The independence graph of a finite group

Abstract

Given a finite group G, we denote by $\Delta (G)$ the graph whose vertices are the elements G and where two vertices x and y are adjacent if there exists a minimal generating set of G containing x and y. We prove that $\Delta (G)$ is connected and classify the groups G for which $\Delta (G)$ is a planar graph.

1 Introduction

The generating graph of a finite group G is the graph defined on the elements of G in such a way that two distinct vertices are connected by an edge if and only if they generate G. It was defined by Liebeck and Shalev in [14], and has been further investigated by many authors: see for example [5, 7, 9, 10, 13, 17, 20,21,22] for some of the range of questions that have been considered. Clearly the generating graph of G is an edgeless graph if G is not 2-generated. We propose and investigate a possible generalization, that gives useful information even when G is not 2-generated.

Let G be a finite group. A generating set X of G is said to be minimal if no proper subset of X generates G. We denote by $\Gamma (G)$ the graph whose vertices are the elements of G and in which two vertices x and y are joined by an edge if and only if $x\ne y$ and there exists a minimal generating set of G containing x and y. Roughly speaking, x and y are adjacent vertices of $\Gamma (G)$ is they are ‘independent’, so we call $\Gamma (G)$ the of G. We will denote by V(G) the set of the non-isolated vertices of $\Gamma (G)$ and by $\Delta (G)$ the subgraph of $\Gamma (G)$ induced by V(G). Our main result is the following.

Theorem 1

If G is a finite group, then the graph $\Delta (G)$ is connected.

We prove a stronger result in the case of finite soluble groups. For a positive integer u, we denote by $\Gamma _u(G)$ the subgraph of $\Gamma (G)$ in which x and y are joined by an edge if and only if there exists a minimal generating set of size u containing x and y. As before, we denote by $\Delta _u(G)$ the subgraph of $\Gamma _u(G)$ induced by the set $V_u(G)$ of its non-isolated vertices. Notice that, even when G is u-generated, the set $V_u(G)$ is in general different from V(G). For example if $G={{\,\mathrm{Sym}\,}}(4)$, then $\{(1,2)(3,4), (1,2), (1,2,3)\}$ is a minimal generating set for G, so $(1,2)(3,4)\in V(G);$ however $(1,2)(3,4)\notin V_2(G)$. If G is a non-cyclic 2-generated group, then $\Gamma _2(G)$ coincides with the generating graph of G and it follows from [10, Theorem 1] that $\Delta _2(G)$ is a connected graph if G is soluble. We generalize this result in the following way.

Theorem 2

If $u\in {\mathbb {N}}$ and G is a finite soluble group, then $\Delta _u(G)$ is connected.

Recall that a graph is said to be embeddable in the plane, or planar, if it can be drawn in the plane so that its edges intersect only at their ends. The 2-generated finite groups whose generating graph is planar have been classified in [18]. Our next result gives a classification of the finite groups G such that $\Gamma (G)$ is a planar graph.

Theorem 3

Let G be a finite group. Then $\Gamma (G)$ is planar if and only either $G\in \{C_2\times C_2, C_2\times C_4, D_4, Q_8, {{\,\mathrm{Sym}\,}}(3)\}$ or $G=C_n$ is cyclic of order n and one of the following occurs:

• (1) n is a prime-power.

• (2) $n=p\cdot q$, where p and q are distinct primes and $p\le 3$.

• (3) $n=4\cdot q$, where q is an odd prime.

Other results, and some related open questions, are presented in Section 5.

2 Proof of Theorem 1

Lemma 4

Let $g\in G$. Then g is isolated in $\Gamma (G)$ if and only if either $G=\langle g\rangle$ or $g\in {{\,\mathrm{Frat}\,}}(G)$.

Proof

Suppose $g\notin {{\,\mathrm{Frat}\,}}(G)$. There exists a maximal subgroup M of G with $g\notin M$. The set $X=\{g\}\cup M$ contains a minimal generating X of G and $g\in X$ (otherwise $G=\langle X\rangle \le M$). If $X\ne \{g\}$, then g is not isolated, otherwise $\langle g \rangle =G$. $\square$

Proposition 5

If G is a finite cyclic group, then $\Delta (G)$ is connected.

Proof

Let $|G|=p_1^{a_1}\ldots p_t^{a_t}$, where $p_1,\ldots ,p_t$ are distinct primes. If $t=1$, then $V(G)=\varnothing$. So assume $t>1$ and, for $1\le i\le t$, let $g_i$ be an element of G of order $|G|/p_i^{a_i}$. The subset $X=\{g_1,\ldots ,g_t\}$ induces a complete subgraph of $\Delta (G)$. Now let $x\in V(G)$. Since $x\notin {{\,\mathrm{Frat}\,}}(G)$, there exists $i\in \{1,\ldots ,t\}$ such that $p_i^{a_i}$ divides |x| and x is adjacent to $g_i$. $\square$

Lemma 6

Let N be a normal subgroup of a finite group G. If $Y=\{y_1,\ldots ,y_t\}$ has the property that $\langle Y, N\rangle =G$, but $\langle Z, N\rangle \ne G$ for every proper subset Z of Y, then there exist $n_1,\ldots ,n_u\in N$ such that $\{y_1,\ldots ,y_t,n_1,\ldots ,n_u\}$ is a minimal generating set of G.

Proof

Since $G=\langle Y, N\rangle$, $Y\cup N$ contains a minimal generating set X of G, and the minimality property of Y implies $Y\subseteq X$. $\square$

Lemma 7

Let N be a normal subgroup of a finite group G. If $x_1N$ and $x_2N$ are joined by an edge of $\Delta (G/N)$, then $x_1n_1$ and $x_2n_2$ are joined by an edge of $\Delta (G)$ for every $n_1, n_2\in N$.

Proof

Let $\{x_1N,x_2N,x_3N,\ldots ,x_tN\}$ be a minimal generating set of G/N. By Lemma 6, for every $n_1, n_2 \in N$, there exists $m_1,\ldots ,m_u\in N$ such that

$$\begin{aligned} \{ x_1n_1,x_2n_2,x_3,\ldots ,x_t,m_1,\ldots ,m_u\} \end{aligned}$$

is a minimal generating set of G. $\square$

We will write $x_1\sim _G x_2$ if $x_1$ and $x_2$ belong to the same connected component of $\Delta (G)$. The following lemma is an immediate consequence of Lemma 7.

Lemma 8

Let N be a normal subgroup of a finite group G and let $x, y \in G$. If $xN, yN \in V(G/N)$ and $xN\sim _{G/N}yN$, then $x\sim _G y$.

Lemma 9

[3, Corollary 1.5] Let G be a finite group with $S:= F^*(G)$ nonabelian simple. If x, y are nontrivial elements of G, then there exists $s\in G$ such that $\langle x, s\rangle$ and $\langle y, s\rangle$ both contain S.

Lemma 10

Let G be a finite monolithic primitive group. Assume that $N={{\,\mathrm{soc}\,}}G$ is non abelian and that $G= \langle x_1, N\rangle =\langle x_2, N\rangle$. Then there exists $m\in N$ such that $\langle x_1, m \rangle =\langle x_2, m \rangle =G$.

Proof

We have $N=S_1\times \cdots \times S_t$, where $t\in {\mathbb {N}}$ and $S_i\cong S$ with S a nonabelian simple group. First consider the case $t=1$. By Lemma 9 there exists $m\in N$ with $\langle x_1, m \rangle =\langle x_2, m \rangle =G$. Assume $t>1$. We have $G\le {{\,\mathrm{Aut}\,}}(S)\wr {{\,\mathrm{Sym}\,}}(t)$ and it is not restrictive to assume $x_1=(h_1,\ldots ,h_t)\sigma$ with $h_1,\ldots ,h_t\in {{\,\mathrm{Aut}\,}}(S)$, $\sigma \in {{\,\mathrm{Sym}\,}}(t)$ and $\sigma (1)=2$. There exists $u\in {\mathbb {Z}}$ such that $x_2^u=(h_1^*,\ldots ,h_2^*)\sigma$, with $h_1^*,\ldots ,h_t^*\in {{\,\mathrm{Aut}\,}}(S)$. Set $l_1:=x_1$, $l_2:=x_2^u$, $k_1:=h_1$, $k_2:=h_1^*$. Let w be an element of S of order 2. By Lemma 9, there exists $s\in S$ such that $\langle w^{k_1},s\rangle =\langle w^{k_2},s\rangle =S$. For $1\le i\le t$, consider the projection $\pi _i: N\rightarrow S_i\cong S$. Let $m=(w,s,1,\ldots ,1)\in N\cong S^t$. For $i\in \{1,2\}$, the subgroup $R_i:=\langle m, x_i\rangle$ contains $\langle m, m^{l_i}\rangle \le N$. Notice that $S=\langle s, w^{k_i}\rangle \le \pi _2(\langle m, m^{l_i}\rangle )$, hence $\pi _2(R_i\cap N)\cong S$. Since $R_iN=G$, we deduce that $\pi _j(R_i\cap N)\cong S$ for each $j\in \{1,\ldots ,t\}$. In particular (see for example [4, Proposition 1.1.39]) either $N\le R_i$ or there exist $k\in \{1,\ldots ,t\}$ and $h \in {{\,\mathrm{Aut}\,}}(S)$ such that $\pi _k(z)=h(\pi _1(z))$ for each $z\in R_i\cap N$. The second possibility cannot occur, since $m=(w,s,1,\ldots ,1)\in R_i\cap N$ and s and w are not conjugate in ${{\,\mathrm{Aut}\,}}S$ ($|w|=2$, while $|s|\ne 2$, otherwise S would be generated by two involutions). So $N\le R_i$ and consequently $R_i=G$. $\square$

Proof of Theorem 1

We prove the theorem by induction on the order of G. If can be easily seen that $x\in V(G)$ if and only if $x{{\,\mathrm{Frat}\,}}(G)\in V(G/{{\,\mathrm{Frat}\,}}(G))$ and that $\Delta (G)$ is connected if and only if $\Delta (G/{{\,\mathrm{Frat}\,}}(G)$ is connected. So if ${{\,\mathrm{Frat}\,}}(G)\ne 1$, the conclusion follows by induction. We may so assume ${{\,\mathrm{Frat}\,}}(G)\ne 1$. Let N be a minimal normal subgroup of G and let $x, y \in V(G)$. If xN and yN are non-isolated vertices of G/N, then by induction $xN \sim _G yN$, so it follows from Lemma 14 that $x\sim _G y$. This means that the set $\Omega _N$ of the elements $g\in V(G)$ such that $gN\in V(G/N)$ is contained in a unique connected component, say $\Gamma _N$, of $\Delta (G)$. Assume now $g\in V(G) \setminus \Omega _N$. If G/N is non-cyclic, then $gN\in {{\,\mathrm{Frat}\,}}(G/N)$. In particular a minimal generating set of G containing g must contain also an element z such that $zN\not \in {{\,\mathrm{Frat}\,}}(G/N)$. But then $z\in \Omega _N$ and, since $z\in \Gamma _N$ and $g\sim _G z$, we conclude $g\in \Gamma _N$. In other words, if G/N is cyclic, then $\Gamma _N=V(G)$. So we may assume that G/N is cyclic for every minimal normal subgroup N of G.

This implies that one of the following occur:

• (1) G is cyclic;

• (2) $G\cong C_p\times C_p$

• (3) G has a unique minimal normal subgroup, say N, and N is not central.

If G is cyclic, then the conclusion follows from Proposition 5. If $G\cong C_p\times C_p$, then $\Delta (G)$ is a complete multipartite graph, with $p+1$ parts of size $p-1$. So we may assume that the third case occurs. First assume that N is abelian. In this case N has a cyclic complement, $H=\langle h \rangle$, acting faithfully and irreducibly on N. We have $\langle n, h \rangle =G$ for every non trivial element n of G, and this implies that there exists a unique connected component $\Lambda$ of $\Delta (G)$ containing all the non trivial elements of N. Let now $g\in G\setminus N$. There is a conjugate $h^*$ of h in G with $g\notin \langle h^* \rangle$. If $1\ne n\in N$, then $G=\langle n, h^*\rangle = \langle g, h^*\rangle$, so $g\sim _G h^*\sim _G n$, hence $g\in \Lambda$. We remain with the case when N is non-abelian. Let $F/N={{\,\mathrm{Frat}\,}}(G/N)$ and set $\Sigma _1=F\setminus \{1\}$, $\Sigma _2=\{g\in G\mid \langle g \rangle N=G\}$, $\Sigma _3=\{g\in G\mid gN\in V(G/N)\}$ (we have $\Sigma _3 = \varnothing$ if and only if |G/N| is a prime power). Notice that V(G) is the disjoint union of $\Sigma _1$, $\Sigma _2$ and $\Sigma _3$. By Lemma 10, all the elements of $\Sigma _2$ belong to the same connected component, say $\Gamma$, of $\Delta (G)$. Assume $\Sigma _3\ne \varnothing$. Fix $y\in \Sigma _2$ and choose n such that $G=\langle y, n\rangle$. Let p be a prime divisor of |G/N| and let $y_1, y_2$ be generators, respectively, of a Sylow p-subgroup and a p-complement of $\langle y \rangle$. Since $\{y_1,y_2,n\}$ is a minimal generating set for G, it follows $y_1, y_2\in \Sigma _3$ and that $y_1\sim _G y_2\sim _G y \sim _G n$. But we noticed in the first part of this proof that all the elements of $\Sigma _3=\Omega _N$ belong to the same connected component, and so $\Sigma _2\cup \Sigma _3 \subseteq \Gamma$. Finally let $g\in \Sigma _1$ and let X be a minimal generating set of G containing g. Certainly $X\cap (\Sigma _2\cup \Sigma _3)\ne \varnothing$, so $g\in \Gamma$. $\square$

3 Soluble groups

Let u be a positive integer and G a finite group. In this section we will use the following notations. We will denote by $\Omega _u(G)$ the set of the minimal generating sets of G of size u, by $\Gamma _n(G)$ the graph whose vertices are the elements of G and in which $x_1$ and $x_2$ are adjacent if and only if there exists $X\in \Omega _n(G)$ with $x_1, x_2 \in X$. Moreover we will denote by $V_n(G)$ the set of the non-isolated vertices of $\Gamma _u(G)$ and by $\Delta _u(G)$ the subgraph of $\Gamma _u(G)$ induces by $V_u(G)$. Finally we will write $x_1\sim _{G,u}x_2$ to indicate that $x_1$ and $x_2$ belong to the same connected component of $\Delta _u(G)$.

We will need a series of preliminary results before giving the proof of Theorem 2. The following is immediate.

Lemma 11

Let G be a finite group. Then $\Delta _u(G)$ is connected if and only if $\Delta _u(G/{{\,\mathrm{Frat}\,}}(G)$ is connected.

Given a subset X of a finite group G, we will denote by $d_X(G)$ the smallest cardinality of a set of elements of G generating G together with the elements of X.

Lemma 12

[10, Lemma 2] Let X be a subset of G and N a normal subgroup of G and suppose that $\langle g_1,\ldots ,g_r, X, N\rangle =G$. If $r\ge d_X(G)$, we can find $n_1,\ldots ,n_r\in N$ so that $\langle g_1n_1,\ldots ,g_rn_r,X\rangle =G$.

Lemma 13

Let N be a normal subgroup of a finite group group G and consider the projection $\pi :G\rightarrow G/N$. Suppose $A\in \Omega _u(G/N)$ and $b \in V_u(G)$ with $bN\in A$. Then there exists $B\in \Omega _u(G)$ such that $b\in B$ and $A=\pi (B)$.

Proof

Let $A=\{bN,z_1N,\ldots ,z_{u-1}N\}$ and $t=d_{\{b\}}(G)$. Since $b\in V_u(G)$, $t\le u-1$. By Lemma 12, there exist $n_1,\ldots ,n_{u-1}\in N$ such that $\langle b,z_1n_1,\ldots ,z_{u-1}n_{u-1}\rangle =G$. The set $B:=\{b,z_1n_1,\ldots ,z_{u-1}n_{u-1}\}$ satisfies the requests of the statement. $\square$

Lemma 14

Let N be a normal subgroup of a finite group G and let $x, y \in V_u(G)$. If $xN, yN \in V_u(G/N)$ and $xN\sim _{{G/N},u}yN$, then there exists $n\in N$ such that $x\sim _{G,u} yn$.

Proof

Since $xN\sim _{G/N,u}yN$, there exists a sequence $A_1,\ldots ,A_t$ of elements of $\Omega _u(G/N)$ such that $xN\in A_1$, $yN\in A_t$ and $A_i\cap A_{i+1}\ne \varnothing$ for $1\le i\le t-1$. We claim that there exists a sequence $B_1,\ldots ,B_t$ of minimal generating sets of G such that $x\in B_1$, $\pi (B_i)=A_i$ for $1\le i\le t$ and $B_i\cap B_{i+1}\ne \varnothing$ for $1\le i\le t-1$. By Lemma 13, there exists a minimal generating set $B_1$ of G with $A_1= \pi (B_1)$ and $x\in B_1$. Suppose that $B_1,\ldots ,B_j$ have been constructed for $j<t$. There exists $g\in B_j$ such that $gN\in A_j\cap A_{j+1}$. Again by Lemma 13, there exists a minimal generating set $B_{j+1}$ of G with $A_{j+1}=\pi (B_{j+1})$ and $g\in B_{j+1}$. $\square$

Denote by d(G) and m(G), respectively, the smallest and the largest cardinality of a minimal generating set of G. A nice result in universal algebra, due to Tarski and known with the name of Tarski irredundant basis theorem (see for example [6, Theorem 4.4]) implies that, for every positive integer k with $d(G)\le k\le m(G)$, G contains an independent generating set of cardinality k. The proof of this theorem relies on a clever but elementary counting argument which implies also the following result:

Lemma 15

For every k with $d(G)\le k<m(G)$ there exists a minimal generating set $\{g_1,\ldots ,g_k\}$ with the property that there are $1\le i\le k$ and $x_1,x_2$ in G such that $\{g_1,\ldots ,g_{i-1},x_1,x_2,g_{i+1},\ldots ,g_k\}$ is again a minimal generating set of G. Moreover $x_1,x_2$ can be chosen with the extra property that $g_i=x_1x_2$.

Recall that for a d-generator finite group G, the swap graph $\Sigma _d(G)$ is the graph in which the vertices are the ordered generating d-tuples and in which two vertices $(x_1,\ldots ,x_d)$ and $(y_1,\ldots ,y_d)$ are adjacent if and only if they differ only by one entry.

Proposition 16

Let G be a finite soluble group. Then $\Delta _{d(G)}(G)$ is connected.

Proof

Let $d=d(G)$. If G is cyclic, then $\Delta _d(G)$ is a null graph, and there is nothing to prove. Assume $d\ge 2$ and let $x, y \in V_d(G)$. Let $X, Y \in \Omega _d(G)$ with $x\in X$ and $y\in Y$. By [12], the swap graph $\Sigma _d(G)$ is connected, so there exists a path in $\Sigma _d(G)$ joining X to Y. Notice that if A, B are adjacent vertices of $\Sigma _d(G)$, then there exists two connected components $\Gamma _A$ and $\Gamma _B$ of $\Delta _d(G)$ containing, respectively, A and B. On the other hand $A\cap B\ne \varnothing$, by the way in which the swap graph is defined. Thus $\Gamma _A\cap \Gamma _B \ne \varnothing$ and consequently $\Gamma _A=\Gamma _B$ and all the elements of $A\cup B$ belong to the same connected component. This implies in particular that if $A_1=X, A_2,\ldots , A_{t-1},A_t=Y$ is a path joining X and Y, then all the elements of $\cup _{1\le i\le t}A_i$ belong to the same connected component. $\square$

Proof of Theorem 2

We may assume $d(G)\le u\le m(G)$, otherwise $V_u(G)$ is empty. If $u=d(G)$, then the results follows from Proposition 16. So we assume $u>d(G)$. We prove the statement by induction on |G|. By Lemma 11, we may assume ${{\,\mathrm{Frat}\,}}(G)=1$.

Let N be a minimal normal subgroup of G. Let K be a complement of N in G. We have $d(K)\le d(G)<u$ and $m(K)=m(G/N)=m(G)-1\ge u-1$ (see [16, Theorem 2]). By the Tarski irredundant basis theorem, K has a minimal generating set $\{k_1,\ldots ,k_{u-1}\}$ of size $u-1$ and $\{k_1,\ldots ,k_{u-1},m\}$ is a minimal generating set of G for every $m\ne 1$. This implies that all the non-trivial elements of N belong to the same connected component, say $\Gamma$, of $\Delta (G)$.

In order to complete our proof, we are going to show that $X\cap \Gamma \ne \varnothing$, for every minimal generating set $X=\{x_1,\ldots ,x_u\}$ of G. For each $i \in \{1,\ldots ,u\}$, there exists $k_i\in K$ and $n_i\in N$ such that $x_i=k_in_i$. We may order the indices in such a way that $Y=\{k_1,\ldots ,k_t\}$ is a minimal generating set for K.

We distinguish two cases.

a) $t < u$. Let $H=\langle x_1,\ldots , x_t\rangle$. Since $G=\langle Y\rangle N=HN$ and $H\ne G$, we deduce that H is a complement for N in G and $\langle H, x_{t+1}\rangle =G$. In particular $t=u-1$ and $\{x_1,\ldots ,x_{t},m\}\in \Omega _u(G)$ for every $1\ne m\in G$. This implies $\{x_1,\ldots ,x_{t}\}\subseteq \Gamma \cap X$.

b) $t=u$. Since $d(K)\le d(G)<u$ and $m(K)\ge u$, by Lemma 15 there exists $\{z_1,\ldots ,z_u\}\in \Omega _u(K)$ with the property that $\{z_1z_2,\ldots ,z_u\}\in \Omega _{u-1}(K)$. We first want to prove that if $n\in N$ and ${{\tilde{z}}}:=z_un\in V_n(G)$, then ${\tilde{z}}\in \Gamma$. First suppose that there exists a complement H on N in G containing ${\tilde{z}}$. There exist $m_1,\ldots ,m_{u-1}\in N$ such that $z_im_i \in H$ for $1\le i\le u-1$. This implies

$$\begin{aligned} H=\langle z_1m_1z_2m_2,z_3m_3,\ldots ,z_{u-1}m_{u-1},{\tilde{z}}\rangle , \end{aligned}$$

but then $\{z_1m_1z_2m_2,z_3m_3,\ldots ,z_{u-1}m_{u-1},{\tilde{z}}, m\}\in \Omega _u(G)$ for every $1\ne m\in M$. Thus ${\tilde{z}}$ and m are adjacent vertices of $\Delta _u(G)$ and consequently ${\tilde{z}} \in \Gamma$. Now assume that no complement of N in G contains ${\tilde{z}}$. If $1\ne m\in N$, then $\langle z_1z_2,z_3,\ldots ,z_u,m\rangle =G$, hence $d_{\{z_u\}}(G)\le u-1$. Since $G=\langle z_1,z_2,\ldots ,z_u,N\rangle$, by Lemma 12 there exist $m_1,\ldots ,m_{u-1}\in M$ such that $\langle z_1m_1,\ldots ,z_{u-1}m_{u-1},z_u\rangle = G$. As before, this implies $z_u\in \Gamma$ and consequently $\{z_1m_1,\ldots ,z_{u-1}m_{u-1}\}\subseteq \Gamma$. On the other hand, since no complement for N in G contains ${\tilde{z}}$, it must be $\langle z_1m_1,\ldots ,z_{u-1}m_{u-1},{\tilde{z}}\rangle = G$. So ${\tilde{z}}$ is adjacent to the vertices $z_1m_1,\ldots ,z_{u-1}m_{u-1}$ of $\Delta _u(G)$ and consequently ${\tilde{z}} \in \Gamma$. Now we can conclude our proof. Since $\{x_1N,\ldots x_uN\}, \{z_1N,\ldots z_uN\} \in \Omega _u(G/N)$, we have $x_1N$, $z_uN\in V_n(G/N)$, so by induction $x_1N \sim _{G/N,u} z_uN$. By Lemma 14 there exists $n\in N$ such that $x_1\sim _{G,u} z_un$. But we proved before that $z_un \in \Gamma$, and this implies $x_1\in \Gamma \cap X$. $\square$

4 Planar graphs

Lemma 17

Let N be a normal subgroup of a finite group G. If $\Gamma (G)$ is planar, then either G/N is cyclic of prime-power order or $|N|\le 2$.

Proof

Assume that G/N is not a cyclic group of prime-power order. Then $\Delta (G/N)$ is not a null-graph. In particular there exist x and y in G such that xN and yN are joined by an edge of $\Gamma (G/N)$. By Lemma 7, the subgraph of $\Gamma (G)$ induced by $xN\cup yN$ is isomorphic to the complete bipartite graph $K_{a,a}$, with $|a|=N$. If $\Gamma (G)$ is planar, then $K_{a,a}$ is planar, and this implies $a\le 2$. $\square$

Proposition 18

$\Gamma (C_n)$ is planar if and only if one of the following occurs:

• (1) n is a prime-power.

• (2) $n=p\cdot q$, where p and q are distinct primes and $p\le 3$.

• (3) $n=4\cdot q$, where q is an odd prime.

Proof

If $n=p^a$ is a prime power, then $\Gamma (C_n)$ is an edgeless graph, and consequently it is planar. Assume that n is not a prime power and let $p < q$ be the two smaller prime divisors of n. We have that $C_n$ contains a normal subgroup N such that $G/N\cong C_{p \cdot q}$, and it follows from Lemma 17 that $|N|\le 2$. If $|N|=1$, then $\Delta (C_n)\cong K_{p-1,q-1}$, and consequently $\Gamma (C_n)$ is planar if and only if $p\le 3$. If $|N|=2$, then $p=2$ and $\Delta (G)\cong K_{2,2(q-1)}$, which is a planar graph. $\square$

Lemma 19

Let G be a finite group. If G is not cyclic, then there exists a normal subgroup N of G with the property that $d(G/N)=2$ but G/M is cyclic for every normal subgroup M of G with $N<M$.

Proof

Let ${\mathcal {M}}$ be the set of the normal subgroups M of G with the property that $d(G/M)=2$. We claim that if G is not cyclic, then ${\mathcal {M}} \ne \varnothing$. Indeed let

$$\begin{aligned} 1=N_t<\cdots <N_0=G \end{aligned}$$

be a chief series of G and let j be the smallest positive integer with the property that $G/N_j$ is not cyclic. By [15, Theorem 1.3], $d(G/N_j)=2$. Once we know that ${\mathcal {M}}$ is not empty, any subgroup in ${\mathcal {M}}$ which is maximal with respect to the inclusion satisfies the requests of the statement. $\square$

Proposition 20

Let G be a finite, non-cyclic group. Then $\Gamma (G)$ is planar if and only if $G\in \{C_2\times C_2, C_2\times C_4, D_4, Q_8, {{\,\mathrm{Sym}\,}}(3)\}$

Proof

Let G be a non-cyclic group. Choose a normal subgroup N of G as described in Lemma 19. It follows from Lemma 7 that $\Gamma (G)$ contains a subgraph isomorphic to $\Delta (G/N)$. So if $\Gamma (G)$ is planar, then $\Gamma (G/N)$ is planar and $|N|\le 2$. By [18] either $G/N\cong C_2\times C_2$ or $G/N\cong {{\,\mathrm{Sym}\,}}(3)$. If $G/N\cong C_2\times C_2$ then either $d(G)=m(G)$ and $G\in \{C_2\times C_2, C_2\times C_4, D_4, Q_8\}$, or $d(G)=m(G)=3$ and $G\cong C_2\times C_2\times C_2$. In the last case $\Delta (G)\cong K_7$ is not planar. In the other cases, $\Gamma (G)$ coincides with the generating graph of G and it is planar. If $G/N \cong {{\,\mathrm{Sym}\,}}(3)$, then $G\cong {{\,\mathrm{Sym}\,}}(3)$, $G\cong D_6$ or $G\cong C_3\rtimes C_4$. If $G\cong S_3$ then $\Gamma (G)$ coincides with the generating graph and it is planar. If $G\cong D_6$, then the six non-central involutions induces a complete subgraph, so $\Gamma (G)$ is not planar. If $G\cong C_3\rtimes C_4$, then the subset $A\cup B$, where A is the set of the six elements of order 4 and B is the set of the four elements with order divisible by 3, induces a non planar graph containing an isomorphic copy of $K_{6,4}$. $\square$

5 Examples and questions

The minimal generating sets for ${{\,\mathrm{Sym}\,}}(4)$ are described in [8]. We have that $d({{\,\mathrm{Sym}\,}}(4))=2$ and $m({{\,\mathrm{Sym}\,}}(4))=3$ and the three graphs $\Gamma _2({{\,\mathrm{Sym}\,}}(4))$, $\Gamma _3({{\,\mathrm{Sym}\,}}(4))$ and $\Gamma ({{\,\mathrm{Sym}\,}}(4))$ are described in the following tables, where the first column contains a representative x of a conjugacy class of ${{\,\mathrm{Sym}\,}}(4)$, the second column describes the set of the elements of ${{\,\mathrm{Sym}\,}}(4)$ adjacent to x in the graph and the third columns gives the degree of x in the graph. We denote by $X_i$ the set of i-cycles (for $2\le i\le 4)$ in ${{\,\mathrm{Sym}\,}}(4)$ and by Y the set of the double transpositions.

$\Gamma _2({{\,\mathrm{Sym}\,}}(4))$
(1,2)(2,3)	$\varnothing$	0
(1,2)	$\{(2,3,4)^{\pm 1},(1,3,4)^{\pm 1}, (1,2,3,4)^{\pm 1},(1,2,4,3)^{\pm 1}\}$	8
(1,2,3)	$X_4\cup \{(1,4), (2,4), (3,4)\}$	9
(1,2,3,4)	$X_3 \cup \{(1,2),(1,4),(2,3),(3,4),(1,3,2,4)^{\pm 1},(1,2,4,3)^{\pm 1}\}$	16

$\Gamma _3({{\,\mathrm{Sym}\,}}(4))$
(1,2)(3,4)	$X_2\cup X_3$	14
(1,2)	$Y\cup \{(1,2,3)^{\pm 1},(1,2,4)^{\pm 1},(1,3),(1,4),(2,3),(2,4),(3,4)\}$	12
(1,2,3)	$Y\cup \{(1,2), (1,3), (2,3), (1,2,4)^{\pm 1},(1,3,4)^{\pm 1},(2,3,4)^{\pm 1}\}$	12
(1,2,3,4)	$\varnothing$	0

$\Gamma ({{\,\mathrm{Sym}\,}}(4))$
(1,2)(3,4)	$X_2\cup X_3$	14
(1,2)	$Y \cup X_3\cup \{(1,3),(1,4),(2,3),(2,4),(3,4),(1,2,3,4)^{\pm 1},(1,2,4,3)^{\pm 1}\}$	20
(1,2,3)	$Y\cup X_2\cup X_4\cup \{(1,2,4)^{\pm 1},(1,3,4)^{\pm 1},(2,3,4)^{\pm 1}\}$	21
(1,2,3,4)	$X_3 \cup \{(1,2),(1,4),(2,3),(3,4), (1,3,2,4)^{\pm 1},(1,2,4,3)^{\pm 1}\}$	16

Denote by $\omega (\Gamma )$ the clique number of a graph $\Gamma$. By [20, Theorem 1.1], we have $\omega (\Gamma _2({{\,\mathrm{Sym}\,}}(4))=4$ and a maximal clique is $\{(1,2,3,4),(1,2,4,3),(1,3,2,4),(1,2,3)\};$ $\omega (\Gamma _3({{\,\mathrm{Sym}\,}}(4)))=7$ and a maximal clique is $X_2\cup \{(1,2)(3,4)\};$ $\omega (\Gamma ({{\,\mathrm{Sym}\,}}(4)))=11$ and a maximal clique is $X_2\cup \{(1,2)(3,4), (1,2,3), (1,2,4), (1,3,4), (2,3,4)\}$. However it is not in general true that $\omega (\Gamma _2({{\,\mathrm{Sym}\,}}(n))\le \omega (\Gamma _{n-1}({{\,\mathrm{Sym}\,}}(n))$. Indeed let n be a sufficiently large odd integer. By [2, Theorem 1], $\omega (\Gamma _2({{\,\mathrm{Sym}\,}}(n))=2^{n-1}$ while by [8, Theorem 2.1] a non-isolated vertex of $\omega (\Gamma _{n-1}({{\,\mathrm{Sym}\,}}(n))$ is either a transposition or a 3-cycle or a double transposition, so $\omega (\Gamma _{n-1}({{\,\mathrm{Sym}\,}}(n))\le \left( {\begin{array}{c}n\\ 2\end{array}}\right) +2\cdot \left( {\begin{array}{c}n\\ 3\end{array}}\right) + 3\cdot \left( {\begin{array}{c}n\\ 4\end{array}}\right)$. The independence number of $\Gamma _2({{\,\mathrm{Sym}\,}}(4))$ is 12 and a maximal independent set is $X_3\cup Y \cup \{id\}$. The independence number of $\Delta _3({{\,\mathrm{Sym}\,}}(4))$ is 8 and a maximal independent set is $X_4\cup \{(1,2),(1,3,4),(1,4,3),id\}$. The independence number of $\Delta ({{\,\mathrm{Sym}\,}}(4))$ is 6 and a maximal independent set is $Y\cup \{(1,2,3,4),(1,4,3,2),id\}$. For $u\in \{1,2\}$, the degree of the vertex of $\Gamma _u({{\,\mathrm{Sym}\,}}(4))$ corresponding to the element g is divisible by the order of g. When $u=2$, this follows from a more general result. Indeed, by [19, Proposition 2.2], if G is a 2-generated group and $g\in G$, then |g| divides the degree of g in the generating graph of G. However this cannot be generalized to $\Gamma _u(G)$ for arbitrary values of u. For example, consider the dihedral group $G=\langle a, b \mid a^6, b^2, (ab)^3\rangle$ of degree 6. Then $\{a^2,a^3,a^ib\}\in \Omega _3(G)$ for $0\le i\le 5$ and there are precisely 7 elements adjacent to $a^2$ in $\Gamma _3(G)$: $a^3$ and $a^ib$ for $0\le i\le 5$. We propose the following question.

Question 21

Let G be a finite group and $g\in G$. Does |g| divide the degree of g in $\Gamma _{d(G)}(G)?$

For a finite group G, let

$$\begin{aligned} W(G)=\bigcap _{d(G)\le u\le m(G)}V_u(G). \end{aligned}$$

We have seen that $W({{\,\mathrm{Sym}\,}}(4))=X_2\cup X_3\ne V({{\,\mathrm{Sym}\,}}(4))$. If $d(G)=m(G)$, then $V(G)=W(G)$ by definition. One may ask whether the converse is true.

Question 22

Does $V(G)=W(G)$ imply $d(G)=m(G)$?

The answer is positive in the soluble case.

Proposition 23

Let G be a finite soluble group. If $V(G)=W(G)$, then $d(G)=m(G)$.

Proof

Let $d=d(G)$, $m=m(G)$. Since $V_u(G/{{\,\mathrm{Frat}\,}}(G))=V_u(G){{\,\mathrm{Frat}\,}}(G)/{{\,\mathrm{Frat}\,}}(G)$, we may assume ${{\,\mathrm{Frat}\,}}(G)=1$. First assume that G is cyclic. Then $|G|=p_1\ldots p_m$, with $p_1,\ldots ,p_m$ distinct primes. Notice that $V_1(G)=\varnothing$, so $V(G)=W(G)\subseteq V_1(G)$ implies $V(G)=\varnothing$ and this is possible only if $m=1$. Now assume that G is not cyclic. By assumption $V_d(G)=V(G)=G\setminus \{1\}$. This is equivalent to say that $d_{\{g\}}(G)=d-1$ for any $1\ne g\in G$. By [9, Corollary 2.20, Theorem 2.21] either G is an elementary abelian p-group or there exist a finite vector space V, a nontrivial irreducible soluble subgroup H of ${{\,\mathrm{Aut}\,}}(V)$ and an integer $d>d(H)$ such that

$$\begin{aligned} G\cong V^{r(d-2)+1} \rtimes H, \end{aligned}$$

where r is the dimension of V over ${{\,\mathrm{End}\,}}_H(V)$ and H acts in the same way on each of the $r(d-2)+1$ factors. In the first case $d=m$, and we are done. In the second case, by [16, Theorem 2], $m=r(d-2)+1+m(H)$. If $d=2$, then $H=\langle h\rangle$ is a cyclic group and $G=V\rtimes H$. Since H is a maximal subgroup, if $h\in \Omega _u(G)$, then $u=2$. On the other hand, by assumption, $h\in V_m(G)$, and therefore $m=2$. Assume $d>2$. This implies $t=r(d-2)+1\ge 2$. We are going to prove that $r=1$. If $r\ne 1$, then there exist $v_1, v_2\in V$ that are ${{\,\mathrm{End}\,}}_G(V)$-linearly independent. This implies that the H-submodule W of $V^t$ generated by $w=(v_1,v_2,1,\ldots ,1)$ is H-isomorphic to $V^2$. As a consequence, if $w\in \Omega _u(G)$, then $u-1\le m(G/W)=m-2$. But then $w\notin \Omega _m(G)$, against the assumption $V_m(G)=V(G)$. So $r=1$, and this implies that H is isomorphic to a subgroup of the multiplicative group of ${{\,\mathrm{End}\,}}_G(V)$, and consequently it is cyclic. Moreover $t=d-1$ and $m(G)=m(H)+d-1$. Let h be a generator for of H. Notice that $h\notin \Omega _u(G)$ if $u-1>t=d-1$. Since, by assumption, $h\in \Omega _u(G)$, it must be $m-1\le d-1$, and consequently $m=d$. $\square$

The finite groups with $d(G)=m(G)$ are described in [1]. All the finite groups with this property are soluble. So Question 22 is equivalent to the following.

Question 24

Does there exist an unsoluble group G with $V(G)=W(G)?$

Another question that we propose is the following.

Question 25

Let G be a finite non-cyclic group. Is the graph $\Delta (G)$ Hamiltonian?

Notice that if G is cyclic, then $\Delta (G)$ is not necessarily Hamiltonian. For example, if $G\cong C_{2\cdot p}$, with p and odd prime, then $\Delta (G)\cong K_{1,p-1}$. In the case of ${{\,\mathrm{Sym}\,}}(4)$ the affirmative answer to the previous question follows from the Dirac’s Theorem, stating that an n-vertex graph in which each vertex has degree at least n/2 must have a Hamiltonian cycle. However it is not in general true that any vertex of $\Delta (G)$ has degree at least |V(G)|/2. Consider for example $G=\langle a, b \mid a^5, b^4, b^{-1}aba^3\rangle$. The graph $\Delta (G)$ has 19 vertices, and the degree of $b^2$ in this graph is 8. In any case, we may use Dirac’s Theorem in the case of finite nilpotent groups.

Theorem 26

If G is a finite non-cyclic nilpotent group, then $\Delta (G)$ is Hamiltonian.

Proof

Let $g\in V(G)$ and let $H=\langle g\rangle {{\,\mathrm{Frat}\,}}(G)$. Let $n=|V(G)|$ and d the degree of $g\in \Delta (G)$. Since $G/{{\,\mathrm{Frat}\,}}(G)$ is a direct product of elementary abelian p-groups, any element of $G\setminus H$ is adjacent to g in $\Delta (G)$. Since G is not cyclic, H is a proper subgroup of G, hence $|G|\ge 2|H|$ and therefore

$$\begin{aligned} d=|G|-|H| \ge \frac{|G|-|{{\,\mathrm{Frat}\,}}(G)|}{2}=\frac{n}{2}. \end{aligned}$$

So the conclusion follows from Dirac’s theorem. $\square$

Finally, a question that remains open is whether Theorem 2 remains true is the solubility assumption is removed.

Question 27

Let G be a finite group and $u\in {\mathbb {N}}$. Is $\Delta _u(G)$ a connected graph?