# Semiregularity and connectivity
of the non-$\mathfrak{F}$ graph of a finite group

Andrea Lucchini  Andrea Lucchini
Università di Padova
Dipartimento di Matematica “Tullio Levi-Civita”
Via Trieste 63, 35121 Padova, Italy
email<EMAIL_ADDRESS>and  Daniele Nemmi  Daniele Nemmi
Università di Padova
Dipartimento di Matematica “Tullio Levi-Civita”
Via Trieste 63, 35121 Padova, Italy
email<EMAIL_ADDRESS>
###### Abstract.

Given a class $\mathfrak{F}$ of finite groups, we consider the graph
$\widetilde{\Gamma}_{\mathfrak{F}}(G)$ whose vertices are the elements of $G$
and where two vertices $g,h\in G$ are adjacent if and only if $\left\langle
g,h\right\rangle\notin\mathfrak{F}$. Moreover we denote by
$\mathcal{I}_{\mathfrak{F}}(G)$ the set of the isolated vertices of
$\widetilde{\Gamma}_{\mathfrak{F}}(G).$ We address the following question: to
what extent the fact that $\mathcal{I}_{\mathfrak{F}}(H)$ is a subgroup of $H$
for any $H\leq G,$ implies that the graph $\Gamma_{\mathfrak{F}}(G)$ obtained
from $\widetilde{\Gamma}_{\mathfrak{F}}(G)$ by deleting the isolated vertices
is a connected graph?

##  1\. Introduction

Let $\mathfrak{F}$ be a class of finite groups and $G$ a finite group.
Consider the graph $\widetilde{\Gamma}_{\mathfrak{F}}(G)$ whose vertices are
the elements of $G$ and where two vertices $g,h\in G$ are adjacent if and only
if $\left\langle g,h\right\rangle\notin\mathfrak{F}$. Moreover denote by
$\mathcal{I}_{\mathfrak{F}}(G)$ the set of isolated vertices of
$\widetilde{\Gamma}_{\mathfrak{F}}(G)$. In [12] we defined the
non-$\mathfrak{F}$ graph $\Gamma_{\mathfrak{F}}(G)$ of $G$ as the subgraph of
$\widetilde{\Gamma}_{\mathfrak{F}}(G)$ obtained by deleting the isolated
vertices. In that paper we concentrated our attention in the particular case
when $\mathfrak{F}$ is a saturated formation. In particular we addressed the
question whether $\mathcal{I}_{\mathfrak{F}}(G)$ is a subgroup of $G$. This is
not always the case, however it occurs for several saturated formations and,
more in general, we called semiregular a class $\mathfrak{F}$ with the
property that $\mathcal{I}_{\mathfrak{F}}(G)$ is a subgroup of $G$ for every
finite group $G$. In the same paper we called connected a class $\mathfrak{F}$
with the property that the graph $\Gamma_{\mathfrak{F}}(G)$ is connected for
any finite group $G$. The results obtained in [12] indicate that often a
semiregular formation is connected. This occurs for example for the formations
of abelian groups, nilpotent groups, soluble groups, supersoluble groups,
groups with nilpotent derived subgroup, groups with Fitting length less or
equal then $t$ for any $t\in\mathbb{N}.$

This stimulated us to investigate to what extent the semiregularity of
$\mathfrak{F}$ implies its connectivity. We addressed the following more
general question. Suppose that $\mathfrak{F}$ is a class containing only
soluble groups and closed under taking subgroups and that a finite group $G$
has the property that $\mathcal{I}_{\mathfrak{F}}(H)$ is a subgroup of $H$ for
any $H\leq G$. Does this implies that $\Gamma_{\mathfrak{F}}(G)$ is connected?

Our main theorem (Theorem 14) says that for a fixed class $\mathfrak{F}$,
either the answer is affirmative, or a minimal counterexample $G$ has a very
peculiar behaviour. Indeed $G$ is soluble, it cannot be generated with 2
elements and there exists an epimorphism $\pi:G\to V^{t}\rtimes H,$ where $H$
is 2-generated, $V$ is a faithful irreducible $H$-module,
$t=1+\dim_{\operatorname{End}_{H}(V)}(V),$ such that the following properties
hold. Denote by $\mathcal{W}$ be the set of the $H$-submodules of $V^{t}$ that
are $H$-isomorphic to $V^{t-1},$ and for any $W\in\mathcal{W}$, let
$M_{W}=\pi^{-1}(WH).$ Then there exists $W\in\mathcal{W}$ such that if
$\overline{W}\neq W,$ then there is no edge in $\Gamma_{\mathfrak{F}}(G)$
connecting two elements of $M_{\overline{W}}$ and one of the two following
situations occurs:

  1. (1)

any egde in $\Gamma_{\mathfrak{F}}(G)$ belongs to the subgraph induced by a
conjugate of $M_{W};$

  2. (2)

$H$ is cyclic of prime order and any egde in $\Gamma_{\mathfrak{F}}(G)$
belongs either to the subgraph induced by a conjugate of $M_{W}$ or to the
subgraph induced by $N=\pi^{-1}(V^{t}).$

We construct two examples in which these two situations occur. In the first
example, $H$ is a quaternion group, $V\cong C_{p}\times C_{p}$, where $p$ is
an arbitrary odd prime, and $\Gamma_{\mathfrak{F}}(G)$ has $p$ connected
components, corresponding to the non-isolated vertices in the subgraphs of
$\Gamma_{\mathfrak{F}}(G)$ induced by the $p$ different conjugates of $M_{W}$.
In the second example, $H\cong C_{3}$, $V\cong C_{2}\times C_{2}$ and
$\Gamma_{\mathfrak{F}}(G)$ has two connected components: one consisting of the
non-isolated vertices of $\Gamma_{\mathfrak{F}}(N),$ the other one consisting
of the union of the sets of non-isolated vertices in the four conjugates of
$M_{W}.$

Despite the presence of these examples, Theorem 14 can be used to prove that
the following classes of finite groups are connected: the class of cyclic
finite groups, the class of finite groups whose order is divisible by only one
prime, the class of finite groups whose order is divisible by at most two
primes.

##  2\. Proof of the main theorem

We begin this section by recalling some known results and proving some
preliminary lemmas, needed in the proof.

######  Proposition 1.

[7] Let $N$ be a normal subgroup of a finite group $G$ and suppose that
$\langle g_{1},\dots g_{k}\rangle N=G$. If $k\geq d(G),$ then there exist
$n_{1},\dots,n_{k}\in N$ so that $\langle g_{1}n_{1},\dots
g_{k}n_{k}\rangle=G$.

Recall that the generating graph of a finite group $G$ is the graph whose
vertices are the elements of $G$ and where $g_{1}$ and $g_{2}$ are adjacent if
and only if $\langle g_{1},g_{2}\rangle=G.$

######  Lemma 2.

Let $G$ be a 2-generated finite soluble group and denote by $\Omega(G)$ the
set of non-isolated vertices of the generating graph of $G$. Assume that $N$
is a minimal normal subgroup of $G$ and that $g_{1},g_{2}$ are two elements of
$G$ with $G=\langle g_{1},g_{2},N\rangle.$ Then there exists $i\in\\{1,2\\}$
with the property that $g_{i}N\subseteq\Omega(G).$

###### Proof.

Suppose $g_{1}n^{*}\notin\Omega(G)$ for some $n^{*}\in N.$ If $n\in N,$ then
$\langle g_{1}n^{*},g_{2}n,N\rangle=G$, while $\langle
g_{1}n^{*},g_{2}n\rangle<G.$ By [11, Lemma 1], there exists $n^{\prime}\in N$
with $\langle g_{1}n^{\prime},g_{2}n\rangle=G.$ Hence $g_{2}n\in\Omega(G).$ ∎

######  Definition 3.

Let $\mathfrak{F}$ be a class of finite groups. We define $\mathfrak{F}_{2}$
as the class of the finite groups $G$ with the property that any 2-generated
subgroup of $G$ is in $\mathfrak{F}.$

Assume that $\mathfrak{F}$ is closed under taking subgroups. We say that
$\mathfrak{F}$ is 2-recognizable whenever a group $G$ belongs to
$\mathfrak{F}$ if all 2-generated subgroups of $G$ belong to $F$. If
$\mathfrak{F}$ is not 2-recognizable, then $\mathfrak{F}_{2}>\mathfrak{F}.$
For example, if $\mathfrak{F}$ is the class of the metabelian finite groups,
and $G$ is a Sylow 2-subgroup of $\operatorname{Sym}(8)$, then
$G\in\mathfrak{F}_{2}\setminus\mathfrak{F}.$ However the following can be
immediately seen.

######  Lemma 4.

If $G$ is a finite group, then
$\Gamma_{\mathfrak{F}}(G)=\Gamma_{\mathfrak{F}_{2}}(G).$ In particular
$\mathcal{I}_{\mathfrak{F}}(G)=\mathcal{I}_{\mathfrak{F}_{2}}(G).$

######  Lemma 5.

Let $\mathfrak{F}$ be a class of finite groups. If
$\mathcal{I}_{\mathfrak{F}}(G)$ is a subgroup of $G$ and
$G=\mathcal{I}_{\mathfrak{F}}(G)\\!\left\langle g\right\rangle$ for some $g\in
G$, then $G\in\mathfrak{F}_{2}$.

###### Proof.

Let $x$ be an arbitrary element of $G$. We have $x=ig^{\alpha}$ for some
$i\in\mathcal{I}_{\mathfrak{F}}(G)$ and $\alpha\in\mathbb{Z}$. Moreover
$\left\langle g,ig^{\alpha}\right\rangle=\left\langle
g,i\right\rangle\in\mathfrak{F}$, since $i\in\mathcal{I}_{\mathfrak{F}}(G)$.
Hence $g\in\mathcal{I}_{\mathfrak{F}}(G)=\mathcal{I}_{\mathfrak{F}_{2}}(G)$,
so $G=\mathcal{I}_{\mathfrak{F}_{2}}(G)$ and therefore $G\in\mathfrak{F}_{2}$.
∎

######  Theorem 6.

[6, Theorem 16.6] Let $L$ and $M$ be inconjugate maximal subgroups of a finite
soluble group $G.$ If $M_{G}\not\leq L_{G},$ then $L\cap M$ is a maximal
subgroup of $L.$

######  Lemma 7.

Let $G$ be a finite soluble group. Suppose that $M$ and $L$ are maximal
subgroups of $G$ with $M,L\notin\mathfrak{F}_{2}$ and that the graphs
$\Gamma_{\mathfrak{F}}(L)$ and $\Gamma_{\mathfrak{F}}(M)$ are connected.
Denote by $\Gamma_{M}$ (resp. $\Gamma_{L}$) the connected component of
$\Gamma_{\mathfrak{F}}(G)$ containing the vertices of
$\Gamma_{\mathfrak{F}}(M)$ (resp. $\Gamma_{\mathfrak{F}}(L)$).

  1. (a)

If $L_{G}\not\leq M_{G},$ then either $\Gamma_{L}=\Gamma_{M}$ or $L\cap
M\subseteq\mathcal{I}_{\mathfrak{F}}(L).$

  2. (b)

If $M_{G}\not\leq L_{G}$ and $L_{G}\not\leq M_{G},$ then
$\Gamma_{L}=\Gamma_{M}.$

  3. (c)

If $M_{G}<L_{G}$ and $\Gamma_{L}\neq\Gamma_{M},$ then $L_{G}/M_{G}$ is the
unique minimal normal subgroup of $G/M_{G}.$

  4. (d)

If $M_{G}<L_{G}$ and $\Gamma_{L}\neq\Gamma_{M},$ then either $L$ is normal in
$G$ or $L/L_{G}$ is cyclic.

###### Proof.

(a) If $\Gamma_{M}\neq\Gamma_{L},$ then $L\cap
M\subseteq\mathcal{I}_{\mathfrak{F}}(M)\cup\mathcal{I}_{\mathfrak{F}}(L).$
Thus either $L\cap M\leq\mathcal{I}_{\mathfrak{F}}(L)$ or $L\cap
M\leq\mathcal{I}_{\mathfrak{F}}(M)$. In the second case, by Theorem 6, $L\cap
M$ is a a maximal subgroup of $M,$ hence $M=\langle L\cap M,m\rangle$ for some
$m\in M.$ But then $M=\mathcal{I}_{\mathfrak{F}}(M)\langle m\rangle.$ By Lemma
5, $M\in\mathfrak{F}_{2},$ against the hypotheses.

(b) If $\Gamma_{M}\neq\Gamma_{L},$ then, by (a), $L\cap
M\leq\mathcal{I}_{\mathfrak{F}}(L)\cap\mathcal{I}_{\mathfrak{F}}(M)$, but then
$L,M\in\mathfrak{F}_{2}$, against the hypotheses.

(c) Since $L_{G}\not\leq M_{G},$ by (b) we have $K:=L\cap
M\leq\mathcal{I}_{\mathfrak{F}}(L).$ Let
$N/M_{G}:=\operatorname{soc}(G/M_{G}).$ Then $N\leq L_{G},$ since $N/M_{G}$ is
the unique minimal normal subgroup of $G/M_{G}$. In particular $L=L\cap
MN=(L\cap M)N=KN$, and therefore $K^{L}=K^{N}.$ If $L_{G}\cap M>M_{G},$ then
$N\leq(L_{G}\cap M)^{N}\leq K^{N}=K^{L},$ and therefore
$L=KN=K^{L}\leq\mathcal{I}_{\mathfrak{F}}(L)$ and $L\in\mathfrak{F}_{2},$
against the assumptions. Hence $L_{G}\cap M=M_{G},$ and consequently
$L_{G}=N.$

(d) Suppose $L$ is not normal in $G$. Let
$U/L_{G}:=\operatorname{soc}(G/L_{G})$. As in the proof of (c), we have
$L=(L\cap M)N=(L\cap M)L_{G}.$ Since $L\notin\mathfrak{F}_{2}$ but, by (a)
$L\cap M\leq\mathcal{I}_{\mathfrak{F}}(L)$, there exist $l\in
L_{G}\setminus\mathcal{I}_{\mathfrak{F}}(L)$ and $s\in L$ such that
$\left\langle l,s\right\rangle\notin\mathfrak{F}$. If $U\\!\left\langle
s\right\rangle<G$, then $U\\!\left\langle s\right\rangle\leq S<G$ for a
maximal subgroup $S$ with $L_{G}<U\leq S_{G}$. Since $\left\langle
l,s\right\rangle\leq L\cap S$, we have $\Gamma_{L}=\Gamma_{S}$ and from (c),
$\Gamma_{M}=\Gamma_{S}=\Gamma_{L}$, against the assumptions. Therefore
$U\\!\left\langle s\right\rangle=G$ and $G/U\cong L/L_{G}$ is cyclic. ∎

The following is immediate.

######  Lemma 8.

Let $\mathfrak{F}$ be a class of groups which is closed under taking subgroups
and epimorphic images. Let $g,h\in G$ and $N\mathrel{\unlhd}G$.

  1. (a)

If $gN$ and $hN$ are adjacent vertices of $\Gamma_{\mathfrak{F}}(G/N)$, then
$g$ and $h$ are adjacent vertices of $\Gamma_{\mathfrak{F}}(G)$.

  2. (b)

If $g\in\mathcal{I}_{\mathfrak{F}}(G),$ then
$gN\in\mathcal{I}_{\mathfrak{F}}(G/N).$

  3. (c)

$\mathcal{I}_{\mathfrak{F}}(G)^{\sigma}=\mathcal{I}_{\mathfrak{F}}(G)$ for
every $\sigma\in\operatorname{Aut}(G)$.

######  Definition 9.

Let $\mathfrak{F}$ be a class of finite groups. We say that a finite group $G$
is $\mathfrak{F}$-semiregular if $\mathcal{I}_{\mathfrak{F}}(H)\leq H$ for
every $H\leq G.$

######  Lemma 10.

Let $\mathfrak{F}$ be a class of finite groups with the following properties:

  1. (1)

All the groups in $\mathfrak{F}$ are soluble.

  2. (2)

$\mathfrak{F}$ is closed under taking subgroups.

Suppose that $G$ is $\mathfrak{F}$-semiregular and $\Gamma_{\mathfrak{F}}(H)$
is connected for any proper subgroup $H$ of $G.$ If there exists a proper
normal subgroup $N$ of $G$ such that $G\setminus N$ is contained in a unique
connected component of $\Gamma_{\mathfrak{F}}(G),$ then either
$\Gamma_{\mathfrak{F}}(G)$ is connected or $N$ is a maximal subgroup of $G.$

###### Proof.

Let $\Omega$ be the connected component of $\Gamma_{\mathfrak{F}}(G)$
containing $G\setminus N.$

If $N\in\mathfrak{F}_{2}$, then every element of
$N\setminus\mathcal{I}_{\mathfrak{F}}(G)$ must be adjacent to an element of
$G\setminus N,$ so $N\setminus\mathcal{I}_{\mathfrak{F}}(G)\subseteq\Omega$.
But this implies $\Omega=G\setminus\mathcal{I}_{\mathfrak{F}}(G)$, and
consequently $\Gamma_{\mathfrak{F}}(G)$ is connected.

If $N\notin\mathfrak{F}_{2}$, then $\mathcal{I}_{\mathfrak{F}}(N)<N.$ Let $H$
be a maximal subgroup of $G$ containing $N$. Since $\Gamma_{\mathfrak{F}}(H)$
is connected, there exists a unique connected component of
$\Gamma_{\mathfrak{F}}(G),$ say $\Delta,$ containing
$H\setminus\mathcal{I}_{\mathfrak{F}}(H).$ Of course
$N\setminus\mathcal{I}_{\mathfrak{F}}(N)\subseteq
H\setminus\mathcal{I}_{\mathfrak{F}}(H),$ so
$N\setminus\mathcal{I}_{\mathfrak{F}}(N)\subseteq\Delta.$ Recall that
$G\setminus N\subseteq\Omega.$ Moreover if
$x\in\mathcal{I}_{\mathfrak{F}}(N)\setminus\mathcal{I}_{\mathfrak{F}}(G),$
then $\langle x,y\rangle\notin\mathfrak{F}$ for some $y\in G\setminus N,$ so
$\mathcal{I}_{\mathfrak{F}}(N)\setminus\mathcal{I}_{\mathfrak{F}}(G)\subseteq\Omega$.
If $\Delta\cap\Omega\neq\emptyset,$ then
$\Delta=\Omega=G\setminus\mathcal{I}_{\mathfrak{F}}(G)$ and
$\Gamma_{\mathfrak{F}}(G)$ is connected. If $\Delta\cap\Omega=\emptyset,$ then
$(H\setminus\mathcal{I}_{\mathfrak{F}}(H))\cap(H\setminus N)=\emptyset$, i.e.
$H=N\cup\mathcal{I}_{\mathfrak{F}}(H).$ Since $H\notin\mathfrak{F}_{2}$,
$\mathcal{I}_{\mathfrak{F}}(H)\neq H,$ and consequently $H=N.$ ∎

######  Lemma 11.

Let $\mathfrak{F}$ be a class of finite groups with the following properties:

  1. (1)

All the groups in $\mathfrak{F}$ are soluble.

  2. (2)

$\mathfrak{F}$ is closed under taking subgroups.

Suppose that $G$ is $\mathfrak{F}$-semiregular and $\Gamma_{\mathfrak{F}}(H)$
is connected for any proper subgroup $H$ of $G.$ If $G$ is not soluble, then
$\Gamma_{\mathfrak{F}}(G)$ is connected.

###### Proof.

Let $R=\operatorname{R}(G)$ be the soluble radical of $G$ and let
$\mathfrak{S}$ be the class of the finite soluble groups. Assume that
$G\neq\operatorname{R}(G).$ By [8, Theorem 6.4], the only isolated vertex of
$\tilde{\Gamma}_{\mathfrak{S}}(G/R)$ is the identity element and the graph
$\Gamma_{\mathfrak{S}}(G/R)$ is connected. By Lemma 8, all the elements of
$G\setminus R$ belong to the same connected component of
$\Gamma_{\mathfrak{S}}(G)$. Since $\mathfrak{F}\subseteq\mathfrak{S},$ there
exists a connected component, say $\Omega$, of $\Gamma_{\mathfrak{F}}(G)$ with
$G\setminus R\subseteq\Omega.$ Since $R$ cannot be a maximal subgroup of $G$,
it follows from Lemma 10 that $\Gamma_{\mathfrak{F}}(G)$ is connected. ∎

######  Lemma 12.

Let $\mathfrak{F}$ be a class of finite groups with the following properties:

  1. (1)

All the groups in $\mathfrak{F}$ are soluble.

  2. (2)

$\mathfrak{F}$ is closed under taking subgroups.

Suppose that $G$ is $\mathfrak{F}$-semiregular and $\Gamma_{\mathfrak{F}}(H)$
is connected for any proper subgroup $H$ of $G.$ If $G$ is a 2-generated
group, then $\Gamma_{\mathfrak{F}}(G)$ is connected.

###### Proof.

By the previous lemma we may assume that $G$ is soluble. Moreover we may
assume that if $g\in G,$ then $\langle g\rangle\in\mathfrak{F}$ (otherwise $g$
would be a universal vertex of $\Gamma_{\mathfrak{F}}(G)$ and consequently
$\Gamma_{\mathfrak{F}}(G)$ would be obviously connected). Finally, we may
assume $G\notin\mathfrak{F}_{2}$ (otherwise $\Gamma_{\mathfrak{F}}(G)$ is the
empty graph). This implies that the subgraph $\Delta(G)$ of the generating
graph of $G$ induced by its non-isolated vertices is a subgraph of
$\Gamma_{\mathfrak{F}}(G).$ By [3, Theorem 1], the graph $\Delta(G)$ is
connected, so there exists a connected component, say $\Gamma$, of
$\Gamma_{\mathfrak{F}}(G)$ containing the set $\Omega(G)$ of the vertices of
$\Delta(G).$ From now on, for any subgroup $H$ of $G,$ we will denote by
$\Omega_{\mathfrak{F}}(H)$ the set $H\setminus\mathcal{I}_{\mathfrak{F}}(H)$
of the vertices of $\Gamma_{\mathfrak{F}}(H).$ Since
$\Gamma_{\mathfrak{F}}(G)$ is disconnected, we must have
$\Omega_{\mathfrak{F}}(G)\neq\Gamma.$

Let $g\in\Omega_{\mathfrak{F}}(G)\setminus\Gamma.$ Then there exists
$\tilde{g}\in G$ such that $\langle g,\tilde{g}\rangle\notin\mathfrak{F}.$
Since $g\notin\Omega(G)$, $\langle g,\tilde{g}\rangle\neq G$ and therefore
there exists a maximal subgroup $M$ of $G$ with
$g\in\Omega_{\mathfrak{F}}(M).$ We have

(2.1) $\Omega_{\mathfrak{F}}(M)\cap\Omega(G)=\emptyset.$

Indeed assume $m\in\Omega_{\mathfrak{F}}(M)\cap\Omega(G)$. Since
$m,g\in\Omega_{\mathfrak{F}}(M),$ then $g$ and $m$ would be contained in the
same connected component of $\Gamma_{\mathfrak{F}}(M),$ and consequently in
the same connected component of $\Gamma_{\mathfrak{F}}(G).$ However, since
$m\in\Omega,$ this connected component would coincide with $\Gamma,$ in
contradiction with $g\notin\Gamma.$

We distinguish two cases:

a) $M$ is a normal subgroup of $G$. In this case
$\mathcal{I}_{\mathfrak{F}}(M)$ is a characteristic subgroup of $M$ and
therefore it is normal in $G.$ Moreover $G=\langle y,M\rangle$ for some $y\in
G.$ Since $G$ is 2-generated, by Proposition 1, there exist $m_{1},m_{2}\in M$
such that $G=\langle ym_{1},m_{2}\rangle.$ Since
$\Omega_{\mathfrak{F}}(M)\neq\emptyset$, we must have
$M\notin\mathfrak{F}_{2}$ and therefore, by Lemma 5,
$M/\mathcal{I}_{\mathfrak{F}}(M)$ is not cyclic. This implies
$m_{2}\notin\mathcal{I}_{\mathfrak{F}}(M).$ Hence
$m_{2}\in\Omega_{\mathfrak{F}}(M)\cap\Omega(G),$ against (2.1).

b) $M$ is not a normal subgroup of $G$. Let $M_{G}$ be the normal core of $M$
in $G$. We have $G/M_{G}\cong A/M_{G}\rtimes M/M_{G},$ where $A/M_{G}$ is a
faithful irreducible $(M/M_{G})$-module. By [5, Theorem 7] there exist
$k_{1},k_{2}\in M$ and $a\in A$ such that $G=\langle k_{1},k_{2}^{a}\rangle
M_{G}.$ By Proposition 1, there exist $m_{1},m_{2}\in M_{G}$ such that
$G=\langle k_{1}m_{1},k_{2}^{a}m_{2}\rangle.$ Hence

$G=\langle x,y^{a}\rangle\text{ with }x=k_{1}m_{1},y=k_{2}m_{2}^{a^{-1}}\in
M\text{ and }a\in A.$

Now we claim that $M_{G}$ is contained in $\mathcal{I}_{\mathfrak{F}}(M)$.
Consider a normal series

$1=U_{0}\leq U_{1}\leq\dots\leq U_{t}=M_{G}$

where, for $1\leq i\leq t,$ $U_{i}/U_{i-1}$ is a chief factor of $G.$ We prove
by induction on $i$ that $U_{i}\leq\mathcal{I}_{\mathfrak{F}}(M).$ Since
$\mathfrak{F}$ contains all cyclic subgroups of $G$,
$U_{0}=1\leq\mathcal{I}_{\mathfrak{F}}(M)$. Let $i<t$ and assume
$U_{i}\leq\mathcal{I}_{\mathfrak{F}}(M)$. It follows from Lemma 2 that there
are two possibilities: either $xuU_{i}\in\Omega(G/U_{i})$ for any $u\in
U_{i+1}$ or $y^{a}uU_{i}\in\Omega(G/U_{i})$ for any $u\in U_{i+1}.$ In the
first case, by Proposition 1, for any $u\in U_{i+1}$ there exists $\bar{u}\in
U_{i}$ such that $xu\bar{u}\in\Omega(G).$ Hence, by (2.1),
$xu\bar{u}\in\mathcal{I}_{\mathfrak{F}}(M).$ So $\langle xu\mid u\in
U_{i+1}\rangle\subseteq\mathcal{I}_{\mathfrak{F}}(M)$ and consequently
$U_{i+1}\subseteq\mathcal{I}_{\mathfrak{F}}(M).$ The same argument can be
applied in the second case. So we have proved
$M_{G}\leq\mathcal{I}_{\mathfrak{F}}(M).$ Since $x\in\Omega(G),$ it follows
from 2.1 that $x\in\mathcal{I}_{\mathfrak{F}}(M).$ But then, since
$M_{G}\leq\mathcal{I}_{\mathfrak{F}}(M),$ $M=\langle
x,y,M_{G}\rangle=\mathcal{I}_{\mathfrak{F}}(M)\langle y\rangle,$ hence
$M\in\mathfrak{F}_{2}$ by Lemma 5, in contradiction with
$g\in\Omega_{\mathfrak{F}}(M).$ ∎

######  Lemma 13.

Let $H$ be a 2-generated finite group and $V$ a faithful irreducible
$H$-module. Consider the semidirect product $X=V^{u}\rtimes H,$ with
$u=\dim_{\operatorname{End}_{H}(V)}(V).$ Suppose $H=\langle
h_{1},h_{2}\rangle.$ Then there exists $w\in V^{u}\rtimes H$ such that
$X=\langle h_{1},h_{2}w\rangle$ if and only if $h_{1}$ acts fixed-point-freely
on $V$.

###### Proof.

Let $F:=\operatorname{End}_{H}(V).$ We may identify $H=\langle
h_{1},h_{2}\rangle$ with a subgroup of the general linear group
$\operatorname{GL}(u,F)$. In this identification, for $i=1,2$, $h_{i}$ becomes
an $u\times u$ matrix $X_{i}$ with coefficients in $F$. Denote by $A_{i}$ the
matrix $I_{u}-X_{i}.$ Let $w=(v_{1},\dots,v_{u})\in V^{u}.$ Then every $v_{j}$
can be viewed as a $1\times u$ matrix. Denote the $u\times u$ matrix with rows
$v_{1},\dots,v_{u}$ by $B$. By [2, Section 4], $\langle h_{1},wh_{2}\rangle=X$
if and only if

(2.2) $\det\begin{pmatrix}A_{1}&A_{2}\\\ 0&B\end{pmatrix}\neq 0.$

A matrix $B$ so that (2.2) is satisfied can be found if and only if $A_{1}$ is
invertible, i.e. if and only if $h_{1}$ acts fixed-point-freely on $V$. ∎

######  Theorem 14.

Let $\mathfrak{F}$ be a class of finite groups with the following properties:

  1. (1)

All the groups in $\mathfrak{F}$ are soluble.

  2. (2)

$\mathfrak{F}$ is closed under taking subgroups.

Suppose that $G$ is a finite group which is minimal with respect to the
following properties: $G$ is $\mathfrak{F}$-semiregular and the graph
$\Gamma_{\mathfrak{F}}(G)$ is not connected. Then $G$ is soluble and there
exists an epimorphism

$\pi:G\to(V_{1}\times\dots\times V_{t})\rtimes H,$

where $H$ is 2-generated and there exists a faithful irreducible $H$-module
$V$ with $V_{i}\cong_{H}V$ for $1\leq i\leq t$ and
$t=1+\dim_{\operatorname{End}_{H}(V)}(V).$

Moreover let $\mathcal{W}$ be the set of the $H$-submodules of
$V_{1}\times\dots\times V_{t}$ that are $H$-isomorphic to $V^{t-1}.$ There
exists one and only one $W\in\mathcal{W}$ with the property that
$M=\pi^{-1}(W\rtimes H)\notin\mathfrak{F}_{2}$. If $g_{1},g_{2}\in G$ and
$\langle g_{1},g_{2}\rangle\notin\mathfrak{F}$, then either $\langle
g_{1},g_{2}\rangle\leq M^{x}$, for some $x\in G,$ or $H$ is cyclic of prime
order and $\langle g_{1},g_{2}\rangle\leq\pi^{-1}(V_{1}\times\dots\times
V_{t}).$

###### Proof.

For a finite group $X$, let $d(X)$ be the smallest cardinality of a generating
set of $X$. By the previous lemmas, $G$ is soluble and $d(G)\geq 3.$ In
particular $G$ contains a normal subgroup $N$ with the property that
$d(G/N)=3$ but $d(G/M)\leq 2$ for every normal subgroup $M$ of $G$ properly
containing $N.$ By [4] $G/N\cong V^{t}\rtimes H$, where $V$ is a faithful
irreducible $H$-module and

$t=1+\dim_{\operatorname{End}_{H}(V)}(V).$

Consider the epimorphism $\pi:G\to V^{t}\rtimes H$ and let $\mathcal{M}$ be
the set of the maximal subgroups of $G$ containing $N$ and with the property
that $M^{\pi}$ does not contain $V^{t}.$ If $M\in\mathcal{M},$ then $M/N\cong
V^{t-1}\rtimes H.$ Let $F:={\operatorname{End}_{H}(V)}.$ The multiplicity of
$V$ in $FH/J(FH)$ (where $FH$ denote the group algebra and $J(FH)$ its
Jacobson radical) is $t-1$ so it follows from [1, Lemma 1] that the smallest
cardinality of a subset of $V^{t}$ generating $V^{t}$ as an $H$-module is
$\lceil t/(t-1)\rceil=2.$ In particular there is no $w\in W$ with the property
that $\langle w\rangle_{H}=V^{t},$ denoting with $\langle w\rangle_{H}$ the
$H$-submodule of $V^{t}$ generated by $w$. If $g\in G,$ then $g^{\pi}=wh$ for
some $w\in V^{t}$ and $h\in H$, so $g^{\pi}\leq\langle w,h\rangle\leq\langle
w\rangle_{H}H,$ and therefore there exists $M\in\mathcal{M}$ with $g\in M.$

Assume that there exists no $M\in\mathcal{M}$ containing two adjacent vertices
of $\Gamma_{\mathfrak{F}}(G).$ Let $T$ be the preimage of $H$ under $\pi$. We
must have $T>N$ (otherwise if $g_{1},g_{2}$ are adjacent vertices, then
$\langle g_{1},g_{2}\rangle N$ is a proper subgroup of $G$ and it is contained
in some $M\in\mathcal{M}).$ In particular $T^{G}=G.$ Fix $l\in T$. For any
$g\in G,$ we have $g^{\pi}=wh,$ with $w\in V^{t}$ and $h\in H.$ Thus $\langle
l,g\rangle^{\pi}=\langle l^{\pi},wh\rangle\leq\langle w\rangle_{H}H.$ Since,
$\langle w\rangle_{H}\neq V^{t},$ there exists $M\in\mathcal{M}$ containing
$\langle l,g\rangle.$ In particular $l$ and $g$ are not adjacent in
$\Gamma_{\mathfrak{F}}(G)$ and therefore $l\in\mathcal{I}_{\mathfrak{F}}(G).$
Thus, $T\leq\mathcal{I}_{\mathfrak{F}}(G).$ But then
$G=T^{G}=\mathcal{I}_{\mathfrak{F}}(G),$ a contradiction.

So we may assume that there exists $K\in\mathcal{M}$ containing two adjacent
vertices $g_{1},g_{2}$ of $\Gamma_{\mathfrak{F}}(G).$ In particular
$K\notin\mathfrak{F}_{2}.$ Let $\tilde{\mathcal{M}}=\\{M\in\mathcal{M}\mid
M\notin\mathfrak{F}_{2}\\}.$ If $M_{1},M_{2}\in\tilde{\mathcal{M}}$ and
$(M_{1})_{G}\neq(M_{2})_{G}$ then if follows from Lemma 7, (b), that
$\Gamma_{M_{1}}=\Gamma_{M_{2}}.$

Suppose that there exists $K^{*}\in\tilde{\mathcal{M}},$ with $K^{*}_{G}\neq
K_{G}$. By the previous paragraph $\Gamma_{K}=\Gamma_{K^{*}}.$ We claim that
in this case $\Gamma_{K}$ contains all the vertices of
$\Gamma_{\mathfrak{F}}(G),$ in contradiction with the assumption that
$\Gamma_{\mathfrak{F}}(G)$ is disconnected. Assume that $g$ is a vertex of
$\Gamma_{\mathfrak{F}}(G)$. There exists $\tilde{g}$ such that $X=\langle
g,\tilde{g}\rangle\notin\mathfrak{F}.$ Let $L$ be a a maximal subgroup of $G$
containing $XN.$ If $L\in\mathcal{M},$ then
$g\in\Gamma_{L}=\Gamma_{K}=\Gamma_{K^{*}}$ by Lemma 7, (b). If
$L\notin\mathcal{M},$ then, up to conjugacy, we may assume $K^{\pi}=W_{1}H$,
$(K^{*})^{\pi}=W_{2}H$, $L^{\pi}=V^{t}X$, with $W_{1}\neq W_{2}$,
$W_{1}\cong_{H}W_{2}\cong_{H}V^{t-1}$ and $X\leq H.$ By Lemma 7, (b), either
$\Gamma_{L}=\Gamma_{K}=\Gamma_{K^{*}}$, or $\langle K\cap L,K^{*}\cap
L\rangle\leq\mathcal{I}_{\mathfrak{F}}(L).$ In the latter case $\langle(K\cap
L)^{\pi},(K^{*}\cap L)^{\pi}\rangle=\langle
W_{1}X,W_{2}X\rangle=V^{t}X=L^{\pi},$ but then
$\mathcal{I}_{\mathfrak{F}}(L)=L,$ a contradiction.

So from now on, we will assume that all the maximal subgroups in
$\tilde{\mathcal{M}}$ are conjugate to $K$. It is not restrictive to assume
$K^{\pi}=(V_{1}\times\dots\times V_{t-1})H$. Suppose now that $H$ is not
cyclic of prime order and, by contradiction, that there exist $g_{1},g_{2}\in
G$ such that $\langle g_{1},g_{2}\rangle\notin\mathfrak{F}$ and $\langle
g_{1},g_{2}\rangle$ is contained in no conjugate of $K.$ There exists a
maximal subgroup $J$ of $G$ with $\langle g_{1},g_{2}\rangle\leq J$ and
$J^{\pi}\geq V_{1}\times\dots\times V_{t}.$ We claim that, for any $g\in G,$
$\Gamma_{J}=\Gamma_{K^{g}}.$ Suppose that this is false. By Lemma 7, (c) and
(d), $J^{\pi}=V^{t}C$ where $C$ is a cyclic maximal subgroup of $H$ and $H=CS$
with $S=\operatorname{soc}H.$ Let $\mathcal{U}$ be the set of the
$H$-submodules $U$ of $V_{1}\times\dots\times V_{t}$ with $U\cong_{H}V$ and
$U\not\leq V_{1}\times\dots\times V_{t-1}.$ If
$\pi^{-1}(U)\leq\mathcal{I}_{\mathfrak{F}}(J)$ for every $U\in\mathcal{U},$
then $R:=\langle\pi^{-1}(U)\mid
U\in\mathcal{U}\rangle=\pi^{-1}(V_{1}\times\dots\times
V_{t})\subseteq\mathcal{I}_{\mathfrak{F}}(J).$ Since $J/R$ is cyclic, by Lemma
5, we would have $J\in\mathfrak{F}_{2}.$ So there exist $U\in\mathcal{U}$ and
$y_{1}\notin\mathcal{I}_{\mathfrak{F}}(J)$ with $y_{1}^{\pi}=u\in U.$ There
exists $y_{2}\in J$ with $\langle y_{1},y_{2}\rangle\notin\mathfrak{F}.$ Let
$y_{2}^{\pi}=wc,$ with $w\in V^{t}$ and $c$ in $C.$ If $\langle c\rangle\neq
C,$ then let $L$ be a maximal subgroup of $G$ containing
$\pi^{-1}((V_{1}\times\dots\times V_{t})S\langle c\rangle).$ Clearly
$\Gamma_{L}=\Gamma_{J}$ since
$\\{y_{1},y_{2}\\}\subseteq\Gamma_{L}\cap\Gamma_{J}.$ However, by Lemma 7 (c),
$\Gamma_{L}=\Gamma_{K^{g}}$. Hence $\Gamma_{J}=\Gamma_{K^{g}},$ independently
on the choice of $g$. Thus $\langle c\rangle=C$. There exists a conjugate
$\bar{c}$ of $c$ in $H$ such that $H=\langle c,\bar{c}\rangle.$ If $c$ acts
fixed-point-freely on $V,$ then, up to conjugacy, we may assume $w\in U$, but
then $\langle y_{1},y_{2}\rangle\leq\pi^{-1}(UC)$ is contained in a maximal
subgroup $M\in\mathcal{M}$, which is not conjugate to $K$ since $U\leq M_{G}$
but $U\not\leq K_{G}.$ Finally assume that $c$ does not act fixed-point-
freely. We have $w=w_{1}+w_{2},$ with $w_{1}\in U$ and
$w_{2}\in(V_{1}\times\dots\times V_{t-1}).$ Since $\bar{c}$, being a conjugate
of $c$, does not act fixed-point-freely of $V,$ by Lemma 13, $\bar{c}$ is an
isolated vertex in $(V_{1}\times\dots\times V_{t-1})H$, so $\langle
w_{2}c,\bar{c}\rangle$ is a proper subgroup of $(V_{1}\times\dots\times
V_{t-1})H$ supplementing $V_{1}\times\dots\times V_{t-1}.$ Thus, there exists
an $H$-submodule $W$ of $V_{1}\times\dots\times V_{t-1}$ such that $W\cong
V^{t-2}$ and $\langle w_{2}c,\bar{c}\rangle\leq WH.$ But then
$M=\pi^{-1}(UWH)\ \in\mathcal{M},$ $M$ is not conjugate to $K$ and $\langle
y_{1},y_{2}\rangle\leq M.$ ∎

######  Definition 15.

Let $\mathfrak{F}$ be a class of finite groups. We say that a finite group $G$
is strongly $\mathfrak{F}$-semiregular if $\mathcal{I}_{\mathfrak{F}}(X/Y)\leq
X/Y$ for every $X\leq G$ and $Y\mathrel{\unlhd}X.$

######  Lemma 16.

Let $\mathfrak{F}$ be a class of finite groups with the following properties:

  1. (1)

All the groups in $\mathfrak{F}$ are soluble.

  2. (2)

$\mathfrak{F}$ is closed under taking subgroups.

  3. (3)

$\mathfrak{F}$ is closed under taking epimorphic images.

Suppose that $G$ is strongly $\mathfrak{F}$-semiregular and
$\Gamma_{\mathfrak{F}}(H)$ is connected for any proper subgroup and any proper
quotient $H$ of $G.$ If $\Gamma_{\mathfrak{F}}(G)$ is disconnected, then
$G/M\in\mathfrak{F}_{2}$ for every $1\neq M\mathrel{\unlhd}G$.

###### Proof.

Suppose that there exists $1\neq M\mathrel{\unlhd}G$ such that
$G/M\notin\mathfrak{F}_{2}.$ Let
$I/M:=\mathcal{I}_{\mathfrak{F}}(G/M)\mathrel{\unlhd}G/M$. If
$g_{1}M,g_{2}M\notin I/M$, then by the hypotheses, they belong to the same
connected component of the graph $\Gamma_{\mathfrak{F}}(G/M)$. Since, by Lemma
8, a path from $g_{1}M$ to $g_{2}M$ in $\Gamma_{\mathfrak{F}}(G/M)$ can be
lifted to a path between $g_{1}$ and $g_{2}$ in $\Gamma_{\mathfrak{F}}(G)$, it
follows that $G\setminus I$ is contained in a unique connected component of
$\Gamma_{\mathfrak{F}}(G)$. By Lemma 10, either $\Gamma_{\mathfrak{F}}(G)$ is
connected or there exists $g\in G$ such that $\left\langle
g\right\rangle\\!I=G$. However, in the second case, $\left\langle
gM\right\rangle\\!I/M=\left\langle
gM\right\rangle\\!\mathcal{I}_{\mathfrak{F}}(G/M)=G/M$ and from Lemma 5 it
would follow $G/M\in\mathfrak{F}_{2},$ against the assumption. ∎

######  Lemma 17.

Let $\mathfrak{F}$ be a class of finite groups with the following properties:

  1. (1)

All the groups in $\mathfrak{F}$ are soluble.

  2. (2)

$\mathfrak{F}$ is closed under taking subgroups.

  3. (3)

$\mathcal{I}_{\mathfrak{F}}(G)$ is a subgroup of $G$ for every finite group
$G.$

  4. (4)

If $G_{1}\in\mathfrak{F}_{2}$ and $|G_{2}|=|G_{1}|,$ then
$G_{2}\in\mathfrak{F}_{2}.$

Then $\Gamma_{\mathfrak{F}}(G)$ is connected for every finite group $G.$

###### Proof.

Let $G$ be a counterexample of minimal order. According to the statement of
Theorem 14, there exists an epimorphism $\pi:G\to(V_{1}\times\dots\times
V_{t})\rtimes H,$ where $H$ is 2-generated and there exists a faithful
irreducible $H$-module $V$ with $V_{i}\cong_{H}V$ for $1\leq i\leq t$ and
$t=1+\dim_{\operatorname{End}_{H}(V)}(V).$ Moreover there exists
$W\in\mathcal{W}$ with $\pi^{-1}(W\rtimes H)\notin\mathfrak{F}_{2}$, while
$\pi^{-1}(W^{*}\rtimes H)\in\mathfrak{F}_{2}$ whenever $W\neq
W^{*}\in\mathcal{W}.$ Since $|\pi^{-1}(W\rtimes H)|=|\pi^{-1}(W^{*}\rtimes
H)|$, this contradicts (4). ∎

##  3\. Applications

######  Proposition 18.

Let $\mathfrak{C}$ be the class of the finite cyclic groups and let $G$ be a
finite group. Denote by $\tilde{\pi}(G)$ be the set of the prime divisors $p$
of $|G|$ with the property that a Sylow $p$-subgroup of $G$ is either cyclic
or a generalized quaternion group. Then
$\mathcal{I}_{\mathfrak{C}}(G)=\prod_{p\in\tilde{\pi}(G)}\operatorname{O}_{p}(Z(G)).$

###### Proof.

Suppose $x\in\mathcal{I}_{\mathfrak{C}}(G).$ Then for any $y\in G,$ $\langle
x,y\rangle$ is cyclic, so in particular $[x,y]=1.$ Hence $x\in Z(G).$ Let
$|x|=n,$ $p$ a prime divisor of $n$ and $\tilde{x}=x^{n/p}.$ Choose
$P\in\operatorname{Syl}_{p}(G)$ and let $y$ be an element of $P$ of order $p$.
Then $\langle\tilde{x},y\rangle$ is contained in $\langle x,y\rangle$, hence
it is cyclic and therefore $\langle\tilde{x}\rangle=\langle y\rangle.$ This
means that $\langle\tilde{x}\rangle$ is the unique minimal subgroup of $P,$
and therefore $P$ is either cyclic or a generalized quaternion group. We have
so proved
$\mathcal{I}_{\mathfrak{C}}(G)\subseteq\prod_{p}\operatorname{O}_{p}(Z(G)).$
Conversely assume $\tilde{\pi}=\\{p_{1},\dots,p_{t}\\}$ and let $x=x_{1}\cdots
x_{t}$ with $x_{i}\in\operatorname{O}_{p_{i}}(Z(G)).$ Let $y\in G$. Since
$x\in Z(G),$ $H=\langle x,y\rangle$ is an abelian subgroup of $G$. For any
prime $p$ dividing $|G|,$ let $Q=\operatorname{O}_{p}(\langle y\rangle)$ and
$P=\operatorname{O}_{p}(H).$ If $p\notin\tilde{\pi}$ then $Q=P$ is cyclic. If
$p\in\tilde{\pi},$ then $P$ is an abelian $p$-group with a unique minimal
subgroup, so again it is cyclic. It follows that $H$ itself is cyclic and
therefore $x\in\mathcal{I}_{\mathfrak{C}}(G).$ ∎

######  Corollary 19.

The graph $\Gamma_{\mathfrak{C}}(G)$ is connected for every finite group $G.$

###### Proof.

Suppose that $G$ is a finite group of minimal order with respect to the
property that $\Gamma_{\mathfrak{C}}(G)$ is not connected. Let $N$ be a
minimal normal subgroup of $G$. By the previous proposition, $G$ is strongly
$\mathfrak{C}$-semiregular, hence it follows from Lemma 16 that $G/N$ is
cyclic. But this implies that $G$ is 2-generated (see [10]), in contradiction
with Lemma 12. ∎

######  Proposition 20.

Let $\mathfrak{P}$ be the class of the finite groups whose order is divisible
by at most a unique prime. Then for any finite group $G$ the graph
$\Gamma_{\mathfrak{P}}(G)$ is connected.

###### Proof.

We may assume that all the elements of $G$ have prime power order (otherwise
$\Gamma_{\mathfrak{F}}(G)$ contains a universal vertex). In this case, let
$H\leq G.$ We have $\mathcal{I}_{\mathfrak{P}}(H)=H$ if $H$ is a $p$-group,
$\mathcal{I}_{\mathfrak{P}}(H)=1$ otherwise. In any case
$\mathcal{I}_{\mathfrak{P}}(H)$ is a subgroup of $H$ and therefore
$\Gamma_{\mathfrak{F}}(G)$ is connected by Lemma 17. ∎

######  Proposition 21.

Let $\mathfrak{D}$ be the class of the finite groups $G$ with the property
that $\pi(G)$ consists of at most two different primes.

  1. (1)

$\mathfrak{D}$ is 2-recognizable.

  2. (2)

For any $g$ in $G$, let $\pi(g)$ be the set of the prime divisors of $|g|.$ If
$|\pi(g)|\geq 3$ for some $g\in G$, then $g$ is adjacent to all the other
vertices of $\tilde{\Gamma}_{\mathfrak{D}}(G)$. If $|\pi(g)|\leq 2$ for every
$g\in G,$ then $\mathcal{I}_{\mathfrak{D}}(G)$ is a subgroup of $G.$

  3. (3)

For any finite group $G$ the graph $\Gamma_{\mathfrak{D}}(G)$ is connected

###### Proof.

(1) Assume that $G\in\mathfrak{D}_{2}$. This implies that $|\pi(g)|\leq 2$ for
any $g$ in $G$ (if $|\pi(x)|>2$ for some $x\in G$, then $\langle
x,y\rangle\notin\mathfrak{D}$ for any $y\in G$). Assume that there exists
$\tilde{g}\in G$ with $\pi(\tilde{g})=\\{p,q\\}.$ Since
$G\in\mathfrak{D}_{2},$ $\langle\tilde{g},g\rangle\in\mathfrak{D}$ for all
$g\in G$. But then $\pi(g)\subseteq\\{p,q\\}$ for all $g\in G$ and therefore
$\pi(G)=\\{p,q\\}$ and $G\in\mathfrak{D}.$ If $G$ is not soluble, then there
exists $H\leq G$ and $K\unlhd H$ such that $H/K\cong S$ is a finite non-
abelian simple group. In particular $H=\langle h_{1},h_{2}\rangle K$ for some
$h_{1},h_{2}\in H.$ But then $S=\langle h_{1},h_{2}\rangle K/K\cong\langle
h_{1},h_{2}\rangle/(K\cap\langle h_{1},h_{2}\rangle),$ so
$3\leq\pi(S)\leq\pi(\langle h_{1},h_{2}\rangle),$ in contradiction with
$\langle h_{1},h_{2}\rangle\in\mathfrak{D}.$ So we may assume that $G$ is a
finite soluble group all of whose elements have prime power order. By [9,
Theorem 1], $\pi(G)\leq 2$, an again we conclude $G\in\mathfrak{D}.$

(2) Suppose that $G$ is a finite group of minimal order with respect to the
properties that $\mathcal{I}_{\mathfrak{D}}(G)\neq\emptyset$ and
$\mathcal{I}_{\mathfrak{D}}(G)$ is not a subgroup of $G$. Let
$x,y\in\mathcal{I}_{\mathfrak{D}}(G)$ such that
$xy\notin\mathcal{I}_{\mathfrak{D}}(G)$. There exists $g\in G$ such that
$\left\langle xy,g\right\rangle\notin\mathfrak{D}$. Notice that the minimality
property of $G$ implies $G=\left\langle x,y,g\right\rangle.$ Let $M$ be a
minimal normal subgroup of $G$ and set
$I/M:=\mathcal{I}_{\mathfrak{D}}(G/M)\mathrel{\unlhd}G/M$. Since
$G=\left\langle x,y,g\right\rangle$, we have $\left\langle gM\right\rangle
I/M=G/M$. By Lemma 5, $G/M\in\mathfrak{D}_{2}=\mathfrak{D}$, so there exist
two primes $p_{1},p_{2}$ with $\pi(G/M)\subseteq\\{p_{1},p_{2}\\}.$ If
$\operatorname{R}(G)=1$, then by [8, Theorem 6.4], for every $1\neq h\in G,$
there exists $\tilde{h}\in G$ such that $\langle h,\tilde{h}\rangle$ is not
soluble. In particular there exists $\tilde{x}$ such that $\langle
x,\tilde{x}\rangle\notin\mathfrak{D},$ in contradiction with
$x\in\mathcal{I}_{\mathfrak{D}}(G).$ Thus we may assume that $M$ is an
elementary abelian $p$-group. We have $\pi(G/M)=\\{p_{1},p_{2}\\}$ and
$p\notin\\{p_{1},p_{2}\\},$ otherwise $G\in\mathfrak{D}.$ So $G=M\rtimes H,$
with $H\cong G/M.$ If $g\in\mathcal{I}_{\mathfrak{D}}(G),$ then $|\pi(g)|\leq
1.$ Suppose that $g$ is a non trivial $p_{1}$-element of $G$. Then there
exists a conjugate $H^{*}$ of $H$ in $G$ and a non-trivial $p_{2}$-element
$g^{*}$ such that $K=\langle g,g^{*}\rangle\leq H^{*}.$ If follows from [5,
Theorem 1] that there exists $m\in M$ such that $\langle g,g^{*}m\rangle$ is
not a complement of $M$ in $MK.$ This implies that $p\cdot p_{1}\cdot p_{2}$
divides $|\langle g,g^{*}m\rangle|$ and consequently
$g\notin\mathcal{I}_{\mathfrak{D}}(G).$ For the same reason if $g$ is a non-
trivial $p_{2}$-element, then $g\notin\mathcal{I}_{\mathfrak{D}}(G).$ Now we
have two possibilities: if $H$ contains no elements of order $p_{1}\cdot
p_{2},$ then $\mathcal{I}_{\mathfrak{D}}(G)=M;$ otherwise
$\mathcal{I}_{\mathfrak{D}}(G)=1.$ In any case,
$\mathcal{I}_{\mathfrak{D}}(G)$ is a subgroup of $G$, against our assumption.

(c) It follows from Lemma 17. ∎

##  4\. Example 1

The statement of Theorem 14 cannot be improved. In this section we exhibit a
class $\mathfrak{F}$ and a finite group $G$ containing a maximal subgroup $M$
with the property that the the connected components of
$\Gamma_{\mathfrak{F}}(G)$ are precisely the subgraphs induced by the
conjugates of $M$ in $G$.

Let $Q$ be the quaternion group and $V\cong C_{3}\times C_{3}$ a faithful
irreducible $Q$-module. Consider the semidirect product $X=(V_{1}\times
V_{2}\times V_{3})\rtimes Q$, with $V_{i}\cong_{Q}V$ for $1\leq i\leq 3.$ Let
$T=V_{1}\times V_{2}\times V_{3}$ and $\mathcal{U}=\\{U_{1},\dots,U_{9}\\}$ be
the set of submodules $U\leq_{Q}T$ with the properties that $T=(V_{1}\times
V_{2})\oplus U.$ Fix $9$ different primes $p_{1},\dots,p_{9}$ with the
property that 12 divides $p_{i}-1$ for $1\leq i\leq 9.$ Then $Y=V^{2}\rtimes
Q$ has a faithful irreducible action on a vector space of dimension 8 over the
field $\mathbb{F}_{p_{i}}$ with $p_{i}$ elements. So for $1\leq i\leq 9,$ we
may consider an irreducible $X$-module, say $W_{i}$, with $W_{i}\cong
C_{p_{i}}^{8}$ and $C_{X}(W_{i})=U_{i}.$ Now consider the semidirect product

$G=(W_{1}\times\dots\times W_{9})\rtimes X.$

Let $W=W_{1}\times\dots\times W_{9}.$ Let $\mathcal{M}$ be the set of the
maximal subgroups $M$ with the property that $W\leq M$ but $T\not\leq M.$ The
conjugacy classes of maximal subgroups in $\mathcal{M}$ are parametrized by
the set $\mathcal{Z}$ of the $Q$-submodules $Z$ of $T$ with $Z\cong_{Q}V^{2}.$
Indeed $\mathcal{M}$ is the disjoint union of the following families:

$\mathcal{M}_{Z}=\\{WZQ^{t}\mid t\in T\\},\text{ with }Z\in\mathcal{Z}.$

Let $Z^{*}=V_{1}\times V_{2}.$ Since $Z^{*}\oplus U_{i}=T,$ it follows that
$W_{i}$ is a faithful irreducible $M/W$-module for any
$M\in\mathcal{M}_{Z^{*}}$ and $1\leq i\leq 9.$ Suppose that $Z^{*}\neq
Z\in\mathcal{Z}.$ There exists $i\in\\{1,\dots,9\\}$ such that $U_{i}\leq Z.$
This implies that if $M\in\mathcal{M}_{Z}$ then $C_{M}(W_{i})>W$ and
consequently the maximal subgroups in $\mathcal{M}_{Z}$ are not isomorphic to
the ones in $\mathcal{M}_{Z^{*}}.$

Consider now the class $\mathfrak{F}$ of the finite groups with no subgroup
isomorphic to $M=WZ^{*}Q.$

Notice that $M$ is 2-generated, while $G$ is not. In particular, if
$g_{1},g_{2}\in G,$ then there is an edge
$(g_{1},g_{2})\in\Gamma_{\mathfrak{F}}(G)$ if and only if $\langle
g_{1},g_{2}\rangle=M^{t}\in\mathcal{M}_{Z^{*}},$ for some $t\in V_{3}.$

For a finite group $Y$, denote by $\Omega(Y)$ the subset of $G$ consisting of
the elements $x$ with the property that $Y=\langle x,y\rangle$ for some $y\in
Y.$

Assume $g=wzh\in\Omega(M^{t})$ with $w\in W,z\in Z^{*}$ and $h\in Q^{t}$. It
must be $h\notin F^{t}$, being $F:=\operatorname{Frat}(Q).$ Moreover it
follows from [12, Proposition 2.2] that $wzh\in\Omega(M^{t})$ if and only if
$zh\in\Omega(\Gamma(Z^{*}Q^{t}))$ and, since the action of $Q^{t}$ on $V$ is
fixed-point-free, by Lemma 13 the condition $h\notin F^{t}$ is sufficient to
ensure $zh\in\Omega(\Gamma(Z^{*}Q))$. In particular
$\mathcal{I}_{\mathfrak{F}}(M^{t})=WZ^{*}F^{t}$. If follows that if $H\leq G,$
then

$\mathcal{I}_{\mathfrak{F}}(H)=\begin{cases}WZ^{*}F^{t}&\text{if $H=M^{t}$ for
some $t\in V_{3},$}\\\ H&\text{otherwise}.\end{cases}$

The set $\Delta_{t}$ of the vertices of $\Gamma_{\mathfrak{F}}(M^{t})$
coincides with $WZ^{*}(Q^{t}\setminus F^{t}).$ Moreover, since the action of
$Q$ on $V$ is fixed-point-free, if $t_{1}\neq t_{2}$, then
$\Delta_{t_{1}}\cap\Delta_{t_{2}}=\emptyset.$ So the connected components of
$\Gamma_{\mathfrak{F}}(G)$ are $WZ^{*}(Q^{t}\setminus F^{t})$, with $t\in
V_{3}.$

######  Remark 22.

In the previous example $\Gamma_{\mathfrak{F}}(G)$ has 9 connected components.
We may repeat a similar construction starting from a faithful irreducible
action of $Q$ on $C_{p}\times C_{p}$, where $p$ is an arbitrary odd prime. In
this way we may construct a group $G$ with the property that
$\Gamma_{\mathfrak{F}}(G)$ has $p^{2}$ connected components.

##  5\. Example 2

In the previous section we have constructed an example of a group $G$
satisfying the assumptions of Theorem 14, in which the graph
$\Gamma_{\mathfrak{F}}(G)$ is disconnected and contains a maximal subgroup $M$
with the property that each arc in the graph belong to the subgraph induced by
a conjugate of $M$. However, as it is indicated in the statement of the
theorem, there is also the possibility that $G$ contains a proper normal
subgroup $N$ in such a way that each arc in the graph belongs either to the
subgraph induced by a conjugate of $M$ or to the subgraph induced by $N$. In
this section we exhibit an example in which this second possibility occur.

Let $A=\langle a_{1}\rangle\times\langle a_{2}\rangle$ with $\langle
a_{1}\rangle\cong\langle a_{2}\rangle\cong C_{8}$ and $Q=\langle
b_{1},b_{2}\rangle\cong Q_{8}.$ Both $A$ and $B$ admit an automorphism of
order 3. So we may consider the semidirect product $X=(A\times Q)\rtimes H$
with $H\cong C_{3}.$ We assume that the map $V_{1}=A/\operatorname{Frat}(A)\to
V_{2}=B/\operatorname{Frat}(B)$ sending $a_{1}\operatorname{Frat}(A)$ to
$b_{1}\operatorname{Frat}(B)$ and $a_{2}\operatorname{Frat}(A)$ to
$b_{2}\operatorname{Frat}(B)$ is an $H$-isomorphism. The group $X$ has
precisely 6 conjugacy classes of maximal subgroups:

  * •

$M_{0}=A\times Q;$

  * •

$M_{1}=(A\times\langle c\rangle)\cong A\rtimes C_{6}$ with
$c=b_{1}^{2}=b_{2}^{2};$

  * •

$M_{2}=(A^{2}\times Q_{8})H$;

  * •

$M_{h}=\langle a_{1}^{h}b_{1},a_{2}^{h}b_{2}\rangle H,$ with $h\in H.$

Moreover $QH$ has an irreducible action on $W\cong C_{7}\times C_{7}$ with the
property that $Q$ acts fixed-point-freely on $W$ while $C_{H}(W)=Z\cong
C_{7}.$ This action can be extended to $X$, with $A\leq C_{X}(W).$ Let
$G=W\rtimes X.$ Let

$B=AH\times C_{7},\quad C=W\rtimes\langle a_{1}b_{1},b_{2}\rangle.$

Moreover let $\mathfrak{F}$ be the class of finite groups with no subgroup
isomorphic to $B$ or $C.$

Suppose that $g_{1}=w_{1}x_{1},g_{2}=w_{2}x_{2}$ have the property that
$Y=\langle g_{1},g_{2}\rangle\notin\mathfrak{F}.$ We have two possibilities.

a) $Y$ has a subgroup isomorphic to $B$. It must be $\langle
x_{1},x_{2}\rangle\leq M_{1}^{q}$ for some $q\in Q$ (since $X$ is not
2-generated and the other maximal subgroups of $X$ with order divisible by 3
do not contain an element of order 8 centralizing a non-trivial element of
$W$). Moreover $(ZAH)^{q}$ is the unique subgroup of $WM_{1}^{q}$ isomorphic
to $B.$

b) $Y$ has a subgroup isomorphic to $C.$ Let $J=\langle x_{1},x_{2}\rangle$
and let $M$ be a maximal subgroup of $X$ containing $J.$ It cannot be that $M$
is a conjugate of $M_{1},$ since all the 2-subgroups of $M_{1}$ are abelian.
It cannot be that $M$ is a conjugate of $M_{2},$ since $M_{2}$ has no element
of order $8$. It cannot be that $M$ is a conjugate of $M_{h},$ since all the
elements of order 2 in $M_{h}$ centralize $W$ while $b_{2}^{2}$ does not. So
$x_{1}=r_{1}s_{1},$ $x_{2}=r_{2}s_{2}$ with $r_{i}\in A,s_{i}\in Q.$ As in the
previous case we want to describe when a subgroup $\tilde{C}$ of $M$ is
isomorphic to $C$. This occurs if and only if $\tilde{C}=\langle
r_{1}s_{1},r_{2}s_{2}\rangle,$ with $\langle s_{1},s_{2}\rangle=Q,$
$r_{1},r_{2}\in A,$ $\langle r_{1},r_{2}\rangle/(\langle
r_{1},r_{2}\rangle\cap A^{2})\cong C_{2}.$

Let $\mathfrak{F}_{B}$ (resp. $\mathfrak{F}_{C}$) the class of finite groups
with no subgroup isomorphic to $B$ (resp. $C$). Let now $K$ be a proper
subgroup of $G$, and assume $K\notin\mathfrak{F}_{2}.$ Since $M_{1}^{q}\cap
M_{0}=A\times\langle c\rangle$ for any $q\in Q$, $K$ cannot contain both a
subgroup isomorphic to $B$ and a subgroup isomorphic to $C$ so we have two
possibilities

a) $(ZAH)^{q}\leq K\leq WM_{1}^{q}$ for some $q\in Q.$ In this case
$\Gamma_{\mathfrak{F}}(K)=\Gamma_{\mathfrak{F}_{B}}(K)$ and
$\mathcal{I}_{\mathfrak{F}}(K)=K\cap(W\operatorname{Frat}(A)\langle
c\rangle).$

b) $K\leq M_{0}$. In this case
$\Gamma_{\mathfrak{F}}(K)=\Gamma_{\mathfrak{F}_{C}}(K)$ and
$\mathcal{I}_{\mathfrak{F}}(K)=K\cap(WA\langle c\rangle).$

In particular

$\Omega_{B}=\cup_{g}WM_{1}^{g}\setminus W\operatorname{Frat}(A)\langle
c\rangle\quad\text{ and }\quad\Omega_{C}=WM_{0}\setminus WA\langle c\rangle$

are, respectively, the set of vertices of $\Gamma_{\mathfrak{F}}(G)$ that are
non isolated in $\Gamma_{\mathfrak{F}}(WM_{1}^{q})$ for some $q$ in $Q$ and
the set of vertices of $\Gamma_{\mathfrak{F}}(G)$ that are non isolated in
$\Gamma_{\mathfrak{F}}(WM_{0}).$ Since $\Omega_{B}\cap\Omega_{C}=\emptyset,$
the graph $\Gamma_{\mathfrak{F}}(G)$ is disconnected (and indeed $\Omega_{B}$
and $\Omega_{C}$ are the two connected components of the graph). Moreover

$\mathcal{I}_{\mathfrak{F}}(G)=(G\setminus\Omega_{B})\cap(G\setminus\Omega_{C})=W(M_{0}\cap(\cap_{q\in
Q}M_{1}^{q}))\cup W\operatorname{Frat}(A)\langle
c\rangle=W\operatorname{Frat}(A)\langle c\rangle.$

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