Non-Solvable Graphs of Groups

Abstract

Let G be a group, and ${{\,\mathrm{Sol}\,}}(G)=\{x \in G : \langle x,y \rangle \text { is solvable for all } y \in G\}$. We associate a graph $\mathcal {NS}_G$ (called the non-solvable graph of G) with G whose vertex set is $G \setminus {{\,\mathrm{Sol}\,}}(G)$ and two distinct vertices are adjacent if they generate a non-solvable subgroup. In this paper, we study many properties of $\mathcal {NS}_G$. In particular, we obtain results on vertex degree, cardinality of vertex degree set, graph realization, domination number, vertex connectivity, independence number and clique number of $\mathcal {NS}_G$. We also consider two groups G and H having isomorphic non-solvable graphs and derive some properties of G and H. Finally, we conclude this paper by showing that $\mathcal {NS}_G$ is neither planar, toroidal, double toroidal, triple toroidal nor projective.

1 Introduction

Let G be a group (finite/infinite), and ${{\,\mathrm{Sol}\,}}(G)=\{x \in G : \langle x,y \rangle \text { is solvable for all } y \in G\}$. Note that ${{\,\mathrm{Sol}\,}}(G)= \underset{x \in G}{\cap } {{\,\mathrm{Sol}\,}}_G(x)$, where ${{\,\mathrm{Sol}\,}}_G(x) = \{g \in G : \langle g,x \rangle \text { is solvable}\}$, is called solvabilizer of x in G. In general, ${{\,\mathrm{Sol}\,}}_G(x)$ is not a subgroup of G. However, it was shown, in [10], that ${{\,\mathrm{Sol}\,}}(G)$ is the solvable radical of G if G is a finite group. That is, if G is a finite group, then ${{\,\mathrm{Sol}\,}}(G)$ is the unique largest solvable normal subgroup of G.

We associate a simple graph $\mathcal {NS}_G$ with G, called the non-solvable graph of G whose vertex set is $G \setminus {{\,\mathrm{Sol}\,}}(G)$ and two distinct vertices x and y are adjacent if and only if $\langle x,y \rangle$ is not solvable. The non-solvable graph of G was introduced in [13] and studied in [4, 13]. The complement of $\mathcal {NS}_G$, known as solvable graph or solubility graph of G, is considered in [5, 6] recently. The concept of non-solvable graph of a group is an extension of non-nilpotent graph and hence non-commuting graph of groups. The non-nilpotent and non-commuting graph of finite groups is studied extensively in [2, 18] and [1, 3, 8, 9, 19], respectively.

We write $V(\Gamma )$ to denote the vertex set of a graph $\Gamma$. The degree of a vertex $x \in V(\Gamma )$ denoted by $\deg _{\Gamma }(x)$ is defined to be the number of vertices adjacent to x, and $\deg (\Gamma ) = \{\deg _{\Gamma }(x) : x \in V(\Gamma )\}$ is the vertex degree set of $\Gamma$. In Sect. 2, we shall study some properties of degree of a vertex and vertex degree set of $\mathcal {NS}_G$. We also obtain some bounds for solvability degree of a finite group one of which is better than an existing bound, in particular [6, Theorem 4.3]. The solvability degree of a finite group G is the probability that a randomly chosen pair of elements of G generate a solvable group (see [6, 12, 22]).

It is known that $\mathcal {NS}_G$ is neither a tree nor a complete graph (see [4, 13]). In Sect. 3, we shall show that $\mathcal {NS}_G$ is not bipartite; more generally, it is not complete multi-partite. We shall also show that $\mathcal {NS}_G$ is Hamiltonian for some classes of finite groups. In Sects. 4–6, we shall obtain several results regarding domination number, vertex connectivity, independence number and clique number of $\mathcal {NS}_G$. In Sect. 7, we shall consider two groups G and H having isomorphic non-solvable graphs and derive some properties of G and H. It is shown in [13] that $\mathcal {NS}_G$ is not planar. In the last section, we shall show that the genus of $\mathcal {NS}_G$ is greater or equal to 4. Hence, $\mathcal {NS}_G$ is neither planar, toroidal, double toroidal nor triple toroidal. We conclude this paper by showing that $\mathcal {NS}_G$ is not projective.

Let U be a non-empty subset of the vertex set of a graph $\Gamma$. The induced subgraph of $\Gamma$ on U is defined to be the graph $\Gamma [U]$ in which the vertex set is U and the edge set consists precisely those edges in $\Gamma$ whose endpoints lie in U. For any non-empty subset S of $V(\Gamma )$, we also write $\Gamma \setminus S$ to denote $\Gamma [V(\Gamma )\setminus S]$.

2 Vertex Degree and Cardinality of Vertex Degree Set

The neighborhood of a vertex x in a graph $\Gamma$, denoted by ${{\,\mathrm{nbd}\,}}_{\Gamma }(x)$, is defined to be the set of all vertices adjacent to x and so $\deg _{\Gamma }(x) = |{{{\,\mathrm{nbd}\,}}}_{\Gamma }(x)|$. It is easy to see that $\deg _{{\mathcal {NS}}_G}(x) = |G|- |{{\,\mathrm{Sol}\,}}_G(x)|$ for any vertex x in the non-solvable graph ${\mathcal {NS}}_G$ of the group G. In [13], Hai-Reuven has shown that

$$\begin{aligned} 6 \le \deg _{{\mathcal {NS}}_G}(x) \le |G|- |{{\,\mathrm{Sol}\,}}(G)|- 2 \end{aligned}$$

(1)

for any $x \in G \setminus {{\,\mathrm{Sol}\,}}(G)$. In this section, we first obtain some bounds for $P_s(G)$ using (1). Recall that $P_s(G)$ is the solvability degree of a finite group G defined by the ratio

$$\begin{aligned} P_s(G) = \frac{|\{(u, v) \in G \times G : \langle u, v\rangle \text{ is } \text{ solvable }\}|}{|G|^2}. \end{aligned}$$

It is worth mentioning that several properties of $P_s(G)$ including some bounds are studied in [6, 12, 22]. The following result gives a connection between $P_s(G)$ and the number of edges in $\mathcal {NS}_G$.

Lemma 1

If G is a finite non-solvable group, then

$$\begin{aligned} \sum _{x\in G\setminus {{\,\mathrm{Sol}\,}}(G)} \deg _{{\mathcal {NS}}_G}(x)= |G|^2(1-P_s(G)). \end{aligned}$$

Proof

Let $U=\{(x,y)\in G\times G : \langle x,y \rangle \text{ is } \text{ not } \text{ solvable }\}$. Then,

$$\begin{aligned} |U|= |G\times G|- |\{(x,y)\in G\times G : \langle x,y \rangle \text{ is } \text{ solvable }\}|= |G|^2-P_s(G)|G|^2. \end{aligned}$$

Note that

$$\begin{aligned} |U|= 2\times \text{ number } \text{ of } \text{ edges } \text{ in } \mathcal {NS}_G=\sum _{x\in G\setminus {{\,\mathrm{Sol}\,}}(G)} \deg _{{\mathcal {NS}}_G}(x). \end{aligned}$$

Hence, the result follows. $\square$

Now we obtain the following bounds for $P_s(G)$.

Theorem 2

If G is a finite non-solvable group, then

$$\begin{aligned} \frac{2(|G|- |{{\,\mathrm{Sol}\,}}(G)|)}{|G|^2} + \frac{2|{{\,\mathrm{Sol}\,}}(G)|}{|G|} - \frac{|{{\,\mathrm{Sol}\,}}(G)|^2}{|G|^2} \le P_s(G) \le 1 - \frac{6(|G|-|{{\,\mathrm{Sol}\,}}(G)|)}{|G|^2}. \end{aligned}$$

Proof

By Lemma 1 and (1), we have

$$\begin{aligned} 6(|G|-|{{\,\mathrm{Sol}\,}}(G)|)\le |G|^2(1-P_s(G)) \le (|G|-|{{\,\mathrm{Sol}\,}}(G)|)(|G|-|{{\,\mathrm{Sol}\,}}(G)|-2), \end{aligned}$$

and hence, the result follows on simplification. $\square$

It was shown in [6, Theorem 4.3] that

$$\begin{aligned} P_s(G) \ge \frac{2(|G|-|{{\,\mathrm{Sol}\,}}(G)|)}{|G|^2} + \frac{|{{\,\mathrm{Sol}\,}}(G)|}{|G|}. \end{aligned}$$

(2)

Note that $\frac{|{{\,\mathrm{Sol}\,}}(G)|}{|G|} - \frac{|{{\,\mathrm{Sol}\,}}(G)|^2}{|G|^2} > 0$ for any finite non-solvable group G. Hence, the lower bound obtained in Theorem 2 for $P_s(G)$ is better than (2).

It was also shown, in [13], that $|\deg (\mathcal {NS}_G)|\ne 2$, where $\deg (\mathcal {NS}_G)$ is the vertex degree set of $\mathcal {NS}_G$. However, we observe that the cardinality of $\deg (\mathcal {NS}_G)$ may be equal to 3. In this section, we shall obtain a class of groups G such that $|\deg (\mathcal {NS}_G)|= 3$. Note that $\deg (\mathcal {NS}_{A_5}) = \{24, 36, 50\}$. More generally, we have the following result.

Proposition 3

Let S be any finite solvable group. Then, $|\deg (\mathcal {NS}_{A_5\times S})|= 3$.

The proof of Proposition 3 follows from the fact that $|\deg (\mathcal {NS}_{A_5})|= 3$ and the result given below.

Lemma 4

Let G be a finite non-solvable group and S be any finite solvable group. Then $|\deg (\mathcal {NS}_G)|= |\deg (\mathcal {NS}_{G \times S})|$.

Proof

Let $(x,s),(y,t)\in G\times S$, then $\langle (x,s),(y,t)\rangle \subseteq \langle x,y\rangle \times \langle s,t\rangle$. Therefore, $\langle (x,s),(y,t)\rangle$ is solvable if and only if $\langle x, y\rangle$ is solvable. Also, ${{\,\mathrm{Sol}\,}}_{G\times S}((x,s)) = {{\,\mathrm{Sol}\,}}_G(x)\times S$, and hence, ${{\,\mathrm{nbd}\,}}_{\mathcal {NS}_{G \times S}}((x, s)) = {{\,\mathrm{nbd}\,}}_{\mathcal {NS}_{G}}(x)\times S$. That is, $\deg _{\mathcal {NS}_{G \times S}}((x,s)) = |S|\deg _{\mathcal {NS}_{G}}(x)$. This completes the proof. $\square$

Now we state the main result of this section.

Theorem 5

If G is a finite non-solvable group such that $G/{{\,\mathrm{Sol}\,}}(G)\cong A_5$, then $|\deg (\mathcal {NS}_G)|= 3$.

To prove this theorem, we need the following results.

Lemma 6

Let H be a subgroup of a finite group G and $x, y \in G$.

• (a) If H is solvable, then $\langle H,{{\,\mathrm{Sol}\,}}(G)\rangle$ is also a solvable subgroup of G.

• (b) If $\langle x, y \rangle$ is solvable, then $\langle xu, yv \rangle$ is also solvable for all $u, v \in {{\,\mathrm{Sol}\,}}(G)$.

• (c) If $\langle x,y \rangle$ is not solvable, then $\langle xu, yv \rangle$ is not solvable for all $u, v \in {{\,\mathrm{Sol}\,}}(G)$.

Proof

Part (a) is proved in [6, Lemma 3.1]. Part (b) follows from part (a). Also, note that parts (b) and (c) are equivalent. $\square$

Lemma 7

Let G be a finite group and $x,y \in G$. Then, $\langle x{{\,\mathrm{Sol}\,}}(G),y{{\,\mathrm{Sol}\,}}(G) \rangle$ is solvable if and only if $\langle x, y \rangle$ is solvable.

Proof

Let $H = \langle x, y \rangle$ and $Z = {{\,\mathrm{Sol}\,}}(G)$. Note that $\langle xZ, yZ \rangle = \frac{HZ}{Z}$. Suppose $\langle xZ, yZ \rangle$ is solvable. Then, $\frac{HZ}{Z}$ is solvable. Therefore,

$$\begin{aligned} {{\,\mathrm{Sol}\,}}_{\frac{HZ}{Z}}(xZ)= \frac{HZ}{Z}. \end{aligned}$$

Let

$$\begin{aligned} \frac{{{\,\mathrm{Sol}\,}}_{HZ}(x)}{Z} := \{ gZ : g \in {{\,\mathrm{Sol}\,}}_{HZ}(x) \} = \{ gZ : \langle g, x \rangle \text { is solvable} \}. \end{aligned}$$

Since $Z\subset {{\,\mathrm{Sol}\,}}(HZ)$ and Z is a normal subgroup of HZ, by [5, Lemma 2.5], we have

$$\begin{aligned} \frac{{{\,\mathrm{Sol}\,}}_{HZ}(x)}{Z} = {{\,\mathrm{Sol}\,}}_{\frac{HZ}{Z}}(xZ). \end{aligned}$$

Therefore, $\frac{{{\,\mathrm{Sol}\,}}_{HZ}(x)}{Z}=\frac{HZ}{Z}$ and so ${{\,\mathrm{Sol}\,}}_{HZ}(x)=HZ$. In particular, ${{\,\mathrm{Sol}\,}}_{H}(x)=H$ and so H is solvable.

If H is solvable then, by Lemma 6(b), ${{\,\mathrm{Sol}\,}}_{HZ}(x)=HZ$ for all $x\in HZ$. Thus, HZ is solvable and so $\frac{HZ}{Z}$ is solvable. Hence, $\langle x_iZ,x_jZ \rangle$ is solvable for $x_i, x_j \in HZ$ and so $\langle xZ, yZ \rangle$ is solvable. $\square$

Proposition 8

Let G be a finite non-solvable group. Then, for all $x \in G \setminus {{\,\mathrm{Sol}\,}}(G)$ we have

$$\begin{aligned} \deg _{\mathcal {NS}_{G}}(x)=\deg _{\mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)}}(x{{\,\mathrm{Sol}\,}}(G))|{{\,\mathrm{Sol}\,}}(G)|. \end{aligned}$$

Proof

Let $y \in {{\,\mathrm{nbd}\,}}_{\mathcal {NS}_{G}}(x)$. By Lemma 6, we have $yz\in {{\,\mathrm{nbd}\,}}_{\mathcal {NS}_{G}}(x)$ for all $z \in {{\,\mathrm{Sol}\,}}(G)$. Thus, ${{\,\mathrm{nbd}\,}}_{\mathcal {NS}_{G}}(x)$ is a union of distinct cosets of ${{\,\mathrm{Sol}\,}}(G)$. Let ${{\,\mathrm{nbd}\,}}_{\mathcal {NS}_{G}}(x)= y_1{{\,\mathrm{Sol}\,}}(G)\cup y_2{{\,\mathrm{Sol}\,}}(G)\cup \dots \cup y_n{{\,\mathrm{Sol}\,}}(G)$. Then, $\deg _{\mathcal {NS}_{G}}(x) = n|{{\,\mathrm{Sol}\,}}(G)|$. By Lemma 7, we have $\langle x{{\,\mathrm{Sol}\,}}(G),y_i{{\,\mathrm{Sol}\,}}(G) \rangle$ is not solvable if and only if $\langle x,y_i \rangle$ is not solvable. Therefore, ${{\,\mathrm{nbd}\,}}_{\mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)}}(x{{\,\mathrm{Sol}\,}}(G)) = \{y_1{{\,\mathrm{Sol}\,}}(G), y_2{{\,\mathrm{Sol}\,}}(G), \dots , y_n{{\,\mathrm{Sol}\,}}(G)\}$ in $\mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)}$. Hence, $\deg _{\mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)}}(x{{\,\mathrm{Sol}\,}}(G)) = n$ and the result follows. $\square$

As a consequence of Proposition 8, we have the following corollary.

Corollary 9

Let G be a finite non-solvable group. Then,

$$\begin{aligned} |\deg (\mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)})|= |\deg (\mathcal {NS}_G)|. \end{aligned}$$

Proof of Theorem 5

Note that $G/{{\,\mathrm{Sol}\,}}(G)\cong A_5$ implies ${\mathcal {NS}}_{G/{{\,\mathrm{Sol}\,}}(G)} \cong {\mathcal {NS}}_{A_5}$. Therefore,

$$\begin{aligned} |\deg \left( \mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)}\right) |= |\deg (\mathcal {NS}_{A_5})|= 3. \end{aligned}$$

Hence, the result follows from Corollary 9. $\square$

We conclude this section with the following upper bound for $|\deg (\mathcal {NS}_G)|$.

Theorem 10

If G is a finite non-solvable group having n distinct solvabilizers, then $|\deg (\mathcal {NS}_G)|\le n - 1$.

Proof

Let $G, X_1, X_2, \dots , X_{n - 1}$ be the distinct solvabilizers of G, where ${{\,\mathrm{Sol}\,}}_G(x_i) = X_i$ for some $x_i \in G \setminus {{\,\mathrm{Sol}\,}}(G)$ and $i = 1, 2, \dots , n - 1$. Then,

$$\begin{aligned} \deg (\mathcal {NS}_G) = \{|G|- |X_1|, |G|- |X_2|, \dots , |G|- |X_{n - 1}|\}. \end{aligned}$$

Hence, the result follows. $\square$

3 Graph Realization

In [13], it was shown that $\mathcal {NS}_G$ is connected with diameter two. It is also shown that $\mathcal {NS}_G$ is not regular and hence not a complete graph. Recently, Akbari [4] have shown that $\mathcal {NS}_G$ is not a tree. In this section, we shall show that $\mathcal {NS}_G$ is not a complete multi-partite graph. We shall also show that $\mathcal {NS}_G$ is Hamiltonian for some groups. The following results are useful in this regard.

Lemma 11

[13] Let G be a finite group. Then, G is solvable if and only if ${{\,\mathrm{Sol}\,}}_G(x)$ is a subgroup of G for all $x\in G$.

Theorem 12

[10] Let G be a finite non-solvable group and $x, y \in G \setminus {{\,\mathrm{Sol}\,}}(G)$. Then, there exists $s \in G \setminus {{\,\mathrm{Sol}\,}}(G)$ such that $\langle x, s \rangle$ and $\langle y, s \rangle$ are not solvable.

Theorem 13

Let G be a finite non-solvable group. Then, $\mathcal {NS}_G$ is not a complete multi-partite graph. In particular, $\mathcal {NS}_G$ is not a complete bipartite graph.

Proof

Suppose $\mathcal {NS}_G$ is a complete multi-partite graph. Let $X_1, X_2, \dots , X_n$ be the partite sets. Let $x \in G \setminus {{\,\mathrm{Sol}\,}}(G)$, then $x \in X_i$ for some i and ${{\,\mathrm{Sol}\,}}_G(x) = {{\,\mathrm{Sol}\,}}(G) \cup X_i$. Let $y, z \in {{\,\mathrm{Sol}\,}}_G(x)$. Then, $\langle y,z \rangle$ is solvable and $yz \in {{\,\mathrm{Sol}\,}}_G(y)={{\,\mathrm{Sol}\,}}_G(x)$. Thus, ${{\,\mathrm{Sol}\,}}_G(x)$ is a subgroup of G. By Lemma 11, G is solvable, a contradiction. Hence, the result follows. $\square$

Theorem 14

Let G be a finite non-solvable group. Then, $\mathcal {NS}_G$ is not a bipartite graph.

Proof

Suppose $\mathcal {NS}_G$ is a bipartite graph. Let X, Y be the partite sets. Let $x \in X$ and $y \in Y$. Then, by Theorem 12, there exists $z \in G\setminus {{\,\mathrm{Sol}\,}}(G)$ such that $\langle x, z \rangle$ and $\langle y, z \rangle$ are not solvable. Therefore, $z \notin X \cup Y$, a contradiction. Hence, the result follows. $\square$

Theorem 15

Let G be a finite non-solvable group such that $|{{\,\mathrm{Sol}\,}}_G(x)|\le \frac{|G|}{2}$ for all $x\in G\setminus {{\,\mathrm{Sol}\,}}(G)$. Then, $\mathcal {NS}_G$ is Hamiltonian.

Proof

Note that $\deg _{\mathcal {NS}_G}(x) = |G|- |{{\,\mathrm{Sol}\,}}_G(x)|$ for all $x \in G \setminus {{\,\mathrm{Sol}\,}}(G)$. Since $|{{\,\mathrm{Sol}\,}}_G(x)|\le \frac{|G|}{2}$ for all $x \in G \setminus {{\,\mathrm{Sol}\,}}(G)$, we have $|G|\ge 2|{{\,\mathrm{Sol}\,}}_G(x)|$. Thus, it follows that

$$\begin{aligned} \deg _{\mathcal {NS}_G}(x) > (|G|- |{{\,\mathrm{Sol}\,}}(G)|)/2. \end{aligned}$$

Therefore, by Dirac’s theorem [7, p. 54], $\mathcal {NS}_G$ is Hamiltonian. $\square$

Corollary 16

The non-solvable graphs of the groups $PSL(3,2) \rtimes {\mathbb {Z}}_2$, $A_6$ and PSL(2, 8) are Hamiltonian.

Proof

The result follows from Theorem 15 using the fact that $|{{\,\mathrm{Sol}\,}}_G(x)|\le \frac{|G|}{2}$ for all $x \in G\setminus {{\,\mathrm{Sol}\,}}(G)$, where $G = PSL(3,2) \rtimes {\mathbb {Z}}_2, A_6$ and PSL(2, 8). $\square$

The following result shows that there is a group G with $|{{\,\mathrm{Sol}\,}}_G(x)|> |G|/2$ for some $x \in G\setminus {{\,\mathrm{Sol}\,}}(G)$ such that $\mathcal {NS}_G$ is Hamiltonian.

Proposition 17

The non-solvable graph of $A_5$ is Hamiltonian.

Proof

For any two vertices a and b, we write $a \sim b$ if a is adjacent to b. It can be verified that

$(1,5,4,3,2)\sim (1,3)(2,5)\sim (2,3,4)\sim (1,4)(3,5)\sim (2,5,4)\sim (1,2)(3,4)\sim (1,5,4)\sim (2,5)(3,4) \sim (1,3,5)\sim (1,4)(2,5)\sim (2,4,3)\sim (1,3)(4,5)\sim (1,2,5)\sim (1,4)(2,3)\sim (3,5,4)\sim (1,5)(2,4)\sim (1,2,3)\sim (1,5)(3,4)\sim (2,3,5)\sim (1,4,2)\sim (2,3)(4,5)\sim (1,5,2)\sim (2,4)(3,5)\sim (1,4,5)\sim (1,2)(3,5)\sim (1,3,4)\sim (1,2)(4,5)\sim (1,5,3)\sim (1,4,2,5,3)\sim (1,3,2)\sim (3,4,5)\sim (1,3)(2,4)\sim (2,5,3)\sim (1,2,4)\sim (1,5)(2,3)\sim (2,4,5)\sim (1,4,3)\sim (1,3,5,2,4)\sim (1,4,5,3,2)\sim (1,2,3,4,5)\sim (1,2,4,3,5) \sim (1,5,3,2,4)\sim (1,4,5,2,3)\sim (1,5,4,2,3)\sim (1,3,4,5,2)\sim (1,5,3,4,2)$ $\sim (1,3,2,4,5) \, \sim \, (1,3,2,5,4) \, \sim \, (1,2,4,5,3) \, \sim \, (1,2,5,4,3) \, \sim \, (1,5,2,3,4)\sim (1,2,3,5,4)\sim (1,4,3,2,5) \sim (1,4,3,5,2)\sim (1,3,4,2,5)\sim (1,4,2,3,5)\sim (1,5,2,4,3)$ $\sim (1,3,5,4,2)\sim (1,2,5,3,4)\sim (1,5,4,3,2)$

is a Hamiltonian cycle of $\mathcal {NS}_{A_5}$. Hence, $\mathcal {NS}_{A_5}$ is Hamiltonian. $\square$

We conclude this section with the following question.

Question 18

Is $\mathcal {NS}_G$ Hamiltonian for any finite non-solvable group G?

4 Domination Number and Vertex Connectivity

For a graph $\Gamma$ and a subset S of the vertex set $V(\Gamma )$, we write $N_\Gamma [S] = S \cup (\cup _{x \in S}{{\,\mathrm{nbd}\,}}_{\Gamma }(x))$. If $N_\Gamma [S] = V(\Gamma )$, then S is said to be a dominating set of $\Gamma$. The domination number of $\Gamma$, denoted by $\lambda (\Gamma )$, is the minimum cardinality of dominating sets of $\Gamma$. In this section, we shall obtain a few results regarding $\lambda (\mathcal {NS}_G)$.

Proposition 19

Let G be a finite non-solvable group. Then, $\lambda (\mathcal {NS}_G) \ne 1$.

Proof

Let $\{x\}$ be a dominating set for $\mathcal {NS}_G$. If ${{\,\mathrm{Sol}\,}}(G)$ contains a non-trivial element z, then xz is adjacent to x, a contradiction. Hence, $|{{\,\mathrm{Sol}\,}}(G)|= 1$.

If $o(x) \ne 2$, then x is adjacent to $x^{-1}$, which is a contradiction. Hence, $o(x) = 2$ and so $x\in P_2$, for some Sylow 2-subgroup $P_2$ of G. Since $|{{\,\mathrm{Sol}\,}}(G)|= 1$ and x is adjacent to all vertices of $\mathcal {NS}_G$, we have ${{\,\mathrm{Sol}\,}}_G(x) =\langle x\rangle$. Also, $P_2\subseteq {{\,\mathrm{Sol}\,}}_G(x)$ and so $P_2=\langle x\rangle$. If $Q_{2}$ is another Sylow 2-subgroup of G, then $|Q_2|= 2$ and so $\langle P_2, Q_2\rangle$ is a dihedral group and hence solvable. That is, x is not adjacent to $y\in Q_2$, $y\ne 1$, which is a contradiction. Thus, it follows that $P_2$ is normal in G. Let $g \in G\setminus P_2$. Then, $gxg^{-1}=x$, that is, $xg=gx$ and so $x\in Z(G)$, which is a contradiction. Hence, the result follows. $\square$

Using GAP [20], it can be seen that $\lambda (A_5) = \lambda (S_5)=4$. In fact, $\{(3,4,5), (1,2,3,4,5), (1,2,4,5,3), (1,5)(2,4)\}$ and $\{(4,5), (1,2)(3,4,5), (1,3)(2,4,5), (1,5)(2,4)\}$ are dominating sets for $A_5$ and $S_5$, respectively. At this point, we would like to ask the following question.

Question 20

Is there any finite non-solvable group G such that $\lambda (\mathcal {NS}_G)=2,3$?

Proposition 21

Let G be a non-solvable group. Then, a subset S of $V(\mathcal {NS}_G)$ is a dominating set if and only if ${{\,\mathrm{Sol}\,}}_G(S) \subset {{\,\mathrm{Sol}\,}}(G)\cup S$.

Proof

Suppose S is a dominating set. If $a \not \in {{\,\mathrm{Sol}\,}}(G)\cup S$ then, by definition of dominating set, there exists $x \in S$ such that $\langle x,a \rangle$ is not solvable. Thus, $a \not \in {{\,\mathrm{Sol}\,}}_G(S)$. It follows that ${{\,\mathrm{Sol}\,}}_G(S) \subset S \cup {{\,\mathrm{Sol}\,}}(G)$.

Now assume that ${{\,\mathrm{Sol}\,}}_G(S) \subset {{\,\mathrm{Sol}\,}}(G) \cup S$. If $a \not \in {{\,\mathrm{Sol}\,}}(G) \cup S$, then by hypothesis, $a \not \in {{\,\mathrm{Sol}\,}}_G(S)$. Therefore, a is adjacent to at least one element of S. This completes the proof. $\square$

The vertex connectivity of a connected graph $\Gamma$, denoted by $\kappa (\Gamma )$, is defined as the smallest number of vertices whose removal makes the graph disconnected. A subset S of the vertices of a connected graph $\Gamma$ is called a vertex cut set, if $\Gamma \setminus S$ is not connected, but $\Gamma \setminus H$ is connected for any proper subset H of S. We conclude this section with the following result on vertex cut set and vertex connectivity of $\mathcal {NS}_G$.

Proposition 22

Let G be a finite non-solvable group, and let S be a vertex cut set of $\mathcal {NS}_G$. Then, S is a union of cosets of ${{\,\mathrm{Sol}\,}}(G)$. In particular, $\kappa (\mathcal {NS}_G)=t|{{\,\mathrm{Sol}\,}}(G)|$, where $t > 1$ is an integer.

Proof

Let $a \in S$. Then, there exist two distinct components $G_1$, $G_2$ of $\mathcal {NS}_G \setminus S$ and two vertices $x \in V(G_1)$, $y \in V(G_2)$ such that a is adjacent to both x and y. By Lemma 6, x and y are also adjacent to az for any $z \in {{\,\mathrm{Sol}\,}}(G)$, and so $a{{\,\mathrm{Sol}\,}}(G) \subset S$. Thus, S is a union of cosets of ${{\,\mathrm{Sol}\,}}(G)$. Hence, $\kappa (\mathcal {NS}_G)= t|{{\,\mathrm{Sol}\,}}(G)|$, where $t \ge 1$ is an integer.

Suppose that $|S|= \kappa (\mathcal {NS}_G)$. It follows from the first part that $\kappa (\mathcal {NS}_G) = t |{{\,\mathrm{Sol}\,}}(G)|$ for some integer $t \ge 1$. If $t = 1$, then $S = b{{\,\mathrm{Sol}\,}}(G)$ for some element $b \in G\setminus {{\,\mathrm{Sol}\,}}(G)$. Therefore, there exist two distinct components $G_1$, $G_2$ of $\mathcal {NS}_G \setminus S$ and $r\in V(G_1)$, $s\in V(G_2)$ such that b is adjacent to both r and s. In other words, $\langle b, r\rangle$ and $\langle b,s\rangle$ are not solvable. Suppose that $o(b) \ne 2$. Then, the number of integers is less than o(b) and relatively prime to it is greater or equal to 2. Let $1 \ne i\in {\mathbb {N}}$ such that $\gcd (i, o(b)) = 1$. Then,

$$\begin{aligned} \langle b^i, r\rangle = \langle b,r\rangle \text { and } \langle b^i, s\rangle =\langle b,s\rangle . \end{aligned}$$

Therefore, $b^i$ is adjacent to both r and s. This is a contradiction since $b^i \notin b{{\,\mathrm{Sol}\,}}(G)$. Hence, $o(b) = 2$.

Suppose $x' \in V(G_1)$ and $y' \in V(G_2)$ are adjacent to bz for some $z \in {{\,\mathrm{Sol}\,}}(G)$. Then, by Lemma 6 (c), b is adjacent to $x'$ and $y'$. Again, by Lemma 7, $x'{{\,\mathrm{Sol}\,}}(G)$ and $y'{{\,\mathrm{Sol}\,}}(G)$ are adjacent to $b{{\,\mathrm{Sol}\,}}(G)$ in the graph $\mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)}$. That is, $g{{\,\mathrm{Sol}\,}}(G)$ and $b{{\,\mathrm{Sol}\,}}(G)$ are adjacent for all $g{{\,\mathrm{Sol}\,}}(G) \in V(\mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)})$. Therefore, $\{b{{\,\mathrm{Sol}\,}}(G)\}$ is a dominating set of $\mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)}$ and so $\lambda (\mathcal {NS}_{G/{{\,\mathrm{Sol}\,}}(G)}) = 1$. Hence, the result follows in view of Proposition 19. $\square$

5 Independence Number

A subset X of the vertices of a graph $\Gamma$ is called an independent set if the induced subgraph on X has no edges. The maximum size of an independent set in a graph $\Gamma$ is called the independence number of $\Gamma$ and it is denoted by $\alpha (\Gamma )$. In this section, we consider the following question.

Question 23

Suppose G is a non-solvable group such that $\mathcal {NS}_G$ has no infinite independent set. Is it true that $\alpha (\mathcal {NS}_G)$ is finite?

It is worth mentioning that Question 23 is similar to [1, Question 2.10] and [17, Question 3.17] where Abdollahi et al. and Nongsiang et al. considered non-commuting and non-nilpotent graphs of finite groups, respectively. Note that the group considered in [17] in order to answer [17, Question 3.17] negatively also gives negative answer to Question 23. However, the next theorem gives affirmative answer to Question 23 for some classes of groups.

Theorem 24

Let G be a non-solvable group such that $\mathcal {NS}_G$ has no infinite independent sets. If ${{\,\mathrm{Sol}\,}}(G)$ is a subgroup and G is an Engel, locally finite, locally solvable, a linear group or a 2-group then G is a finite group. In particular, $\alpha (\mathcal {NS}_G)$ is finite.

Note that Theorem 24 and its proof are similar to [1, Theorem 2.11] and [17, Theorem 3.18].

Proposition 25

Let G be a group. Then, for every maximal independent set S of $\mathcal {NS}_G$ we have

$$\begin{aligned} \underset{x\in S}{\cap }{{\,\mathrm{Sol}\,}}_G(x)=S\cup {{\,\mathrm{Sol}\,}}(G). \end{aligned}$$

Proof

The result follows from the fact that S is maximal and $S\cup {{\,\mathrm{Sol}\,}}(G)\subset {{\,\mathrm{Sol}\,}}_G(x)$ for all $x \in S$. $\square$

Remark 1

Let $R=\{(3,4,5),(1,4)(3,5),(2,5,3)\}\subset A_5$. Then, R is an independent set of $\mathcal {NS}_{A_5}$ and $\langle R\rangle \cong A_5$. This shows that a subgroup generated by an independent set may not be solvable. Also there exists a maximal independent set S, such that $R\subseteq S$. Since the edge set of $\mathcal {NS}_{A_5}$ is non-empty, we have $S\ne A_5\setminus {{\,\mathrm{Sol}\,}}(A_5)$, showing that $S\cup {{\,\mathrm{Sol}\,}}(A_5)$ is not a subgroup of $A_5$. Thus, for a finite non-solvable group G and a maximal independent set S of $\mathcal {NS}_G$, $S\cup {{\,\mathrm{Sol}\,}}(G)$ need not be a subgroup of G.

We conclude this section with the following result.

Theorem 26

The order of a finite non-solvable group G is bounded by a function of the independence number of its non-solvable graph. Consequently, given a non-negative integer k, there are at the most finitely many finite non-solvable groups whose non-solvable graphs have independence number k.

Proof

Let $x \in G$ such that $x, x^2 \not \in {{\,\mathrm{Sol}\,}}(G)$. Then, $x{{\,\mathrm{Sol}\,}}(G)\cup x^2{{\,\mathrm{Sol}\,}}(G)$ is an independent set of $\mathcal {NS}_G$. Thus, $|{{\,\mathrm{Sol}\,}}(G)|\le \frac{k}{2}$. Let P be a Sylow subgroup of G, then P is solvable. Thus, it follows that $P\setminus {{\,\mathrm{Sol}\,}}(G)$ is an independent set of $\mathcal {NS}_G$. Hence, $|P\setminus {{\,\mathrm{Sol}\,}}(G)|\le k$, that is, $|P |\le \frac{3k}{2}$. Since, the number of primes less or equal to $\frac{3k}{2}$ is at most $\frac{3k}{4}$, we have $|G|\le (\frac{3k}{2})^{\frac{3k}{4}}$. This completes the proof. $\square$

6 Clique Number of Non-Solvable Graphs

A subset of the vertex set of a graph $\Gamma$ is called a clique of $\Gamma$ if it consists entirely of pairwise adjacent vertices. The least upper bound of the sizes of all the cliques of $\Gamma$ is called the clique number of $\Gamma$ and is denoted by $\omega (\Gamma )$. Note that $\omega ({\tilde{\Gamma }}) \le \omega (\Gamma )$ for any subgraph ${\tilde{\Gamma }}$ of $\Gamma$.

Proposition 27

Let G be a finite non-solvable group.

• (a) If H is a non-solvable subgroup of G, then $\omega (\mathcal {NS}_H)\le \omega (\mathcal {NS}_G)$.

• (b) If N is a normal subgroup of G and $\frac{G}{N}$ is non-solvable, then $\omega (\mathcal {NS}_{\frac{G}{N}}) \le \omega (\mathcal {NS}_G)$. The equality holds when $N = {{\,\mathrm{Sol}\,}}(G)$.

Proof

Part (a) follows from the fact that $\mathcal {NS}_H$ is a subgraph of $\mathcal {NS}_G$. For part (b), we shall show that $\mathcal {NS}_{\frac{G}{N}}$ is isomorphic to a subgraph of $\mathcal {NS}_G$.

Let $V(\mathcal {NS}_{\frac{G}{N}})=\{x_1N,x_2N,\dots ,x_nN\}$ and $K=\{x_1,x_2,\dots ,x_n\}$. Then, for $x_iN \in V(\mathcal {NS}_{\frac{G}{N}})$, there exists $x_jN \in V(\mathcal {NS}_{\frac{G}{N}})$ such that $\langle x_iN,x_jN\rangle$ is not solvable. Let $H = \langle x_i, x_j\rangle$. Then,

$$\begin{aligned} \langle x_iN, x_jN \rangle = \frac{H N}{N}. \end{aligned}$$

Suppose H is solvable. Then, there exists a sub-normal series $\{1\} = H_0 \le H_1 \le H_2 \le \cdots \le H_n = H$, where $H_i$ is normal in $H_{i+1}$ and $\frac{H_{i+1}}{H_i}$ is abelian for all $i = 0, 1, \dots , n - 1$. Consider the series $N = H_0N/N \le H_1N/N \le \cdots \le H_nN/N = HN/N$. We have $H_iN/N$ is normal in $H_{i+1}N/N$, for if $aN\in H_iN/N$ and $bN\in H_{i+1}N/N$, where $a \in H_iN$ and $b \in H_{i+1}N$, then $bNaN(bN)^{-1}=bab^{-1}N \in H_iN/N$ since $bab^{-1}\in H_iN$ by [16, Theorem C.2.]. Therefore, $H_iN/N$ is normal in $H_{i+1}N/N$. Also,

$$\begin{aligned} \frac{H_{i+1}N/N}{H_iN/N} \cong \frac{H_{i+1}N}{H_iN}, \end{aligned}$$

which is abelian by [16, Theorem C.2.]. Therefore, $HN/N = \langle x_iN,x_jN \rangle$ is solvable, which is a contradiction. Therefore, H is non-solvable.

Let L be a graph such that $V(L)= K$ and two vertices x, y of L are adjacent if and only if xN and yN are adjacent in $\mathcal {NS}_{\frac{G}{N}}$. Then, L is a subgraph of $\mathcal {NS}_G[K]$ and hence a subgraph of $\mathcal {NS}_G$. Define a map $\phi : V(\mathcal {NS}_{\frac{G}{N}}) \rightarrow V(L)$ by $\phi (x_iN)=x_i$. Then, $\phi$ is one to one and onto. Also two vertices $x_iN$ and $x_jN$ are adjacent in $\mathcal {NS}_{\frac{G}{N}}$ if and only if $x_i$ and $x_j$ are adjacent in L. Thus, $\mathcal {NS}_{\frac{G}{N}} \cong L$.

If $N = {{\,\mathrm{Sol}\,}}(G)$ then, by Lemma 7, it follows that $\{x_1{{\,\mathrm{Sol}\,}}(G), x_2{{\,\mathrm{Sol}\,}}(G),\dots$, $x_t{{\,\mathrm{Sol}\,}}(G)\}$ is a clique of $\mathcal {NS}_\frac{G}{{{\,\mathrm{Sol}\,}}(G)}$ if and only if $\{x_1, x_2, \dots , x_t\}$ is a clique of $\mathcal {NS}_G$. Hence, $\omega (\mathcal {NS}_\frac{G}{{{\,\mathrm{Sol}\,}}(G)})= \omega (\mathcal {NS}_G)$. $\square$

Theorem 28

For any non-solvable group G and a solvable group S, we have

$$\begin{aligned} \omega (\mathcal {NS}_G)=\omega (\mathcal {NS}_{G\times S}). \end{aligned}$$

Proof

Suppose C is a clique of $\mathcal {NS}_G$. Let $a,b \in C$, then $\langle a,b \rangle$ is not solvable. Now, $\langle (a,e_s),(b,e_s)\rangle \cong \langle a,b \rangle$, where $e_s$ is the identity element of S, and so $\langle (a,e_s),(b,e_s)\rangle$ is not solvable. Thus, $C\times \{e_s\}$ is a clique of $\mathcal {NS}_{G\times S}$. Now suppose D is a clique of $\mathcal {NS}_{G\times S}$. Let $(x,s_1),(y,s_2)\in D$, where $x \ne y$. Then $\langle (x,s_1),(y,s_2)\rangle \subseteq \langle x,y\rangle \times \langle s_1,s_2\rangle$. Since S is solvable, we have $\langle s_1,s_2 \rangle$ is solvable. Since $\langle (x,s_1),(y,s_2)\rangle$ is not solvable, we have $\langle x,y \rangle$ is not solvable. Thus, $E=\{x : (x,s)\in D\}$ is a clique of $\mathcal {NS}_G$. Hence, the result follows noting that $|D|= |E|$. $\square$

The following lemma is useful in obtaining a lower bound for $\omega (\mathcal {NS}_G)$.

Lemma 29

Let G be a finite non-solvable group. Then, there exists an element $x \in G\setminus {{\,\mathrm{Sol}\,}}(G)$ such that o(x) is a prime greater or equal to 5.

Proof

Suppose that $1 \ne o(x) = 2^{\alpha }3^{\beta }$ for all $x \in G\setminus {{\,\mathrm{Sol}\,}}(G)$, where $\alpha$ and $\beta$ are non-zero integers. Then, $|G/{{\,\mathrm{Sol}\,}}(G)|= 2^{m}3^{n}$ for some non-zero integers m, n. Therefore, $G/{{\,\mathrm{Sol}\,}}(G)$ is solvable and so, by Lemma 7, G is solvable, a contradiction. This proves the existence of an element $x \in G\setminus {{\,\mathrm{Sol}\,}}(G)$ such that o(x) is a prime greater or equal to 5. $\square$

Proposition 30

Let G be a finite non-solvable group. Then, $\omega (\mathcal {NS}_G)\ge 6$.

Proof

By Lemma 29, we have an element $x \in G\setminus {{\,\mathrm{Sol}\,}}(G)$ such that o(x) is a prime greater or equal to 5. Let $y \in G\setminus {{\,\mathrm{Sol}\,}}(G)$ such that x is adjacent to y. Then, $\{x, y, xy, x^2y, x^3y, x^4y\}$ is a clique of $\mathcal {NS}_G$ and so $\omega (\mathcal {NS}_G)\ge 6$. $\square$

The following program in GAP [20] shows that

$$\begin{aligned} \omega (\mathcal {NS}_{A_5}) = \omega (\mathcal {NS}_{SL(2,5)}) = \omega (\mathcal {NS}_{{\mathbb {Z}}_2 \times A_5}) = 8 \text { and } \omega (\mathcal {NS}_{S_5}) = 16. \end{aligned}$$

Note that $A_5 = \text {SmallGroup}(60, 5)$, $SL(2,5) \, = \, \text {SmallGroup}(120, 5), \,\quad S_5=$

$\text {SmallGroup}(120, 34)$ and ${\mathbb {Z}}_2 \times A_5 = \text {SmallGroup}(120, 35)$. Also $G/{{\,\mathrm{Sol}\,}}(G) \cong A_5$ for $G = A_5, SL(2,5)$ and ${\mathbb {Z}}_2 \times A_5$.

The following program in GAP [20] shows that the clique number of $\mathcal {NS}_G$ for groups of order less or equal to 360 with $G/{{\,\mathrm{Sol}\,}}(G) \not \cong A_5$ is greater or equal to 9.

We conclude this section with the following conjecture.

Conjecture 31

Let G be a non-solvable group such that $\omega (\mathcal {NS}_G) = 8$. Then, $G/{{\,\mathrm{Sol}\,}}(G)\cong A_5$.

7 Groups with the Same Non-Solvable Graphs

In [15], Moghaddamfar et al. conjectured that if G and H are two non-abelian finite groups such that their non-commuting graphs are isomorphic then $|G|= |H|$. This conjecture was verified for several classes of finite groups in [1, 15]. However, the conjecture was refuted by Moghaddamfar [14] in the year 2006. Recently, Nongsiang and Saikia [17] posed similar conjecture for non-nilpotent graphs of finite groups. In this section, we consider the following problem.

Problem 32

Let G and H be two non-solvable groups such that $\mathcal {NS}_G \cong \mathcal {NS}_H$. Determine whether $|G|= |H|$.

We begin the section with the following theorem.

Theorem 33

Let G and H be two non-solvable groups such that $\mathcal {NS}_G \cong \mathcal {NS}_H$. If G is finite, then H is also finite. Moreover, $|{{\,\mathrm{Sol}\,}}(H)|$ divides

$$\begin{aligned} \gcd (|G|- |{{\,\mathrm{Sol}\,}}(G)|,|G|- |{{\,\mathrm{Sol}\,}}_G(x)|, |{{\,\mathrm{Sol}\,}}_G(g)|- |{{\,\mathrm{Sol}\,}}(G)|), \end{aligned}$$

where $g \in G\setminus {{\,\mathrm{Sol}\,}}(G)$, and hence, $|H|$ is bounded by a function of $|G|$.

Proof

Since $\mathcal {NS}_G \cong \mathcal {NS}_H$, we have $|H\setminus {{\,\mathrm{Sol}\,}}(H)|= |G\setminus {{\,\mathrm{Sol}\,}}(G)|$ and so $|H\setminus {{\,\mathrm{Sol}\,}}(H)|$ is finite. If $h \in H\setminus {{\,\mathrm{Sol}\,}}(H)$ then $\{aha^{-1} : a \in H\} \subset H\setminus {{\,\mathrm{Sol}\,}}(H)$, since ${{\,\mathrm{Sol}\,}}(H)$ is closed under conjugation. Thus, every element in $H\setminus {{\,\mathrm{Sol}\,}}(H)$ has finitely many conjugates in H. It follows that $K = C_H (H\setminus {{\,\mathrm{Sol}\,}}(H))$ has finite index in H. Since $\mathcal {NS}_H$ has no isolated vertex, there exist two adjacent vertices u and v in $\mathcal {NS}_H$. Now, if $s \in K$, then $s \in C_H (\{u,v\})$ and so $\langle su,v \rangle$ is not solvable since $\langle su, v \rangle \cong \langle u, v \rangle \times \langle s \rangle$. Therefore, $Ku \subset H\setminus {{\,\mathrm{Sol}\,}}(H)$ and so K is finite. Hence, H is finite.

It follows that ${{\,\mathrm{Sol}\,}}(H)$ is a subgroup of H and so $|{{\,\mathrm{Sol}\,}}(H)|$ divides $|H|- |{{\,\mathrm{Sol}\,}}(H)|$. Since $|H|- |{{\,\mathrm{Sol}\,}}(H)|= |G|- |{{\,\mathrm{Sol}\,}}(G)|$, we have $|{{\,\mathrm{Sol}\,}}(H)|$ divides $|G|- |{{\,\mathrm{Sol}\,}}(G)|$. Let $x' \in H \setminus {{\,\mathrm{Sol}\,}}(H)$ and $y \in {{\,\mathrm{Sol}\,}}_H(x')$. Then, by Lemma 6, $\langle x', yz \rangle$ is solvable for all $z \in {{\,\mathrm{Sol}\,}}(H)$. Thus, ${{\,\mathrm{Sol}\,}}_H(x') = {{\,\mathrm{Sol}\,}}(H)\cup y_1{{\,\mathrm{Sol}\,}}(H) \cup \dots \cup y_n {{\,\mathrm{Sol}\,}}(H)$, for some $y_i \in H$. Therefore, $|{{\,\mathrm{Sol}\,}}(H)|$ divides $|{{\,\mathrm{Sol}\,}}_H(x')|$ and so $|{{\,\mathrm{Sol}\,}}(H)|$ divides $|H|- |{{\,\mathrm{Sol}\,}}_H(x')|$. We have $\deg (\mathcal {NS}_G) = \deg (\mathcal {NS}_H)$ since $\mathcal {NS}_G \cong \mathcal {NS}_H$. Also $\deg _{\mathcal {NS}_G}(g) = |G|- |{{\,\mathrm{Sol}\,}}_G(g)|$ for any $g \in V(\mathcal {NS}_G)$ and $\deg _{\mathcal {NS}_H}(h) = |H|- |{{\,\mathrm{Sol}\,}}_H(h)|$ for any $h \in V(\mathcal {NS}_H)$. Therefore, $|{{\,\mathrm{Sol}\,}}(H)|$ divides $|H|- |{{\,\mathrm{Sol}\,}}_H(h)|$ and hence $|G|- |{{\,\mathrm{Sol}\,}}_G(g)|$ for any $g \in G \setminus {{\,\mathrm{Sol}\,}}(G)$. Since $|{{\,\mathrm{Sol}\,}}(H)|$ divides $|G|- |{{\,\mathrm{Sol}\,}}(G)|$ and $|G|- |{{\,\mathrm{Sol}\,}}_G(g)|$, it divides $|G|- |{{\,\mathrm{Sol}\,}}(G)|- (|G|- |{{\,\mathrm{Sol}\,}}_G(g)|)= |{{\,\mathrm{Sol}\,}}_G(g)|- |{{\,\mathrm{Sol}\,}}(G)|$. This completes the proof. $\square$

Proposition 34

Let G be a non-solvable group such that $\mathcal {NS}_G$ is finite. Then, G is a finite group.

Proof

It follows directly from the first paragraph of the proof of the above theorem. $\square$

Proposition 35

Let G be a group such that $\mathcal {NS}_G\cong \mathcal {NS}_{A_5}$, then $G\cong A_5$.

Proof

Since $\mathcal {NS}_G\cong \mathcal {NS}_{A_5}$, we have G is a finite non-solvable group and

$$\begin{aligned} |G\setminus {{\,\mathrm{Sol}\,}}(G)|= |A_5\setminus {{\,\mathrm{Sol}\,}}(A_5)|= 59. \end{aligned}$$

Therefore, $|G|= |{{\,\mathrm{Sol}\,}}(G)|+ 59$. Since ${{\,\mathrm{Sol}\,}}(G)$ is a subgroup of G, we have $|{{\,\mathrm{Sol}\,}}(G)|\le \frac{|G|}{2}$ and so $|G|\le 118$. Hence, the result follows. $\square$

Remark 2

Using the following program in GAP [20], one can see that the non-solvable graphs of SL(2, 5) and ${\mathbb {Z}}_2\times A_5$ are isomorphic. It follows that non-solvable graphs of two groups are isomorphic which need not implies that their corresponding groups are isomorphic.

Proposition 36

Let G and H be two finite non-solvable groups. If $\mathcal {NS}_G \cong \mathcal {NS}_H$, then $\mathcal {NS}_{G \times A} \cong \mathcal {NS}_{H \times B}$, where A and B are two solvable groups having equal order.

Proof

Let $\varphi :\mathcal {NS}_G \rightarrow \mathcal {NS}_H$ be a graph isomorphism and $\psi :A \rightarrow B$ be a bijective map. Then, $(g, a) \mapsto (\varphi (g),\psi (a))$ defines a graph isomorphism between $\mathcal {NS}_{G \times A}$ and $\mathcal {NS}_{H \times B}$. $\square$

A non-solvable group G is called an Fs-group if for every two elements $x , y \in G\setminus {{\,\mathrm{Sol}\,}}(G)$ such that ${{\,\mathrm{Sol}\,}}_G(x)\ne {{\,\mathrm{Sol}\,}}_G(y)$ implies ${{\,\mathrm{Sol}\,}}_G(x) \not \subset {{\,\mathrm{Sol}\,}}_G(y)$ and ${{\,\mathrm{Sol}\,}}_G(y) \not \subset {{\,\mathrm{Sol}\,}}_G(x)$.

Proposition 37

Let G be an Fs-group. If H is a non-solvable group such that $\mathcal {NS}_G \cong \mathcal {NS}_H$, then H is also an Fs-group.

Proof

Let $\psi : \mathcal {NS}_H \rightarrow \mathcal {NS}_G$ be a graph isomorphism. Let $x, y \in H\setminus {{\,\mathrm{Sol}\,}}(H)$ such that ${{\,\mathrm{Sol}\,}}_H(x) \subseteq {{\,\mathrm{Sol}\,}}_H(y)$. Then $\psi ({{\,\mathrm{Sol}\,}}_H(x)\setminus {{\,\mathrm{Sol}\,}}(H)) \subseteq \psi ({{\,\mathrm{Sol}\,}}_H(y)\setminus {{\,\mathrm{Sol}\,}}(H))$. We have

$$\begin{aligned} \psi ({{\,\mathrm{Sol}\,}}_H(z)\setminus {{\,\mathrm{Sol}\,}}(H)) = {{\,\mathrm{Sol}\,}}_G(\psi (z))\setminus {{\,\mathrm{Sol}\,}}(G) \text { for all } z\in H\setminus {{\,\mathrm{Sol}\,}}(H). \end{aligned}$$

Therefore, ${{\,\mathrm{Sol}\,}}_G(\psi (x))\setminus {{\,\mathrm{Sol}\,}}(G) \subseteq {{\,\mathrm{Sol}\,}}_G(\psi (y))\setminus {{\,\mathrm{Sol}\,}}(G)$. Since G is an Fs-group, we have

$$\begin{aligned} {{\,\mathrm{Sol}\,}}_G(\psi (x))\setminus {{\,\mathrm{Sol}\,}}(G) = {{\,\mathrm{Sol}\,}}_G(\psi (y))\setminus {{\,\mathrm{Sol}\,}}(G). \end{aligned}$$

It follows that ${{\,\mathrm{Sol}\,}}_H(x)\setminus {{\,\mathrm{Sol}\,}}(H) = {{\,\mathrm{Sol}\,}}_H(y)\setminus {{\,\mathrm{Sol}\,}}(H)$ and so ${{\,\mathrm{Sol}\,}}_H(x) = {{\,\mathrm{Sol}\,}}_H(y)$. Hence, H is an Fs-group. $\square$

8 Genus of Non-Solvable Graph

The genus of a graph $\Gamma$, denoted by $\gamma (\Gamma )$, is the smallest non-negative integer g such that the graph can be embedded on the surface obtained by attaching g handles to a sphere. Clearly, if ${\tilde{\Gamma }}$ is a subgraph of $\Gamma$, then $\gamma ({\tilde{\Gamma }}) \le \gamma (\Gamma )$. Graphs having genus zero are called planar graphs, while those having genus one are called toroidal graphs. Graphs having genus two are called double-toroidal graphs and those having genus three are called triple-toroidal graphs. In [13, Corollary 3.14], it was shown that $\mathcal {NS}_G$ is not planar for finite non-solvable group G. In this section, we extend [13, Corollary 3.14] and show that $\mathcal {NS}_G$ is neither planar, toroidal, double toroidal nor triple toroidal.

It is well known (see [21, Theorem 6-39]) that $\gamma (K_{n})=\left\lceil \frac{(n-3)(n-4)}{12}\right\rceil$, where $n\ge 3$ and $K_n$ is the complete graph on n vertices. Also, if $m,n\ge 2$, then

$$\begin{aligned}\gamma (K_{m,n})=\left\lceil \frac{(m-2)(n-2)}{4} \right\rceil \text { and } \gamma (K_{m, m, m})= \frac{(m - 2)(m - 1)}{2},\end{aligned}$$

where $K_{m,n}$, $K_{m,m, m}$ are complete bipartite and tripartite graphs, respectively.

Proposition 38

Let G be a finite non-solvable group. Then,

$$\begin{aligned} |{{\,\mathrm{Sol}\,}}(G)|\le \sqrt{2\gamma (\mathcal {NS}_G)} + 2. \end{aligned}$$

Proof

Assume that $Z={{\,\mathrm{Sol}\,}}(G)$. By Proposition 30, we have $\omega (\mathcal {NS}_G) \ge 3$. So, there exist $u,v,w \in G\setminus Z$ such that they are adjacent to each other. Then, by Lemma 6, $\mathcal {NS}_G[uZ\cup vZ \cup wZ]$ is isomorphic to $K_{|Z|, |Z|, |Z|}$. We have

$$\begin{aligned} \gamma (\mathcal {NS}_G) \ge \gamma ( K_{|Z|, |Z|, |Z|})=\frac{(|Z|- 2)(|Z|- 1)}{2} \ge \frac{(|Z|- 2)(|Z|- 2)}{2}, \end{aligned}$$

and hence, the result follows. $\square$

Theorem 39

Let G be a finite non-solvable graph. Then, $\gamma (\mathcal {NS}_G)\ge 4$. In particular, $\mathcal {NS}_G$ is neither planar, toroidal, double toroidal nor triple toroidal.

Proof

By Lemma 29, we have an element $x \in G\setminus {{\,\mathrm{Sol}\,}}(G)$ such that o(x) is a prime greater or equal to 5. Clearly, ${{\,\mathrm{nbd}\,}}_{\mathcal {NS}_G}(x)\ne \emptyset$. Assume that $o(y) = 2$ for all $y\in {{\,\mathrm{nbd}\,}}_{\mathcal {NS}_G}(x)$. Then, $xy \in {{\,\mathrm{nbd}\,}}_{\mathcal {NS}_G}(x)$ and so $o(xy)=2$. Thus, $\langle x, y\rangle = \langle y, xy\rangle$ is isomorphic to a dihedral group, which is a contradiction. Therefore, there exists $y \in {{\,\mathrm{nbd}\,}}_{\mathcal {NS}_G}(x)$ such that $o(y)\ge 3$. Let $1 \ne j \in {\mathbb {N}}$ and $\gcd (j, o(x)) = 1$. Consider the subsets $H = \{x, x^2, x^3, x^4\}$, $K = \{y^ix^j : i = 1, 2, j= 0, 1, 2, 3, 4\}$ of $G\setminus {{\,\mathrm{Sol}\,}}(G)$ and the induced graph $\mathcal {NS}_G[H\cup K]$. Notice that $\mathcal {NS}_G[H\cup K]$ has a subgraph isomorphic to $K_{4,10}$ and hence

$$\begin{aligned} \gamma (\mathcal {NS}_G) \ge \gamma (\mathcal {NS}_G[H\cup K]) \ge \gamma (K_{4,10}) = 4. \end{aligned}$$

This completes the proof. $\square$

Remark 3

By GAP [20], using the following program, we see that $\mathcal {NS}_{A_5}$ has 1140 edges and 59 vertices. Thus, by [21, Corollary 6-14], we have

$$\begin{aligned} \gamma (\mathcal {NS}_{A_5})\ge \frac{1140}{6}-\frac{59}{2}+1=161.5 \end{aligned}$$

and so $\gamma (\mathcal {NS}_{A_5}) \ge 162$.

Similarly $\mathcal {NS}_{S_5}, \mathcal {NS}_{SL(2,5)}$ and $\mathcal {NS}_{{\mathbb {Z}}_2\times A_5}$ have 4560 edges and 119 vertices. So their genera are at least 732.

A compact surface $N_k$ is a connected sum of k projective planes. The number k is called the crosscap of $N_k$. A simple graph which can be embedded in $N_k$ but not in $N_{k-1}$ is called a graph of crosscap k. The notation ${\bar{\gamma }}(\Gamma )$ stands for the crosscap of a graph $\Gamma$. It is easy to see that ${\bar{\gamma }}({\tilde{\Gamma }})\le {\bar{\gamma }}(\Gamma )$ for all subgraphs ${\tilde{\Gamma }}$ of $\Gamma$. Also, a graph $\Gamma$, such that ${\bar{\gamma }}(\Gamma )=1$ is called a projective graph. It is shown in [11] that $2K_5$ is not projective. Hence, any graph containing a subgraph isomorphic to $2K_5$ is not projective. We conclude this paper with the following result.

Theorem 40

Let G be a finite non-solvable group. Then, $\mathcal {NS}_G$ is not projective.

Proof

As shown in the proof of Theorem 39, there exist $x, y \in G\setminus {{\,\mathrm{Sol}\,}}(G)$ such that o(x) is a prime greater or equal to 5, $o(y) \ge 3$ and they are adjacent. Let $1 \ne j \in {\mathbb {N}}, \gcd (j, o(y))=1$. Consider the subsets $H = \{y, xy, x^2y, x^3y, x^4y\}$ and $K = \{y^j, xy^j,$ $x^2y^j, x^3y^j, x^4y^j\}$ of $G\setminus {{\,\mathrm{Sol}\,}}(G)$. Then, $H\cap K = \emptyset$ and $\mathcal {NS}_G[H]\cong \mathcal {NS}_G[K]\cong K_5$. It follows that $\mathcal {NS}_G$ has a subgraph isomorphic to $2K_5$. Hence, $\mathcal {NS}_G$ is not projective. $\square$