Assume to the contrary that |C(L)| ≥ 8. Then the induced subgraph of Γ₂(L) with vertex set C(L) is isomorphic to K₈, which is a contradiction.

By Lemma 2.1., it is sufficient for us to probe the toroidality of the graph Γ₂(L) in the cases in which the size of C(L) is 3, 4, 5, 6 or 7. In this paper, we discuss on the case that |C(L)| = 3. First we begin with the following notation.

Set n:=∣∪t=13St∣. Then we have the following cases:

• Case 1. |S_i| ≠ n − 2, for i = 1, 2, 3. Since the contraction of Γ₂(L) contains a subgraph isomorphic to K_3,7 or K_4,5, one can conclude that the graph Γ₂(L) is not toroidal.

• Case 2. There exists 1 ≤ i ≤ 3, such that |S_i| = n − 2. If S_jk is an empty set, for j, k ∉ {i} with 1 ≤ i, j, k ≤ 3, then Γ₂(L) is planar, which is not a toroidal graph. If S_jk ≠ Ø, for j, k ∉ {i} with 1 ≤ i, j, k ≤ 3, then we can find a copy of K_3,8 in the contraction of Γ₂(L), and thus the graph Γ₂(L) is not toroidal.

Now, by Lemma 2.4., we state necessary and sufficient conditions for toroidality of the graph Γ₂(L), when |C(L)| = 3. It should be noted that in the proof of following theorem, according to Remark 2.3., the cases where the graph Γ₂(L) is planar is ignored.

If one of the above statements holds, then one can easily check that Γ₂(L) is a toroidal graph.

Conversely, let Γ₂(L) be toroidal. By Lemma 2.4., 5≤ ∣∪t=13St∣ ≤9. Thus we have the following situations:

(i) ∣∪t=13St∣ =5.

Assume that there is some i, say i = 1, such that |S_i| = 3. If S₂₃ is non-empty, then the contraction of Γ₂(L) contains a subgraph isomorphic to K_3,3, and so it is not planar. Also when |S₂₃| ≥ 5, we have a copy of K_3,7 in the contraction of Γ₂(L) with vertex set {a₁, a₂, a₃} ∪ {b, c, s₁, s₂, …, s₅}, where a₁, a₂, a₃ ∈ S₁, b ∈ S₂, c ∈ S₃ and s₁, s₂, …, s₅ ∈ S₂₃. It is clear that the graph Γ₂(L) is not toroidal. Therefore, we may assume that 1 ≤ |S₂₃| ≤ 4. In this situation, the complement of Γ₂(L) contains C603, one of the listed graphs in [4], which is pictured in Figure 1. In Figure 1, we replace x₁, x₂, …, x₉ by a₁, a₂, a₃, b, s₁, s₂, s₃, s₄, c, respectively, which a₁, a₂, a₃ ∈ S₁, b ∈ S₂, c ∈ S₃ and s₁, s₂, s₃, s₄ ∈ S₂₃. Therefore, Γ₂(L) is a toroidal graph (see Figure 2).

Now, if there is a unique i, say i = 1, such that |S_i| = 1, and the sets S₁₂ and S₁₃ are non-empty, then we can find a subdivision of K₅ in the structure of the contraction of Γ₂(L) as it is shown in Figure 3, where a ∈ S₁, b₁, b₂ ∈ S₂, c₁, c₂ ∈ S₃, s₁₂ ∈ S₁₂ and s₁₃ ∈ S₁₃. Thus Γ₂ (L) is toroidal.

(ii) ∣∪t=13St∣ =6.

Suppose that there exists only one i, say i = 1, such that |S_i| = 4. If the size of S₂₃ is at least 3, then one can easily observe that the contraction of Γ₂(L) contains a subgraph isomorphic to K_4,5. Hence the graph Γ₂(L) is not toroidal. So, for toroidality, the size of S₂₃ is necessarily 1 or 2. In this case, Γ₂(L) is contained in K₈ (K₃ ∪ K₂) (cf. [4, p.55]). Hence Γ₂(L) is a toroidal graph (see Figure 4). In Figure 4, we assume that a₁, a₂, a₃, a₄ ∈ S₁, b ∈ S₂, c ∈ S₃ and s₂₃, s23′∈S23.

Now, assume that there exists a unique i, say i = 1, such that |S_i| = 3. If S₂₃ has at least 4 elements, then the contraction of Γ₂(L) contains a copy of K_3,7, and hence Γ₂(L) is not a toroidal graph. Also if |S₂₃| = 3 and S₁₃ is non-empty, then the complement of the contraction of Γ₂(L) is contained in U6.6b, one of the listed graphs in [4]. So the graph Γ₂(L) is not toroidal (see Figure 5). In Figure 5, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂ ∈ S₂, c ∈ S₃, s₁₃ ∈ S₁₃ and s₂₃, s23′,s23″∈S23.

Therefore, for toroidality of Γ₂(L), when |S₂₃| = 3, necessarily, S₁₃ = Ø. In this situation, the complement of Γ₂(L) contains C610, one of the listed graphs in [4] (see Figure 6). In Figure 6, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂ ∈ S₂, c ∈ S₃ and s₂₃, s23′,s23″∈S23. In this situation, the embedding of the graph Γ₂(L) in the torus is pictured in Figure 7.

In addition, if |S₂₃| ≤ 2, then the contraction of Γ₂(L) contains a copy of K_3,3, and so Γ₂(L) is not planar. In this situation, Γ₂(L) is contained in K₈ (K₃ ∪ K₂), (cf. [4, p.55]). Therefore, Γ₂(L) is a toroidal graph (see Figure 8). In Figure 8, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂ ∈ S₂, c ∈ S₃, s₁₃ ∈ S₁₃ and s₂₃, s23′∈S23.

Finally, suppose that |S_i| = 2, for all 1 ≤ i ≤ 3 and only one of the sets S₁₂, S₁₃ or S₂₃ is non-empty. Then it is easy to check that the contraction of Γ₂(L) contains K_3,3 and so it is not planar. Now we assume that at least one of the sets S₁₂, S₁₃ or S₂₃ are non-empty sets. Obviously, we can find a subdivision of K₆ in Γ₂(L) as it is shown in Figure 9, where a₁, a₂ ∈ S₁, b₁, b₂ ∈ S₂, c₁, c₂ ∈ S₃, s₁₂ ∈ S₁₂, s₁₃ ∈ S₁₃ and s₂₃ ∈ S₂₃. Thus Γ₂ (L) is a toroidal graph.

(iii) ∣∪t=13St∣ =7.

First, suppose that |S_i| = 5, for some 1 ≤ i ≤ 3, without loss of generality, we may assume that i = 1. If S₂₃ has exactly 1 element, then Γ₂(L) is contained in K₈ (K₃∪K₂), (cf. [4, p.55]). Hence the graph Γ₂(L) is toroidal (see Figure 10). In Figure 10, we have the vertices a₁, a₂, a₃, a₄, a₅ ∈ S₁, b ∈ S₂, c ∈ S₃ and s₂₃ ∈ S₂₃.

We may assume that S₂₃ has at least 2 elements. In this case, the vertices of the set {a1,a2,…,a5}∪{b,c,s23,s23′} form a subgraph isomorphic to K_4,5 in the contraction of Γ₂(L), where a₁, a₂, …, a₅ ∈ S₁, b ∈ S₂, c ∈ S₃ and s₂₃, s23′∈S23. Therefore, Γ₂(L) is not a toroidal graph.

Suppose that there is i with 1 ≤ i ≤ 3, say 1, such that |S_i| = 4. When S₂₃ is a singleton set, the graph Γ₂(L) is contained in K₈ (K₃ ∪ K₂) (cf. [4, p.55]). So the graph Γ₂(L) is toroidal (see Figure 11). In Figure 11, we have the vertices a₁, a₂, a₃, a₄ ∈ S₁, b₁, b₂ ∈ S₂, c ∈ S₃ and s₂₃ ∈ S₂₃.

As |S₂₃| ≥ 2, the contraction of Γ₂(L) contains a copy of K_4,5, which implies that Γ₂(L) is not a toroidal graph. If S₁ and S₂ have exactly 3 elements, and S₁₃ or S₂₃ has at least 3 elements, then we have a subgraph isomorphic to K_3,7 in the contraction of Γ₂(L), and so Γ₂(L) is not a toroidal graph. Therefore, we assume that |S₁₃| ≤ 2, S₂₃ = Ø or |S₂₃| ≤ 2, S₁₃ = Ø. Then the complement of Γ₂(L) contains C603, one of the listed graphs in [4] (see Figure 1). In Figure 1, we replace vertices x₁, x₂, …, x₉ by b₁, b₂, b₃, a₁, s₁₃, c, a₂, a₃, s13′, respectively, where a₁, a₂, a₃ ∈ S₁, b₁, b₂, b₃ ∈ S₂, c ∈ S₃ and s₁₃, s13′∈S13. Hence the graph Γ₂(L) is toroidal (see Figure 12).

In addition, we may assume S₁₃ and S₂₃ have 1 element, exactly. Then the complement of Γ₂(L) contains C402, one of the listed graphs in [4] (see Figure 13). In Figure 13, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂, b₃ ∈ S₂, c ∈ S₃, s₁₃ ∈ S₁₃ and s₂₃ ∈ S₂₃. So Γ₂(L) is a toroidal graph, which is pictured in Figure 14.

If one of the sets S₁₃ or S₂₃ has exactly 2 elements and the other one has only 1 element, then Γ₂(L) contains a subgraph isomorphic to G₃, one of the listed graphs in [12]. So Γ₂(L) is not a toroidal graph. To do this, in Figure 15, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂, b₃ ∈ S₂, c ∈ S₃, s₁₃, s13′∈S13 and s₂₃ ∈ S₂₃.

Now, consider the case that there is a unique i, say 1, such that |S_i| = 3. If S₂₃ has at least 3 elements, then the contraction of Γ₂(L) contains a copy of K_3,7, and so the graph Γ₂(L) is not toroidal. When |S₂₃| = 2, and also S₁₂ or S₁₃ is non-empty, the complement of the contraction of Γ₂(L) is contained in U6.6b, one of the listed graphs in [4] (see Figure 16). To do this, in Figure 16, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂ ∈ S₂, c₁, c₂ ∈ S₃, s₂₃, s23′∈S23 and s₁₂ ∈ S₁₂. So the graph Γ₂(L) is not toroidal.

Hence we assume that |S₂₃| = 2 and S₁₂ = S₁₃ = Ø. Then the complement of Γ₂(L) contains C603, one of the listed graphs in [4] (see Figure 1). In Figure 1, we replace the vertices x₁, x₂, …, x₉ by a₁, a₂, a₃, b₁, s₂₃, c₁, b₂, s23′, c₂, respectively, where a₁, a₂, a₃ ∈ S₁, b₁, b₂ ∈ S₂, c₁, c₂ ∈ S₃ and s₂₃, s23′∈S23. Hence Γ₂(L) is a toroidal graph, which is pictured in Figure 17.

When S₂₃ is a singleton set, Γ₂(L) is contained in K₈ (K₃∪K₂), (cf. [4, p.55]). Thus the graph Γ₂(L) is toroidal (see Figure 18). In Figure 18, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂ ∈ S₂, c₁, c₂ ∈ S₃, s₁₂ ∈ S₁₂, s₁₃ ∈ S₁₃ and s₂₃ ∈ S₂₃.

(iv) ∣∪t=13St∣ =8.

Suppose that there exists only one i, say i = 1, such that |S_i| = 6. If the size of S₂₃ is at least 2, then one can easily see that the contraction of Γ₂(L) contains a subgraph isomorphic to K_4,6. Hence Γ₂(L) is not a toroidal graph. So the size of S₂₃ is necessarily 1. In this situation, the complement of Γ₂(L) contains C603, one of the listed graphs in [4] (see Figure 1). To do this, in Figure 1, we replace the vertices x₁, x₂, …, x₉ by b, s₂₃, c, a₁, a₃, a₅, a₂, a₄, a₆, respectively, where a₁, a₂, …, a₆ ∈ S₁, b ∈ S₂, c ∈ S₃ and s₂₃ ∈ S₂₃. Thus Γ₂(L), which is pictured in Figure 19, is a toroidal graph.

Now, suppose that there exists some i, say i = 1, such that |S_i| = 5. If S₂₃ is non-empty, then the contraction of Γ₂(L) contains a copy of K_4,5, and so Γ₂(L) is not a toroidal graph. Otherwise, S₂₃ = Ø, and so Γ₂(L) is contained in K₈ (K₃ ∪ K₂), (cf. [4, p.55]). Therefore, Γ₂(L) is a toroidal graph (see Figure 20). In Figure 20, we have the vertices a₁, a₂, …, a₅ ∈ S₁, b₁, b₂ ∈ S₂ and c ∈ S₃.

Suppose that there exist unique i and j with 1 ≤ i, j ≤ 3, say i = 1 and j = 2, such that |S_i| = 4, |S_j| = 3. If S₁₃ is non-empty, then the complement of Γ₂(L) is contained in S5.5, one of the listed graphs in [4] (see Figure 21). In fact, in Figure 21, we have the vertices a₁, a₂, a₃, a₄ ∈ S₁, b₁, b₂, b₃ ∈ S₂, c ∈ S₃ and s₁₃ ∈ S₁₃. And so the graph Γ₂(L) is not toroidal.

Also if S₂₃ is non-empty, then one can easily see that the contraction of Γ₂(L) contains a copy of K_4,5. Hence Γ₂(L) is not toroidal. So the size of the sets S₁₃ and S₂₃ is 0. In this case, Γ₂(L) is contained in K₈ (K₃ ∪ K₂) (cf. [4, p.55]), which is a toroidal graph (see Figure 22). In Figure 22, we have the vertices a₁, a₂, a₃, a₄ ∈ S₁, b₁, b₂, b₃ ∈ S₂ and c ∈ S₃.

Suppose that S₁ and S₂ have exactly 3 elements. If |S₁₃| ≥ 2 or |S₂₃| ≥ 2, then it is easy to see that the contraction of Γ₂(L) contains a copy of K_3,7. Hence Γ₂(L) is not a toroidal graph. Also, if S₁₃ and S₂₃ have only 1 element, then Γ₂(L) contains a subgraph isomorphic to G₃, one of the listed graphs in [12]. So the graph Γ₂(L) not toroidal. To do this, in Figure 23, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂, b₃ ∈ S₂, c₁, c₂, ∈ S₃, s₁₃ ∈ S₁₃ and s₂₃ ∈ S₂₃.

So we may assume that S₁₂ is non-empty and also S₁₃ or S₂₃ has only 1 element. Then the complement of the contraction of Γ₂(L) is contained in S5.6, one of the listed graphs in [4] (see Figure 24). Hence the graph Γ₂(L) is not toroidal. In Figure 24, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂, b₃ ∈ S₂, c₁, c₂ ∈ S₃, s₁₂ ∈ S₁₂ and s₁₃ ∈ S₁₃.

Therefore if the size of the set S₁₃ or S₂₃ is 1, then necessarily S₁₂ must be an empty set. Since the complement of Γ₂(L) contains C402, one of the listed graphs in [4]. To do this, in Figure 25, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂, b₃ ∈ S₂, c₁, c₂ ∈ S₃, s₁₃ ∈ S₁₃. Hence Γ₂(L) is a toroidal graph (see Figure 26).

Consequently, two sets S₁₃ and S₂₃ are both empty. In this situation, Γ₂(L) is contained in K₈ (K₃ ∪ K₂) (cf. [4, p.55]), which is a toroidal graph. To do this, in Figure 27, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂, b₃ ∈ S₂, c₁, c₂ ∈ S₃ and s₁₂ ∈ S₁₂.

Now, suppose that S₁ and S₂ have 2 elements. If S₁₂ is non-empty, then the contraction of Γ₂(L) contains a copy of K_4,5, and thus the graph Γ₂(L) is not toroidal. So the set S₁₂ is necessarily empty. In this case, Γ₂(L) is contained in K₈ (K₃ ∪ K₂), (cf. [4, p.55]), which is a toroidal graph (see Figure 28). In Figure 28, we have the vertices a₁, a₂ ∈ S₁, b₁, b₂ ∈ S₂ and c₁, c₂, c₃, c₄ ∈ S₃.

(v) ∣∪t=13St∣ =9.

First, suppose that |S_i| = 7, for some 1 ≤ i ≤ 3. Without loss of generality, we may assume that i = 1. If S₂₃ is non-empty, then the contraction of Γ₂(L) contains a copy of K_3,7. Hence the graph Γ₂(L) is not toroidal.

Also suppose that there is some i with 1 ≤ i ≤ 3, say 1, such that |S_i| = 6 and S₂₃ is non-empty. Then we can find a copy of K_4,6 in the contraction of Γ₂(L). Therefore, Γ₂(L) is not a toroidal graph. So, for toroidality of Γ₂(L), it is sufficient S₂₃ = Ø, because in this situation the complement of Γ₂(L) contains C603, one of the listed graphs in [4]. To do this, in Figure 1, we replace vertices x₁, x₂, …, x₉ by b₁, b₂, c, a₁, a₃, a₅, a₂, a₄, a₆, respectively, where a₁, a₂, …, a₆ ∈ S₁, b₁, b₂ ∈ S₂, c ∈ S₃. The embedding of Γ₂(L) in the torus is pictured in Figure 29.

Now, suppose that all of the sets S₁, S₂ and S₃ have 3 elements, exactly. If S₁₂, S₁₃ and S₂₃ are empty, then the complement of Γ₂(L) contains C315, one of the listed graphs in [4] (see Figure 30). In Figure 30, we have the vertices a₁, a₂, a₃ ∈ S₁, b₁, b₂, b₃ ∈ S₂, c₁, c₂, c₃ ∈ S₃. Therefore, the graph Γ₂(L), which is pictured in Figure 31, is toroidal.

Otherwise, we may assume that at least one of the sets S₁₂, S₁₃ or S₂₃ is nonempty. Then the contraction of Γ₂(L) contains a copy of K_3,7. Thus Γ₂ (L) is not a toroidal graph. Otherwise, there exists some i, with 1 ≤ i ≤ 3, such that |S_i| = 4 or |S_i| = 5. In these situations, the contraction of Γ₂(L) contains a copy of K_4,5. Therefore, Γ₂(L) is not a toroidal graph.