## 3 Edge Decomposition of Bipartite Circulant Graphs \(C_{2 nm,nm}\) Based on Cartesian Products
These novel chains represent the edge decomposition of \(C_{2 nm,nm}\), i.e., \(E(\mathbb{C}_{\sigma}t)=\{\ (e_{0}+\sigma,\pi(f_{0,\sigma})),(e_{1}+\sigma,\pi(f_{1, \sigma})),\ldots,\)\((e_{nm-1}+\sigma,\pi(f_{nm-1,\sigma}))\ \}\), \(\sigma\in\mathbb{Z}_{mn}\) and \(\cup_{\sigma\in\mathbb{Z}_{nm}}E(\mathbb{C}_{\sigma})=E(C_{2 nm,nm})\) that can be proved as follows. Let \(\lambda_{i}=f_{i}-e_{i}\), which is unique for every pair \((e_{i},f_{i});i\in\mathbb{Z}_{nm}\). For \(u\
eq v\in\mathbb{Z}_{nm}\), let \((e_{i}+u,f_{i}+u)\in E(\mathbb{C}_{u})\) and \((e_{i}+v,f_{i}+v)\in E(\mathbb{C}_{v})\). Suppose that \(|E(\mathbb{C}_{u})\cap\ E(\mathbb{C}_{v})|\
eq 0\) such that there is at least one edge \((e_{i},f_{i})\in\{E(\mathbb{C}_{u})\cap E(\mathbb{C}_{v})\}\). From the translation definition, we have \((e_{i}+u-u,f_{i}+u-u)=(e_{i},f_{i}),(e_{i}+v-v,f_{i}+v-v)=(e_{i},f_{i})\), and \(\lambda_{i}=f_{i}-e_{i}\) is unique for every edge in \(\mathbb{C}_{0}\). Hence, \(|E(\mathbb{C}_{u})\cap\ E(\mathbb{C}_{v})|\
eq 0\). Then, the addition of \(nm\) to the second component of the ordered pairs \((e_{0}+\sigma,f_{0}+\sigma),(e_{1}+\sigma,f_{1}+\sigma),\ldots,\)\((e_{nm-1}+\sigma,f_{nm-1}+\sigma),\sigma\in\mathbb{Z}_{nm}\) constructs the edge decomposition of \(C_{2 nm,nm}\). Hence, \(\cup_{\sigma\in\mathbb{Z}_{nm}}E(\mathbb{C}_{\sigma})=E(C_{2 nm,nm})\). \(\square\)  
**Theorem 1**. : _There is an edge decomposition of \(C_{8n,An}\) by the graph \(nK_{2,2}\) and an edge decomposition of \(C_{2m,m}\) by the graph \(K_{1,m}\), then there is an edge decomposition of \(C_{8 nm,Anm}\) by the graph \(\mathbb{C}\cong nK_{2,2m};m\geq 1,gcd(n,3)=1\)._  
**Proof. ** The edge set of \(nK_{2,2}\) can be represented by \((a_{0},b_{0}+4n),(a_{1},b_{1}+4n),\ldots,\)\((a_{4n-1},b_{4n-1}+4n)\), where  
\[\begin{array}{c}a_{i}=\left\{ \begin{array}{cc}i&if\ i<2n\\ 2n+i&if\ i\geq 2n\\ b_{i}=a_{i}+i.\end{array} \right.,\\ \mbox{Hence, }\{b_{i}-a_{i}:i\in\mathbb{Z}_{4n}\}=\mathbb{Z}_{4n}.\\ \mbox{The edge set of the second graph $K_{1,m}$ can be represented by $(c_{0},d_{0}+m)$, $(c_{1},d_{1}+m)$, }\ldots,(c_{m-1},d_{m-1}+m)\), where  
\[ \begin{array}{c}c_{i}=0;i\in\mathbb{Z}_{m},\\ d_{i}=c_{i}+i.\end{array}  \tag{8}\]  
Hence, \(\left\{d_{j}-c_{j}:j\in\mathbb{Z}_{m}\right\}=\mathbb{Z}_{m}\). Suppose that the first graph \(nK_{2,2}\) is represented by the vector \(v(nK_{2,2})=(a_{0},a_{1},a_{2},\ldots,\)\(a_{4n-1})\), and the second graph \(K_{1,m}\) is represented by the vector \(w(K_{1,m})=(c_{0},c_{1},c_{2},\ldots,\)\(c_{m-1})\), then \(v(nK_{2,2})\times w(K_{1,m})=(a_{0}c_{0},a_{0}c_{1},\ldots,a_{p}c_{q},\ldots,a_ {4n-1}c_{m-1},\) where \(p\in\mathbb{Z}_{4n}\) and \(q\in\mathbb{Z}_{m}\). Then, construct the ordered pairs \((a_{0}c_{0},b_{0}d_{0}),(a_{0}c_{1},b_{0}d_{1}),\ldots,(a_{p}c_{q},b_{p}d_{q}),\ldots,\)\((a_{4n-1}c_{m-1},b_{4n-1}d_{m-1})\) where \(p\in\mathbb{Z}_{4n}\) and \(q\in\mathbb{Z}_{m}\). Hence, from (7) and (8), we can conclude that  
\[\left\{(b_{i}-a_{i})\big{(}d_{j}-c_{j}\big{)}:i\in\mathbb{Z}_{4n},j\in\mathbb{ Z}_{m}\right\}=\mathbb{Z}_{4n}\times\mathbb{Z}_{m}. \tag{9}\]  
Then, let \(\varphi:\mathbb{Z}_{4n}\times\mathbb{Z}_{m}\rightarrow\mathbb{Z}_{4 nm}\) be a one-one mapping defined by  
\[\varphi\big{(}a_{i}c_{j}\big{)}=e_{mi+j}=ma_{i}+c_{j},i\in\mathbb{Z}_{4n},j\in \mathbb{Z}_{m}, \tag{10}\]  
and \(\psi:\mathbb{Z}_{4n}\times\mathbb{Z}_{m}\rightarrow\mathbb{Z}_{4 nm}\) be a one-one mapping defined by  
\[\psi\big{(}b_{i}d_{j}\big{)}=f_{mi+j}=mb_{i}+d_{j},i\in\mathbb{Z}_{4n},j\in \mathbb{Z}_{m}. \tag{11}\]  
Hence, we can obtain the following chain \((e_{0},f_{0}),(e_{1},f_{1}),\ldots,(e_{4 nm-1},f_{4 nm-1})\). The translation for the previous chain is \((e_{0}+\sigma,f_{0}+\sigma),(e_{1}+\sigma,f_{1}+\sigma),\ldots,\)\((e_{4 nm-1}+\sigma,f_{4 nm-1}+\sigma),\sigma\in\mathbb{Z}_{4 nm}\), where the additions are calculated modulo \(4 nm\).