On the connectivity of the non-generating graph

Abstract

Given a 2-generated finite group G, the non-generating graph of G has as vertices the elements of G and two vertices are adjacent if and only if they are distinct and do not generate G. We consider the graph $\Sigma (G)$ obtained from the non-generating graph of G by deleting the universal vertices. We prove that if the derived subgroup of G is not nilpotent, then this graph is connected, with diameter at most 5. Moreover, we give a complete classification of the finite groups G such that $\Sigma (G)$ is disconnected.

1 Introduction

Let G be a finite group. The generating graph of G,  written $\Gamma (G),$ is the graph in which the vertices are the elements of G and there is an edge between $g_1$ and $g_2$ if G is generated by $g_1$ and $g_2.$ If G is not 2-generated, then there will be no edge in this graph. Thus, it is natural to assume that G is 2-generated. Quite a lot is known about this graph when G is a non-Abelian simple group; for example, Guralnick and Kantor [9] showed that there is no isolated vertex in $\Gamma (G)$ but the identity, and Breuer, Guralnick, Kantor [2] showed that the diameter of the subgraph of $\Gamma (G)$ induced by non-identity elements is 2 for all G. If G is an arbitrary finite group, then $\Gamma (G)$ could contain many isolated vertices. Let $\Delta (G)$ be the subgraph of $\Gamma (G)$ that is induced by all the vertices that are not isolated. In [5] and [11], it is proved that if G is a 2-generated soluble group, then $\Delta (G)$ is connected and ${{\,\mathrm{diam}\,}}(\Delta (G))\le 3.$ The situation is different if the solubility assumption is dropped. It is an open problem whether or not $\Delta (G)$ is connected, but even when $\Delta (G)$ is connected, its diameter can be arbitrarily large. For example, if G is the largest 2-generated direct power of ${{\,\mathrm{SL}\,}}(2,2^p)$ and p is a sufficiently large odd prime, then $\Delta (G)$ is connected but ${{\,\mathrm{diam}\,}}(\Delta (G))\ge 2^{p-2}-1$ (see [6, Theorem 5.4]).

The aim of this paper is to investigate the connectivity of the complement graph of $\Delta (G),$ denoted by $\Sigma (G).$ This graph can be described as follows: we take the non-generating graph of G,  i.e. the graph whose vertices are the elements of G and where there is an edge between $g_1$ and $g_2$ if $\langle g_1, g_2\rangle \ne G$ and we remove the universal vertices (corresponding to the isolated vertices of the generating graph). We prove that $\Sigma (G)$ is connected, except for some families that can be completely described. In any case, if $\Sigma (G)$ is disconnected, then G is soluble and its derived subgroup is nilpotent.

Theorem 1

Let G be a 2-generated finite group. Then $\Sigma (G)$ is connected if and only if none of the following occurs:

• (1) G is cyclic;

• (2) G is a p-group;

• (3) $G/{{\,\mathrm{Frat}\,}}(G)\cong (V_1\times \dots \times V_t)\rtimes H,$ where $H\cong C_p$ for some prime p,  and $V_1,\dots ,V_t$ are pairwise non-H-isomorphic non-trivial irreducible H-modules.

• (4) $G/{{\,\mathrm{Frat}\,}}(G)\cong (V_1\times \dots \times V_t)\rtimes H,$ where $H\cong C_p\times C_p$ for some prime p,  $V_1,\dots ,V_t$ are pairwise non-H-isomorphic non-trivial irreducible H-modules and $C_H(V_1\times \dots \times V_t)\cong C_p.$

Moreover, if $\Sigma (G)$ is connected, then ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 5$, and ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 3$ under the additional assumption that G is soluble.

We do not know whether the bound on the diameter of $\Sigma (G)$ is the best possible. In any case, if ${{\,\mathrm{B}\,}}$ is the Baby Monster, then ${{\,\mathrm {diam}\,}}(\Sigma (\mathrm {B}))\ge 4$ (see the end of Section 3). On the other hand, for soluble groups, the bound ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 3$ is the best possible. Consider for example $G=\langle a \rangle \times \langle b \rangle \times \langle c \rangle ,$ with $|a|=|b|=2$ and $|c|=3.$ In this case, $1, c, c^2$ are the isolated vertices of the generating graph $\Gamma (G)$. Moreover, if g is adjacent to ac in $\Sigma (G)$, then $g\in \langle ac\rangle$, hence $g\in \{a, ac^2\}.$ Similarly b and $bc^2$ are the only vertices of $\Sigma (G)$ that are adjacent to bc. Hence a shortest path in $\Sigma (G)$ between ac and bc is (ac, a, b, bc).

When $\Sigma (G)$ is disconnected, it is possible that it contains some isolated vertices. However this occurs only in few particular cases.

Proposition 2

Let G be a 2-generated finite group. Then $\Sigma (G)$ has an isolated vertex if and only if

• (1) G is cyclic;

• (2) $G\cong C_2\times C_2$;

• (3) $G\cong D_p$ is a dihedral group with 2p elements for p an odd prime.

The structure of the paper is as follows. In Section  2, we study $\Sigma (G)$ in the particular case when G is a primitive soluble group. In Section  3, we investigate $\Sigma (G)$ when G is a monolithic group whose socle is non-Abelian. Thanks to the fact that if N is a proper normal subgroup of G and $\Sigma (G/N)$ is connected, then $\Sigma (G)$ is also connected (see Proposition 15), using the results from Sections 2 and 3, we prove in Section  4 that if $\Sigma (G)$ is disconnected, then the derived subgroup of G is nilpotent. The case when the derived subgroup is nilpotent is analysed in Section  5. In the last section, we combine the partial results obtained in the previous sections to complete the proof of Theorem 1. Although it is elementary and independent from the proof of Theorem 1, the proof of Proposition 2 is given at the end of Section  6.

2 Primitive soluble groups

Definition 3

Let G be a finite group. We denote by V(G) the subset of G consisting of the elements x with the property that $G=\langle x, y\rangle$ for some y.

Lemma 4

([14, Proposition 2.2]). Let G be a primitive soluble group. Let $N={{\,\mathrm{soc}\,}}(G)$ and let H be a core-free maximal subgroup of G. Given $1\ne h\in H$ and $n\in N$, $hn \in V(G)$ if and only if $h\in V(H).$ In particular, $V(H)\setminus \{1\}\subseteq V(G).$

Proposition 5

Let G be a 2-generated primitive soluble group. Then $\Sigma (G)$ is disconnected if and only if either $G\cong C_p$ or $G/{{\,\mathrm{soc}\,}}(G)\cong C_p$. Moreover, if G is connected, then ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 3.$

Proof

If G is nilpotent, then $G\cong C_p.$ In this case, $\Sigma (G)$ is a disconnected graph with p vertices and no edges. So we may assume that G is not nilpotent. Thus $G\cong V \rtimes H,$ where $V={{\,\mathrm{soc}\,}}(G)$ is a faithful irreducible H-module.

If $|H|=p,$ with p a prime, then $\langle g, v \rangle =G$ for any $1\ne v \in V$ and $g\notin V.$ In particular, $V\setminus \{1\}$ is a connected component of $\Sigma (G)$.

So we may assume that |H| is not a prime. Suppose that $g_1=v_1h_1, g_2=v_2h_2$ are two different elements of V(G),  with $v_1, v_2 \in V, h_1, h_2 \in H.$ By Lemma 4, for $i\in \{1,2\},$ either $h_i=1$ (and therefore H is cyclic) or $h_i \in V(G).$ If neither $h_1$ nor $h_2$ is a generator of H, then $(v_1h_1,h_1,h_2,v_2h_2)$ is a path in $\Sigma (G)$. So we may assume $H=\langle h_1 \rangle .$ In this case, $h_1$ and $g_1$ are conjugate in G, so it is not restrictive to assume $v_1=1.$ Since |H| is not a prime, we can choose $1\ne h \in H$ with $|h|<|H|.$ Since all the complements of V in G are conjugate, there exists $v\in V$ such that $\langle g_2 \rangle \le \langle h_1^v\rangle .$ But then $(h_1,h,h^v,g_2)$ is a path in $\Sigma (G).$ $\square$

3 Monolithic groups with non-Abelian socle

Let G be a 2-generated finite monolithic group, with a non-Abelian socle. The aim of this section is to prove that the graph $\Sigma (G)$ is connected, with diameter at most 5.

Assume $A={{\,\mathrm{soc}\,}}(G)\cong S^n$, with S a finite non-Abelian simple group and $n\in {\mathbb {N}}.$ We may identify G with a subgroup of $\mathrm {Aut}(S^n)={{\,\mathrm{Aut}\,}}(S) \wr \mathrm {Sym}(n)$, the wreath product of ${{\,\mathrm{Aut}\,}}(S)$ with the symmetric group of degree n. So the elements of G are of the kind $g=(\alpha _1, \dots , \alpha _n)\sigma$, with $\alpha _i \in {{\,\mathrm{Aut}\,}}(S)$ and $\sigma \in \mathrm {Sym}(n)$. For this section, we will refer to this identification and we will denote by $\pi$ the homomorphism $\pi : {{\,\mathrm{Aut}\,}}(S)\wr \mathrm {Sym}(n) \rightarrow \mathrm {Sym}(n)$ mapping $(\alpha _1, \dots , \alpha _n)\sigma$ to $\sigma$.

We begin with two lemmas concerning some properties of ${{\,\mathrm{Aut}\,}}(S).$

Lemma 6

Let S be a finite non-Abelian simple group. There exist a subgroup H of ${{\,\mathrm{Aut}\,}}(S)$ and a prime divisor r of the order of S with the following properties:

• (1) $H \cap S < S;$

• (2) $HS={{\,\mathrm{Aut}\,}}(S);$

• (3) for every $h \in H$, we can find an element $s \in S\cap H$ such that $|h|_r \ne |h s|_r$.

Proof

First suppose that S is an alternating group of degree n (with $n\ne 6$) or a sporadic simple group. We claim that in this case we can take $r=2$ and $H \in {{\,\mathrm{Syl}\,}}_2({{\,\mathrm{Aut}\,}}(S)).$ This is equivalent to saying that any coset $(S \cap H)h$ of $S\cap H$ in H contains x and y with $|x|_2\ne |y|_2.$ If $h\in S$, then it suffices to take $x=1$ and y a non-trivial element of $H\cap S$. If $h\notin S,$ then, since $|{{\,\mathrm{Aut}\,}}(S):S|\le 2$, we must have $|H:H\cap S|=|{{\,\mathrm{Aut}\,}}(S):S|=2$ and $(H\cap S)h=H{\setminus } H\cap S.$ It is known that in this case ${{\,\mathrm{Aut}\,}}(S) \setminus S$ (and consequently $H\setminus H\cap S$) contains both an involution and an element of order 4 (see, for example, [1, Theorem 2]).

Now assume that S is a simple group of Lie type, defined over a field of characteristic p. The proof of the Lemma in [13, Section 2] implies that, except when $S=\mathrm {PSL}(2,q)$ and q is odd, we can take $r=p$ and $H=N_{{\,\mathrm {Aut}\,}(S)}(P)$ for $P \in {{\,\mathrm{Syl}\,}}_p(S).$ We remain with the case when $S=\mathrm {PSL}(2,q)$ and q is odd. In this case, let $r=2,$ $P\in {{\,\mathrm {Syl}\,}}_2(S)$, and $H=N_{{\,\mathrm {Aut}\,}(S)}(P).$ We may choose P so that the Frobenius automorphism $\sigma$ belongs to H. Let $h \in H.$ Up to multiplying with a suitable element of $S\cap H$, we may assume $h=y\sigma _1\sigma _2$, with $y \in H$, $|y|=2$, $\sigma _1,\sigma _2 \in \langle \sigma \rangle ,$ $|\sigma _1|$ a 2-power, $|\sigma _2|$ odd, and $[y,\sigma ]=1.$ Let ${\tilde{q}}$ be the size of the subfield of GF(q) centralized by $\sigma _2.$ By [1, Theorem 2], there exists $t \in \mathrm {PSL}(2,{\tilde{q}})\langle y\sigma _1\rangle \cap H$ such that $|ty\sigma _1|_2>|y\sigma _1|_2.$ We have $|th|_2>|h|_2.$ $\square$

Lemma 7

Let S and H be as in the previous lemma. For any $h_1, h_2\in H,$ there exist $s \in S,$ $t \in H\cap S$ such that $\langle h_1s, h_2t\rangle = \langle h_1, h_2\rangle S.$

Proof

It follows from Lemma 6 (3) that there exists $t \in H \cap S$ such that $\langle h_2t\rangle \cap S \ne 1.$ Indeed, if $\langle h_2\rangle \cap S \ne 1,$ then we can take $t=1.$ Otherwise take $t\in H \cap S$ such that $|h_2|_r \ne |h_2t|_r;$ then $\langle h_2t \rangle \cap S$ contains a non-trivial r-element. Now let $1\ne u \in \langle h_2t \rangle \cap S.$ By [3, Theorem 1], there exists $s \in S$ such that $\langle h_1s, u\rangle =\langle h_1, S\rangle .$ This implies $\langle h_1s, h_2t\rangle = \langle h_1, h_2\rangle S.$ $\square$

In the next two lemmas, let ${\tilde{H}}=\{(\alpha _1,\dots ,\alpha _n)\sigma \in G \mid \alpha _1,\dots ,\alpha _n \in H\},$ where H is the subgroup of ${{\,\mathrm{Aut}\,}}(S)$ introduced in the statement of Lemma 6. Clearly, since $H\cap S < S,$ ${\tilde{H}}$ is a proper subgroup of G. Moreover, in the proof of the next lemma, we will use a quasi-ordering relation on the set of the cyclic permutations which belong to the group $\mathrm {Sym}(n)$, whose definition depends on the choice of the prime r appearing in the statement of Lemma 6. Let $\sigma _1$, $\sigma _2 \in \mathrm {Sym}(n)$ be two cyclic permutations (including cycles of length 1); we define $\sigma _1 \le \sigma _2$ if either $|\sigma _1|_r < |\sigma _2|_r$ or $|\sigma _1|_r =|\sigma _2|_r$ and $|\sigma _1| \le |\sigma _2|$.

Lemma 8

Suppose $\langle g_1, g_2 \rangle A=G$ and that one of the following holds:

• (1) $g_1^\pi$ has a fixed point;

• (2) $g_2^\pi$ has a fixed point;

• (3) $(g_1g_2^i)^\pi$ is fixed-point-free for every $i \in {\mathbb {Z}}.$

Then there exist $u_1, u_2$ in A such that $\langle u_1g_1, u_2g_2 \rangle =G$ and $u_1g_1 \in {\tilde{H}}$.

Proof

Since $SH={{\,\mathrm{Aut}\,}}(S),$ it is not restrictive to assume $g_1, g_2 \in {\tilde{H}}$. By [13, Theorem 1.1] and its proof, there exist $u_1, u_2 \in A$ such that $\langle u_1g_1, u_2 g_2\rangle =G.$ With a careful revision of the proof, it can be shown that $u_1, u_2$ can be chosen with the additional property that $u_1 \in A \cap H^n$. The proof of [13, Theorem 1.1] is constructive, but unfortunately long (7 pages) and intricate. Our aim here is not to repeat all the proof, but only to give a short sketch and indicate which parts of the proof require some adjustment in order to achieve the stronger formulation.

First we fix our notation. Since they are quite technical and it could be difficult to understand immediately their meaning, at any step we will indicate how our definitions work in the particular case when $g_1^\pi$ has a fixed point.

We assume $g_1=(\alpha _1,\dots ,\alpha _n)\rho$, $g_2=(\beta _1,\dots ,\beta _n)\sigma$, with $\alpha _i, \beta _j \in {{\,\mathrm{Aut}\,}}(S)$ and $\rho , \sigma \in \mathrm {Sym}(n)$. Then we write $\rho =\rho _1 \cdots \rho _{s(\rho )}$ as products of disjoint cycles (including possibly cycles of length 1) in such a way that

$$\begin{aligned} \rho _1 \le \dots \le \rho _{s(\rho )}. \end{aligned}$$

For $1 \le i \le s(\rho ),$ let $\rho _i=(m_{i,1},\dots ,m_{i,|\rho _i|})$ and set $m=m_{1,1}.$ Notice that in the particular case when $g_1^\pi =\rho$ has at least one fixed point, m is one of these fixed points (i.e. $\rho _1=(m)$).

The permutation $\sigma$ has an orbit (possibly of length 1) containing m and acts on this orbit as a cycle $(m,n_2,\dots ,n_t).$

For a given pair $v_1=(x_1,\dots ,x_n)$, $v_2=(y_1,\dots ,y_n)$ of elements of $A=S^n$, we define

$$\begin{aligned} {\bar{\alpha }}_r= & {} x_r\alpha _r,\quad {\bar{\beta }}_r=y_r\beta _r,\quad 1 \le r \le n,\\ a_i= & {} \ {\bar{\alpha }}_{m_{i,1}} \cdots {\bar{\alpha }}_{m_{i,|\rho _i|}}, \ 1 \le i \le s(\rho ). \end{aligned}$$

Moreover, let $a =\alpha _{m} {\bar{\alpha }}_{m_{1,2}} \cdots {\bar{\alpha }}_{m_{1,|\rho _1|}}$, $b =\beta _{m} {\bar{\beta }}_{n_{2}} \cdots {\bar{\beta }}_{n_{t}}$ and consider $K=\langle a, b, S \rangle$. In the particular case when m is a fixed point of $\rho$, we just have $a=\alpha _m$.

Now we say that the 2n-tuple $(x_1,\dots ,x_n,y_1,\dots ,y_n)\in S^{2n}$ is good if the following two conditions are satisfied:

• (1) $\langle a_1, b_1 \rangle = \langle x_ma, y_mb \rangle =K$.

• (2) If $2 \le i \le s(\rho )$, then $a_i^{|\rho _1 \cdots \rho _i|/|\rho _i|}$ is not conjugate to $a_1^{|\rho _1 \cdots \rho _i|/|\rho _1|}$ in ${{\,\mathrm{Aut}\,}}(S).$

The reason for this definition is the following. The first part of the proof of [13, Theorem 1.1] (ending with the last paragraph of page 177) shows that conditions (1) and (2) ensure that there exists a partition $\Phi$ of $\{1,\dots ,n\}$ invariant for the action of $\langle \rho , \sigma \rangle$ such that $\langle v_1g_1, v_2g_2\rangle \cap S^n=\prod _{B \in \Phi }D_B$, where, for every block $B \in \Phi$, $D_B$ is a full diagonal subgroup of $\prod _{j \in B}S_j$. Moreover, if B is the block of $\Phi$ containing m, then $B \subseteq \mathrm {supp}(\rho _1).$ In particular, if m is fixed by $\rho$, then $B=\{m\}$, hence $A=S^n \le \langle v_1g_1, v_2g_2\rangle$ and therefore $G=\langle v_1g_1, v_2g_2\rangle .$

We claim that we can find a good 2n-tuple with the additional property that $x_1,\dots ,x_n\in H.$ To construct such a 2n-tuple, we start by choosing arbitrarily $x_1,\ldots ,x_n,$ $y_1,\ldots ,y_n\in S\cap H,$ and then we modify the elements $x_m, y_m$ and the elements $x_i$ for $i \in \{m_{i,1}\mid 2 \le i \le s(\rho )\}$. This can be done in two consecutive steps.

• (i) First, by Lemma 7, we can find $x \in H\cap S$ and $y \in S$ such that $\langle x a, yb \rangle =K$. We substitute the original $x_m, y_m$ by $xx_m$ and $yy_m.$

• (ii) Then let $i\in \{2,\dots ,s(\rho )\}.$ Since $\rho _1 \le \dots \le \rho _i$, $|\rho _1 \cdots \rho _i|/|\rho _i|$ is coprime with r and therefore, by Lemma 6, there exists $s_i \in S\cap H$ such that $(s_ia_i)^{|\rho _1 \cdots \rho _i|/|\rho _i|}$ is not conjugate to $a_1^{|\rho _1 \cdots \rho _i|/|\rho _1|}$ in ${{\,\mathrm{Aut}\,}}(S)$. We substitute $x_{m_{i,1}}$ by $s_i x_{m_{i,1}}.$

Our first conclusion is that if $\rho$ has at least a fixed point and $(x_1,\dots ,x_n.y_1,\dots ,y_n)$ is a good 2n-tuple with the additional property ensured by the previous paragraph, then $(x_1,\dots ,x_n)$ and $(y_1,\dots ,y_n)$ satisfy the request of our statement.

The case when $\sigma$ has a fixed point can be dealt with a similar argument, working with the ordered pair $(g_2,g_1)$ instead of $(g_1,g_2)$. However, if we repeat precisely the same procedure as above, we obtain $v_1, v_2 \in A$ such that $\langle v_1g_2, v_2g_1\rangle = G$ and $v_1g_2 \in {\tilde{H}}$, while our request is to have $v_2g_1 \in {\tilde{H}}.$ On the other hand, when in step (i) we apply Lemma 7, we are free to choose x, y either with the property that $x \in H\cap S$ and $y\in S$ or with the property that $x \in S$ and $y \in H\cap S$, and the second choice solves the problem.

We remain with the case when $\rho \sigma ^i$ is fixed-point-free for every $i \in {\mathbb {Z}}.$ As we have seen before, we can find $v_1=(x_1,\dots ,x_n)$, $v_2=(y_1,\dots ,y_n) \in S^n$ with the properties that $(x_1,\dots ,x_n,y_1,\dots ,y_n)$ is a good 2n-tuple with $x_1,\dots ,x_n\in H\cap S.$ This implies that $\langle v_1g_1, v_2g_2\rangle \cap S^n=\prod _{B \in \Phi }D_B$ for a suitable partition $\Phi$ of $\{1,\dots ,n\}.$ Clearly this is not enough to reach our conclusion and indeed this is the case for which the proof of [13, Theorem 1.1] requires more work. The crucial observation is that the conditions that make $(x_1,\dots ,x_n,y_1,\dots ,y_n)$ a good 2n-tuple involve only the elements $x_1,\dots ,x_n$ and the value of the product $y_m\beta _{m} y_{n_2}\beta _{n_{2}} \cdots y_{n_{t}}\beta _{n_{t}}$. This leaves a large freedom in the choice of elements $y_1,\dots ,y_n.$ Indeed, the second part of the proof of [13, Theorem 1.1], from page 177 to page 181, shows that we may choose $y_1,\dots ,y_n$ so that $\prod _{B \in \Phi }D_B$ is normalized by $\langle v_1g_1, v_2g_2\rangle$ only if $\Phi =\{\{1\},\dots , \{n\}\}.$ With this choice, $\langle v_1g_1, v_2g_2\rangle =G$, as required. $\square$

Corollary 9

Let g be a vertex of $\Sigma (G).$ If $G/{{\,\mathrm{soc}\,}}(G)$ is not cyclic, then there exists a vertex ${\tilde{g}} \in \Sigma (G)$ such that ${\tilde{g}} \in {\tilde{H}}$ and the distance between g and ${\tilde{g}}$ in $\Sigma (G)$ is at most 2.

Proof

Since $g \in \Sigma (G),$ there exists $g_2 \in G$ such that $\langle g, g_2\rangle =G.$ If g and $g_2$ satisfy one of the three conditions in the statement of Lemma 8, then there exists $u, u_2$ in ${{\,\mathrm{soc}\,}}(G)=A$ such that $\langle ug, u_2g_2 \rangle =G$ and ${\tilde{g}}= ug \in {\tilde{H}}.$ In particular, ${\tilde{g}} \in \Sigma (G)$ and $\langle g, {\tilde{g}} \rangle \le \langle g \rangle A < G$ since we are assuming that G/A is not cyclic, and therefore g and ${\tilde{g}}$ are adjacent vertices of $\Sigma (G).$ Now assume that $g^\pi$ and $g_2^\pi$ are fixed-point-free but there exists $i\in {\mathbb {Z}}$ such that $g_3=gg_2^i$ has a fixed point. If $\langle g, g_3 \rangle =G,$ then we repeat the previous argument using $g_3$ instead of $g_2$ and we find an element ${\tilde{g}}$ with the required properties. Suppose $\langle g,g_3 \rangle \ne G.$ In any case, $\langle g_2, g_3\rangle = \langle g_2, g\rangle =G$, so $g_3$ is a vertex of $\Sigma (G)$ which is adjacent to g. Moreover, we may apply Lemma 8 to the generating pair $(g_3,g_2)$ in order to find ${\tilde{g}}_3 \in {\tilde{H}}$ which is adjacent in $\Sigma (G)$ to $g_3$. So $(g, g_3, {\tilde{g}}_3)$ is a path in $\Sigma (G)$ and we may take ${\tilde{g}}={\tilde{g}}_3.$ $\square$

Proposition 10

The graph $\Sigma (G)$ is connected, with diameter at most 5.

Proof

We distinguish two cases:

a) $G/{{\,\mathrm{soc}\,}}(G)$ is not cyclic. Suppose that $g_1, g_2$ are two different vertices of $\Sigma (G).$ Choose ${\tilde{g}}_1, {\tilde{g}}_2$ as in the statement of Corollary 9. Since $\langle {\tilde{g}}_1, {\tilde{g}}_2 \rangle \le {\tilde{H}} < G,$ ${\tilde{g}}_1, {\tilde{g}}_2$ are adjacent vertices of $\Sigma (G)$. So the distance in $\Sigma (G)$ between $g_1$ and $g_2$ is at most 5.

b) $G/{{\,\mathrm{soc}\,}}(G)$ is cyclic. Recall that the intersection graph ${{\,\mathrm{I}\,}}(G)$ of G is the graph whose vertices are the non-trivial proper subgroups of G and in which two vertices H and K are adjacent if and only if $H\cap K\ne 1.$ By [3, Theorem 1], when $G/{{\,\mathrm{soc}\,}}(G)$ is cyclic, the vertex set V(G) of $\Sigma (G)$ coincides with the set of the non-trivial elements of G. In particular, as explained in [4, Section 12], ($\Sigma (G), {{\,\mathrm{I}\,}}(G)$) is a dual pair of graphs, and therefore there is a natural bijection between connected components of $\Sigma (G)$ and connected components of ${{\,\mathrm{I}\,}}(G)$ with the property that corresponding components have diameters which are either equal or differ by 1. If G is neither soluble nor simple, then ${{\,\mathrm{I}\,}}(G)$ is connected with ${{\,\mathrm{diam}\,}}({{\,\mathrm{I}\,}}(G))\le 4$ (see [16, Lemma 5]) hence ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 5$. Freedman recently proved that also when G is a finite non-Abelian simple group, the graph $\Sigma (G)$ is connected with diameter at most 5 (see the remark after [4, Proposition 12]). $\square$

We do not know whether the bound in the previous proposition is the best possible. However, Freedman [7] proved that ${{\,\mathrm{diam}\,}}({{\,\mathrm{I}\,}}(G))\le 5$ for any finite non-Abelian simple group G and that the upper bound is attained only by the Baby Monster ${{\,\mathrm{B}\,}}$ and some unitary groups. In particular, since $({{\,\mathrm{I}\,}}({{\,\mathrm{B}\,}}),\Sigma ({{\,\mathrm{B}\,}}))$ is a dual pair of graphs, it follows that ${{\,\mathrm{diam}\,}}(\Sigma ({{\,\mathrm{B}\,}}))\ge 4.$

4 A reduction to the case when the derived subgroup is nilpotent

The main result of this section is the following proposition, which says in particular that if $\Sigma (G)$ is disconnected, then G is soluble and the derived subgroup of G is nilpotent.

Proposition 11

Let G be a 2-generated group. If $\Sigma (G)$ is disconnected, then G is soluble. Moreover if G is soluble, then at least one of the following occurs.

• (1) $\Sigma (G)$ is connected and ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 3;$

• (2) The derived subgroup of G is nilpotent and $G/{{\,\mathrm{Frat}\,}}(G)$ has the following structure:

  $$\begin{aligned} G/{{\,\mathrm{Frat}\,}}(G)\cong (V_1\times \dots \times V_t)\rtimes H, \end{aligned}$$

  where H is Abelian and $V_1,\dots ,V_t$ are pairwise non-H-isomorphic non-trivial irreducible H-modules (including the possibility $t=0$).

The proof requires some preliminary easy results. First we need to notice that $\Sigma (G)$ is disconnected if G is cyclic or has prime power order.

Lemma 12

If G is a non-trivial cyclic group, then $\Sigma (G)$ is disconnected.

Proof

If $\langle g \rangle =G,$ then g is an isolated vertex of $\Sigma (G).$ $\square$

Lemma 13

If G is a 2-generated finite p-group, then $\Sigma (G)$ is disconnected.

Proof

By the previous lemma, we may assume that G is not cyclic. Let $F={{\,\mathrm{Frat}\,}}(G)$. Then G has precisely $p+1$ maximal subgroups $M_1,\dots ,M_{p+1}$ and $V(G)=G{\setminus } F.$ Moreover, two distinct vertices x and y of $\Sigma (G)$ are adjacent if and only if $x, y \in M_i$ for some $1\le i \le p+1.$ This implies that $\Sigma (G)$ is the union of $p+1$ complete graphs, with vertex sets $M_1\setminus F,\dots ,M_{p+1}\setminus F.$ $\square$

A crucial role in our proof will also be played by the following result, due to Gaschütz.

Proposition 14

([8]). Let N be a normal subgroup of a finite group G and suppose that $\langle g_1,\dots g_k\rangle N=G$. If $k \ge d(G),$ then there exist $n_1,\dots ,n_k \in N$ so that $\langle g_1n_1,\dots g_kn_k\rangle =G$.

Using the previous proposition, we may prove the following basic observation.

Proposition 15

Let N be a proper normal subgroup of G. If $\Sigma (G/N)$ is connected, then $\Sigma (G)$ is connected and ${{\,\mathrm{diam}\,}}(\Sigma (G))\le {{\,\mathrm{diam}\,}}(\Sigma (G/N)).$

Proof

Let $g_1,g_2$ be two different vertices of $\Sigma (G).$ If $g_1N=g_2N,$ then $\langle g_1, g_2 \rangle \le \langle g_1, g_2\rangle N < G$ since, by Lemma 12, G/N is not cyclic. In this case, $g_1,g_2$ are adjacent vertices of $\Sigma (G).$ So we may assume $g_1N\ne g_2N.$ In this case, there exists a path $(g_1N,y_1N,\dots ,y_rN,g_2N)$ in $\Sigma (G/N).$ By Proposition 14, for any $1\le i\le r$, $y_in_i \in V(G)$ for some $n_i \in N.$ Thus $(g_1,y_1n_1,\dots ,y_rn_r,g_2)$ is a path in $\Sigma (G).$ $\square$

Corollary 16

If G is non-soluble, then $\Sigma (G)$ is connected and ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 5.$

Proof

A non-soluble group G has an epimorphic image which is monolithic with non-Abelian socle. So the conclusion follows combining the previous proposition with Proposition 10. $\square$

A chief factor $A=X/Y$ of a finite group G is said to be a non-Frattini chief factor if $X/Y \not \le {{\,\mathrm{Frat}\,}}(G/Y).$

Corollary 17

Let G be a 2-generated soluble group. If there exists a non-Frattini and non-central chief factor A of G such that $G/C_G(A)$ is not cyclic of prime order, then G is connected and ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 3.$

Proof

If A is a non-Frattini chief factor of G, then G admits as an epimorphic image the semidirect product $A\rtimes G/C_G(A),$ so the conclusion follows combining Propositions  5 and 15. $\square$

Proof of Proposition 11

By Corollary 16, we may assume that G is soluble. Denote by W and F, respectively, the Fitting subgroup and the Frattini subgroup of G. By [10, Theorem 10.6 (c)], W/F has a complement in G/F and it is a direct product of minimal normal subgroups of G/F. In particular, $G/F \cong (V_1\times \dots \times V_u) \rtimes K,$ where $V_j$ is an irreducible K-module for $1\le j \le u$ and $\bigcap _{1\le j \le u}C_K(V_j)=1.$ Now let ${\mathcal {A}}$ be the set of the non-trivial irreducible G-modules that are G-isomorphic to a non-Frattini chief factor. By Corollary 17, we may assume that $G/C_G(A)$ is cyclic of prime order for every $A\in {\mathcal {A}}.$ This implies in particular that, for $1\le j\le u,$ either $C_K(V_j)=K$ or $K/C_K(V_j)\cong C_{p_j}$ for some prime $p_j.$ Moreover, $K \le \prod _{1\le j\le u}K/C_K(V_j)$ is Abelian. This implies in particular that G/W is Abelian, hence the derived subgroup $G^\prime$ of G is contained in W and therefore is nilpotent. We may assume that $V_j \le Z(G)$ if and only if $j>t.$ So we have $G/F\cong (V_1\times \dots \times V_t) \rtimes H$, with $H=(V_{t+1}\times \dots \times V_u)\times K.$ If $1\le j_1 < j_2\le t,$ then $(V_{j_1}\times V_{j_2})\rtimes H,$ being an epimorphic image of G, must be 2-generated, and this implies $V_{j_1}\not \cong _H V_{j_2}.$

$\square$

5 Groups with nilpotent derived subgroup

In this section, we investigate the connectivity of the graph $\Sigma (G)$ under the additional assumption that G is a finite soluble group whose derived subgroup is nilpotent. First we consider the particular case when G itself is nilpotent. The analysis of this case relies on Lemma 19, whose proof requires the following preliminary observation.

Lemma 18

Let A and B be soluble groups without common epimorphic images. Then two elements $(x_1,y_1)$, $(x_2,y_2)$ of $A\times B$ generate $A\times B$ if and only if $\langle x_1, x_2\rangle =A$ and $\langle y_1, y_2\rangle =B$.

Proof

Suppose $A=\langle x_1, x_2\rangle$ and $B=\langle y_1, y_2\rangle .$ Since H is a subdirect product of A and B,  by Goursat’s lemma (see, for example, [15, 4.3.1]), we can have $H < G$ only if A and B have a common non-trivial epimorphic image. $\square$

Lemma 19

If $G=A \times B$ is a non-cyclic soluble group and A and B have no common Abelian epimorphic image, then $\Sigma (G)$ is connected and ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 3,$ with equality if and only one of the following occurs:

• (1) A is cyclic and either $\Sigma (B)$ is disconnected or ${{\,\mathrm{diam}\,}}(\Sigma (B))>2.$

• (2) B is cyclic and either $\Sigma (A)$ is disconnected or ${{\,\mathrm{diam}\,}}(\Sigma (A))>2.$

Proof

By Lemma 18, $V(G)=V(A)\times V(B).$ We may assume that A is not cyclic (if A and B are both cyclic, then either they have a common epimorphic image or $A\times B$ is cyclic). Suppose that $g_1=(a_1,b_1)$, $g_2=(a_2,b_2)$ are two different vertices of $\Sigma (G).$ If $B\ne \langle b_1 \rangle$, then $((a_1,b_1),(a_2,b_1),(a_2,b_2))$ is a path in $\Sigma (G).$ Similarly, if $B\ne \langle b_2 \rangle$, then $((a_1,b_1),(a_1,b_2),(a_2,b_2))$ is a path in $\Sigma (G).$ If $B= \langle b_1 \rangle =\langle b_2 \rangle$, then $((a_1,b_1),(a_1,1),$ $(a_2,1),(a_2,b_2))$ is a path in $\Sigma (G)$. In the last case, if $(a_1,a,a_2)$ is a path in $\Sigma (A),$ then $((a_1,b_1), (a,b_1), (a_2,b_2))$ is also a path in $\Sigma (G).$

Finally assume that $B=\langle b \rangle$ is cyclic and there exist $a_1, a_2 \in V(A)$ without a common neighbour in $\Sigma (A).$ Then $(a_1,b)$ and $(a_2,b)$ do not have a common neighbour in $\Sigma (G)$ and therefore ${{\,\mathrm{diam}\,}}(\Sigma (G))\ge 3.$ $\square$

Corollary 20

Let G be a 2-generated finite nilpotent group. Then $\Sigma (G)$ is disconnected if and only if G is either a cyclic group or a p-group. Moreover, if G is neither a cyclic group nor a p-group, then ${{\,\mathrm{diam}\,}}(\Sigma (G))=3$ if G has only one non-cyclic Sylow subgroup, ${{\,\mathrm{diam}\,}}(\Sigma (G))=2$ otherwise.

Proof

We decompose $G=P_1\times \cdots \times P_t\times Q_1\times \cdots \times Q_u$ where $P_1,\dots ,P_t$ are the cyclic Sylow subgroups of G and $Q_1,\dots ,Q_u$ are the remaining Sylow subgroups. If $u=0$, then $\Sigma (G)$ is disconnected by Lemma 12. So we may assume $u\ge 1$. If $u=1$ and $t=0,$ then $\Sigma (G)$ is disconnected by Lemma 13. If $u=1$ and $t>0,$ then $G=X \times Q_1$ with $X=P_1\times \dots \times P_t$. So by Lemma 19, $\Sigma (G)$ is connected, and, since X is cyclic and $\Sigma (Q_1)$ is disconnected, ${{\,\mathrm{diam}\,}}(\Sigma (G))=3.$ If $u\ge 2,$ then $G=K \times Q_u$ with $K=P_1\times \cdots \times P_t\times Q_1\times \cdots \times Q_{u-1}.$ Neither K nor $Q_u$ is cyclic, so, again by Lemma 19, $\Sigma (G)$ is connected and ${{\,\mathrm{diam}\,}}(\Sigma (G))=2.$ $\square$

It remains to investigate the case when G is as described in Proposition 11 (2) and $t>0.$ By the following lemma it is not restrictive to assume ${{\,\mathrm{Frat}\,}}(G)=1.$

Lemma 21

Let G be a 2-generated finite group. Then $\Sigma (G)$ is connected if and only if $\Sigma (G/{{\,\mathrm{Frat}\,}}(G))$ is connected.

Proof

Let $F={{\,\mathrm{Frat}\,}}(G).$ Since G is cyclic if and only if G/F is cyclic, by Lemma 12, we may assume that G is not cyclic. By Proposition 15, we only have to prove that if $\Sigma (G/F)$ is disconnected, then $\Sigma (G)$ is also disconnected. So assume that $\Omega$ is a connected component of $\Sigma (G/F)$ and that there exists $yF \in V(G/F) {\setminus } \Omega .$ Since $\langle g_1, g_2 \rangle =G$ if and only if $\langle g_1F, g_2F \rangle =G/F,$ it follows immediately that $\Omega ^*=\{x \in G\mid xF \in \Omega \}$ is a connected component of $\Sigma (G)$ and $y \in V(G) \setminus \Omega ^*.$ $\square$

Assume now that ${{\,\mathrm{Frat}\,}}(G)=1$ and that G is a non-nilpotent group satisfying the description given in Proposition 11 (2). We have:

($*$):

$G\cong (V_1\times \dots \times V_t)\rtimes H$, where H is Abelian and $V_1,\dots ,V_t$ are pairwise non-H-isomorphic non-trivial irreducible H-modules.

Notice that if G satisfies $(*)$ and M is a maximal subgroup of G, then either $M=(V_1\times \dots \times V_t)K$ for some maximal subgroup K of H or M contains a conjugate of H (and consequently it contains $H\cap Z(G)$). In particular, $Z(G)\cap {{\,\mathrm{Frat}\,}}(H)\le {{\,\mathrm{Frat}\,}}(G)=1.$

The following lemma describes the possibilities that can occur if G satisfies $(*).$

Lemma 22

Assume that G satisfies $(*).$ Then one of the following occurs:

• (1) $\Sigma (G)$ is connected and ${{\,\mathrm{diam}\,}}(G)\le 3;$

• (2) $H \cong C_p$ for a suitable prime p;

• (3) $H\cong C_p^2$ for a suitable prime p.

Proof

Let $C_i=C_H(V_i)$ and $H_i=H/C_i.$ The primitive soluble group $K_i=V_i \rtimes H_i$ is an epimorphic image of G. If $|H_i|$ is not a prime, then by Propositions 5 and 15, $\Sigma (G)$ is connected and ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 3.$ So we may assume $|H_i|=p_i,$ with $p_i$ a prime for $1\le i\le t.$ Suppose $p_i\ne p_j$ for some $1\le i<j\le t.$ Then $K_i \times K_j$ is an epimorphic image of G and it can be easily seen that $V(K_1 \times K_2)=V(K_1)\times V(K_2)$. Arguing as in the proof of Lemma 19, it can be deduced that $\Sigma (K_1\times K_2),$ and consequently $\Sigma (G),$ is connected with diameter at most 3. So $p_i=p$ for $1\le i \le t.$ Moreover, $\bigcap _{1\le i\le t} C_i=Z(G)$, and therefore H/Z(G) is an Abelian group of exponent p. Let A be a $p^\prime$-Hall subgroup of Z(G). Then A has a complement, say B,  in H and $G=((V_1\times \dots \times V_t) \rtimes B) \times A$. Since A and $(V_1\times \dots \times V_t) \rtimes B$ have no common Abelian epimorphic image, it follows from Lemma 19 that $\Sigma (G)$ is connected with diameter at most 3. So H is a p-group and since $H^p\le Z(G),$ it follows that $H^p \le {{\,\mathrm{Frat}\,}}(H)\cap Z(G)=1.$ Since G is 2-generated, we conclude that $H\cong C_p^d$ with $d\le 2.$ $\square$

In the notation introduced in $(*)$, let $F_i={{\,\mathrm{End}\,}}_H(V_i).$ Since H is Abelian, $\dim _{F_i}V_i=1$. We may identify $V_i$ with the additive group of the field $F_i.$ Moreover, if $h \in H,$ then there exists $\alpha _i(h) \in F_i^*$ such that $v^h=\alpha _i(h)v$ for every $v\in V_i.$ The following holds:

Lemma 23

Assume that $g_1=(v_{1,1},\dots ,v_{1,t})h_1$ and $g_2=(v_{2,1},\dots ,v_{2,t})h_2$ are elements of G. For $1\le j\le t$, consider the matrix

$$\begin{aligned} A_j:=\begin{pmatrix} 1-\alpha _j(h_1)&{}1-\alpha _j(h_2)\\ v_{1,j}&{}v_{2,j} \end{pmatrix}. \end{aligned}$$

Then $\langle g_1,g_2\rangle =G$ if and only if the following hold:

• (1) $\langle h_1,h_2\rangle =H;$

• (2) $\det (A_j)\ne 0$ for every $1\le j\le t.$

Proof

See [12, Proposition 2.1 and Proposition 2.2]. $\square$

With the help of the previous lemma, we may analyse the connectivity of $\Sigma (G)$ when G is as described in (2) or (3) of Lemma 22. There are three situations that can occur, as described in the following three lemmas.

Lemma 24

Assume that G satisfies $(*).$ If $H \cong C_p,$ then $\Sigma (G)$ is disconnected.

Proof

Let $W=V_1\times \dots \times V_t.$ By Lemma 23, if $w=(v_1,\dots ,v_t) \in W$, then $w \in V(G)$ if and only if $v_i \ne 0$ for $1\le i \le t.$ Consider $\Omega =W \cap V(G).$ If $w_1, w_2$ are two different elements of $\Omega ,$ then they are adjacent in $\Sigma (G).$ Again by Lemma 23, if $g\in V(G){\setminus } \Omega ,$ then $G=\langle g, w\rangle$ for any $w \in \Omega .$ This implies that $\Omega$ is a proper connected component of $\Sigma (G).$ $\square$

Lemma 25

Assume that G satisfies $(*).$ If $H \cong C_p\times C_p$ and $Z(G)\ne 1,$ then $\Sigma (G)$ is disconnected.

Proof

In this case, $Z(G)=\langle h\rangle$ is a subgroup of G of order p. Let $W=V_1\times \dots \times V_t.$ By Lemma 23, if $x=(v_1,\dots ,v_t)h^j \in W\langle h\rangle$, then $x \in V(G)$ if and only if $h^j\ne 1$ and $v_i \ne 0$ for $1\le i \le t.$ Consider $\Omega =W\langle h\rangle \cap V(G).$ If $x_1, x_2$ are two different elements of $\Omega ,$ then they are adjacent in $\Sigma (G).$ Again by Lemma 23, if $g\in V(G) {\setminus } \Omega ,$ then $G=\langle g, x\rangle$ for any $x \in \Omega .$ This implies that $\Omega$ is a proper connected component of $\Sigma (G).$ $\square$

Lemma 26

Assume that G satisfies $(*).$ If $H \cong C_p\times C_p$ and $Z(G)= 1,$ then $\Sigma (G)$ is connected and ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 2.$

Proof

For $h \in H,$ let $\Delta (h)=\{i \in \{1,\dots ,t\}\mid h \in C_H(V_i)\}.$ Let $g=(v_1,\dots ,v_t)h \in G.$ By Lemma 23, $g\in V(G)$ if and only if $h\ne 1$ and $v_j\ne 0$ for any $j\in \Delta (h).$ Suppose that $g_1=(x_1,\dots ,x_t)h_1$, $g_2=(y_1,\dots ,y_t)h_2$ are two distinct vertices of $\Sigma (G).$ We may assume $\langle g_1, g_2\rangle =G$, otherwise $g_1,g_2$ are adjacent vertices of $\Sigma (G).$ Up to reordering, we may assume $\Delta (h_1)=\{1,\dots ,r\}$ for some $r \in \{0,\dots ,t\}.$ Since $H=\langle h_1, h_2\rangle$ and $Z(G)=\bigcap _{1\le j \le t}C_H(V_j)=1,$ we must have $\Delta (h_2)\subseteq \{r+1,\dots ,t\}.$ Up to reordering, we may assume $\Delta (h_2) = \{r+1,\dots ,r+s\}$ for some $s \in \{0,\dots ,t-r\}.$ Moreover, up to conjugation with a suitable element of $V_1\times \dots \times V_t,$ we may assume $x_j=0$ if $j > r$ and $y_k=0$ if $k\le r.$

If $r+s< t,$ then $g=(0,\dots ,0,y_{r+1},\dots ,y_{r+s},0,\dots ,0)h_2\in V(G).$ On the other hand, $\langle g_1, g\rangle$ is contained in $(V_1\times \dots \times V_{t-1})\rtimes H$ and $\langle g_2, g\rangle$ is contained in $(V_1\times \dots \times V_{t})\rtimes \langle h_2 \rangle ,$ so $(g_1,g,g_2)$ is a path in $\Sigma (G).$

Finally assume $r+s=t.$ In this case, $\Delta (h_1h_2)=\emptyset$, so $h_1h_2\in V(G)$. Moreover, $r>0,$ otherwise $\Delta (h_2)=\{1,\dots ,t\}$ and $h_2 \in Z(G),$ and $r<t$, otherwise $\Delta (h_1)=\{1,\dots ,t\}$ and $h_1 \in Z(G).$ Thus $\langle g_1, h_1h_2\rangle \le (V_1\times \dots \times V_{t-1})\rtimes H$ and $\langle g_2, h_1h_2\rangle \le (V_2\times \dots \times V_{t})\rtimes H$. But then $(g_1,h_1h_2,g_2)$ is a path in $\Sigma (G).$ $\square$

6 Proofs of Theorem 1 and Proposition 2

Proof of Theorem 1

Suppose that G is a 2-generated finite group and that $\Sigma (G)$ is disconnected. By Proposition 11, G is soluble with a nilpotent derived subgroup. If G is nilpotent, then, by Corollary 20, $\Sigma (G)$ is disconnected if and only if G is a cyclic group or a p-group. Suppose that G is not nilpotent. By Lemma 21, $\Sigma (G)$ is connected if and only if $\Sigma (G/{{\,\mathrm{Frat}\,}}(G))$ is connected. Combining Proposition 11 and Lemma  22, it follows that if $\Sigma (G)$ is disconnected, then $G/{{\,\mathrm{Frat}\,}}(G) \cong (V_1\times \dots \times V_t)\rtimes H,$ where either $H\cong C_p$ or $H\cong C_p\times C_p$ for some prime p,  and $V_1,\dots ,V_t$ are pairwise non-H-isomorphic non-trivial irreducible H-modules. If $H\cong C_p,$ then $\Sigma (G)$ is disconnected by Lemma 24. If $H \cong C_p\times C_p,$ then, by Lemmas 25 and 26, $\Sigma (G)$ is disconnected if and only if $Z(G/{{\,\mathrm{Frat}\,}}(G))\ne 1,$ or, equivalently, if and only if $C_H(V_1\times \dots \times V_t)\cong C_p.$

Finally, suppose $\Sigma (G)$ is connected. If G is non-soluble, Corollary 16 implies ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 5$. If G is soluble, then ${{\,\mathrm{diam}\,}}(\Sigma (G))\le 3$: indeed, when G is nilpotent, the conclusion follows from Corollary 20 and when G is non-nilpotent, it follows by combining Proposition 11, Lemma 22, and Lemma 26. $\square$

Proof of Proposition 2

First notice that generators of a cyclic group and involutions in $C_2\times C_2$ and $D_p$ are isolated vertices in the corresponding graphs. Conversely, let G be a 2-generated finite group and suppose that $g\in G$ is an isolated vertex of $\Sigma (G)$. If $g\ne g^{-1}$, then $G=\left\langle g,g^{-1}\right\rangle =\left\langle g\right\rangle$, so G is cyclic. Otherwise g is an involution. Suppose that this is the case and assume that G is not cyclic. Since $g \in V(G),$ the set $V_g=\{h \in G\mid \langle h, g\rangle =G\}$ is non-empty. Suppose $h \in V_g.$ Then $G=\langle g,h\rangle = \langle gh^i, h\rangle$ for any $i \in {\mathbb {Z}}.$ Hence $gh^i \in V(G)$ and therefore either $h^i=1$ or $\langle g, gh^i\rangle =\langle g, h^i\rangle =G$. In other words, if g generates G together with h, then it generates G together with any non-trivial power of h. If $\langle g,h \rangle$ is Abelian, this is possible only if |h| is a prime, and we must have $|h|=2$ otherwise G would be cyclic. So if G is Abelian, then $G\cong C_2\times C_2.$ If G is non-Abelian and $h \in V_g$, then $g\ne g^h \in V(G),$ thus $G=\left\langle g,g^h\right\rangle \cong D_n$ is a dihedral group of order 2n with $n=|gg^h|.$ In particular, g generates G together with any non-trivial power of $gg^h$ and this is possible only if n is a prime. $\square$