In this section, by considering the graph \mathrm{\Gamma}({G}_{i}), 1\le i\le 5, we mainly deal with some graph properties, namely diameter, maximum and minimum degrees, girth, chromatic number, clique number, domination number, degree sequence and irregularity index of \mathrm{\Gamma}({G}_{i}).
2.1 Case 1: the graph \mathrm{\Gamma}({G}_{1}), where {G}_{1}={\mathbb{Z}}_{k}\ast ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{n})
If we consider the graph of the group {G}_{1}, then we have a subgraph of Figure 1(a) with vertices {v}_{1}={a}^{m}, {v}_{2}=a, {v}_{4}={x}^{k}, {v}_{5}=x, {v}_{9}=b and {v}_{10}={b}^{n}. In this graph the edge set depends on the orders of factor groups of {G}_{1}. If we take k,m,n=2, then by the edge definition, we have the edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{15} in this subgraph of Figure 1(a). In the case k,m,n>2, we do not have any edges. In the remaining case, i.e., one or two orders of factor groups of {G}_{1} are equal to two, we have one or two edges among {\alpha}_{1}, {\alpha}_{4} and {\alpha}_{15}.
If we reconsider that the graph in Figure 1(a) depends on the group {G}_{1} with above facts, then the picture will be shown as an unconnected graph which is not related to the numbers k, m and n. Thus the first result is the following.
Theorem 1 diam(\mathrm{\Gamma}({G}_{1}))=\mathrm{\infty}.
Theorem 2
The maximum and minimum degrees of the graph
\mathrm{\Gamma}({G}_{1})
are
\mathrm{\Delta}(\mathrm{\Gamma}({G}_{1}))=\{\begin{array}{ll}1;& \mathit{\text{at least one of}}k,m,n=2,\\ 0;& k,m,n>2\end{array}
and
\delta (\mathrm{\Gamma}({G}_{1}))=\{\begin{array}{ll}1;& k,m,n=2,\\ 0;& \mathit{\text{at most two of}}k,m,n=2\mathit{\text{or}}k,m,n>2,\end{array}
respectively.
Proof If we take that at least one of k, m, n is two, then we have at least one of the edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{15}. Thus we get \mathrm{\Delta}(\mathrm{\Gamma}({G}_{1}))=1. If we consider k=m=n=2, then since we have edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{15}, this gives us that all vertices in the graph \mathrm{\Gamma}({G}_{1}) have degree one. So, \mathrm{\Delta}(\mathrm{\Gamma}({G}_{1}))=\delta (\mathrm{\Gamma}({G}_{1}))=1. Now we consider the case k,m,n>2. In this case, since we have no edges in the graph \mathrm{\Gamma}({G}_{1}), we obtain \mathrm{\Delta}(\mathrm{\Gamma}({G}_{1}))=\delta (\mathrm{\Gamma}({G}_{1}))=0. If we take that at most two of k, m, n are equal to two, then we get four vertices having degree one and two vertices having degree zero. Therefore, in this case, \delta (\mathrm{\Gamma}({G}_{1}))=0. □
Theorem 3 For any k,m,n\ne 1, the girth of the graph \mathrm{\Gamma}({G}_{1}) is equal to infinity.
Proof Since we just have the edges {\alpha}_{1}, {\alpha}_{4} and {\alpha}_{15} depending on the numbers k, m, n, we do not have any cycle in the graph \mathrm{\Gamma}({G}_{1}). So, gr(\mathrm{\Gamma}({G}_{1}))=\mathrm{\infty}. □
Theorem 4
The chromatic number of
\mathrm{\Gamma}({G}_{1})
is equal to
\chi (\mathrm{\Gamma}({G}_{1}))=\{\begin{array}{ll}2;& \mathit{\text{at least one of}}k,m,n=2,\\ 1;& k,m,n>2.\end{array}
Proof If we take that at least one of k, m, n is two, then we have at least one of the edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{15}. Thus we use two different colors since there exist neighbor vertices. By the edge definition of \mathrm{\Gamma}({G}_{1}), we do not have any edges between generators and elements of three factor groups of {G}_{1}. Thus we obtain \chi (\mathrm{\Gamma}({G}_{1}))=2. If we consider k,m,n>2, then since we do not have any edges in the graph, we can label all vertices with the same color. Therefore \chi (\mathrm{\Gamma}({G}_{1}))=1. □
Theorem 5
The clique number of
\mathrm{\Gamma}({G}_{1})
is equal to
\omega (\mathrm{\Gamma}({G}_{1}))=\{\begin{array}{ll}2;& \mathit{\text{at least one of}}k,m,n=2,\\ 1;& k,m,n>2.\end{array}
Proof The proof of this theorem is similar to the proof of Theorem 4. If we take that at least one of k, m, n is 2, then we have at least one of the edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{15}, i.e., we have a disconnected graph which has at least three complete subgraphs. Since these complete subgraphs have two vertices, we get \omega (\mathrm{\Gamma}({G}_{1}))=2. If we consider k,m,n>2, then since we do not have any edges in the graph, the number of vertices in maximal clique is one. □
Theorem 6 The domination number of \mathrm{\Gamma}({G}_{1}) is equal to infinity.
Proof For all cases of k, m, n, since the graph \mathrm{\Gamma}({G}_{1}) is disconnected, we get \gamma (\mathrm{\Gamma}({G}_{1}))=\mathrm{\infty}. □
Theorem 7
The degree sequence and irregularity index of
\mathrm{\Gamma}({G}_{1})
are given by
DS(\mathrm{\Gamma}({G}_{1}))=\{\begin{array}{ll}(1,1,1,1,1,1);& k,m,n=2,\\ (0,0,0,0,0,0);& k,m,n>2,\\ ({i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5},{i}_{6});& \mathit{\text{otherwise}},\end{array}
where {i}_{j}=0,1 for 1\le j\le 6, and
t(\mathrm{\Gamma}({G}_{1}))=\{\begin{array}{ll}1;& k,m,n\ge 2,\\ 2;& \mathit{\text{at least one of}}k,m,n\mathit{\text{is equal}}2,\end{array}
respectively.
Proof By the graph \mathrm{\Gamma}({G}_{1}), if k, m, n are equal to two and greater than two, then the degrees of the vertices are one and zero, respectively. But if at least one of k, m, n is equal to two, then some vertices have degree one and some of them have degree zero. Hence, by the definition of a degree sequence, we clearly obtain the set DS(\mathrm{\Gamma}({G}_{1})), as depicted. Nevertheless, it is easily seen that the irregularity index t(\mathrm{\Gamma}({G}_{1}))=1 and 2, as required. □
2.2 Case 2: the graph \mathrm{\Gamma}({G}_{2}), where {G}_{2}={\mathbb{Z}}_{k}\times ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{n})
If we consider the graph of the group {G}_{2}, then we have a subgraph of Figure 1(a) with vertices {v}_{1}={a}^{m}, {v}_{2}=a, {v}_{3}={x}^{k}a, {v}_{4}={x}^{k}, {v}_{5}=x, {v}_{6}=x{a}^{m}, {v}_{7}=xa, {v}_{8}={x}^{k}b, {v}_{9}=b, {v}_{10}={b}^{n}, {v}_{11}=x{b}^{n} and {v}_{12}=xb. In this graph the edge set depends on the orders of factor groups of {G}_{2}. If we take k,m,n=2 then, by the edge definition, we have the edges {\alpha}_{j}, 1\le j\le 23. In the case k,m,n>2, we do not have edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{9}, {\alpha}_{12}, {\alpha}_{15}, {\alpha}_{20} and {\alpha}_{23} in \mathrm{\Gamma}({G}_{2}).
Theorem 8 The diameter of the graph \mathrm{\Gamma}({G}_{2}) is equal to four.
Proof By considering the graph of the group {G}_{2}, we say that the diameter of the graph \mathrm{\Gamma}({G}_{2}) does not depend on the numbers k, m, n. For any k, m, n, in the graph \mathrm{\Gamma}({G}_{2}) the vertices {v}_{7}=xa and {v}_{12}=xb are adjacent to vertices {v}_{1}, {v}_{2}, {v}_{3}, {v}_{4}, {v}_{5}, {v}_{6} and {v}_{4}, {v}_{5}, {v}_{8}, {v}_{9}, {v}_{10}, {v}_{11}, respectively. If we connect any two vertices, except {v}_{4} and {v}_{5}, via the shortest path, we need to pass through the vertices {v}_{7} and {v}_{12}. For instance, we need the edges {\alpha}_{7}, {\alpha}_{11}, {\alpha}_{18} and {\alpha}_{22} to connect two vertices {v}_{1}=a and {v}_{10}=b. This gives us diam(\mathrm{\Gamma}({G}_{2}))=4. □
Theorem 9
The maximum and minimum degrees of the graph
\mathrm{\Gamma}({G}_{2})
are
\mathrm{\Delta}(\mathrm{\Gamma}({G}_{2}))=\{\begin{array}{ll}6;& k,m,n=2,\\ 4;& k,m,n>2\end{array}
and
\delta (\mathrm{\Gamma}({G}_{2}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 2;& k,m,n>2,\end{array}
respectively.
Proof For k,m,n=2, in the graph of {G}_{2} the vertices {v}_{7}=xa and {v}_{12}=xb are adjacent to vertices {v}_{1}, {v}_{2}, {v}_{3}, {v}_{4}, {v}_{5}, {v}_{6} and {v}_{4}, {v}_{5}, {v}_{8}, {v}_{9}, {v}_{10}, {v}_{11}, respectively. Since these vertices have the largest degrees in \mathrm{\Gamma}({G}_{2}), we get \mathrm{\Delta}(\mathrm{\Gamma}({G}_{2}))=6. The other vertices {v}_{1}, {v}_{2}, {v}_{3}, {v}_{6}, {v}_{8}, {v}_{9}, {v}_{10} and {v}_{12} have degree three and the remaining vertices {v}_{4} and {v}_{5} have degree five. So, the minimum degree of the graph \mathrm{\Gamma}({G}_{2}) is \delta (\mathrm{\Gamma}({G}_{2}))=3. Now we take k,m,n>2. In this case, we do not have edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{9}, {\alpha}_{12}, {\alpha}_{15}, {\alpha}_{20} and {\alpha}_{23}. Thus the vertices {v}_{4}, {v}_{5}, {v}_{7}, {v}_{12} have degree four and the remaining vertices have degree two. So, \mathrm{\Delta}(\mathrm{\Gamma}({G}_{2}))=4 and \delta (\mathrm{\Gamma}({G}_{2}))=2. □
Theorem 10
The girth of the graph
\mathrm{\Gamma}({G}_{2})
is equal to
gr(\mathrm{\Gamma}({G}_{2}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 4;& k,m,n>2.\end{array}
Proof By the considering the graph of the group {G}_{2}, we have twelve triangles and five squares for k=m=n=2 and k,m,n>2, respectively. By the definition of girth, this gives us the required result. □
Theorem 11
The chromatic number of the graph
\mathrm{\Gamma}({G}_{2})
is equal to
\chi (\mathrm{\Gamma}({G}_{2}))=\{\begin{array}{ll}3;& \mathit{\text{at least one of}}k,m,n=2,\\ 2;& k,m,n>2.\end{array}
Proof If the graph \mathrm{\Gamma}({G}_{2}) has one of the following forms: {\mathbb{Z}}_{2}\times ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{n}), {\mathbb{Z}}_{k}\times ({\mathbb{Z}}_{2}\ast {\mathbb{Z}}_{n}), {\mathbb{Z}}_{k}\times ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{2}), {\mathbb{Z}}_{2}\times ({\mathbb{Z}}_{2}\ast {\mathbb{Z}}_{n}), {\mathbb{Z}}_{2}\times ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{2}), {\mathbb{Z}}_{k}\times ({\mathbb{Z}}_{2}\ast {\mathbb{Z}}_{2}) or {\mathbb{Z}}_{2}\times ({\mathbb{Z}}_{2}\ast {\mathbb{Z}}_{2}), then we have similar neighbors for the graphs of each group. So, we can label the vertices with three different colors. If k,m,n\ne 2, then in the graph of {G}_{2} each vertex has two or four neighbors. In this graph, since the opposite vertices, which have an edge between them, can be labeled with the same color, we have two different colors. Hence \chi (\mathrm{\Gamma}({G}_{2}))=2. □
Theorem 12
The domination number of the graph
\mathrm{\Gamma}({G}_{2})
is
\gamma (\mathrm{\Gamma}({G}_{2}))=\{\begin{array}{ll}2;& k,m,n=2,\\ 4;& k,m,n>2.\end{array}
Proof Firstly, we consider the case k=m=n=2. Since the vertices {v}_{7}=xa and {v}_{12}=xb are connected with all other vertices in the graph of {G}_{2}, we can take the dominating set as \{{v}_{7},{v}_{12}\}. Thus \gamma (\mathrm{\Gamma}({G}_{2}))=2. In the case k,m,n\ne 2, since the number of edges is decreasing, the number of connected edges is decreasing as well. In this case, let us choose the dominating set as \{{v}_{4},{v}_{5},{v}_{7},{v}_{12}\}. Every vertex, except the vertices in the dominating set, is joined to at least one vertex of this dominating set by an edge. Therefore we have \gamma (\mathrm{\Gamma}({G}_{2}))=4. □
Theorem 13
The clique number of
\mathrm{\Gamma}({G}_{2})
is equal to
\omega (\mathrm{\Gamma}({G}_{2}))=\{\begin{array}{ll}3;& \mathit{\text{at least one of}}k,m,n=2,\\ 2;& k,m,n>2.\end{array}
Proof In the graph \mathrm{\Gamma}({G}_{2}), for k=m=n=2, we have twelve complete subgraphs. These are obtained by the vertices {v}_{1}{v}_{2}{v}_{7}, {v}_{2}{v}_{3}{v}_{7}, {v}_{3}{v}_{4}{v}_{7}, {v}_{4}{v}_{5}{v}_{7}, {v}_{5}{v}_{6}{v}_{7}, {v}_{4}{v}_{5}{v}_{12}, {v}_{4}{v}_{8}{v}_{12}, {v}_{8}{v}_{9}{v}_{12}, {v}_{9}{v}_{10}{v}_{12}, {v}_{10}{v}_{11}{v}_{12} and {v}_{5}{v}_{11}{v}_{12}. Hence \omega (\mathrm{\Gamma}({G}_{2}))=3. If k,m,n\ne 2, then we can find the smallest complete subgraphs as edges obtained by any two vertices in the graph \mathrm{\Gamma}({G}_{2}). So, \omega (\mathrm{\Gamma}({G}_{2}))=2. □
Theorem 14
The degree sequence and irregularity index of
\mathrm{\Gamma}({G}_{2})
are given by
DS(\mathrm{\Gamma}({G}_{2}))=\{\begin{array}{ll}(3,3,3,5,5,3,6,3,3,3,3,6);& k,m,n=2,\\ (2,2,2,4,4,2,4,2,2,2,2,4);& k,m,n>2\end{array}
and
t(\mathrm{\Gamma}({G}_{2}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 2;& k,m,n>2,\end{array}
respectively.
Proof It is easily seen by the graph \mathrm{\Gamma}({G}_{2}). □
Example 1 Let us consider the group G={\mathbb{Z}}_{3}\times ({\mathbb{Z}}_{3}\ast {\mathbb{Z}}_{3}) presented by
{\mathcal{P}}_{G}=\u3008x,a,b;{x}^{3}=1,{a}^{3}=1,{b}^{3}=1,xa=ax,xb=bx\u3009,
and x>a>b, the graph \mathrm{\Gamma}(G) as drawn in Figure 1(b), with the vertex set
V(\mathrm{\Gamma}(G))=\{x,a,b,{x}^{3},{a}^{3},{b}^{3},{x}^{3}a,{x}^{3}b,x{a}^{3},x{b}^{3},xa,xb\}.
By the result of Theorems 8 and 14, we have diam(\mathrm{\Gamma}(G))=4, \mathrm{\Delta}(\mathrm{\Gamma}(G))=4, \delta (\mathrm{\Gamma}(G))=2, gr(\mathrm{\Gamma}(G))=4, \chi (\mathrm{\Gamma}(G))=2, \gamma (\mathrm{\Gamma}(G))=4, \omega (\mathrm{\Gamma}(G))=2, DS(\mathrm{\Gamma}(G))=(2,2,2,4,4,2,4,2,2,2,2,4) and t(\mathrm{\Gamma}(G))=2.
2.3 Case 3: the graph \mathrm{\Gamma}({G}_{3}), where {G}_{3}={\mathbb{Z}}_{k}\ast ({\mathbb{Z}}_{m}\times {\mathbb{Z}}_{n})
If we consider the graph of the group {G}_{3}, then we have a subgraph of Figure 1(a) with vertices {v}_{1}={a}^{m}, {v}_{2}=a, {v}_{3}=a{b}^{n}, {v}_{4}={b}^{n}, {v}_{5}=b, {v}_{6}={a}^{m}b, {v}_{7}=ab, {v}_{9}=x and {v}_{10}={x}^{k}. If we take k,m,n=2, then by the edge definition, we have the edges {\alpha}_{j}, 1\le j\le 12 and {\alpha}_{15} in this subgraph of Figure 1(a). In the case k,m,n>2, we do not have edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{9}, {\alpha}_{12}, {\alpha}_{15} in the graph \mathrm{\Gamma}({G}_{3}).
Theorem 15
The maximum and minimum degrees of the graph
\mathrm{\Gamma}({G}_{3})
are
\mathrm{\Delta}(\mathrm{\Gamma}({G}_{3}))=\{\begin{array}{ll}6;& k,m,n=2,\\ 4;& k,m,n>2\end{array}
and
\delta (\mathrm{\Gamma}({G}_{3}))=\{\begin{array}{ll}1;& k,m,n=2,\\ 0;& k,m,n>2,\end{array}
respectively.
Proof Let us consider the graph \mathrm{\Gamma}({G}_{3}) and take k,m,n=2. In this case, the vertex {v}_{7} has the maximum degree six and the vertices {v}_{9} and {v}_{10} have the minimum degree one. But if we take k,m,n>2, then since there do not exist the edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{9}, {\alpha}_{12} and {\alpha}_{15} in the graph \mathrm{\Gamma}({G}_{3}), we obtain the maximum degree four by the vertex {v}_{7} and the minimum degree zero by the vertices {v}_{9} and {v}_{10}. □
Theorem 16
The girth of the graph
\mathrm{\Gamma}({G}_{3})
is
gr(\mathrm{\Gamma}({G}_{3}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 4;& k,m,n>2.\end{array}
Proof Firstly, we take account of the case k=m=n=2. In this case, we have six triangles which have the edges {\alpha}_{1}{\alpha}_{7}{\alpha}_{8}, {\alpha}_{2}{\alpha}_{8}{\alpha}_{9}, {\alpha}_{3}{\alpha}_{9}{\alpha}_{10}, {\alpha}_{4}{\alpha}_{10}{\alpha}_{11}, {\alpha}_{5}{\alpha}_{11}{\alpha}_{12}, {\alpha}_{6}{\alpha}_{7}{\alpha}_{12} in the graph \mathrm{\Gamma}({G}_{3}). Thus gr(\mathrm{\Gamma}({G}_{3}))=3. Now we consider the case k,m,n>2. In this case, since we do not have the edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{9}, {\alpha}_{12} and {\alpha}_{15}, we have two squares which have the edges {\alpha}_{1}{\alpha}_{6}{\alpha}_{5}{\alpha}_{7} and {\alpha}_{2}{\alpha}_{3}{\alpha}_{4}{\alpha}_{7} in the graph \mathrm{\Gamma}({G}_{3}). Therefore gr(\mathrm{\Gamma}({G}_{3}))=4. □
Theorem 17
The chromatic number of the graph
\mathrm{\Gamma}({G}_{3})
is
\chi (\mathrm{\Gamma}({G}_{3}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 2;& k,m,n>2.\end{array}
Proof Let us take k=m=n=2. In the graph \mathrm{\Gamma}({G}_{3}), since the vertex {v}_{7} is connected with all vertices except the vertices {v}_{9} and {v}_{10}, this vertex must be labeled by a different color than other vertices. In addition, since other vertices are connected with each other doubly, they can be labeled by two different colors. This gives us \chi (\mathrm{\Gamma}({G}_{3}))=3. In the case k,m,n>2, since we have two squares, as in the previous proof, in the graph \mathrm{\Gamma}({G}_{3}), it is enough to label two adjacent vertices by different colors. Hence \chi (\mathrm{\Gamma}({G}_{3}))=2. □
Theorem 18 The domination number of the graph \mathrm{\Gamma}({G}_{3}) is equal to infinity.
Proof For all cases of k, m, n, since the graph \mathrm{\Gamma}({G}_{3}) is disconnected, we get \gamma (\mathrm{\Gamma}({G}_{1}))=\mathrm{\infty}. □
Theorem 19
The clique number of the graph
\mathrm{\Gamma}({G}_{3})
is equal to
\omega (\mathrm{\Gamma}({G}_{3}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 2;& k,m,n>2.\end{array}
Proof For the case k=m=n=2, we have six maximal complete subgraphs of the graph \mathrm{\Gamma}({G}_{3}) which are triangles as in the proof of Theorem 16. Thus the largest number of the vertices in any maximal complete subgraph is three. If we take k,m,n>2, then we get eight maximal complete subgraphs, namely {\alpha}_{2}, {\alpha}_{3}, {\alpha}_{5}, {\alpha}_{6}, {\alpha}_{7}, {\alpha}_{8}, {\alpha}_{10} and {\alpha}_{11}, having two vertices. So, \omega (\mathrm{\Gamma}({G}_{3}))=2. □
Theorem 20
The degree sequence and the irregularity index of
\mathrm{\Gamma}({G}_{3})
are given by
DS(\mathrm{\Gamma}({G}_{3}))=\{\begin{array}{ll}(3,3,3,3,3,3,6,1,1);& k,m,n=2,\\ (2,2,2,2,2,2,4,0,0);& k,m,n>2\end{array}
and
t(\mathrm{\Gamma}({G}_{3}))=3;\phantom{\rule{1em}{0ex}}k,m,n\ge 2,
respectively.
Proof It is easily seen by the graph of the group {G}_{3}. □
2.4 Case 4: the graph \mathrm{\Gamma}({G}_{4}), where {G}_{4}=({\mathbb{Z}}_{k}\ast {\mathbb{Z}}_{l})\ast ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{n})
If we consider the graph of the group {G}_{4}, then we get a subgraph \mathrm{\Gamma}({G}_{4}) of the graph in Figure 1(a) with vertices {v}_{1}={a}^{m}, {v}_{2}=a, {v}_{4}={x}^{k}, {v}_{5}=x, {v}_{9}=b, {v}_{10}={b}^{n}, {v}_{14}={y}^{l} and {v}_{15}=y. If we take k,l,m,n=2, then by Section 1.1, we obtain the edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{15} and {\alpha}_{26} in \mathrm{\Gamma}({G}_{4}). For the case k,l,m,n>2, we do not have any edges. On the other hand, since at most three orders of factor groups of {G}_{4} are equal to two, we have at most three edges between {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{15} and {\alpha}_{26}.
Since the proof of each condition of the next result is quite similar to the related results over the group {G}_{1}={\mathbb{Z}}_{k}\ast ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{n}) in Case 1, we omit it.
Theorem 21 Let us consider the group {G}_{4}=({\mathbb{Z}}_{k}\ast {\mathbb{Z}}_{l})\ast ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{n}) with its subgraph \mathrm{\Gamma}({G}_{4}) as defined in the first paragraph of this case.

(i)
The maximum and minimum degrees of the graph
\mathrm{\Gamma}({G}_{4})
are
\mathrm{\Delta}(\mathrm{\Gamma}({G}_{4}))=\{\begin{array}{ll}1;& \mathit{\text{at least one of}}k,m,n=2,\\ 0;& k,m,n>2\end{array}
and
\delta (\mathrm{\Gamma}({G}_{4}))=\{\begin{array}{ll}1;& k,m,n=2,\\ 0;& k,m,n>2,\end{array}
respectively.

(ii)
For any k, l, m, n different from one by considering a subgraph of Figure 1(a), the girth of the graph \mathrm{\Gamma}({G}_{4}) is gr(\mathrm{\Gamma}({G}_{4}))=\mathrm{\infty}.

(iii)
The chromatic number of the graph
\mathrm{\Gamma}({G}_{4})
is equal to
\chi (\mathrm{\Gamma}({G}_{4}))=\{\begin{array}{ll}2;& \mathit{\text{at least one of}}k,m,n=2,\\ 1;& k,m,n>2.\end{array}

(iv)
The clique number of
\mathrm{\Gamma}({G}_{4})
is equal to
\omega (\mathrm{\Gamma}({G}_{4}))=\{\begin{array}{ll}2;& k,m,n=2,\\ 1;& k,m,n>2.\end{array}

(v)
The domination number of \mathrm{\Gamma}({G}_{4}) is equal to infinity.

(vi)
The degree sequence and the irregularity index of
\mathrm{\Gamma}({G}_{4})
are given by
DS(\mathrm{\Gamma}({G}_{4}))=\{\begin{array}{ll}(1,1,1,1,1,1,1,1);& k,m,n=2,\\ (0,0,0,0,0,0,0,0);& k,m,n>2,\\ ({i}_{1},{i}_{2},{i}_{3},{i}_{4},{i}_{5},{i}_{6},{i}_{7},{i}_{8});& \mathit{\text{otherwise}},\end{array}
where {i}_{j}=0,1 (1\le j\le 8) and
t(\mathrm{\Gamma}({G}_{4}))=\{\begin{array}{ll}1;& k,m,n\ge 2,\\ 2;& \mathit{\text{otherwise}},\end{array}
respectively.
2.5 Case 5: the graph \mathrm{\Gamma}({G}_{5}), where {G}_{5}=({\mathbb{Z}}_{k}\ast {\mathbb{Z}}_{l})\times ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{n})
Similarly as in Case 4, for the group {G}_{5}, we obtain a subgraph \mathrm{\Gamma}({G}_{5}) of the graph in Figure 1(a) having vertices {v}_{1}={a}^{m}, {v}_{2}=a, {v}_{3}={x}^{k}a, {v}_{4}={x}^{k}, {v}_{5}=x, {v}_{6}=x{a}^{m}, {v}_{7}=xa, {v}_{8}={x}^{k}b, {v}_{9}=b, {v}_{10}={b}^{n}, {v}_{11}=x{b}^{n}, {v}_{12}=xb, {v}_{13}={y}^{l}b, {v}_{14}={y}^{l}, {v}_{15}=y, {v}_{16}=y{b}^{n}, {v}_{17}=yb, {v}_{18}={y}^{l}a, {v}_{19}={v}_{2}=a, {v}_{20}={v}_{1}={a}^{m}, {v}_{21}=y{a}^{m} and {v}_{22}=ya. In this graph, the edge set depends on the orders of factor groups of {G}_{5}. If we take k,l,m,n=2, then, by the adjacency definition in Section 1.1, we have the edges {\alpha}_{j}, 1\le j\le 45 with {\alpha}_{1}={\alpha}_{37}. For the case k,l,m,n>2, we do not have any edges {\alpha}_{1}, {\alpha}_{4}, {\alpha}_{9}, {\alpha}_{12}, {\alpha}_{15}, {\alpha}_{20}, {\alpha}_{23}, {\alpha}_{26}, {\alpha}_{31}, {\alpha}_{34}, {\alpha}_{37}={\alpha}_{1}, {\alpha}_{42}, {\alpha}_{45} in \mathrm{\Gamma}({G}_{5}).
In the following result (Theorem 22 below), we again omit the proof of it as in Theorem 21 since it is quite similar to the related results over the group {G}_{2}={\mathbb{Z}}_{k}\times ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{n}) in Case 2.
Theorem 22 Let us consider the group {G}_{5}=({\mathbb{Z}}_{k}\ast {\mathbb{Z}}_{l})\times ({\mathbb{Z}}_{m}\ast {\mathbb{Z}}_{n}) with its related graph \mathrm{\Gamma}({G}_{5}) as defined in the first paragraph of Case 5.

(i)
The maximum and minimum degrees of the graph
\mathrm{\Gamma}({G}_{5})
are equal to
\mathrm{\Delta}(\mathrm{\Gamma}({G}_{5}))=\{\begin{array}{ll}6;& k,m,n=2,\\ 4;& k,m,n>2\end{array}
and
\delta (\mathrm{\Gamma}({G}_{5}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 2;& k,m,n>2,\end{array}
respectively.

(ii)
The girth of the graph
\mathrm{\Gamma}({G}_{5})
is
gr(\mathrm{\Gamma}({G}_{5}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 4;& k,m,n>2.\end{array}

(iii)
The chromatic number of the graph
\mathrm{\Gamma}({G}_{5})
is
\chi (\mathrm{\Gamma}({G}_{5}))=\{\begin{array}{ll}3;& \mathit{\text{at least one of}}k,m,n=2,\\ 2;& k,m,n>2.\end{array}

(iv)
The domination number of the graph
\mathrm{\Gamma}({G}_{5})
is
\gamma (\mathrm{\Gamma}({G}_{5}))=\{\begin{array}{ll}4;& k,m,n=2,\\ 6;& k,m,n>2.\end{array}

(v)
The clique number of the graph
\mathrm{\Gamma}({G}_{5})
is
\omega (\mathrm{\Gamma}({G}_{5}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 2;& k,m,n>2.\end{array}

(vi)
The degree sequence and the irregularity index of
\mathrm{\Gamma}({G}_{5})
are given by
DS(\mathrm{\Gamma}({G}_{5}))=\{\begin{array}{ll}(5,5,3,5,5,3,6,3,5,5,3,6,3,5,5,3,6,3,5,5,3,6);& k,m,n=2,\\ (4,4,2,4,4,2,4,2,4,4,2,4,2,4,4,2,4,2,4,4,2,4);& k,m,n>2\end{array}
and
t(\mathrm{\Gamma}({G}_{5}))=\{\begin{array}{ll}3;& k,m,n=2,\\ 2;& k,m,n>2,\end{array}
respectively.