vertex of C. We use cyindeg(C) to denote the indegree of a cycle vertex of C.


## 4. The Main Result Whenn=pe

In this section we studyG(pe,k), where p is prime. It is clear that G2(pe,k) consists of a single component. We begin with the following Lemma.Lemma 16.
Letp be an odd prime and e, k two positive integers. Suppose that gcd((p-1)pe-1,k)≥pe-1. Then (10)Gpe,k≃O1pe-1×Gp,k.Proof.
LetCi denote the cyclic group of order i. Recall that for an abelian group H, G(H,k) is the digraph whose vertex set is H. There exists a directed edge from a to b if and only if b=ak. Note that the unit group Zpe is a cyclic group of order (p-1)pe-1. Thus, G1(pe,k)≃G(C(p-1)pe-1,k)≃G(Cp-1,k)×G(Cpe-1,k) for any e≥1.
By hypothesis we havepe-1∣k, and hence G(Cpe-1,k)≃O1pe-1. Since G(p,k)≃O1∪G(Cp-1,k), we have (11)Gpe,k≃G1pe,k∪G2pe,k≃GCp-1,k×GCpe-1,k∪O1pe-1≃O1pe-1×GCpe-1,k∪O1pe-1×O1≃O1pe-1×O1∪GCp-1,k≃O1pe-1×Gp,k.Recall that a digraphG is called semiregular if there exists a positive integer d such that each vertex of G has either indegree d or 0. In this case, G is called d-semiregular.Lemma 17 ([10] Theorem 4.4).
Letp be a prime and e, k≥2 two positive integers.(1)
Ifp is odd, then we can obtain that G2(pe,k) is semiregular if and only if e≤k or gcd(p-1,k)=1 and e≤k+r+1, where pr||k. In the first case G2(pe,k) is pe-1-semiregular, and in the second case G2(pe,k) is pe-2-semiregular. Thus, G(pe,k) is semiregular if and only if gcd(λ(pe),k)=pe-1. (2)
Ifp=2, then G2(2e,k) is semiregular if and only if one of the following conditions hold:
(a)
k = 2 and e∈{1,2,3,4,6}; (b)
k = 4 and 1≤e≤9; (c)
k ≥ 6 and 1≤e≤k+r+2, where 2r||k. Moreover,G(2e,k) is semiregular if and only if one of the following conditions hold: (a)
k = 2 and e∈{1,2,4}; (b)
k = 4 and 1≤e≤5; (c)
k ≥ 6 and 1≤e≤r+2, where 2r∥k.Lemma 18.
Letp be a prime. Suppose that gcd(p-1,k)=1. Then for any positive integers e≥1, k≥2, we have indegpek(0)≥indegpek(1), except the case (p,e,k)=(2,3,2).Proof.
In fact, fore≤k, we have indegpek(0)=pe-1 and indegpek(1)=δgcd(λ(pe),k), where δ=2 if 2∣k and 8∣pe, and δ=1 otherwise. It is clear that indegpek(0)≥indegpek(1), since gcd(p-1,k)=1.
Now, lete>k. Then e≥3. Assume that p is odd. Suppose that pr||k. Let s=⌈e/k⌉. Then (s-1)k<e≤sk. If s≥3, then e-s≥(s-1)k-s+1=(s-1)(k-1)≥2(k-1)≥k≥r. Then indeg(0)=pe-s≥pr=indeg(1). If s=2, we still have e-s=e-2≥r. Otherwise we have pe-1∣k. Since e≥3, then e>k≥pe-1, which is a contradiction.
Suppose thatp=2. By Lemmas 5 and 7, we know that 2=indeg82(0)<indeg82(1)=4. In the following, we assume that e≥4. Suppose that 2r||k. Let s=⌈e/k⌉. Then (s-1)k<e≤sk. If s≥3, then e-s≥(s-1)k-s+1=(s-1)(k-1)≥2(k-1)≥k≥r+1. Then indeg(0)=2e-s≥2r+1≥indeg(1). If s=2, by easy computations we still have e-s=e-2≥r+1 since e>k.We use the following notation. Letp be a fixed prime. Let the mapping vp be defined by vp:N∗→N such that vp(n)=max{m≥0:pm∣n} for n∈N∗, where N is the set of nonnegative integers and N∗ is the set of positive integers. Furthermore, we define a function Vp from the set of digraphs to the set of subsets of N as follows: (12)VpG≔vpindegc:c∈G,indegc>0,for any digraph G. Let A and B be two nonempty subsets of N. We denote A+B={a+b:a∈A,b∈B}. Since vp(mn)=vp(m)+vp(n), we have Vp(G×H)=Vp(G)+Vp(H) for any digraphs G and H.Lemma 19.
Letp be a prime and e, 2≤k2≤k1 be positive integers. Suppose that Vp(G2(pe,k1))=Vp(G2(pe,k2)) and vp(gcd(λ(pe),k1))=vp(gcd(λ(pe),k2))=r. Then e≤min{k1,k2} or k1=k2.Proof.
Ife≤min{k1,k2}, then Vp(G2(n,k1))=Vp(G2(n,k2))={e-1} since ak1≡ak2≡0(modpe) for any a∣p. Suppose that k2≤e<k1. Then e-1∈Vp(G2(pe,k2)). But e-1∉Vp(G2(pe,k1)). It contradicts the fact that Vp(G2(n,k1))=Vp(G2(n,k2)). In the following we always assume that e>k1≥k2.
Case 1. First we assume that p is an odd prime. We show that ⌈e/k1⌉=⌈e/k2⌉. Let ⌈e/ki⌉=si. We need to show that max{Vp(G2(pe,ki))}=e-si. By Lemmas 5, 6, and 7, we see that(13)VpG2pe,ki=ki-1t+min⁡e-kit-1,r:t=1,2,…,si-1∪0,e-si.We have (si-1)ki<e≤siki, and then e-kit-1≥ki>r for any t<si-1. Hence (14)ki-1si-1-ki-1t-min⁡e-kit-1,r=ki-1si-t-1-r≥0for any 1≤t<si-1. It implies that (15)max⁡VpG2pe,ki=max⁡ki-1si-1+min⁡e-kisi-1-1,r,e-si.But (ki-1)(si-1)+min{e-ki(si-1)-1,r}≤(ki-1)(si-1)+(e-ki(si-1)-1)=e-si. So max{Vp(G2(pe,ki))}=e-si. We get s1=s2.
Now ifs1=s2≥3, by (13) we have min{Vp(G2(pe,ki))\{0}}=ki-1+r. Then k1-1+r=k2-1+r, k1=k2. If s1=s2=2, then min{Vp(G2(pe,ki))\{0}}=ki-1+min{e-ki-1,r}. We have k1-1+min{e-k1-1,r}=k2-1+min{e-k2-1,r}.
Subcase 1. If r≤e-k2-1≤e-k1-1, then k1-1+r=k2-1+r and k1=k2.
Subcase 2. If e-k2-1≤r≤e-k1-1, then k1-1+r=k2-1+e-k2-1=e-2 and e=k1+r+1. But we have e>k2 and pr∣ki, i=1,2. Let k1=prm1, k2=prm2. We have(16)k2-k1=prm2-m1<r+1.We get m2=m1 from this equation and thus k1=k2.
Subcase 3. If e-k2-1≤e-k1-1≤r, we have k2<e≤k1+r+1. Using the same notation we still have (16); thus k1=k2.
Case 2. Now we treat the case p=2. The case e≤3 is trivial. Assume that e≥4; in this case we still have ⌈e/k1⌉=⌈e/k2⌉. Let ⌈e/ki⌉=si. It suffices to show that max⁡V2(G2(2e,ki))=e-si. By Lemmas 5, 6, and 7, we see that (17)V2G2e,ki=v2δ2ki-1tgcdλ2e-kit,ki:t=1,2,…,si-1∪0,e-si,where δ=2 if 2∣ki and e-kit≥3, and δ=1 otherwise. If r=0, then max{V2(G2(pe,ki))}=max{e-si,(ki-1)(si-1)}=e-si. We have s1=s2. Suppose that r>0. Then 2∣ki. If t<si-1, e-kit>(si-1)ki-kit=(si-t-1)ki≥max{r,3}. So(18)V2G2e,ki=1+ki-1t+r:t=1,2,…,si-2∪0,e-si,ϵ+ki-1si-1+v2gcdλ2e-kisi-1,ki,where ϵ=1 if e-ki(si-1)≥3, and ϵ=0 otherwise. We have (19)ϵ+v2gcdλ2e-kisi-1,ki≤e-kisi-1-1.
Thusmax{V2(G2(pe,ki))}=e-si, s1=s2.
Now ifs1=s2≥3, we have min{V2(G2(2e,ki))\{0}}=1+ki-1+r=ki+r. Then k1+r=k2+r, k1=k2. If s1=s2=2, then min{V2(G2(2e,ki))\{0}}=ϵi+ki-1+v2(gcd(λ(2e-ki),2r)), where ϵi=1 if e-ki≥3, and ϵ=0 otherwise. Assume that k1≤k2.
Subcase 1. 3≤e-k2≤e-k1. Then, we have k1+min{e-k1-2,r}=k2+min{e-k2-2,r}.
(i) Ifr≤e-k2-2≤e-k1-2, then k1+r=k2+r, k1=k2.
(ii) Ife-k2-2≤r≤e-k1-2, then e-2=k1+r. We have k2<e=k1+r+2, let ki=2rmi, i=1,2. Then(20)2rm2-m1<r+2.We get r=0,1, m2=m1+1. If r=0, then k2=k1+1. That is, one of k1, k2 is odd and the other one is even. It is contrary to gcd(λ(2e),k1)=gcd(λ(2e),k2)=2r. If r=1, we have 4∣k1 or 4∣k2 and gcd(2e-2,k1)=gcd(2e-2,k2)=2. It implies that e-2=1, e=3. But it is contrary to 3≤e-k2.
(iii) Ife-k2-2≤e-k1-2≤r, then k2<k1+r+2. Using the same notations we still have (20). It is impossible.
Subcase 2. e-k2<3≤e-k1. Then we have k1+min{e-k1-2,r}=k2+min{e-k2-1,r}.
(i) Ifr≤e-k2-1 and r≤e-k1-2, then k1+r=k2+r, k1=k2.
(ii) Ife-k2-1≤r and r≤e-k1-2, then e-1=k1+r. We have k2<e=k1+r+1. Let ki=2rmi, i=1,2. Then (21)2rm2-m1<r+1.We must have m2=m1, k2=k1.
(iii) Ife-k2-1≤r and e-k1-2≤r, then e-1=e-2. It is a contradiction.
Subcase 3. e-k2≤e-k1<3. We have k1+min{e-k1-1,r}=k2+min{e-k2-1,r}.
(i) Ife=k1+2, then k1≤k2<e=k1+2, k2=k1+1. One of k1 and k2 is odd and the other one is even. It is contrary to gcd(λ(2e),k1)=gcd(λ(2e),k2)=2r.
(ii) Ife=k1+1, then k1≤k2<e=k1+1, and we still have k2=k1. The lemma is proved.Corollary 20.
Letp be a prime and e be a positive integer. Let k1≥2, k2≥2 be two positive integers. Then G(pe,k1)≃G(pe,k2) if and only if G1(pe,k1)≃G1(pe,k2) and G2(pe,k1)≃G2(pe,k2).Proof.
Suppose thatG(pe,k1)≃G(pe,k2). It suffices to show that G2(pe,k1)≃G2(pe,k2). If e≤2, then G2(pe,k1)≃G2(pe,k2)≃O1p. We assume that e≥3 and p≥3. If G2(pe,k1)≄G2(pe,k2), by Theorem 3 we have G2(pe,k1)≃C2 and G2(pe,k2)≃C1 where Ci is the component of G(pe,ki) containing the vertex 1. By Lemma 17 we have gcd((p-1)pe-1,k1)=pe-2 or pe-1, gcd((p-1)pe-1,k2)=pe-2 or pe-1. It implies that pe-2∣ki and therefore e≤pe-2≤ki since e≥3. We have G2(pe,k1)≃G2(pe,k2)≃O1pe-1, which is a contradiction.
Ifp=2, e=3, one of k1, k2 is 2. Then by easy computations we know that G(n,k1)≃G(n,k2) if and only if k1=k2. Now let e≥4; we have that C1, C2, G2(n,k1) and G2(n,k2) are semiregular. Then we have (22)indeg2ek10=indeg2ek21andindeg2ek20=indeg2ek11. By Lemma 18, (23)indeg2ek10≥indeg2ek11andindeg2ek20≥indeg2ek21.Combining these formulas, we obtain d=indeg2ek1(0)=indeg2ek1(1)=indeg2ek2(0)=indeg2ek1(1) and G(n,ki) is d-semiregular, i=1,2.
Ife≤min⁡{k1,k2}, then G2(2e,k1)≃G2(2e,k2)≃O1d. If e>min⁡{k1,k2}, by Lemma 17 we know that k1=k2. We still have G2(2e,k1)≃G2(2e,k2). The other direction is trivial.Theorem 21.
Letp be an odd prime and e, k1≥2, k2≥2 positive integers. Then G(pe,k1)≃G(pe,k2) if and only if one of the following conditions holds:
(i) Ife>min⁡{k1,k2}, then k1=k2.
(ii) Ife≤min⁡{k1,k2}, then gcd(λ(pe),k1)=gcd(λ(pe),k2), and we have a factorization of λ(pe)=uv, where u is the maximal divisor of λ(pe) relatively prime to k1 and is also the maximal divisor of λ(pe) relatively prime to k2. Moreover, for any divisor d of u, orddk1=orddk2.Proof.
By Corollary20, we know that G(pe,k1)≃G(pe,k2) if and only if Gi(pe,k1)≃Gi(pe,k2), i=1,2. But G1(pe,ki)≃G(C(p-1)pe-1,ki) if p is odd, and G1(2e,ki)≃G(C2⊕C2e-2,ki), where we use Cm to denote the cyclic group of order m.
So we haveVp(G2(pe,k1))=Vp(G2(pe,k2)) and pr||gcd(λ(pe),ki) for some r. The proof is complete by Lemma 18 and Theorem 3.



---

*Source: 9304645-2018-12-02.xml*