# A Theory of Cartesian Product and Factorization

Authors: V. Vilfred
Journal: Journal of Discrete Mathematics
 (2013)

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## Abstract

We determine when the Cartesian product of two circulant graphs is also a circulant graph. This leads to a theory of factorization of circulant graphs.


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## Body

## 1. Introduction

Just as integers can be factored into prime numbers, there are many results on decomposition of structures throughout mathematics [1]. The standard products—Cartesian, lexicographic, tensor, and strong—all belong to a class of products introduced by Imrich and Izbicki [2] and called B-products [3]. Properties of circulant graphs are extensively studied by many authors [2–19] and products of graphs have been studied for almost 50 years now. Sabidussi [20] proved that every (nonnull) graph is the unique product of prime graphs. Broere and Hattingh [3] established that the lexicographic product of two circulant graphs is again circulant. They and Sanders and George [12] established that this is not the case with other products. Alspach and Parsons [5] introduced metacirculant graphs as a generalization of circulant graphs and characterized metacirculant graphs in terms of their automorphism groups. Sanders [11] established that any B-product of two circulant graphs is always a metacirculant graph with parameters that are easily described in terms of the product graphs and also established that any metacirculant graph with the appropriate structure is isomorphic to the B-product of a pair of circulant graphs. After a graph is identified as a circulant graph, its properties can be derived easily. This paper gives a detailed study of Cartesian product and factorization of circulant graphs similar to the theory of product and factorization of natural numbers. For more details on circulant graphs, see [9, 10].Letn be a positive integer and let R be a subset of {1,2,…,[n/2]}. The circulant graph Cn(R) has vertices v1,v2,…,vn=v0 with vi adjacent to vi+r for each r∈R, subscript addition taken modulo n. When discussing circulant graphs, we will often assume, without further comment, that the vertices are the corners of a regular n-gon, labeled clockwise. Circulant graphs C7(1,3) and C8(2,4) are shown in Figures 1(a) and 1(b).(a)C7(1,3). (b) C8(2,4).
(a)
(b)Whenn/2∈R, edge vivi+n/2 is taken as a single edge while considering the degree of a vertex, but as a double edge while counting number of edges or cycles in Cn(R)  [3, 6–10, 13, 14, 17, 18, 21]. We generally write Cn for Cn(1) and Cn() for Cn(Φ)=Knc, the null graph on n vertices. Note that if a graph G is circulant, then its adjacency matrix A(G) is circulant. It follows that if the first row of the adjacency matrix of a circulant graph is [a1,a2,…,an], then a1=0 and ai= an-i+2, 2≤i≤n.Throughout this paper, for a setR={r1,r2,…,rk},Cn(R) denotes Cn(r1,r2,…,rk) where 1≤r1<r2<⋯<rk≤[n/2]; we consider only connected circulant graphs of finite order greater than 2 and allcycles have length at least 3, unless otherwise specified.Letn, r be positive integers, n≥2 and r<n/2. Then, Cn(r) consists of a collection of disjoint cycles, v0vrv2r,…,v0; v1v1+rv1+2r,…,v1;…; vr-1v2r-1v3r-1,…,vr-1. If d=gcd(n,r), then there are d such disjoint cycles and each has length n/d. We say that each of these cycles has period  r,length  n/d, and rotation  r/d.Ifr=n/2, then obviously Cn(r) is just a 1-factor. Since Cn(R) is just the union of the cycles of Cn(r) for r∈R, we have a decomposition of Cn(R).Theorem 1 (see [7]).
Circulant graphCn(R) for a set R={r1,r2,…,rk} is connected if and only if gcd(n,r1,r2,…,rk)=1.Theorem 2 (see [15, 18]).
Letr∈R. Then, in Cn(R), the length of a cycle of period r is n/gcd(n,r) and the number of vertex-disjoint periodic cycles of period r is gcd(n,r).In the circulant graphC10(1,4,5),  (v1,v2,…,v10) is a cycle of period 1 and length 10; cycles (v1,v5,v9,v3,v7) and (v2,v6,v10,v4,v8) are of period 4 and length 5=10/gcd(10,4) each; by considering the edges v1v16, v2v7, v3v8, v4v9,  and v5v10 as double edges, they are cycles of period 5 and of length 2=10/gcd(10,5), each. See Figure 2.Figure 2
C
10
(
1,4
,
5
).Theorem 3 (see [15, 18]).
IfCn(R)≅Cn(S), then there is a bijection f from R to S so that for all r∈R,   gcd(n,r)=gcd(n,f(r)).Proof.
The proof is by induction on the order ofR.Remark 4.
Let|R|=k. Then, the circulant graph Cn(R) for a set R={r1,r2,…,rk} is (2k-1)-regular if n/2∈R and 2k-regular otherwise.Definition 5.
The cross product or Cartesian product of two simple graphsG(V,E) and H(W,F) is the simple graph G□H with vertex set V×W in which two vertices u=(u1,u2) and v=(v1,v2) are adjacent if and only if either u1=v1 and u2v2∈F or u2=v2 and u1v1∈E.Definition 6 (see [4, 13–18]).
Circulant graphsCn(R) and Cn(S) for R={r1,r2,…,rk} and S={s1,s2,…,sk} are said to be Adam’s isomorphic if there exists a positive integer x relatively prime to n with {s1,s2,…,sk}={xr1,xr2,…,xrk}n* where 〈ri〉n*, the reflexive modular reduction of a sequence 〈ri〉, is the sequence obtained by reducing each ri modulo n to yield ri′ and then replacing all resulting terms ri′ which are larger than n/2 by n-ri′.
The following lemmas are useful to obtain one-to-one mappings.Lemma 7.
LetA and B be two nonempty sets. Let f:A→B be a mapping. Then, f is one-to-one if and only if f/A′ is one-to-one for every nonempty subset A′ of A.Lemma 8.
LetA and B be nonempty sets and let A1,A2,…,Ak be a partition of A (each Ai being nonempty), i=1,2,…,k. Let f:A→B be a mapping. Then f is one-to-one if and only if f/Ai is one-to-one for every i,i=1,2,…,k.The motivation to do this research work is to develop a theory among circulant graphs similar to the product and factorization of natural numbers.


## 2. Main Results

Throughout this section, the following notation is used. Letc∈ℕ and A⊆ℕ. Then cA={ca:a∈A}. Let [r,s] denote {r,r+1,…,s}, r∈ℤ, and let nG denote disjoint union of n copies of the graph G. G□H denotes the Cartesian product of graphs G and H.In this paper, Cartesian product and factorization ofconnected circulant graphs is studied. Moreover C4=P2□P2 and for n≠4 and nontrivial graphs G and H, Cn≠G□H and in particular, C2n+1(R)≠P2□G for any R⊆[1,n].In any circulant graphCn(R) the length of a periodic cycle of period r is n/gcd(n,r). And 2, 3, 4, and 6 are the only numbers such that in any circulant graph Cn(R) periodic cycle of length 2, 3, 4, or 6, if it exists, occurs without rotation always. This follows from the fact that for any natural number m greater than 6, there exists at least one natural number s such that 1<s<[m/2] and gcd(s,m)=1 so that Cm≅Cm(s) and rotation of the periodic cycle of period s is s/gcd(m,s). When m=5, s=2 is the required value. Thus, we call the periodic cycles of length 2, 3, 4, and 6 as the irrotational periodic cycles.It is easy to verify the following:(i)
G1□(G2∪G3)≅(G1□G2)∪(G1□G3) for graphs G1, G2 and G3. Here union is meant to be disjoint union only;(ii)
G1□(Cpm(pR))≅G1□(p·Cm(R))≅p(G1□Cm(R));(iii)
Cpm(pR)□Cqn(qS)≅pq(Cm(R)□Cn(S)) for sets R⊆[1,[m/2]],  S⊆[1,[n/2]],  p,q∈ℕ;(iv)
in any circulant graphCn(R) removal or addition of one or more jump sizes, if possible, will not change the property of being circulant;(v)
ifG and H are nontrivial graphs of order m and n, respectively, m,n≥2, then G□H contains n number of disjoint copies of G and m number of disjoint copies of H.Theorem 9.
Letn>2. Then P2□Cn is circulant if and only if n is odd. Furthermore, in that case, P2□Cn≅C2n(2,n).Proof.
Forn>2, P2□Cn is a 3-regular connected graph of order 2n. When n is even, C2n(2,n)≅2·Cn(1,n/2) which is a disconnected graph, whereas P2□Cn is a connected one and hence P2□Cn≠C2n(2,n) when n is even. For n>2,  P2□Cn contains two disjoint copies of Cn, say Cn(1)(1) and Cn(2)(1), so that Cn(1)∪Cn(2)≅2·Cn(1)≅C2n(2) and n disjoint copies of P2. Let Cn(j)=(u0,ju1,ju2,j,…,un-1,j) and ui,1ui,2∈E(P2□Cn), i=0,1,2,…,n-1, and j=1,2. Thus P2□Cn≅Cn(1)∪Cn(2)∪{ui,1ui,2/i=0,1,2,…,n-1}≅C2n(2)∪{ui,1ui,2/i=0,1,2,…,n-1} and V(P2□Cn)={ui,j/i=0,1,2,…,n-1 and j=1,2}.
LetC2n(R) be a circulant graph with n∈R. In C2n(R), cycle of period n is of length 2 and irrotational. Using Remark 4, for n>2, 3-regular connected graph P2□Cn is a circulant of the form C2n(S) for some S⊆[1,n] if and only if each edge ui,1ui,2 acts as a cycle of period n (double edge) and equal number of vertices of Cn(1) (and of Cn(2)) are on each sides of ui,1ui,2, i=0,1,2,…,n-1. This is possible only when n is odd, in which case the circulant graph is nothing but C2n(2,n). The following transformation gives the required circulant graph representation of P2□Cn when n is odd.
Letn=2m+1, m∈ℕ. Let V(K2(2m+1))={v0,v1,v2,…,v2(2m+1)-1} where the vertices v0,v1,v2,…,v2(2m+1)-1 are considered as the corners of a regular 2(2m+1)-gon, assumed to be located in the plane proceeding cyclically clockwise. Define the mapping ϕ2,2m+1:V(P2□C2m+1)→V(K2(2m+1)) such that Φ2,2m+1(ui,j)=v2i+(j-1)(2m+1) and Φ2,2m+1((u,v))=(Φ2,2m+1(u),Φ2,2m+1(v)) for every (u,v)∈E(P2□C2m+1), using subscript arithmetic modulo 2(2m+1),  j=1,2 and i=0,1,…,2m. Under this mapping, Φ2,2m+1(C2m+1(j))=Φ2,2m+1((u0,ju1,ju2,j…u2m,j))=(Φ2,2m+1(u0,j)Φ2,2m+1(u1,j)Φ2,2m+1(u2,j)…Φ2,2m+1(u2m,j))=(v(j-1)(2m+1)v2+(j-1)(2m+1)v4+(j-1)(2m+1)…v2(2m)+(j-1)(2m+1)), which is a periodic cycle of period 2 and of length 2m+1 in C2(2m+1)(1,2,…,2m+1)=K2(2m+1),  Φ2,2m+1((ui,1,ui+1,1))=(v2i,v2i+2), Φ2,2m+1((ui,2,ui+1,2))=(v2i+2m+1,v2i+2m+3) and Φ2,2m+1((ui,1,ui,2))=(v2i,v2i+2m+1), using subscript arithmetic modulo 2(2m+1), i=0,1,2,…,2m and j=1,2. This implies that the mapping is a one-to-one mapping and preserves adjacency and the transformed graph Φ2,2m+1(P2□C2m+1) is C2(2m+1)(2,2m+1)=C2n(2,n), n=2m+1 and m∈ℕ. Hence the result.Note 1.
ConsiderP2□P2=C4  which is a connected circulant graph. When n is even and greater than 2, C2n(2,n)≅2·Cn(1,[n/2]), whereas  P2□Cn  is connected but not circulant and hence  C2n(2,n)≠P2□Cn when n is even and greater than 2.Note 2.
Fork≥3 and n≥2,Pk□Cn is not circulant since it is not a regular graph.Note 3.
Form>2  and  connected circulant  Cm(R), P2□Cmn(nR)≅n(P2□Cm(R)) since H□(nG)≅n(H□G) for any connected graphs  H and G,  n∈ℕ.Theorem 10.
Forn∈ℕ,  P2□C2n+1(R)≅C2(2n+1)(2R∪{2n+1})≅C2(2n+1)(2dR∪{2n+1}), where gcd(2(2n+1),d)=1.Proof.
P
2
□
C
2
n
+
1
(
R
) contains two disjoint copies of C2n+1(R), say G1 and G2, and n+1 disjoint copies of P2 so that G1∪G2≅2·C2n+1(R)≅C2(2n+1)(2R) and P2□C2n+1(R)≅C2(2n+1)(2R)∪n·P2. Let V(Gj)=(u0,ju1,ju2,j,…,u2n,j) and ui,1ui,2∈E(P2□C2n+1(R)), i=0,1,2,…,2n and j=1,2 so that P2□C2n+1(R)≅C2(2n+1)(2R)∪{ui,1ui,2/i=0,1,2,…,2n} and V(P2□C2n+1(R))={ui,j/i=0,1,2,…,2n and j=1,2}. At first let us assume that C2n+1(R) is connected, n∈ℕ. Then, for every r∈R, the length of each periodic cycle of period r in C2n+1(R) is (2n+1)/gcd(2n+1,r), an odd number. Thus using the proof similar to that of Theorem 9, equal number of vertices of each periodic cycle, starting from ui,1 or ui,2 of copies of C2n+1(R), can be made on each sides of ui,1ui,2 in P2□C2n+1(R), using the one-to-one mapping Φ2,2n+1:V(P2□C2n+1(R))→V(K2(2n+1)) defined by Φ2,2n+1(ui,j)=v2i+(j-1)(2n+1)  and Φ2,2n+1((u,v))=(Φ2,2n+1(u),Φ2,2n+1(v)) for every (u,v)∈E(P2□C2n+1(R)), using subscript arithmetic modulo 2(2n+1),  i=0,1,…,2n, and j=1,2 where V(K2(2n+1))={v0,v1,v2,…,v2(2n+1)-1} and the vertices v0,v1,v2,…,v2(2n+1)-1 are considered as the corners of a regular 2(2n+1)-gon assumed to be located in the plane proceeding cyclically clockwise. Under this transformation Φ2,2n+1((ui,1,ui+r,1))=(v2i,v2i+2r),  Φ2,2n+1((ui,2,ui+r,2))=(v2i+2n+1,v2(i+r)+2n+1) and Φ2,2n+1((ui,1,ui,2))=(v2i,v2i+2n+1), using subscript arithmetic modulo  2(2n+1),  r∈R and i=0,1,2,…,2n. And the transformed graph is C2(2n+1)(2R∪{2n+1}), n∈ℕ. The above result is true also when C2n+1(R) is disconnected, n∈ℕ. Hence P2□C2n+1(R)≅C2(2n+1)(2R∪{2n+1}). When gcd(2(2n+1),d)=1,  d is an odd number, hence d(2n+1)=2n+1 under arithmetic modulo 2(2n+1). Also C2(2n+1)(2R∪{2n+1}) and C2(2n+1)(2dR∪{d(2n+1)})=C2(2n+1)(2dR∪{2n+1}) are Adam’s isomorphic graphs. Hence the result.Corollary 11.
For nonempty subsetR of [1,n], if either S≠2dR∪{2n+1} for any d relatively prime to 2(2n+1) under arithmetic modulo 2(2n+1) or C2n+1(R) is disconnected, then connected circulant C2(2n+1)(S)≠P2□C2n+1(R).Proof.
WhenΦ≠R⊆[1,n] and S=2dR∪{2n+1} for any d relatively prime to 2(2n+1), then C2(2n+1)(S)=P2□C2n+1(R), using Theorem 10. When C2n+1(R) is disconnected, then the product graph P2□C2n+1(R) is also disconnected whereas it is given that C2(2n+1)(S) is connected. Hence the result.Theorem 12.
Ford,n∈ℕ and gcd(4(2n+1),d)=1,  C4□C2n+1(S)≅C4(2n+1)(4S∪{2n+1})≅C4(2n+1)(4dS∪{2n+1}).Proof.
At first let us assume thatC2n+1(S) is connected. Let G1, G2, G3, and G4 be the four copies of C2n+1(S) in C4□C2n+1(S) so that G1∪G2∪G3∪G4≅4·C2n+1(S)≅C4(2n+1)(4S). Let V(Gi)={ui,1,ui,2,…,ui,2n+1},  (u1,ju2,ju3,ju4,j) be the jth copy of C4 in C4□C2n+1(S) and V(C4□C2n+1(S))={ui,j:i=1,2,3,4 and j=1,2,…,2n+1}, i=1,2,3,4 and j=1,2,…,2n+1.
LetV(K4(2n+1))={v1,v2,…,v4(2n+1)=v0} and the vertices v0,v1,v2,…,v4(2n+1)-1 are considered as the corners of a regular 4(2n+1)-gon assumed to be located in the plane proceeding cyclically clockwise. Define one-to-one mapping Φ4,2n+1:V(C4□C2n+1(S))→V(K4(2n+1)) such that Φ4,2n+1(ui,j)=v(i-1)(2n+1)+4(j-1) using subscript arithmetic modulo 4(2n+1)   and  Φ4,2n+1((u,v))=(Φ4,2n+1(u),Φ4,2n+1(v)) for every (u,v)∈E(C4□C2n+1(S)),  i=1,2,3,4 and j=1,2,…,2n+1. Then Φ4,2n+1((ui,1ui,2ui,3…ui,2n+1))=(Φ4,2n+1(ui,1)Φ4,2n+1(ui,2)Φ4,2n+1(ui,3)…Φ4,2n+1(ui,2n+1))=(v0+(i-1)(2n+1)v4+(i-1)(2n+1)v8+(i-1)(2n+1)…v4(2n)+(i-1)(2n+1))=C2n+1(i), say, and Φ4,2n+1((u1,ju2,ju3,ju4,j))=(v4(j-1)v(2n+1)+4(j-1)v2(2n+1)+4(j-1)v3(2n+1)+4(j-1)) which is a periodic cycle of period 2n+1 and of length 4 in C4(2n+1)(1,2,…,2(2n+1))=K4(2n+1), using subscript arithmetic modulo 4(2n+1), i=1,2,3,4 and j=1,2,…,2n+1. Also, C2n+1(i) is a periodic cycle of period 4 and of length 2n+1 in C4(2n+1)(1,2,…,2(2n+1))=K4(2n+1) since v(i-1)(2n+1)+4(j-1)=v(i-1)(2n+1)+4(k-1) implies that k=j,  1≤k,j≤2n+1, and i=1,2,3,4. Moreover, each cycle of period s of C2n+1(S) becomes periodic cycle of period 4s in the transformed graph. Thus the transformed graph Φ4,2n+1(C4□C2n+1(S)) is nothing but C4(2n+1)(R) where R=4S∪{2n+1}. Also when gcd(4(2n+1),d)=1, C4(2n+1)(4S∪{2n+1}) and C4(2n+1)(4dS∪{d(2n+1)})=C4(2n+1)(4dS∪{2n+1}) are Adam’s isomorphic. Hence the result.Theorem 13.
LetC2n(R) be connected, n∈ℕ. Then C4□C2n(R)≠C8n(S) for any S⊆[1,4n].Proof.
Using Theorem9, C4□C2=C4□P2 is not circulant. For n≥2, if C2n(R) is connected, then C4□C2n(R) is also connected and it contains 4 disjoint copies of C2n(R) and 2n number of disjoint copies of C4. And hence if C2n(R) is connected and C4□C2n(R)≅C8n(S) for some S⊆[1,4n], then S=4kR∪{2nd} for some k and d such that C8n(4kR)=4·C2n(kR)≅4·C2n(R),  C8n(2nd)=2n·C4(d) and gcd(2n,k)=1=gcd(4,d). This implies C4□C2n(R)≅C8n(4kR∪{2dn})≅2·C4n(2kR∪{dn}) which is disconnected, a contradiction when C2n(R) is connected. Hence the result.Lemma 14.
Letm,n≥2, r∈R, and s∈S. If Cm(r)□Cn(s) is not circulant, then Cm(R)□Cn(S) is not circulant.Proof.
Letr∈R and s∈S. Now, Cm(r)□Cn(s)≠Cmn(T) for any T⊆[1,mn/2] implies that Cm(R)□Cn(s)≠Cmn(U) for any U⊆[1,mn/2] (if not, let r∈R and Cm(R)□Cn(s)=Cmn(U) for at least one U⊆[1,mn/2]. Then by removing all other jump sizes other than r from Cm(R), the resultant graph, Cm(r)□Cn(s), remains circulant, a contradiction to the given condition). Similarly we can prove that for s∈S if Cm(R)□Cn(s) is not circulant, then Cm(R)□Cn(S) is not circulant. Combining the above arguments, we get the result.Theorem 15.
LetC2(2n+1)(R) be connected, n∈ℕ. Then P2□C2(2n+1)(R) is circulant if and only if C2(2n+1)(R)≅P2□C2n+1(S) for some S⊆[1,n]. Furthermore, in that case, P2□C2(2n+1)(R)≅C4□C2n+1(S)≅C4(2n+1)(4S∪{2n+1})≅C4(2n+1)(4dS∪{2n+1}), where gcd(d,4(2n+1))=1.Proof.
LetC2(2n+1)(R)≅P2□C2n+1(S). Then C2(2n+1)(R)≅C2(2n+1)(2S∪{2n+1}), using Theorem 10. This implies that P2□C2(2n+1)(R)≅C4□C2n+1(S)≅C4(2n+1)(4S∪{2n+1})≅C4(2n+1)(4dS∪{2n+1}), where gcd(d,4(2n+1))=1, using Theorem 12.
Conversely, letP2□C2(2n+1)(R) be circulant. If C2(2n+1)(R)≠P2□C2n+1(S) for any connected C2n+1(S), then P2□C2(2n+1)(R) has P2 as a factor and not C4=P2□P2. Let R=2S∪T⊆[1,2n+1] where T⊆[1,2n+1]⊂2ℕ+1 and T≠Φ since C2(2n+1)(R) is connected. Let 2r+1∈T, r∈ℕ∪{0}. Then using Theorem 2 the length of the periodic cycle of period 2r+1 in C2(2n+1)(R) is 2(2n+1)/gcd(2(2n+1),2r+1), an even number. Therefore P2□C2(2n+1)(2r+1) cannot be a circulant graph (see the proof of Theorem 9) and using Lemma 14, P2□C2(2n+1)(R) cannot be a circulant graph, a contradiction to P2□C2(2n+1)(R) is circulant. This implies that there exists a connected circulant C2n+1(S) such that C2(2n+1)(R)≅P2□C2n+1(S). And thereby P2□C2(2n+1)(R)≅C4□C2n+1(S)≅C4(2n+1)(4dS∪{2n+1}) using Theorem 12 where gcd(d,4(2n+1))=1. Hence the result.Corollary 16.
Letn≥2 and Cn(R) be connected. Then C4□Cn(R) is circulant if and only if n is odd. Furthermore, in that case, C4□Cn(R)≅C4n(4R∪{n})≅C4n(4dR∪{n}) where gcd(d,4n)=1.Proof.
Whenn is odd C4□Cn(R)≅C4n(4dR∪{n}), using Theorem 12 where gcd(d,4n)=1. Conversely, let C4□Cn(R) be circulant, say C4n(S) for some S⊆[1,2n]. When n is odd, the result is true by Theorem 12. When n is even, using Theorem 13, C4□Cn(R) is not circulant. Hence the result.Corollary 17.
Letn≥2 and Cn(R) be connected. Then C4□Cn(R) is circulant if and only if n is odd if and only if C4n(4R∪{n}) is connected. Furthermore, in that case, C4□C2n+1(R)≅C4(2n+1)(4R∪{2n+1})≅C4(2n+1)(4dR∪{2n+1}) and C4(1,2)□C2n+1(R)≅C4(2n+1)(4dR∪{2n+1,2(2n+1)}), where gcd(d,4(2n+1))=1.Theorem 18.
Letn>2 and Cn(R) be connected. Then P2□Cn(R) is circulant if and only if n is odd or Cn(R)≅P2□C2m+1(S) for some connected C2m+1(S),m∈ℕ.Proof.
Whenn is odd, P2□Cn(R)≅C2n(2dR∪{n}), using Theorem 10 where gcd(d,2n)=1. When Cn(R)≅P2□C2m+1(S) for some connected C2m+1(S), P2□Cn(R)≅C4□C2m+1(S)≅C4(2m+1)(4dS∪{2m+1}), using Theorem 12 where gcd(d,4(2m+1))=1.
Conversely, letP2□Cn(R) be circulant. Consider the following two cases of Cn(R).
Case i.  Cn(R)≅P2□Ck(S) for some connected Ck(S).
C
n
(
R
)
≅
P
2
□
C
k
(
S
) is connected and P2□Cn(R) is circulant implies that P2□Cn(R)≅P2□P2□Ck(S)≅C4□Ck(S) is a connected circulant. This implies that k is odd, using Corollary 16. Let k=2m+1 so that Cn(R)≅P2□Ck(S)=P2□C2m+1(S) for some connected C2m+1(S), m∈ℕ.
Case ii. Cn(R)≠P2□C2m+1(S) for any connected C2m+1(S).
Our aim is to prove thatn is odd in this case. If not, let n=2m, m∈ℕ. Then P2□Cn(R)=P2□C2m(R) is a connected circulant graph such that Cn(R)=C2m(R)≠P2□C2k+1(S) for any connected C2k+1(S). This implies that 2m/gcd(2m,r), the length of a periodic cycle of period r in C2m(R), is odd for all r∈R since each copy of P2 in the connected circulant graph P2□C2m(R) acts as a periodic cycle of length 2 (see the proof of Theorem 9). This is possible only when r is even for every r∈R. This implies that R=2R′⊆[1,m] and thereby Cn(R)=C2m(R)=C2m(2R′)=2·Cm(R′) which is not connected, a contradiction. This implies that n must be odd. Hence the result.Corollary 19.
Forn>2, the products P2□Cn and P2□Kn=P2□Cn(1,2,…,[n/2]) are circulant if and only if n is odd. Furthermore, in that case, P2□C2n+1≅C2(2n+1)(2,2n+1) and P2□K2n+1≅C2(2n+1)(2,4,…,2n,2n+1)⊂K2(2n+1),  n∈ℕ.Proof.
Forn>2, P2□Cn is circulant if and only if n is odd, follows from Theorem 9. Using Theorem 10, P2□K2m+1=P2□C2m+1(1,2,…,m)≅C2(2m+1)(2{1,2,…,m}∪{2m+1})=C2(2m+1)(2,4,…,2m,2m + 1) which is a proper subgraph of C2(2m+1)(1,2,…,2m+1)=K2(2m+1), m∈ℕ. Now consider the case of P2□Kn when n is even. Let n=2m, m∈ℕ. For m≥2, using Theorem 9, P2□C2m=P2□C2m(1) is not a circulant graph and hence using Lemma 14, P2□K2m is not a circulant graph. Hence the result.Corollary 20.
Forn>2,  Kn≠P2□Cm(S) and Kn≠C4□Cm(S) for any connected circulant Cm(S).Proof.
Clearly, forn>2,  Kn=Cn(1,2,…,[n/2]), a connected circulant graph. If connected circulant graph Kn=Cn(1,2,…,[n/2])≅P2□Cm(S), then, using Theorem 18, m is odd or Cm(S)≅P2□C2k+1(R) for some connected circulant C2k+1(R), k∈ℕ. And in both cases, using Theorems 10 and 12, P2□Cm(S) is a proper subgraph of K2m=Kn and ≠Kn, a contradiction for m>1. This implies that Kn≠P2□Cm(S) for any connected circulant Cm(S), n>2. Similarly if connected circulant graph Cn(1,2,…,[n/2])=Kn≅C4□Cm(S), then, using Corollary 16, m is odd and C4□Cm(S)=C4□C2k+1(S)≅C4(2k+1)(4dS∪{2k+1})=C4m(4dS∪{m}) which is a proper subgraph of K4m=Kn and ≠Kn, a contradiction where m=4k+1 and gcd(4m,d)=1,   k,d∈N. Hence the result.So far we could find out when the cross product ofP2 or C4 with another circulant is also circulant. It is interesting to know, in general, whether the cross product of any two connected circulant graphs is circulant or not. If so, when is it circulant? The following give some positive results in this direction.Theorem 21.
Form,n>2, Cm□Cn≅Cmn(m,n) if and only if gcd(m,n)=1.Proof.
Letm,n>2. At first, assume that Cm□Cn≅Cmn(m,n). Cm□Cn is a connected graph which implies that Cmn(m,n) is also connected. This implies, using Theorem 1, gcd(mn,m,n)=1 which implies gcd(m,n)=1.
Conversely, letgcd(m,n)=1. Without loss of generality, let us assume that 2<m<n. Now Cm□Cn is a connected 4-regular graph containing m disjoint copies of Cn, (n disjoint copies of Cm) and through all the m isomorphic images of each vertex of Cn, there exists a cycle of length m. And for all possible values of m and n, n·Cm≅Cmn(n) and m·Cn≅Cmn(m) are the two edge disjoint spanning circulant subgraphs of Cm□Cn.
Claim. Cm□Cn≅Cmn(m,n).
C
m
□
C
n contains m copies of Cn and Cmn(m)≅m·Cn(1) is a spanning subgraph of Cm□Cn. Let Cn(i)=(ui,1ui,2…ui,n) be the ith copy of Cn in Cm□Cn where ui,j is the isomorphic image of u1,j of Cn(1) in Cn(i) so that (u1,ju2,j…um,j) is a cycle of length m in Cm□Cn and V(Cm□Cn)={ui,j:i=1,2,…,m, and j=1,2,…,n}, j=1,2,…,n and i=1,2,…,m. Let V(Kmn)={v1,v2,…,vmn=v0}. Vertices v0,v1,v2,…,vmn-1 are considered as the ordered vertices of a regular mn-gon, assumed to be located in the plane, proceeding cyclically clockwise. Define a mapping Φm,n:V(Cm□Cn)→V(Cmn(1,2,…,mn/2))=V(Kmn) such that Φm,n(ui,j)=v(j-1)m+(i-1)n and Φm,n((u,v))=(Φm,n(u),Φm,n(v)) for every (u,v)∈E(Cm□Cn), using subscript arithmetic modulo mn, 1≤i≤m and 1≤j≤n. Under this transformation Φm,n(Cn(i))=Φm,n((ui,1ui,2ui,3…ui,n))=(Φm,n(ui,1)Φm,n(ui,2)Φm,n(ui,3)…Φm,n(ui,n))=1(v(i-1)nv(i-1)n+mv(i-1)n+2m…v(i-1)n+(n-1)m) which is a periodic cycle of (period m) length mn/gcd(mn,m)=n in Cmn(1,2,…,mn/2), i=1,2,…,m and these m cycles are vertex disjoint periodic cycles in Cmn(1,2,…,mn/2). Similarly, ϕm,n((u1,ju2,ju3,j⋯ui,j…um,j))=(v(j-1)mv(j-1)m+nv(j-1)m+2n…v(j-1)m+(m-1)n) which is a periodic cycle of period n and length m in Cmn(1,2,…,mn/2), j=1,2,…,n, and these n cycles are vertex disjoint periodic cycles in Cmn(1,2,…,mn/2) (transformation Φm,n is similar to θn,r,t defined in [13–18] to define type-2 isomorphism of circulant graphs). And the edges of Cm□Cn are the edges of n disjoint copies of Cm and m disjoint copies of Cn only. Clearly, Φm,n is a one-to-one mapping and preserves adjacency and Φm,n(Cm□Cn)=Cmn(m,n). Here the mapping Φm,n is one-to-one that follows from Lemma 8 once by considering each Cn(i)=Ai and A=V(Cm□Cn) and in another by considering each Cm(j)=Aj and A=V(Cm□Cn), i=1,2,…,m and j=1,2,…,n. The transformation Φm,n defined on C5□C6 is illustrated in Figures 3 and 4. C5□C6 is given in Figure 3 and Φ5,6(C5□C6)=C30(5,6) is given in Figure 4.Figure 3
C
5
×
C
6.Figure 4
Transformed graph ofC5×C6≅C30(5,6).Corollary 22.
Letm,n>2. Then Cm□Cn is circulant if and only if gcd(m,n)=1 if and only if Cm□Cn≅Cmn(dm,dn), where gcd(mn,d)=1.Proof.
Using Theorem21, for m,n>2, Cm□Cn≅Cmn(m,n) if and only if gcd(m,n)=1. Moreover, if Cn(p,q) and Cn(r,s) are isomorphic, then they are Adam’s isomorphic only [18, 19] and, when gcd(n,d)=1, Cn(p,q) and Cn(dp,dq) are Adam’s isomorphic circulant graphs. Hence Cm□Cn≅Cmn(m,n)≅Cmn(dm,dn), where gcd(m,n)=1=gcd(mn,d).Note 4.
For any integerd relatively prime to mn and for a set R⊆[1,mn/2], Cmn(R) and Cmn(dR) are Adam’s isomorphic. Thus when gcd(m,n)=1=gcd(mn,d), Cm□Cn≅Cmn(m,n)≅Cmn(dm,dn), d∈ℕ. Thus P2□C2n+1≅C2(2n+1)(2,2n+1), C5□C6≅C30(5,6), and C7□C10≅C70(7,10). Also C5□C6≅C30(5,6)≅C30(7×5,7×6)≅C30(5,12)≅C30(5,7×6) and C7□C10≅C70(7,10)≅C70(9×7,9×10)≅C70(7,20)≅C70(7,9×10).Corollary 23.
Leta,b,m,n∈ℕ, 1≤2r+1≤2m+1 and 1≤2s+1≤2n+1. Then C2a(2m+1)(2r+1)□C2b(2n+1)(2s+1) is not circulant.Proof.
Letgcd(2m+1,2r+1)=2x+1 and gcd(2n+1,2s+1)=2y+1, x,y∈ℕ∪{0}. This implies gcd((2m+1)/(2x+1),(2r+1)/(2x+1))=1=gcd(2a(2m+1)/(2x+1),(2r+1)/(2x+1)), and gcd((2n+1)/(2y+1),(2s+1)/(2y+1))=1=gcd(2b(2n+1)/2y+1,2s+1/2y+1) which implies that C2a(2m+1)(2r+1)≅(2x+1)·C2a(2m+1)/(2x+1)((2r+1)/(2x+1))≅(2x+1)·C2a(2m+1)/2x+1≅(2x+1)·C2a(2m′+1) and similarly C2b(2n+1)(2s+1)≅(2y+1)·C2b(2n′+1) for some m′,n′∈ℕ where (2m+1)/(2x+1)=2m′+1 and (2n+1)/(2y+1)=2n′+1. This implies that C2a(2m+1)(2r+1)□C2b(2n+1)(2s+1)≅(2x+1)(2y+1)·C2a(2m′+1)□C2b(2n′+1), which is not circulant using Theorem 21 since gcd(2a(2m′+1),2b(2n′+1))≥2, m′,n′∈ℕ. Hence the result.Corollary 24.
LetCkam(R) and Ckbn(S) be connected, m,n>2, k≥1, gcd(m,n)=1=gcd(mn,k) and a,b∈ℕ∪{0}. Then Cm(R)□Ckbn(S) and C2a+1m(R)□C2b+1n(S) are circulants when C2a+1m(R)=C2m(R)≅P2□Cm(T1) and C2b+1m(R)=C2n(S)≅P2□Cn(T2) for some T1 and T2.Corollary 25.
Fora,b,k,m,n∈ℕ, m,n>2, k≥2, gcd(m,n)=1, ka+b≠4 and connected circulants Cn(R), Cn(S), Ckam(R), and Ckbn(S), Cn(R)□Cn(S) and Ckam(R)□Ckbn(S) are not circulant.In the proof of Theorem21, we have, when m,n>2 and gcd(m,n)=1, Cm□Cn≅Cmn(m,n)≅Cmn(dm,dn) where gcd(mn,d)=1. In the transformation Φm,n defined on the graph Cm□Cn each copy of the cycle Cn is rotated uniformly (and thereby the relative positions of the vertices of each copy of Cn are not changed) with respect to the regular mn-gon so as to make uniform rotation of each copy of Cm with respect to the regular mn-gon. Here any vertex ui,j→v(j-1)m+(i-1)n using subscript arithmetic modulo mn, i=1,2,…,m and j=1,2,…,n. Thereby, for m,n>2, gcd(m,n)=1 and connected circulant Cm(R) and Cn(S), Φm,n(Cm(R)□Cn)≅Φn,m(Cn□Cm(R))≅Cmn(nR∪{m})≅Cmn(dnR∪{dm}) and Φm,n(Cm□Cn(S))≅Cmn(mS∪{n})≅Cmn(dmS∪{dn}) where gcd(mn,d)=1. And if we introduce jump sizes r1,r2,…,rj instead of jump size 1 in Cm and s1,s2,…,sk instead of 1 in Cn and making the above transformation on Cm(R)□Cn(S), we get Φm,n(Cm(R)□Cn(S))=Cmn(nR∪mS)≅Cmn(dnR∪dmS), where Cm(R) and Cn(S) are connected, m,n>2, gcd(m,n)=1, gcd(mn,d)=1, R={r1,r2,…,rj}, and S={s1,s2,…,sk}. Thus, we obtain the following result.Theorem 26.
LetCm(R) and Cn(S) be connected, m,n>2 and gcd(m,n)=1=gcd(mn,d). Then Cm(R)□Cn(S)≅Cmn(mS∪nR)≅Cmn(dmS∪dnR).Theorem 27.
Letm,n∈ℕ and C2m+1(R) and C2n+1(S) be connected circulant graphs. Then C2m+1□C2n+1 is circulant if and only if K2m+1□K2n+1 is circulant if and only if C2m+1(R)□C2n+1(S) is circulant if and only gcd(2m+1,2n+1)=1. Furthermore, in that case, there exists natural number d relatively prime to (2m+1)(2n+1) such that C2m+1(R)□C2n+1(S)≅C(2m+1)(2n+1)((2n+1)R∪(2m+1)S)≅C(2m+1)(2n+1)(d(2n+1)R∪d(2m+1)S).Theorem 28.
LetG be a connected graph of order n, n>2. Then P2□G is circulant if and only if G≅H or P2□H where H is a connected circulant graph of odd order.Proof.
LetG≅H or P2□H for some connected circulant graph H of odd order. Then, by Theorem 18, P2□G is circulant.
Conversely, letP2□G≅C2n(S) be a connected circulant of order 2n for some S⊆[1,n], n>2. Clearly, P2□G≅C2n(S)⊆P2□Kn, n>2. Now, consider different cases of G.
Case i.  O(G)=n=2m+1, m∈N.
Clearly,G⊆K2m+1 and, using Corollary 19, P2□G≅C2n(S)=C2(2m+1)(S)⊆P2□K2m+1≅C2(2m+1)(2,4,…,2m,2m+1). This implies that S⊆2T∪{2m+1}, where T=[1,m] and S≠ϕ. S={2m+1} and S⊆2T are not possible since C2(2m+1)(S) is connected. When S=2T′∪{2m+1}, P2□G≅C2(2m+1)(2T′∪{2m+1})≅P2□C2m+1(T′), using Theorem 10 where ϕ≠T′⊆T=[1,m]. Thus in this case, we get G≅C2m+1(T′) for some connected circulant C2m+1(T′).
Case ii. G≅P2□H where H is a connected circulant graph of order 2m+1,  m∈ℕ.
Then,P2□G is circulant which implies that C4□H is circulant. Clearly, using Corollary 17, P2□G≅C4□H≅C4(2m+1)(S)⊆C4□K2m+1≅C4(2m+1)(4d,8d,…,4md,2m+1) for some S⊆[1,2(2m+1)] and every d relative prime to 4(2m+1). This implies that S⊆4dT∪{2m+1}, d∈ℕ, gcd(4(2m+1),d)=1, T=[1,m], and S≠ϕ. S={2m+1} and S⊆4dT are not possible since C4(2m+1)(S) is connected. When S=4dT′∪{2m+1} for some T′ and d∈ℕ such that gcd(4(2m+1),d)=1, Φ≠T′⊆T, then P2□G≅C4□H≅C4(2m+1)(S)≅C4(2m+1)(4dT′∪{2m+1})≅C4□C2m+1(T′)  using Theorem 15. This implies G≅P2□C2m+1(T′) for some connected circulant C2m+1(T′).
Also,P2□G≅C2n(S), a circulant graph implies that 2G and n·P2 are two edge-disjoint spanning subgraphs of C2n(S). This implies if we remove all the n copies of circulant subgraph P2=C2 from the circulant graph C2n(S)≅P2□G, then the resultant graph 2G must be a circulant graph (in any circulant graph removal or addition of one or more jump sizes, if possible, will not change the property of being circulant) which implies that G is circulant. Now G and P2□G are circulants and using Theorem 18 the graph G is either of odd order or product of P2 and an odd order circulant graph. Hence the result.Theorem 29.
LetG be a connected graph of order n≥2. Then C4□G is circulant if and only if G is circulant of odd order.Proof.
WhenG is a connected circulant of odd order, then using Corollary 19, C4□G is circulant. Conversely, let C4□G be a connected circulant, say C4n(R). In any circulant graph C4n(R) periodic cycles, each of length 4 occur without rotation. The spanning subgraph n·C4 is also a circulant subgraph in C4n(R)≅C4□G. And hence if we remove all the edges of the spanning circulant subgraph n·C4 from the circulant graph  C4n(S)≅C4□G, then the resultant graph 4G must be a circulant graph which implies that G is circulant. When G and C4□G are connected circulant graphs, then, using Corollary 19, the order of G is odd. Hence the result.Theorem 30.
IfG and H are connected graphs and G□H is circulant, then G and H are circulants.Proof.
LetG and H be connected graphs of order m and n, respectively, m,n∈ℕ. Then the order of the graph G□H is mn. Let G□H be circulant. For m or n=1 or m=n=2, the result is true. Now let 2≤m≤n. When at least one of the two graphs, say G, is circulant, then nG is a spanning subgraph which is also a circulant subgraph of the circulant graph G□H. If we remove all the edges of spanning circulant subgraph nG from the circulant graph G□H the resultant graph mH is circulant which implies that H is circulant. When both the graphs G and H are not circulant, then let Cm(R1) be a spanned subgraph of G obtained from G by removal of minimum number of edges and Cm(R2) be a circulant graph which is obtained from G by adding minimum number of edges in G. Similarly let Cn(S1) and Cn(S2) be the corresponding circulant graphs obtained with respect to the graph H. This implies that Cm(Φ)□Cn(Φ)⊆Cm(R1)□Cn(S1)⊂G□H⊂Cm(R2)□Cn(S2)⊆Km□Kn which implies, from the construction, that G□H≠ product of a circulant graph of order m and a circulant graph of order n which implies G□H is not a circulant graph, a contradiction to the given condition that the graph G□H is circulant. This implies G and H are circulant graphs. Hence the result.Theorem 31.
LetG and H be connected graphs, each of order >2. Then G□H is circulant if and only if G and H are circulants and satisfy one of the following conditions:   (i)
G≅Cm(R); H≅Cn(S) and gcd(m,n)=1;(ii)
G≅C2m+1(R); H≅C2n+1(S), P2□C2n+1(S),  C4□C2n+1(S) or C2k(2n+1)(S) and gcd(2m+1,2k(2n+1))=1,  k∈ℕ;(iii)
G≅P2□C2m+1(R); H≅C2n+1(S) or P2□C2n+1(S) and gcd(2m+1,2n+1)=1;(iv)
G≅C2k(2m+1)(R)≠P2□C2k-1(2m+1)(T) for any C2k-1(2m+1)(T); H≅C2n+1(S) and gcd(2k(2m+1),2n+1)=1, k∈ℕ;(v)
G≅C4□C2m+1(R); H≅C2n+1(S) and gcd(2m+1,2n+1)=1;(vi)
G≅C2k(2m+1)(R)≠C4□C2k-2(2m+1)(T),P2□C2k-1(2m+1)(U) for any C2k-2(2m+1)(T) and C2k-1(2m+1)(U);  H≅C2n+1(S) and gcd(2k(2m+1),2n+1)=1, k∈ℕ, k≥2.Proof.
LetG and H be of order m and n, respectively, m,n>2. Then G□H is a connected graph of order mn, mn>4. Graph G□H is circulant if and only if G and H are circulants that follow from Theorem 30. Let G≅Cm(R) and H≅Cn(S) for some R⊆[1,m/2] and S⊆[1,n/2]. Then the rest of the result follows from Theorems 10, 12, and 26 and Corollaries 17, 24, and 25.Corollary 32.
P
2
□
C
4
□
G and C4□P2□H are not circulant for any graphs G and H.Proof.
P
2
□
C
4 is not circulant using Theorem 18. Then the result follows from Theorem 31.Definition 33 (see [21]).
A nontrivial graphG is said to be prime if G=G1□G2 implies that G1 or G2 is trivial; G is composite if it is not prime.
Sabidussi [20] proved that every nontrivial graph is the unique product of prime graphs. In view of this and of the result that m·Cn(R)≅Cmn(mR), let us look into the definition of prime circulant graphs.Definition 34.
IfCm(R), Cn(S) and Cmn(T) are circulant graphs such that Cm(R)□Cn(S)≅Cmn(T), then we say that Cm(R) and Cn(S) are divisors or factors of Cmn(T).
Thus for any connected circulant graph, the graph andC1()=K1 are always divisors and so we call them as improper divisors of the circulant graph. Divisors which are integer multiple of improper divisors also are called as improper divisors of the circulant graph. This does not arise since we consider divisors of connected graphs only. Divisor(s) other than improper divisors is called proper divisor(s) of the circulant graph.Definition 35.
A circulant graph whose only divisors are improper is called a prime circulant graph. Other circulant graphs are called composite circulant graphs.
In view of Corollary32, for any connected circulant, both P2 and C4m(R) cannot be proper divisors, m∈N; C4=P2□P2=C2□C2 and we consider C2=P2 as a prime circulant.
It is easy to verify the following result using Theorems9, 10, 15, 18, 27, 28, 29, and 31, and Corollaries 17, 24, and 25.Theorem 36 (factorization theorem on circulant graphs).
Letm and n be relatively prime integers. If R⊆[1,m/2], S⊆[1,n/2], and T⊆[1,mn/2] with T=dnR∪dmS for some d such that gcd(mn,d)=1, then Cmn(T)≅Cm(R)□Cn(S).Note 5.
From the factorization theorem, we get, in particular, whengcd(2R(2n+1),d)=1 and T=2dR∪{2n+1}, C2(2n+1)(T)≅P2□C2n+1(R) and when gcd(4R(2n+1),d)=1 and T=4dR∪{2n+1} or 4dR∪{2n+1,2(2n+1)}, C4(2n+1)(T)≅C4□C2n+1(R) or C4(1,2)□C2n+1(R), respectively.Theorem 37.
Ifn≠4 and 1∈R, then Cn(R) is a prime circulant.Proof.
Forn=4,  Cn=C4≅C2□C2=P2□P2. Let 1∈R. For n≠4, if possible, assume that Cn(R)≅Cn1(S)□Cn2(T) where R, S, and T are sets of positive integers such that R⊆[1,n/2], S⊆[1,n1/2], T⊆[1,n2/2], n1,n2∈ℕ∖{1}, and n=n1n2. Then, using Theorems 26 and 27, there exist integer d with gcd(n1n2,d)=1 such that Cn1(S)□Cn2(T)≅Cn1n2(dn2S∪dn1T). Now the length of periodic cycles of Cn1n2(dn2S∪dn1T) is either n1n2/gcd(n1n2,dn2s)=n1/gcd(n1,s) or n1n2/gcd(n1n2,dn1t)=n2/gcd(n2,t), s∈S and t∈T. This implies that no periodic cycle of Cn1n2(dn2S∪dn1T) is of length n=n1n2, whereas Cn(R) has periodic cycle (of period 1 and) of length n. This is a contradiction to Cn(R)≅Cn1n2(dn2S∪dn1T)≅Cn1(S)□Cn2(T) for some d such that gcd(n1n2,d)=1.Remark 38.
The converse of the above theorem is not true. That is, ifCn(R) is a prime circulant graph, then R need not contain 1. For example,C30(3,4,5) is a prime circulant graph, using the factorization theorem, but it is not isomorphic to C30(1,r,s) for all possible values of r and s.Corollary 39.
Ifn≠4 and if R contains an integer relatively prime to n, then Cn(R) is prime circulant.Proof.
Letr∈R be relatively prime to n. Hence there exists d∈ℤn* such that dr≡1(modn). Clearly, Cn(dR) is Adam’s isomorphic to Cn(R) and contains jump size 1 corresponding to dr which is ≡1(modn)  [22]. Now the result follows from Theorem 37.Corollary 40.
Ifn is a prime power other than 4 and Cn(R) is connected, then Cn(R) is prime circulant for all R≠ϕ.Proof.
ClearlyCp(R) with R={r1,r2,…,rk}≠Φ is prime circulant for any prime number p. Let n=pm≠22 where p is a prime number and m∈ℕ. Given that Cpm(R) is connected which implies that gcd(pm,r1,r2,…,rk)=1. Since p is prime, the above relation implies that there exists at least one ri such that gcd(pm,ri)=1, 1≤i≤k. Now the result follows from Corollary 39.Theorem 41 (fundamental theorem of circulant graphs).
Every connected circulant graph is the unique product of prime circulant graphs (uniqueness up to isomorphism).
(It follows that if a connected circulantCn(T)≅Ck1(R1)□Ck2(R2)□⋯□Ckm(Rm)≅Cl1(S1)□Cl2(S2)□⋯□Clh(Sh), where Cki(Ri) and Clj(Sj) are connected prime circulant graphs, then m=h and there exists a permutation Π on the symbols 1,2,…,m  such that Π(i)=j, Clj(Sj)≅Cki(Ri),  gcd(kα,kβ)=1 and gcd(la,lb)=1,  α≠β,  a≠b and kα,kβ,la,lb≠2, each prime circulant and prime circulant of even order occur at the most once in the prime factorization except C2 which may occur at the most twice, n=k1k2,…,  km=l1l2,…,lh, and i,j,α,β,a,b=1,2,…,m. Here we have if A□G≅A□H,  then G≅H for any graph A.)Proof.
Using the factorization theorem, we can write any circulant graph as a product of prime circulant graphs. The number of even order prime factors (circulant) of any given connected circulant is at the most two. Furthermore, in the case of having two even order prime factors, each prime factor isC2 only. The uniqueness follows from Theorems 10, 15, 18, and 26–31, and Corollaries 17 and 25.Note 6.
In the above theorem we have ifA□G≅A□H, then G≅H for any graph A. But the converse need not be true always. That is if G and H are isomorphic connected simple graphs and A is any connected simple graph, then A□G and A□H need not be isomorphic. For example, (i) C16(1,2,7)□C16(1,2,7) is not isomorphic to C16(1,2,7)□C16(3,4,5), even though C16(1,2,7)≅C16(2,3,5). Another example is (ii) C16(1,2,7)□C27(1,3,8,10) is not isomorphic to C16(1,2,7)□C27(3,4,5,13), even though C27(1,3,8,10)≅C27(3,4,5,13). The product graphs in example (i) are not circulant, whereas in example (ii) the product graphs are circulant graphs.Remark 42.
IfG is a connected graph such that G≅G1□G2□⋯□Gk,  then the diameter of G,  dia(G)=∑i=1k‍dia(Gi) [23].
Thus, we can find the diameter of any given circulant graph, provided that diameters of its prime circulant graphs are known. Also the above relation helps to generate (circulant) graphs of bigger diameters.Concluding Remarks
(1)
In prime factorization of connected circulantsC1()=K1 and C2=P2 act similar to 1 and 2 among the set of all natural numbers, respectively. Thus,  C1() is a unit, like 1 in number theory.
(2)
There exist  two types of prime circulant graphs of ordern, one with periodic cycle(s) of length n and the other without periodic cycle of length n. See Theorem 37, Corollary 39, and Remark 38.
(3)
The theory of factorization of circulants is similar to the theory of factorization of natural numbers and one of the very few well-known mathematical structures so vividly classified (expressed) in terms of prime factors. It can be applied in cryptography.
(4)
POLY315.exe is a VB program developed by us to show visually how the transformationΦm,n acts on Cm□Cn for different values of m and n, m,n∈ℕ.
(5)
An interesting problem is, for a given integern, finding the number of prime (composite) circulant graphs of order either equal to n or less than or equal to n.
(6)
One can develop theories similar to the theory of Cartesian product and factorization of circulant graphs to the other standard products of circulant graphs.



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