The diameter of the generating graph of a finite soluble group

TL;DR

This paper investigates the diameter of the generating graph of finite soluble groups, establishing conditions under which certain generating pairs exist and providing a counterexample for a specific case.

Contribution

It proves new conditions for the existence of common generators in finite soluble groups and constructs a counterexample demonstrating limitations of these conditions.

Findings

Existence of a common generator under certain conditions

Counterexample of a 2-generated soluble group where the condition fails

Weaker universal result for all finite soluble groups

Abstract

Let $G$ be a finite 2-generated soluble group and suppose that $\langle a_1,b_1\rangle=\langle a_2,b_2\rangle=G$. If either $G^\prime$ is of odd order or $G^\prime$ is nilpotent, then there exists $b \in G$ with $\langle a_1,b\rangle=\langle a_2,b\rangle=G.$ We construct a soluble 2-generated group $G$ of order $2^{10}\cdot 3^2$ for which the previous result does not hold. However a weaker result is true for every finite soluble group: if $\langle a_1,b_1\rangle=\langle a_2,b_2\rangle=G$, then there exist $c_1, c_2$ such that $\langle a_1, c_1\rangle = \langle c_1, c_2\rangle =\langle c_2, a_2\rangle=G.$

Equations

$${\varphi }_{G,N}(X,k)=|N{|}^k{P}_{G,N}(X,k)$$

$$P_{G,N}(X,k)=\sum _{\begin{array}{c}X\subseteq H\le G\\ HN=G\end{array}}\frac{\mu (H,G)}{|G:H{|}^k}.$$

$$P_{G,N}(X,k)=1-\frac{c}{|N{|}^k},$$

$$P_{G,N}(X,k)=1-\frac{|{\mathrm{Der}}_X(H,N)|}{|N{|}^k}.$$

$$n={\dim }_{{\mathbb{F}}_q}N,a={\dim }_{{\mathbb{F}}_q}{\mathrm{Der}}_X(H,N).$$

$$P_{G,N}(X,k)=1-\frac{q^a}{q^{nk}}.$$

$$P_{G,N}(X,k)=1-\frac{q^a}{q^{nk}}\ge 1-\frac{1}{q}=\frac{q-1}{q}.\text{qed}$$

$${\Omega }_i=\{n\in N\mid ⟨a_i,gn⟩=G\}.$$

$$|{\Omega }_i|=|N|P_{G,N}(\{a_i\},1)\ge |N|\frac{q-1}{q}\ge \frac{2|N|}{3}.$$

$$\left(\begin{array}{cc}1-k_1& 1-k_2\\ v_1& v_3\\ v_2& v_4\end{array}\right)\ne 0.$$

$$(v_1,v_2,v_3,v_4{)}^{(x,y)}=(v_1^x,v_2^x,v_3^y,v_4^y)$$

$$G=H⋉W.$$

$$N_1:=$$

$$N_2:=$$

$$(x_1,y_1)(v_{11},v_{12},v_{13},v_{14})=g_1,(x_2,y_2)(v_{21},v_{22},v_{23},v_{24})=g_2.$$

$$⟨g_1,g_2⟩R_G(Z)=G,⟨g_1,g_2⟩R_G(U_1)=G,⟨g_1,g_2⟩R_G(U_2)=G$$

$$⟨g_1,g_2⟩R_G(V_1)=G\text{ if and only if }⟨x_1(v_{11},v_{12}),x_2(v_{21},v_{22})⟩=(V_1\times V_2)⋊\mathrm{GL}(2,2),$$

$$⟨g_1,g_2⟩R_G(V_3)=G\text{ if and only if }⟨y_1(v_{31},v_{32}),y_2(v_{41},v_{42})⟩=(V_3\times V_4)⋊\mathrm{GL}(2,2).$$

$$⟨g_1,g_2⟩=G$$

$$(1)$$

$$(2)$$

$$(3)$$

$$x:=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right),y:=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right),z:=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right),$$

$$0=(0,0),e_1=(1,0),e_2=(0,1).$$

$$a_1:=$$

$$b_1:=$$

$$⟨(x,x),(y,z)⟩=H.$$

$$\det \left(\begin{array}{cc}1-x& 1-y\\ 0& e_1\\ e_2& 0\end{array}\right)=\det \left(\begin{array}{cccc}0& 0& 0& 1\\ 1& 0& 1& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\end{array}\right)=1,$$

$$\det \left(\begin{array}{cc}1-x& 1-z\\ 0& e_1\\ e_2& 0\end{array}\right)=\det \left(\begin{array}{cccc}0& 0& 0& 1\\ 1& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 1& 0& 0\end{array}\right)=1,$$

$$\det \left(\begin{array}{cc}1-x& 1-y\\ e_1& 0\\ e_2& 0\end{array}\right)=\det \left(\begin{array}{cccc}0& 0& 0& 1\\ 1& 0& 1& 1\\ 1& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right)=1,$$

$$\det \left(\begin{array}{cc}1-x& 1-z\\ e_1& e_1\\ e_2& 0\end{array}\right)=\det \left(\begin{array}{cccc}0& 0& 0& 1\\ 1& 0& 0& 0\\ 1& 0& 1& 0\\ 0& 1& 0& 0\end{array}\right)=1,$$

$$⟨a_1,b_1⟩=⟨a_2,b_2⟩=G.$$

Full text

1. Introduction

Let $G$ be a finite group. The generating graph for $G,$ written $\Gamma(G),$ is the graph where the vertices are the nonidentity 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 edges in this graph. Thus, it is natural to assume that $G$ is 2-generated when looking at this graph. There could be many isolated vertices in this graph. For example, all of the elements in the Frattini subgroup will be isolated vertices. We can also find isolated vertices outside the Frattini subgroup (for example the nontrivial elements of the Klein subgroup are isolated vertices in $\Gamma(\operatorname{Sym}(4)).$ Let $\Delta(G)$ be the subgraph of $\Gamma(G)$ that is induced by all of the vertices that are not isolated. In [4] it is proved that if $G$ is a 2-generated soluble group, then $\Delta(G)$ is connected. In this paper we investigate the diameter $\operatorname{diam}(\Delta(G))$ of this graph.

Theorem 1**.**

If $G$ is a 2-generated finite soluble group, then $\Delta(G)$ is connected and $\operatorname{diam}(\Delta(G))\leq 3.$

The situation is completely 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 $\operatorname{SL}(2,2^{p})$ and $p$ is a sufficiently large odd prime, then $\Delta(G)$ is connected but $\operatorname{diam}(\Delta(G))\geq 2^{p-2}-1$ (see [2, Theorem 5.4]).

For soluble groups, the bound $\operatorname{diam}(\Delta(G))\leq 3$ given in Theorem 1 is best possible. In Section 3 we construct a soluble 2-generated group $G$ of order $2^{10}\cdot 3^{2}$ with $\operatorname{diam}(\Delta(G))=3.$ However we prove that $\operatorname{diam}(\Delta(G))\leq 2$ in some relevant cases.

Theorem 2**.**

Suppose that a finite 2-generated soluble group $G$ has property that $|\operatorname{End}_{G}(V)|>2$ for every nontrivial irreducible $G$-module which is $G$-isomorphic to a complemented chief factor of $G$. Then $\operatorname{diam}(\Delta(G))\leq 2$, i.e. if $\langle a_{1},b_{1}\rangle=\langle a_{2},b_{2}\rangle=G,$ then there exists $b\in G$ with $\langle a_{1},b\rangle=\langle a_{2},b\rangle=G.$

Corollary 3**.**

Let $G$ be a 2-generated finite group. If the derived subgroup of $G$ has odd order, then $\operatorname{diam}(\Delta(G))\leq 2.$

Corollary 4**.**

Let $G$ be a 2-generated finite group. If the derived subgroup of $G$ is nilpotent, then $\operatorname{diam}(\Delta(G))\leq 2.$

2. Proof of Theorem 2

We could prove Theorem 2 with the same approach that will be used in the proof of Theorem 1. However we prefer to give in this particular case an easier and shorter proof. Before doing that, we briefly recall some necessary definitions and results. 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 $X.$ The following generalizes a result originally obtained by W. Gaschütz [7] for $X=\varnothing.$

Lemma 5** ([4] Lemma 6).**

Let $X$ be a subset of $G$ and $N$ a normal subgroup of $G$ and suppose that $\langle g_{1},\dots,g_{k},X\rangle N=G.$ If $k\geq d_{X}(G),$ then there exist $n_{1},\dots,n_{k}\in N$ so that $\langle g_{1}n_{1},\dots,g_{k}n_{k},X\rangle=G.$

It follows from the proof of [4, Lemma 6] that the number, say $\phi_{G,N}(X,k),$ of $k$-tuples $(g_{1}n_{1},\dots,g_{k}n_{k})$ generating $G$ with $X$ is independent of the choice of $(g_{1},\dots,g_{k}).$ In particular

[TABLE]

where $P_{G,N}(X,k)$ is the conditional probability that $k$ elements of $G$ generate $G$ with $X,$ given that they generate $G$ with $XN$.

Proposition 6** ([9] Proposition 16).**

If $N$ is a normal subgroup of a finite group $G$ and $k$ is a positive integer, then

[TABLE]

where $\mu$ is the Möbius function associated with the subgroup lattice of $G.$

Corollary 7**.**

Let $N$ be a minimal normal subgroup of a finite group $G$. Assume that $N$ is abelian and let $q=|\operatorname{End}_{G}(N)|.$ For every $X\subseteq G$, if $k\geq d_{X}(G)$ and $P_{G,N}(X,k)\neq 0,$ then $P_{G,N}(X,k)\geq\frac{q-1}{q}.$

Proof.

We may assume that $N$ is not contained in the Frattini subgroup of $G$ (otherwise $P_{G,N}(X,k)=1$). In this case, if $H$ is a proper supplement of $N$ in $G,$ then $H$ is a maximal subgroup of $G$ and complements $N$. Therefore $\mu(H,G)=-1$ and $|G:H|=|N|$. It follows from Proposition 6 that

[TABLE]

where $c$ is the number of complements of $N$ in $G$ containing $X.$ If $c=0,$ then $P_{G,N}(X,k)=1.$ Assume $c\neq 0$ and fix a complement $H$ of $N$ in $G$ containing $X.$ Let $\operatorname{Der}_{X}(H,N)$ be the set of derivations $\delta$ from $H$ to $N$ with the property that $x^{\delta}=1$ for every $x\in X.$ The complements of $N$ in $G$ containing $X$ are precisely the subgroups of $G$ of the kind $H_{\delta}=\{hh^{\delta}\mid h\in H\}$ with $\delta\in\operatorname{Der}_{X}(H,N)$, hence

[TABLE]

Now let $\mathbb{F}_{q}=\operatorname{End}_{G}(N)$. Both $N$ and $\operatorname{Der}_{X}(H,N)$ can be viewed as vector spaces over $\mathbb{F}_{q}.$ Let

[TABLE]

We have

[TABLE]

Since $P_{G,N}(X,k)\neq 0,$ we have $a<kq$ and

[TABLE]

Proof of Theorem 2.

We prove the theorem by induction on the order of $G.$ We may assume that $G$ is not cyclic and that the Frattini subgroup of $G$ is trivial. We distinguish two cases:

a) All the minimal normal subgroups of $G$ have order 2. In this case $G$ is an elementary abelian group of order 4 and $a_{1}$ and $a_{2}$ are nontrivial elements of $G.$ If $b\notin\{1,a_{1},a_{2}\},$ then $\langle a_{1},b\rangle=\langle a_{2},b\rangle=G.$

b) $G$ contains a minimal normal subgroup $N$ with $|N|\geq 3.$ By assumption $q=|\operatorname{End}_{G}(N)|\geq 3.$ Assume $\langle a_{1},b_{1}\rangle=\langle a_{2},b_{2}\rangle=G.$ By induction there exists $g\in G$ such that $\langle a_{1},g\rangle N=\langle a_{2},g\rangle N=G.$ For $i\in\{1,2\},$ let

[TABLE]

Since $\langle a_{i},b_{i}\rangle=G,$ we have $d_{\{a_{i}\}}(G)\leq 1,$ hence, by Lemma 5, $P_{G,N}(\{a_{i}\},1)\neq 0$ and consequently we deduce from Corollary 7 that

[TABLE]

But then $\Omega_{1}\cap\Omega_{2}\neq\varnothing.$ Let $b=gn$ with $n\in\Omega_{1}\cap\Omega_{2}.$ Then $G=\langle a_{1},b\rangle=\langle a_{2},b\rangle.$ ∎

Proof of Corollary 3.

Let $G^{\prime}$ be the derived subgroup of $G.$ If $|G^{\prime}|$ is odd, then $G^{\prime}$ is soluble by the Feit-Thompson Theorem, and consequently $G$ is also soluble. Moreover if $X$ and $Y$ are normal subgroups of $G$ such that $A=X/Y$ is a nontrivial irreducible $G$-module then $|A|$ is a power of a prime divisor $p$ of $|G^{\prime}|$ and $F=\operatorname{End}_{G}(A)$ is a finite field of characteristic $p$. Hence $|F|\geq p\geq 3$ and we may apply Theorem 2. ∎

Proof of Corollary 4.

We may assume $\operatorname{Frat}(G)=1.$ This means that $G=M\rtimes H$ where $H$ is abelian and $M=V_{1}\times\cdots\times V_{u}$ is the direct product of $u$ irreducible non trivial $H$-modules $V_{1},\dots,V_{u}.$ Let $F_{i}=\operatorname{End}_{H}(V_{i})=\operatorname{End}_{G}(V_{i})$: for each $i\in\{1,\dots,u\},$ $V_{i}$ is an absolutely irreducible $F_{i}H$-module so $\dim_{F_{i}}V_{i}=1.$ Now assume that $A$ is a nontrivial irreducible $G$-module $G$-isomorphic to a complemented chief factor of $G$: it must be $A\cong_{G}V_{i}$ for some $i$, so $|\operatorname{End}_{G}(A)|=|F_{i}|=|V_{i}|=|A|.$ It cannot be $|A|=2,$ otherwise $A$ would be a trivial $G$-module. Again we may apply Theorem 2. ∎

Remark 8**.**

Let $N$ be a noncentral and complemented minimal normal subgroup of a 2-generated soluble group $G$ and assume that $\mathbb{F}_{2}=\operatorname{End}_{G}(N).$ It follows from [9, Lemma 18] and [5, Lemma 18] that if $P_{G,N}(X,k)\leq 1/2,$ then there are $\dim_{\mathbb{F}_{2}}N$ different complemented factors $G$-isomorphic to $N$ in every chief series of $G.$ This means that, with the same arguments used in the proof of Theorem 2, a little bit stronger result can be proved:* Suppose that a finite 2-generated soluble group $G$ has the following property: if $V$ is a nontrivial irreducible $G$-module with $\operatorname{End}_{G}(V)=\mathbb{F}_{2},$ then a chief series of $G$ does not contain $\dim_{\mathbb{F}_{2}}V$ different complemented factors $G$-isomorphic to $V.$ Then $\operatorname{diam}(\Delta(G))\leq 2$.*

3. A finite soluble group $G$ with $\operatorname{diam}(\Delta(G))>2$

Let first recall some results that we will be applied in the discussion of our example. Let $G$ be a finite soluble group, and let $\mathcal{V}_{G}$ be a set of representatives for the irreducible $G$-groups that are $G$-isomorphic to a complemented chief factor of $G$. For $V\in\mathcal{V}_{G}$ let $R_{G}(V)$ be the smallest normal subgroup contained in $C_{G}(V)$ with the property that $C_{G}(V)/R_{G}(V)$ is $G$-isomorphic to a direct product of copies of $V$ and it has a complement in $G/R_{G}(V)$. The factor group $C_{G}(V)/R_{G}(V)$ is called the $V$-crown of $G$. The non-negative integer $\delta_{G}(V)$ defined by $C_{G}(V)/R_{G}(V)\cong_{G}V^{\delta_{G}(V)}$ is called the $V$-rank of $G$ and it coincides with the number of complemented factors in any chief series of $G$ that are $G$-isomorphic to $V$. If $\delta_{G}(V)\neq 0$, then the $V$-crown is the socle of $G/R_{G}(V)$. The notion of crown was introduced by Gaschütz in [8]. We have (see for example [10, Proposition 2.4]):

Proposition 9**.**

Let $G$ and $\mathcal{V}_{G}$ be as above. Let $x_{1},\ldots,x_{u}$ be elements of $G$ such that $\langle x_{1},\dots,x_{u},R_{G}(V)\rangle=G$ for any $V\in\mathcal{V}_{G}$. Then $\langle x_{1},\dots,x_{u}\rangle=G$.

Now let $V$ be a finite dimensional vector space over a finite field of prime order. Let $K$ be a $d$-generated linear soluble group acting irreducibly and faithfully on $V$ and fix a generating $d$-tuple $(k_{1},\dots,k_{d})$ of $K.$ For a positive integer $u$ we consider the semidirect product $G_{u}=V^{u}\rtimes K$ where $K$ acts in the same way on each of the $u$ direct factors. Put $F=\mathrm{End}_{K}(V)$. Let $n$ be the dimension of $V$ over $F$. We may identify $K=\langle k_{1},\dots,k_{d}\rangle$ with a subgroup of the general linear group $\operatorname{GL}(n,F)$. In this identification $k_{i}$ becomes an $n\times n$ matrix $X_{i}$ with coefficients in $F$; denote by $A_{i}$ the matrix $I_{n}-X_{i}.$ Let $w_{i}=(v_{i,1},\dots,v_{i,u})\in V^{u}.$ Then every $v_{i,j}$ can be viewed as a $1\times n$ matrix. Denote the $u\times n$ matrix with rows $v_{i,1},\dots,v_{i,u}$ by $B_{i}$. The following result is proved in [3, Section 4].

Proposition 10**.**

The group $G_{u}=V^{u}\rtimes K$ can be generated by $d$ elements if and only if $u\leq n(d-1).$ Moreover

1. (1)

$\mathrm{rank}\begin{pmatrix}A_{1}&\dots&A_{d}\end{pmatrix}=n.$** 2. (2)

$\langle k_{1}w_{1},\dots,k_{d}w_{d}\rangle=V^{u}\rtimes K$* if and only if $\mathrm{rank}\begin{pmatrix}A_{1}&\cdots&A_{d}\\ B_{1}&\cdots&B_{d}\end{pmatrix}=n+u.$*

In this section we will use in particular the following corollary of the previous proposition:

Corollary 11**.**

Let $V=\mathbb{F}_{2}\times\mathbb{F}_{2},$ where $\mathbb{F}_{2}$ is the field with 2 elements and let $\Gamma=\operatorname{GL}(2,2)\ltimes V^{2}.$ Assume that $\langle k_{1},k_{2}\rangle=\operatorname{GL}(2,2)$ and let $\gamma_{1}=k_{1}(v_{1},v_{2}),$ $\gamma_{2}=k_{2}(v_{3},v_{4})$ in $\Gamma.$ We have that $\Gamma=\langle\gamma_{1},\gamma_{2}\rangle$ if and only if

[TABLE]

Now we are ready to start the construction of a finite 2-generated soluble $G$ with $\operatorname{diam}(\Delta(G))>2.$ Let $H=\operatorname{GL}(2,2)\times\operatorname{GL}(2,2)$ and let $W=V_{1}\times V_{2}\times V_{3}\times V_{4}$ be the direct product of four 2-dimensional vector spaces over the field $\mathbb{F}_{2}$ with two elements. We define an action of $H$ on $W$ by setting

[TABLE]

and we consider the semidirect product

[TABLE]

Let

[TABLE]

A set of representatives for the $G$-isomorphism classes of the complemented chief factor of $G$ contains precisely 5 elements:

• •

$Z,$ a central $G$-module of order 2, with $R_{G}(Z)=G^{\prime}=\operatorname{SL}(2,2)^{2}\ltimes W.$

• •

$U_{1},$ a non central $G$-module of order 3, with $R_{G}(U_{1})=N_{2}\ltimes W.$

• •

$U_{2},$ a non central $G$-module of order 3, with $R_{G}(U_{2})=N_{1}\ltimes W.$

• •

$V_{1}$, with $R_{G}(V_{1})=V_{3}\times V_{4}\times N_{2}.$

• •

$V_{3}$, with $R_{G}(V_{3})=V_{1}\times V_{2}\times N_{1}.$

Let

[TABLE]

We want to apply Proposition 9 to check whether $\langle g_{1},g_{2}\rangle=G$. The three conditions

[TABLE]

are equivalent to $\langle g_{1},g_{2}\rangle W=G$, i.e. to $\langle(x_{1},y_{1}),(x_{2},y_{2})\rangle=H.$ Moreover

[TABLE]

[TABLE]

Applying Corollary 11 we conclude that

[TABLE]

if and only if the following conditions are satisfied:

[TABLE]

Consider the following elements of $\operatorname{GL}(2,2)$:

[TABLE]

and the following elements of $\mathbb{F}_{2}^{2}$:

[TABLE]

Let

[TABLE]

It can be easily checked that

[TABLE]

Moreover

[TABLE]

[TABLE]

[TABLE]

[TABLE]

so either $a_{1},b_{1}$ as $a_{2},b_{2}$ satisfy the three conditions (1), (2) (3) and therefore

[TABLE]

Now we want to prove that there is no $b\in G$ with $\langle a_{1},b\rangle=\langle a_{2},b\rangle=G.$ Let $b=(h_{1},h_{2})(v_{1},v_{2},v_{3},v_{4})$, and assume by contradiction that $\langle a_{1},b\rangle=\langle a_{2},b\rangle=G.$ We must have in particular that condition (1) holds, i.e. $\langle(x,x),(h_{1},h_{2})\rangle=H.$ Since $(x,x)$ has order 2 and $H$ cannot be generated by two involutions (otherwise it would be a dihedral group) at least one of the two elements $h_{1},h_{2}$ must have order 3: it is not restrictive to assume $h_{1}=y.$ Let $v_{1}=(\alpha,\beta),$ $v_{2}=(\gamma,\delta).$ Conditions (2) and (3) must be satisfied, hence we must have

[TABLE]

However

[TABLE]

[TABLE]

However, since $\alpha\in\mathbb{F}_{2}$ either $\alpha=0$ or $\alpha+1=0,$ so there is no $b\in G$ with $\langle a_{1},b\rangle=\langle a_{2},b\rangle=G.$

4. A problem in linear algebra

Before to prove Theorem 1, we need to collect a series of results in linear algebra. Denote by $M_{r\times s}(F)$ the set of the $r\times s$ matrices with coefficients over the field $F.$

Lemma 12**.**

[4, Lemma 3]** Let $V$ be a finite dimensional vector space over the field $F$. If $W_{1}$ and $W_{2}$ are subspaces of $V$ with $\dim W_{1}=\dim W_{2}$, then $V$ contains a subspace $U$ such that $V=W_{1}\oplus U=W_{2}\oplus U.$

Lemma 13**.**

Let $v_{1},\dots,v_{n},w_{1},\dots w_{n}\in F^{n}$, where $F$ is a finite field and either $|F|>2$ or $n\neq 1.$ There exist $z_{1},\dots,z_{n}\in F^{n}$ so that the two sequences

[TABLE]

[TABLE]

are both basis of $F^{n}.$

Proof.

Equivalently, we want to prove that for every pair of matrices $A,B\in M_{n\times n}(F),$ there exists $C\in M_{n\times n}(F),$ such that $\det(A+C)\neq 0$ and $\det(B+C)\neq 0.$ Since either $|F|>2$ or $n\neq 1,$ every element of $M_{n\times n}(F)$ can be expressed as the sum of two units [11]. In particular $A-B=U-V$ with $U,V\in\operatorname{GL}(n,F).$ We may take $C=U-A=B-V.$ ∎

Lemma 14**.**

Let $F$ be a finite field and assume $r\leq n.$ Given $R\in M_{r\times n}(F)$ and $S\in M_{r\times r}(F)$ consider the matrix $\begin{pmatrix}R&S\end{pmatrix}\in M_{r\times(n+r)}$. Assume $\operatorname{rank}\begin{pmatrix}R&S\end{pmatrix}=r$ and let $\pi_{R,S}$ be the probability that a matrix $Z\in M_{r\times n}(F)$ satisfies the condition $\operatorname{rank}(R+SZ)=r.$ Then

[TABLE]

Proof.

There exist $m\leq r,$ $X\in\operatorname{GL}(r,F)$ and $Y\in\operatorname{GL}(r,F)$ such that

[TABLE]

where $I_{m}$ is the identity element in $M_{m\times m}(F).$ Since

[TABLE]

and

[TABLE]

it is not restrictive (replacing $R$ by $XR$, $S$ by $XSY$ and $Z$ by $Y^{-1}Z$) to assume

[TABLE]

Denote by $v_{1},\dots,v_{r}$ the rows of $R$ and by $z_{1},\dots,z_{r}$ the rows of $Z.$ The fact that the rows of $(R\ S)$ are linearly independent implies that $v_{m+1},\dots,v_{r}$ are linearly independent vectors of $F^{n}.$ The condition $\operatorname{rank}(R+SZ)=r$ is equivalent to ask that

[TABLE]

are linearly independent. The probability that $z_{1},\dots,z_{m}$ satisfy this condition is

[TABLE]

Hence

[TABLE]

Lemma 15**.**

Assume that $F$ is a finite field and that $A$, $B_{1},$ $B_{2},$ $D_{1}$ and $D_{2}$ are elements of $M_{n\times n}(F)$ with the property that

[TABLE]

[TABLE]

Moreover assume that at least one of the following conditions holds:

1. (1)

$|F|>2$; 2. (2)

$\det A=0$; 3. (3)

$n\geq 2$* and $(\det B_{1},\det B_{2})\neq(0,0).$*

Then there exists $C\in M_{n\times n}(F)$ such that

[TABLE]

Proof.

Let $r=\operatorname{rank}(A).$ There exist $X,Y\in\operatorname{GL}(n,F)$ such that

[TABLE]

where $I_{r}$ is the identity element in $M_{r\times r}(F).$ Let $B_{11},B_{21}\in M_{r\times n}(F)$ and $B_{12},B_{22}\in M_{(n-r)\times n}(F)$ such that

[TABLE]

For $i\in\{1,2\},$ since

[TABLE]

it must be $\operatorname{rank}(B_{i2})=n-r$. In particular there exists $Z_{i}\in\operatorname{GL}(n,F)$ such that

[TABLE]

with $B_{i1}^{*}\in M_{r\times r}(F),$ $B_{i2}^{*}\in M_{r\times(n-r)}(F).$ Notice that

[TABLE]

This means that it is not restrictive to assume

[TABLE]

with $B_{i1}^{*}\in M_{r\times r}(F),$ $B_{i2}^{*}\in M_{r\times(n-r)}(F).$ Let $C_{1},D_{i1}\in M_{n\times r}(F)$ and $C_{2},D_{i2}\in M_{n\times(n-r)}(F)$ such that $\begin{pmatrix}C_{1}&C_{2}\end{pmatrix}=C$ and $\begin{pmatrix}D_{i1}&D_{i2}\end{pmatrix}=D.$ Notice that

[TABLE]

Assume that we can find $C_{1}$ such that

[TABLE]

and let $W_{1},W_{2}$ be the subspaces of $F^{n}$ spanned, respectively, by the columns of the two matrices $D_{11}-C_{1}B_{11}^{*}$ and $D_{21}-C_{1}B_{21}^{*}$. By Lemma 12, there exists a subspace $U$ of $F^{n}$ such that $F^{n}=W_{1}\oplus U=W_{2}\oplus U.$ If $C_{2}$ is a matrix whose columns are a basis for $U,$ then

[TABLE]

and $C=(C_{1}\ C_{2})$ is a matrix with the request property. Set

[TABLE]

The previous observation implies that a matrix $C$ with the requested properties exists if and only if there exists $Z\in M_{r\times n}(F)$ such that

[TABLE]

Notice that $R_{1},R_{2}\in M_{r\times n}(F)$, $S_{1},S_{2}\in M_{r\times r}(F)$ have the property that

[TABLE]

First assume that either $|F|=q>2$ or $r<n:$ by Lemma 14, we have

[TABLE]

and this is sufficient to ensure that a matrix $Z$ with the requested property exists. Therefore we may assume $r=n$ (i.e. $\det A\neq 0$) and $q=2.$ In this case, we assume also that at least one of the two matrices $B_{1}$ and $B_{2}$ is invertible. Let for example $\det B_{1}\neq 0.$ This implies $\det S_{1}\neq 0.$ There exist $m\leq n$ and $X,Y\in\operatorname{GL}(n,F)$ such that

[TABLE]

Notice that $R_{2}+S_{2}Z$ is invertible if and only if $XR_{2}Y+XS_{2}YY^{-1}ZY$ is invertible. Moreover $R_{1}+S_{1}Z$ is invertible if and only if $R_{1}Y+S_{1}YY^{-1}ZY$ is invertible, if and only if $(S_{1}Y)^{-1}R_{1}Y+Y^{-1}ZY$ is invertible. This means that (replacing $R_{1}$ by $(S_{1}Y)^{-1}R_{1}Y$, $R_{2}$ by $XR_{2}Y$ and $Z$ by $Y^{-1}ZY$) we may assume

[TABLE]

Let $v_{1},\dots,v_{n}$ be the rows of $R_{1},$ $w_{1},\dots,w_{n}$ the rows of $R_{2}$ and $z_{1},\dots,z_{n}$ the rows of $Z.$ Our request on $Z$ is equivalent to ask that the sequences

[TABLE]

[TABLE]

are both linearly independent. Notice that the condition $\operatorname{rank}(R_{2}\ S_{2})=n$ implies in particular that $w_{m+1},\dots,w_{n}$ are linearly independent. First assume $m\neq 1$. For $j>m,$ let $z_{j}=v_{j}+w_{j}$ so that $z_{j}+v_{j}=w_{j}$ and let $W=\langle w_{m+1},\dots,w_{n}\rangle.$ We then work in the vector space $F^{n}/W$ of dimension $m$ and our request is that the vectors $v_{1}+z_{1}+W,\dots,v_{m}+z_{m}+W$ and the vectors $w_{1}+z_{1}+W,\dots,w_{m}+z_{m}+W$ are linearly independent: Lemma 13 ensures that this request is fulfilled for a suitable choice of $z_{1},\dots,z_{m}.$ Finally assume $m=1$. As before for $j>2,$ let $z_{j}=v_{j}+w_{j}$ and let $W=\langle w_{3},\dots,w_{n}\rangle.$ We want to find $z_{1}$ and $z_{2}$ so that the two vectors $v_{1}+z_{1}+W,v_{2}+z_{2}+W$ and the two vectors $w_{1}+z_{1}+W,w_{2}+W$ are linearly independent. This is always possible. First choose $z_{1}$ so that $\langle w_{1}+z_{1}+W\rangle\notin\langle w_{2}+W\rangle$ and $v_{1}+z_{1}\notin W.$ Once $z_{1}$ has been fixed, choose $z_{2}$ so that $\langle v_{2}+z_{2}+W\rangle\notin\langle v_{1}+z_{1}+W\rangle.$ ∎

Remark 16**.**

Notice that when $|F|=2$ and $\det A\neq 0,$ we cannot drop the assumption $(\det B_{1},\det B_{2})\neq(0,0).$ Consider for example

[TABLE]

Then, as we noticed at the end of Section 3, there is no $C\in M_{2\times 2}(F)$ with

[TABLE]

This restriction in the statement of Lemma 15 is indeed the reason why we cannot have $\operatorname{diam}(\Delta(G))=2$ for every 2-generated finite soluble group $G.$**

Proposition 17**.**

Let $F$ be a finite field and let $n$ be a positive integer. Assume that either $n\geq 2$ or $|F|>2.$ Assume that $A_{0}$, $A_{1},$ $A_{2},$ $A_{3},$ $B_{0}$ and $B_{3}$ are elements of $M_{n\times n}(F)$ with the property that

[TABLE]

and

[TABLE]

Then there exist $B_{1},B_{2}\in M_{n\times n}(F)$ such that

[TABLE]

Proof.

Set $i=1$ if either $|F|>2$ or $|F|=2$ and $\det A_{1}=0,$ $i=2$ otherwise.

a) Assume $i=1.$ First choose $B_{2}$ so that

[TABLE]

Then, since either $\det A_{1}=0$ or $|F|>2,$ by Lemma 15 there exists $B_{1}$ such that

[TABLE]

b) Assume $i=2.$ First choose $B_{1}$ so that

[TABLE]

Then, since $\det A_{1}\neq 0$ or $|F|>2,$ by Lemma 15 there exists $B_{2}$ such that

[TABLE]

Corollary 18**.**

Let $K$ be a non-trivial 2-generated linear soluble group acting irreducibly and faithfully on $V$ and consider the semidirect product $G=V^{\delta}\rtimes K$ with $\delta\leq n=\dim_{\operatorname{End}_{G}(V)}V.$ Assume that there exists $x_{0},x_{1},x_{2},x_{3},$ in $K$ such that

1. (1)

$x_{0}w_{0}$* and $x_{3}w_{3}$ are non isolated vertices in the generating graph of $G,$* 2. (2)

$\langle x_{0},x_{1}\rangle=\langle x_{1},x_{2}\rangle=\langle x_{2},x_{3}\rangle=K.$**

Then there exist $w_{1},w_{2}\in G$ with

[TABLE]

Proof.

Since $V^{\delta}\rtimes K$ is an epimorphic image of $V^{n}\rtimes K$, it suffices to prove the statement in the particular case $G=V^{n}\times K.$ We may identify $x_{0},x_{1},x_{2},x_{3}$ with $X_{0},X_{1},X_{2},X_{3}\in\operatorname{GL}(n,F)$, where $F=\operatorname{End}_{G}(V)$ and $w_{0},w_{1},w_{2},w_{3}\in V^{n}$ with four matrices $B_{0},B_{1},B_{2},B_{3}$ in $M_{n\times n}(F).$ We now apply Proposition 10. Let

[TABLE]

Conditions (1) and (2) implies that

[TABLE]

and

[TABLE]

Moreover the statement is equivalent to say that there exist $B_{1},B_{2}\in M_{n\times n}(F)$ with

[TABLE]

The existence of $B_{1}$ and $B_{2}$ is ensured by Lemma 15 (notice that the fact that $K$ is a non-trivial subgroup of $\operatorname{GL}(n,F)$ implies that $n\geq 2$ if $|F|=2$). ∎

5. Proof of Theorem 1

At the beginning of Section 3 we recalled some properties of the crowns of a finite soluble group. In the proof of Theorem 1, we will use other two related results.

Lemma 19**.**

[1, Lemma 1.3.6]** Let $G$ be a finite solvable group with trivial Frattini subgroup. There exists a crown $C/R$ and a non trivial normal subgroup $U$ of $G$ such that $C=R\times U.$

Lemma 20**.**

[6, Proposition 11]** Assume that $G$ is a finite soluble group with trivial Frattini subgroup and let $C,R,U$ as in the statement of Lemma 19. If $HU=HR=G,$ then $H=G.$

Proof of Theorem 1.

We prove the theorem making induction on the order of $G$. Choose two non-isolated vertices $x$ and $y$ in the generating graph of $G$. Let $F=\operatorname{Frat}(G)$ be the Frattini subgroup of $G$. Clearly $xF$ and $yF$ are non-isolated vertices of the generating graph of $G/F.$ If $F\neq 1,$ then by induction there exists a path

[TABLE]

in the graph $\Gamma(G/F)$, with $n\leq 3.$ For every $0\leq i\leq n-1,$ we have $G=\langle g_{i},g_{i+1}\rangle F=\langle g_{i},g_{i+1}\rangle,$ hence $x=g_{0},\dots,g_{n}=y$ is a path in $\Gamma(G).$ Therefore we may assume $F=1.$ In this case, by Lemma 19, there exists a crown $C/R$ of $G$ and a normal subgroup $U$ of $G$ such that $C=R\times U.$ We have $R=R_{G}(A)$ where $A$ is an irreducible $G$-module and $U\cong_{G}A^{\delta}$ for $\delta=\delta_{G}(A).$ By induction the graph $\Gamma(G/U)$ contains a path $xU=g_{0}U,g_{1}U,\dots,g_{n-1}U,g_{n}U=yU$ with $n\leq 3.$ We may assume $n=3:$ indeed if $n=1$ we may consider the path $g_{0}U,g_{1}U,g_{0}U,g_{1}U$ and if $n=2$ we may consider the path $g_{0}U,g_{0}g_{1}U,g_{1}U,g_{2}U.$ So we are assuming

[TABLE]

We work in the factor group $\bar{G}=G/R.$ We have $\bar{C}=C/R=UR/R\cong U\cong A^{\delta}$ and either $A\cong C_{p}$ is a trivial $G$-module and $\bar{G}\cong(C_{p})^{\delta}$ or $\bar{G}=\bar{U}\rtimes\bar{H}\cong A^{\delta}\rtimes K$ where $K\cong\bar{H}$ acts in the same say on each of the $\delta$ factors of $A^{\delta}$ and this action is faithful and irreducible. Since $\bar{G}$ is 2-generated, we have $\delta\leq 2$ if $A$ is a trivial $G$-module, $\delta\leq n:=\dim_{{\operatorname{End}_{G}(A)}}A$ otherwise. By Theorem 2 in the first case (we are working in the nilpotent group $A^{\delta}$) and by Proposition 18 in the second case, there exist $u_{1},u_{2}\in U$ with $\langle\bar{x},\bar{g}_{1}\bar{u}_{1}\rangle=\langle\bar{g}_{1}\bar{u}_{1},\bar{g}_{2}\bar{u}_{2}\rangle=\langle\bar{g}_{2}\bar{u}_{2},\bar{y}\rangle=\bar{G}.$ i.e.

[TABLE]

By Lemma 20, from (5.1) and (5.2), we deduce

[TABLE]