# The Engel graph of a finite group

Eloisa Detomi  E. Detomi, Dipartimento di Ingegneria dell’Informazione - DEI,
Università di Padova, Via G. Gradenigo 6/B, 35121 Padova, Italy
<EMAIL_ADDRESS>,  Andrea Lucchini  A. Lucchini, Università di Padova,
Dipartimento di Matematica “Tullio Levi-Civita”, Via Trieste 63, 35121 Padova,
Italy<EMAIL_ADDRESS>and  Daniele Nemmi  D. Nemmi, Università di
Padova, Dipartimento di Matematica “Tullio Levi-Civita”, Via Trieste 63, 35121
Padova, Italy<EMAIL_ADDRESS>
###### Abstract.

For a finite group $G,$ we investigate the direct graph $\Gamma(G),$ whose
vertices are the non-hypercentral elements of $G$ and where there is an edge
$x\mapsto y$ if and only if $[x,_{n}y]=1$ for some $n\in\mathbb{N}.$ We prove
that $\Gamma(G)$ is always weakly connected and is strongly connected if
$G/Z_{\infty}(G)$ is neither Frobenius nor almost simple.

##  1\. Introduction

Let $G$ be a finite group. The commuting graph of $G$ has as vertices the non-
central elements of $G$ and two distinct vertices are adjacent if and only if
they commute in $G$. Commuting graphs arise naturally in many different
contexts and they have been intensively studied by various authors in recent
years (see in particular [5], [8], [10], [12], [14]).

The commuting graph of a group can be viewed through a different lens, which
leads naturally to some interesting generalisations. To do this, first let
$\mathcal{A}$ be the class of abelian groups and let
$\Lambda_{\mathcal{A}}(G)$ be the graph with vertex set $G$ so that $x$ and
$y$ are adjacent if and only if the subgroup $\langle x,y\rangle$ of $G$ is
abelian. Clearly, every vertex in the center $Z(G)$ is connected to every
other vertex in this graph (i.e. it is a universal vertex of the graph), so it
makes sense to consider the more restrictive graph $\Gamma_{\mathcal{A}}(G)$,
which is only defined on the non-central elements of $G$. Then
$\Gamma_{\mathcal{A}}(G)$ is the commuting graph of $G$ defined above. More in
general, let $\mathcal{F}$ be a family of groups and consider the graph
$\Lambda_{\mathcal{F}}(G)$ on the elements of $G$, where distinct vertices $x$
and $y$ are adjacent if and only if $\langle x,y\rangle$ is in $\mathcal{F}$.
As before, we would like to define the related graph $\Gamma_{\mathcal{F}}(G)$
on $G\setminus I_{\mathcal{F}}(G)$, where $I_{\mathcal{F}}(G)$ is the set of
universal vertices of $\Lambda_{\mathcal{F}}(G)$. If $\mathcal{F}=\mathcal{N}$
is the class of nilpotent groups, then $I_{\mathcal{F}}(G)$ coincides with the
hypercenter $Z_{\infty}(G)$ of $G$ which is the final term in the upper
central series of $G$ (see [1, Proposition 2.1]) and $\Gamma_{\mathcal{F}}(G)$
is called the nilpotent graph. Similarly, if $\mathcal{F}=\mathcal{S}$ is the
class of soluble groups, then [9, Theorem 1.1] implies that
$I_{\mathcal{F}}(G)=R(G)$ is the soluble radical and $\Gamma_{\mathcal{F}}(G)$
is the soluble graph of $G$. Notice that if $Z(G)=1,$ then the nilpotent graph
$\Gamma_{\mathcal{N}}(G)$ is connected if and only if the abelian graph
$\Gamma_{\mathcal{A}}(G)$ is connected and
$\operatorname{diam}(\Gamma_{\mathcal{A}}(G))\leq
2\cdot\operatorname{diam}(\Gamma_{\mathcal{N}}(G)).$ Indeed if $x$ and $y$ are
non-trivial elements of $G$ and $\langle x,y\rangle$ is nilpotent, then
$\langle x,y\rangle$ contains a non-trivial central element $z$ and $(x,z,y)$
is a path in $\Gamma_{\mathcal{A}}(G).$

The notion of commuting graph can be also generalized in a different
direction. Let $w$ be a word in the free group $F_{2}$ of rank 2. We may
define a direct graph $\Lambda_{w}(G)$ on the elements of $G,$ where there is
an edge $x\mapsto y$ if and only if $w(x,y)=1.$ Set

$\displaystyle I_{r,w}(G)$ $\displaystyle=\\{g\in G\mid w(g,x)=1\text{ for all
}x\in G\\},$ $\displaystyle I_{l,w}(G)$ $\displaystyle=\\{g\in G\mid
w(x,g)=1\text{ for all }x\in G\\},$ $\displaystyle I_{w}(G)$
$\displaystyle=I_{r,w}(G)\cap I_{l,w}(G).$

Noticing that $I_{{w}}(G)$ is the set of universal vertices of
$\Lambda_{w}(G)$, we define the related graph $\Gamma_{w}(G)$ on $G\setminus
I_{w}(G)$.

Set $[x,{}_{0}y]=x$ and $[x,{}_{i+1}y]=[[x,{}_{i}y],y]$ for $i\geq 0$. The
word $[x,{}_{n}y]$ is called the $n$-th Engel word. We will use the symbols
$\Lambda_{n}(G),\Gamma_{n}(G),$ $I_{r,n}(G),I_{l,n}(G)$ to denote the graphs
$\Lambda_{w}(G)$ and $\Gamma_{w}(G)$ and the sets $I_{r,w}(G),I_{l,w}(G),$
when $w=[x,_{n}y]$ is the $n$-Engel word. Notice that $\Gamma_{1}(G)$
coincides with the commuting graph of $G.$

In this paper we want to study the graph
$\Lambda(G)=\cup_{n>0}\Lambda_{n}(G),$ i.e. the graph whose vertices are the
elements of $G$ and where there is an edge $x\mapsto y$ if and only if
$[x,_{n}y]=1$ for some $0<n\in\mathbb{N}.$

The union $I_{l}(G)=\cup_{n}I_{l,n}(G)$ is the set of the left Engel elements
and coincides with the Fitting subgroup $F(G)$ of $G.$ The union
$I_{r}(G)=\cup_{n}I_{r,n}(G)$ is the set of the right Engel elements and
coincides with the hypercenter $Z_{\infty}(G)$ of $G$ (see e.g. [13, 12.3.7]).
In particular the elements of $G$ that are connected to every other vertex
(both as starting and ending vertex of some edge) are the elements of
$I_{l}(G)\cap I_{r}(G)=Z_{\infty}(G).$ So it makes sense to consider the more
restrictive graph $\Gamma(G)$, which is only defined on the non-hypercentral
elements of $G$. We will call $\Gamma(G)$ the Engel graph of $G.$

Our aim is to study the connectivity properties of the Engel graph. Since
$\Gamma(G)$ is a direct graph we may consider the strong connectivity and the
weak connectivity. A directed graph is strongly connected  if it contains a
directed path from $x$ to $y$ (and from $y$ to $x$) for every pair of vertices
$(x,y).$ The strong components are the maximal strongly connected subgraphs. A
directed graph is weakly connected if the undirected underlying graph obtained
by replacing all directed edges of the graph with undirected edges is a
connected graph. Our first main result is the following.

######  Theorem 1.1.

If $G$ is a finite group, then $\Gamma(G)$ is weakly connected and its
undirected diameter is at most $10.$

The study of the strong connectivity of $\Gamma(G)$ is more complicated. It
can be easily proved that $\Gamma(G)$ is strongly connected if and only if
$\Gamma(G/Z_{\infty}(G))$ is strongly connected (see Lemma 2.1) and that the
Engel graph of a Frobenius group is not strongly connected (Lemma 3.1). Our
main result is the following.

######  Theorem 1.2.

Suppose that $G/Z_{\infty}(G)$ is not an almost simple group. Then $\Gamma(G)$
is strongly connected if and only if $G/Z_{\infty}(G)$ is not a Frobenius
group.

The case where $G/Z_{\infty}(G)$ is an almost simple group is much more
difficult to handle. Indeed, there are infinitely many simple groups whose
Engel graph is strongly connected, for example the alternating groups $A_{n}$
if $n\geq 6$ (see Subsection 5.1), but also infinitely many simple groups
whose Engel graph is not strongly connected, for example
$\operatorname{PSL}(2,q)$ when $q$ is even and the Suzuki groups
$\operatorname{Sz}(q)$ (see Subsection 5.3). An immediate application of Lemma
3.2 is that if $G$ is almost simple and $\Gamma(\operatorname{soc}(G))$ is
strongly connected, then $\Gamma(G)$ is also strongly connected. The converse
is not always true: for example $\Gamma(S_{5})$ is strongly connected, but
$\Gamma(A_{5})$ is not.

In the case where $G$ is soluble and $G/Z_{\infty}(G)$ is not a Frobenius
group, we can also bound the diameter of $\Gamma(G)$.

######  Theorem 1.3.

Suppose that $G$ is soluble and $G/Z_{\infty}(G)$ is not a Frobenius group.
Then $\operatorname{diam}\Gamma(G)\leq 4$. Furthermore, there exists a soluble
group $G$ such that $\operatorname{diam}\Gamma(G)=4$.

It is interesting to note that if $Z_{\infty}(G)=1$ and $G$ is not almost
simple, then $\Gamma(G)$ is strongly connected if and only if $\Gamma_{2}(G)$
is strongly connected (see Theorem 3.12). This is no more true if the
assumption $Z_{\infty}(G)=1$ is removed. For example
$\Gamma_{3}(\operatorname{GL}(2,3))$ is strongly connected, but
$\Gamma_{2}(\operatorname{GL}(2,3))$ is not (see Remark 3.13).

The nilpotent graph and the Engel graph of $G$ have the same vertex-set,
$G\setminus Z_{\infty}(G).$ As all elements of a nilpotent group are (left and
right) Engel, if $g_{1}$ and $g_{2}$ are adjacent in the nilpotent graph, then
$g_{1}\mapsto g_{2}$ and $g_{2}\mapsto g_{1}$ in the Engel graph. So if the
nilpotent graph of $G$ is connected, then the Engel graph $\Gamma(G)$ is
strongly connected. The converse is false, indeed if $G$ is $S_{4}$, and more
in general a 2-Frobenius group, or $A_{p}$, when $p$ is a prime with $p>5$,
then the commuting graph of $G$ is not connected (see [12]), and the same
holds for its nilpotent graph, as mentioned above. On the other hand, it
follows from Theorem 1.2 and Subsection 5.1 that the Engel graph of $G$ is
strongly connected.

The paper is organized as follows. In Section 2 we collect some preliminary
results about the Engel graph and prove Theorem 1.1. In Section 3 we prove
Theorem 1.2. The proof of Theorem 1.3 will be completed in Section 4. In
Section 5 we analyze the Engel graph of some families of simple and almost
simple groups. Finally, in Section 6, with the help of some computer
calculations, we prove that the Engel graph of a sporadic simple group is
strongly connected.

### Acknowledgements

We would like to thank Pablo Spiga for fruitful discussions and his precious
help in the computations with some sporadic simple groups.

##  2\. Preliminary results and proof of Theorem 1.1

In this and in the following sections we will use the symbols $x\mapsto y$ and
$x\mapsto_{n}y$ to say there there is an edge from $x$ to $y$ in the graph
$\Lambda(G)$, or respectively $\Lambda_{n}(G)$, and the symbols
$x\twoheadrightarrow y$ and $x\twoheadrightarrow_{n}y$ to say that there is a
direct path in the graph $\Gamma(G)$, or respectively $\Gamma_{n}(G)$, joining
$x$ to $y.$ Recall that $\Gamma(G)$ and $\Gamma_{n}(G)$ are the subgraphs
defined on the set of non-universal vertices of $\Lambda(G)$ and
$\Lambda_{n}(G)$, respectively.

######  Lemma 2.1.

$\Gamma(G)$ is weakly or strongly connected if and only if
$\Gamma(G/Z_{\infty}(G))$ is weakly or strongly connected.

###### Proof.

If $y$ is adjacent to $x$ in $\Gamma(G)$, then its image $\bar{y}$ in
$G/Z_{\infty}(G)$ is adjacent to $\bar{x}$ in $\Gamma(G/Z_{\infty}(G))$.
Conversely, if $\bar{y}\in G/Z_{\infty}(G)$ is adjacent to $\bar{x}$ in
$\Gamma(G/Z_{\infty}(G))$, then $[y,_{n}x]\in Z_{\infty}(G)=Z_{m}(G)$ for some
$n$ and $m$. Thus $[y,_{n+m}x]=[[y,_{n}x],_{m}x]=1$ and $y$ is adjacent to $x$
in $\Gamma(G)$. ∎

######  Lemma 2.2.

Let $x,y\in G.$

  1. (1)

If $[y,_{n}x]=[y,_{m}x]\neq 1$ for some $0\leq n<m$, then $y$ is not adjacent
to $x$ in $\Gamma(G).$

  2. (2)

If $y\in F(G)$, then $x\mapsto y$ for every $x\in G$.

  3. (3)

If $y\in N_{G}(\langle x\rangle)$, then $y\mapsto_{2}x$.

###### Proof.

  1. (1)

Assume $1\neq[y,_{n}x]=[y,_{m}x]$ for some $0\leq n<m$ and let $z=[y,_{n}x]$.
Then $z=[z,_{m-n}x]$ and so $z=[z,_{l(m-n)}x]$ for every $l\geq 0$. If
$[y,_{N}x]=1$, then $1=[y,_{N}x,_{N(m-n-1)}x]=[y,_{N(m-n)}x]=z,$ a
contradiction.

  2. (2)

It follows from the fact that the Fitting subgroup $F=F(G)$ coincides with the
set of the left Engel elements of $G$.

  3. (3)

If $y\in N_{G}(\langle x\rangle)$, then $[y,x]\in\langle x\rangle$ and it
commutes with $x$, so $[y,_{2}x]=1$.∎

The proof of Theorem 1.1 relies on the main result of [2], which states that
the soluble graph of an insoluble group is connected and its diameter is at
most $5$.

###### Proof of Theorem 1.1.

By Lemma 2.1, we may assume that $Z_{\infty}(G)=1.$ If the soluble radical
$R(G)$ is non-trivial, then $F(G)\neq 1$ and $\Gamma(G)$ is weakly connected
by Lemma 2.2 (2). So we may assume $R(G)=1.$ Let $x_{1},x_{2}$ be two distinct
non-identical elements of $G$. If $\langle x_{1},x_{2}\rangle$ is soluble then
there exists $1\neq f\in F(\langle x_{1},x_{2}\rangle)$ and $x_{1}\mapsto
f,x_{2}\mapsto f.$ Since the soluble graph of $G$ is connected with diameter
at most $5$ (see [2]), we deduce that $\Gamma(G)$ is weakly connected and its
undirected diameter is at most $10$, as claimed. ∎

Note that if $\Omega$ is a strong component of $\Gamma(G)$ and a non-
hypercentral element $x\in G$ commutes with some element of $\Omega$, then
$x\in\Omega$. Thus the following result will be relevant for our proofs.

######  Theorem 2.3.

[4, Theorem 10.2.1] If a finite group $G$ admits a fixed-point-free
automorphism of prime order, then $G$ is nilpotent.

We will also need the following result concerning Frobenius complements.

######  Lemma 2.4.

[10, Lemma 2.1] Suppose $X$ is a Frobenius complement. Then every Sylow
subgroup of $X$ is cyclic or generalized quaternion. If $X$ has odd order then
any two elements of prime order commute and $X$ is metacyclic. If $X$ has even
order, then $X$ contains a unique involution.

##  3\. Proof of Theorem 1.2

######  Lemma 3.1.

If $G=K\rtimes H$ is a Frobenius group with kernel $K$ and complement $H$,
then $\Gamma(G)$ is not strongly connected. In particular for every nontrivial
$k\in K$ there is no element $g\in G\setminus K$ with $k\mapsto g$.

###### Proof.

Let $k$ be a nontrivial element of $K$ and assume that there exists $g\in
G\setminus K$ with $k\mapsto g$. Thus $[k,_{n}g]=1$ for some integer $n.$ Take
$g\in G\setminus K$ such that $n$ is minimal. Since $g\in G\setminus
K=\cup_{k\in K}H^{k}$, without loss of generality we can assume that $g$
belongs to $H$. So $z=[k,_{n-1}g]$ belongs to $K$, since $K$ is normal, and it
belongs to $H$, since it centralizes $g\in H$ and $G$ is Frobenius. Therefore
$z=1$, a contradiction to the minimality of $n$. ∎

For the remaining part of this section our aim is to prove that if $G$ is
neither Frobenius nor almost simple, then $\Gamma(G)$ is strongly connected.
In the following we will denote by $\Delta(G)$ the subgraph of $\Lambda(G)$
induced by the non-trivial elements of $G.$ Clearly $\Gamma(G)=\Delta(G)$ if
$Z_{\infty}(G)=1.$

######  Lemma 3.2.

Let $N$ be a non-nilpotent normal subgroup of a finite group $G$. If
$\Delta(N)$ is strongly connected, or, more generally, if $N\setminus\\{1\\}$
is contained in a strong component of $\Delta(G)$, then $\Delta(G)$ is
strongly connected.

###### Proof.

Let $\Omega$ be the strong component of $\Delta(G)$ containing
$N\setminus\\{1\\}.$ Let $x$ be a non-trivial element of $G$ and let
$\tilde{x}$ be an element of prime order in $\langle x\rangle$. Since $N$ is
not nilpotent, it follows from Theorem 2.3 that there exists $y\in
N\setminus\\{1\\}$ such that $[y,\tilde{x}]=1.$ Clearly $x$, $\tilde{x}$ and
$y$ form a clique in the commuting graph of $G$, and consequently
$x\in\Omega.$ ∎

######  Lemma 3.3.

Suppose that $N=G_{1}G_{2}$ is the central product of two non-trivial finite
groups. Then $\Delta(N)$ is strongly connected and $\operatorname{diam}(N)\leq
3.$

###### Proof.

Let $g_{1}=x_{1}y_{1},g_{2}=x_{2}y_{2}$ be two non trivial elements of $N$
with $x_{1},x_{2}\in G_{1},y_{1},y_{2}\in G_{2}.$ We may assume $x_{1}\neq 1.$
If $y_{2}\neq 1$, then $g_{1}=x_{1}y_{1},x_{1},y_{2},x_{2}y_{2}=g_{2}$ is a
path in $\Delta(N).$ Suppose $y_{2}=1$. This implies $x_{2}\neq 1.$ If
$y_{1}\neq 1,$ then $g_{1}=x_{1}y_{1},y_{1},x_{2},x_{2}y_{2}=g_{2}$ is a path
in $\Delta(N).$ We remain with the case $y_{1}=y_{2}=1.$ Then, let $1\neq z\in
G_{2}.$ In this case $g_{1}=x_{1},z,x_{2}=g_{2}$ is a path in $\Delta(N).$ ∎

In the following we will denote by $F^{*}(G)$ the generalized Fitting subgroup
of $G$.

######  Lemma 3.4.

If $G$ is not an almost simple group and $F^{*}(G)>F(G)$, then $\Delta(G)$ is
strongly connected.

###### Proof.

Recall that $F^{*}(G)=E(G)F(G)$ is the central product of the Fitting subgroup
of $G$ with the layer $E(G)$ of $G$. Since $E(G)$ is the central product of
the components, either $F^{*}(G)$ is a non-abelian simple group or
$\Delta(F^{*}(G))$ is strongly connected by Lemma 3.3. In the first case $G$
is almost simple, in the second $\Delta(G)$ is strongly connected by Lemma
3.2. ∎

In the following, given two vertices $x,y$ of the graph $\Gamma(G),$ we will
denote by $d(x,y)$ the direct distance from $x$ to $y$ in $\Gamma(G).$

######  Lemma 3.5.

Let $x,y$ be two distinct non-trivial elements of a finite group $G$. If
$Z_{\infty}(G)=1$, $F(G)=F^{*}(G)$ and $G$ is not a Frobenius group, then the
following hold:

  1. (1)

If $d(x,y)>3$, then $N_{G}(\langle y\rangle)F$ (and consequently
$C_{G}(\langle y\rangle)F$) is a Frobenius group.

  2. (2)

If $d(x,y)>4$ and $y_{r}$ is a power of $y$ of order a prime $r$, then
$C_{G}(y_{r})$ has odd order and it is metacyclic.

###### Proof.

Assume $d(x,y)>3$. If $F(G)\cap N_{G}(\langle y\rangle)\neq 1$, then there
exists a non trivial element $g\in F(G)\cap N_{G}(\langle y\rangle)$, so by
Lemma 2.2 (2) and (3),

$x\mapsto g\mapsto y$

a contradiction. So $F(G)\cap N_{G}(\langle y\rangle)=1$. Assume, by
contradiction, that $N_{G}(\langle y\rangle)F(G)$ is not a Frobenius group.
Then there exists an element $1\neq g\in C_{F(G)}(k)$ for some $1\neq k\in
N_{G}(\langle y\rangle)$ and

$x\mapsto g\mapsto k\mapsto y$

against $d(x,y)>3.$

Now assume $d(x,y)>4$. In particular, $d(x,y_{r})>3$. By (1), $C_{G}(y_{r})$
is a Frobenius complement. Assume that $|C_{G}(y_{r})|$ is even and take an
involution $z\in C_{G}(y_{r})$. As every non-trivial element of $C_{G}(y_{r})$
acts fixed-point-freely on $F,$ $z$ is the unique involution of
$C_{G}(y_{r})$. But then, for every $g\in C_{G}(y_{r})$, we have $z^{g}\in
C_{G}(y_{r})$, hence $z^{g}=z$. Since $z$ acts fixed-point-freely on $F(G)$,
it inverts the elements of $F(G)$ and $F(G)$ is abelian. In particular
$F(G)=[F(G),z]$ as a set. We claim:

$(*)\quad G=C_{G}(z)F(G).$

Indeed let $g\in G$. As $z^{g}$ acts as involution on $F(G)$, then $[z,g]$
centralizes $F$. Since $C_{G}(F^{*}(G))\leq F^{*}(G)=F(G),$ it follows
$[z,g]\in F(G)=[z,F(G)]$. Thus there exists an element $f\in F(G)$ such that
$[z,g]=[z,f]$. So $z^{g}=z^{f}$, hence $g\in C_{G}(z)F(G)$. It follows that
$G=C_{G}(z)F(G)$ is not a Frobenius group. Since $C_{G}(z)\cap F(G)=1,$ there
exists a non-trivial element $g\in C_{G}(z)$ such that $C_{F}(g)\neq 1$. Thus,
for some $1\neq f\in C_{F}(g)$, since $y\in C_{G}(y_{r})$ centralizes $z$, we
have

$x\mapsto f\mapsto g\mapsto z\mapsto y$

a contradiction. Thus $C_{G}(y_{r})$ is a Frobenius complement with odd order
and our conclusion follows from Lemma 2.4. ∎

######  Corollary 3.6.

Assume that $Z_{\infty}(G)=1$ and $G$ is neither a Frobenius group nor an
almost simple group. Then all the elements whose centralizer has even order
belong to the same strong component of $\Gamma(G).$

###### Proof.

It suffices to prove that if $|x|=|y|=2,$ then $x\twoheadrightarrow y.$ This
follows from Lemmas 3.4 and 3.5. ∎

######  Lemma 3.7.

Let $X$ be a finite group. Assume that $F^{*}(X)=F(X)$ and that $X/F(X)\cong
S^{t},$ where $S$ is a finite non-abelian simple group. Then $\Delta(X)$ is
strongly connected.

###### Proof.

The statement is trivially true if $Z_{\infty}(G)\neq 1.$ So we may assume
$Z_{\infty}(G)=1.$

It follows from the Brauer-Suzuki theorem that there is no simple group with
generalized quaternion Sylow 2-subgroups. In particular, by Lemma 2.4,
$X/F(X)$ cannot be isomorphic to a Frobenius complement and consequently $X$
is not a Frobenius group. By Corollary 3.6, all the involutions of $X$ belong
to the same strong component, say $\Omega,$ of $X$. Moreover if $f$ is a non-
trivial element of $F(X)$ and $z$ is an involution of $X$, then
$f\twoheadrightarrow z$ by Lemma 3.5 and $z\mapsto f$ by Lemma 2.2, so
$F(X)\setminus\\{1\\}\subseteq\Omega.$

We claim that for any odd prime $p$ dividing $|S|,$ $\Omega$ contains all the
elements of $X$ with order $p.$ If $t>1,$ then a Sylow $p$-subgroup of $S^{t}$
cannot act fixed-point-freely on $F(X)$, so any Sylow $p$-subgroup of $X$
contains an element in $\Omega,$ and, consequently, all the elements of this
Sylow subgroup belong to $\Omega$ (using the fact that the center of a Sylow
subgroup is non-trivial). Assume $t=1$ and that, by contradiction, there
exists no element of order $p$ in $\Omega$ and let $p$ be the smallest (odd)
prime with this property. Let $P$ be a Sylow $p$-subgroup of $S$. Since $P$
acts fixed-point-freely on $F(X)$, $P=\langle u\rangle$ is cyclic. By
Burnside’s Theorem [13, Theorem 10.1.8], as $S$ is not $p$-nilpotent, we have
that $P\not\leq Z(N_{S}(P))$, so there exists an element $v\in
N_{S}(P)\setminus C_{S}(P)$. In particular, $vC_{S}(P)$ is a nontrivial
element of $N_{S}(P)/C_{S}(P)\leq{\rm Aut}(P)$, so its order divides
$\varphi(p^{n})=p^{n-1}(p-1)$ and is coprime to $p$ (since $P\leq C_{G}(P)$).
But then there is a prime $q$ dividing $(|v|,p-1)$ and so the minimality of
$p$ implies that $v$ is contained in $\Omega$. Since $v$ normalizes $\langle
u\rangle$, $v\to u.$ On the other hand, $u\to f$ for any $f\in F(X),$ which
means that $u\in\Omega$ and we have reached a contradiction. ∎

######  Lemma 3.8.

Assume $Z_{\infty}(G)=1,$ $F(G)=F^{*}(G)$ and there exist $x,y\in G$ with
$d(x,y)>4.$ Let $J/F(G)=F(G/F).$ Then $|J/F(G)|$ is prime to $|F(G)|$.

###### Proof.

Let $s$ be a prime divisor of $|F(G)|$ and let $S$ be a Sylow $s$-subgroup of
$J$. Take a power of $y$ of prime order, say $y_{r}$. As $F(G)S$ is normal in
$G$, either $y_{r}$ acts fixed-point-freely on $F(G)S$, and so $F(G)S$ is
nilpotent hence contained in $F(G)$, or $C_{F(G)S}(y_{r})\neq 1$. Let $g\in
C_{F(G)S}(y_{r})$ be a non-trivial element of prime power order. Thus either
$g\in F(G)$, and

$x\mapsto g\mapsto y_{r}\mapsto y$

or $g$ is an $s$-element and it is contained in a Sylow $s$-subgroup
containing $S\cap F(G)\neq 1$. Thus $g$ is contained in a nilpotent subgroup
having nontrivial intersection with $F(G)$, hence there exists a nontrivial
element $f\in F(G)$ such that $[f,_{n}g]=1$. It follows that

$x\mapsto f\mapsto g\mapsto y_{r}\mapsto y$

against the assumptions. ∎

######  Lemma 3.9.

Assume $Z_{\infty}(G)=1,$ $F(G)=F^{*}(G)$ and there exist $x,y\in G$ with
$d(x,y)>4.$ Let $J/F(G)=F(G/F(G)),$ $J^{*}/F(G)=F^{*}(G/F(G))$ and
$Z/F(G)=Z(J/F(G))$. Let $y_{r}$ be a power of $y$ of prime order $r.$ If $G$
is not Frobenius and $J=J^{*}$, then $y_{r}\in Z$.

###### Proof.

Suppose $y_{r}\notin Z$. As $J/F(G)=F^{*}(G/F(G))$, then $y_{r}\notin
C_{G/F(G)}(J/F(G))$, and so $y_{r}$ does not centralize a Sylow $t$-subgroup
of $J/F(G)$.

Assume by contradiction $t=r$ and let $R$ be a Sylow $r$-subgroup of $J\langle
y_{r}\rangle$ containing $y_{r}$. As $y_{r}$ does not centralize an
$r$-subgroup of $J/F(G)$, $y_{r}$ is not central in $R$ and the center of $R$
provides another element of order $r$ that centralizes $y_{r}$. Since, by
Lemma 3.5, the Sylow subgroups of $C_{G}(y_{r})$ are cyclic, we get a
contradiction. Thus we have $t\neq r$.

We now claim that there exists a Sylow $t$-subgroup $T$ of $J$ which is
$y_{r}$-invariant. Notice that $y_{r}$ acts on $J=\prod_{p}T_{p}F(G)$ and
centralizes the $r$-Sylow subgroups. If $r$ does not divide $|F(G)|$ then the
action is coprime, so there exists an invariant $t$-Sylow. If $r$ divides
$|F(G)|$, then $y_{r}$ belongs to a $r$-Sylow subgroup of $F(G)\langle
y_{r}\rangle$, and there exists a nontrivial element $f$ of $F(G)$ such that
$[f,_{n}y_{r}]=1$. Thus

$x\mapsto f\mapsto y_{r}\mapsto y,$

a contradiction.

Let $T$ be a Sylow $t$-subgroup of $J$ which is $y_{r}$-invariant. Note that
$y_{r}$ acts faithfully on $T$.

Let $V$ be a minimal normal subgroup of $G$. Thus $V\leq F(G)$. We claim that
$C_{G}(V)\leq F(G)$. Otherwise $y_{r}$ could not act fixed-point-freely on
$C_{G}(V)$, and consequently there would exist $g\in C_{G}(V)\cap
C_{G}(y_{r})$ and thus, for any $1\neq v\in V$,

$x\mapsto v\mapsto g\mapsto y_{r}\mapsto y.$

Moreover, $V$ is a faithful $Q\langle y_{r}\rangle$-module for every $\langle
y_{r}\rangle$-invariant subgroup $Q$ of $T$. Indeed $C_{G}(V)\leq F(G)$ and
$|F(G)|$ is prime to the order of $T\langle y_{r}\rangle$.

Since $C_{V}(\langle y_{r}\rangle)$ is trivial, by [12, Lemma 2.5] we deduce
that $T$ is a $2$-group and that every abelian characteristic subgroup of $T$
is centralized by $y_{r}$. Since $C_{G}(y_{r})$ has odd order, we get a
contradiction. ∎

######  Proposition 3.10.

Assume $Z_{\infty}(G)=1,$ $F(G)=F^{*}(G)$ and there exist $x,y\in G$ with
$d(x,y)>4.$ Let $J/F(G)=F(G/F(G))$ and $J^{*}/F(G)=F^{*}(G/F(G))$. If
$J=J^{*}$, then $G$ is a Frobenius group.

###### Proof.

We use the notation introduced in the previous lemma. Since $y_{r}\in Z$, in
particular $y_{r}$ belongs to the center of a Sylow $r$-subgroup $R$ of $J$.
Thus $R\leq C_{G}(y_{r})$ is cyclic, and $\langle y_{r}\rangle$ is a
characteristic subgroup of $R$. Now, $RF(G)$ is normal in $G$ and $\langle
y_{r}\rangle F(G)$ is a characteristic subgroup of $RF(G)$. Thus $\langle
y_{r}\rangle F(G)$ is normal in $G$. By the Frattini argument, as $y_{r}$ has
order prime to $F(G)$,

$G=N_{G}(\langle y_{r}\rangle)\left(\langle y_{r}\rangle
F(G)\right)=N_{G}(\langle y_{r}\rangle)F(G)$

hence $G$ is Frobenius by Lemma 3.5. ∎

The first part of Theorem 1.3 follows directly from Lemma 2.1 and Proposition
3.10.

######  Corollary 3.11.

If $G$ is a soluble and $G/Z_{\infty}(G)$ is not a Frobenius group, then
$\Gamma(G)$ is strongly connected and has diameter at most $4$.

Now we are ready to prove our main result.

###### Proof of Theorem 1.2.

Let $G$ be a finite group such that the factor $G/Z_{\infty}(G)$ is not an
almost simple group. By Lemma 2.1, we can assume that $Z_{\infty}(G)=1$. In
Lemma 3.1 we have seen that if $G$ is a Frobenius group then $\Gamma(G)$ is
not strongly connected. So we are left to prove that if $G$ is neither almost
simple nor Frobenius, then $\Gamma(G)$ is strongly connected. By Lemma 3.4 and
Proposition 3.10 we are reduced to the case where $F^{*}(G)=F(G)$ and $J\neq
J^{*}$. If $J=F(G),$ then $\Delta(J^{*})$ is connected by Lemma 3.7. In this
case $\Gamma(G)$ is strongly connected by Lemma 3.2.

So we are left with the case $J>F(G).$ Since $J$ is not nilpotent, by Lemma
3.2 it is sufficient to prove that $J\setminus\\{1\\}$ is contained in a
strong component of $\Gamma(G)$.

If $J$ is not a Frobenius group, then $\Gamma(J)$ is strongly connected, by
Proposition 3.10, and we are done. So $J$ is a Frobenius group. Let $P/J$ be a
Sylow 2-subgroup of $J^{*}/J.$ Note that $P/F$ is a central product of a Sylow
$2$-subgroup of $E(G/F(G))$ and $J/F(G)$.

We distinguish two cases:

a) $2$ divides $|F(G)|.$ In this case $P$ is neither Frobenius nor almost
simple, so, by Proposition 3.10, $\Gamma(P)$ is strongly connected. As a
consequence $J\setminus\\{1\\}$ is contained in a strong component of
$\Gamma(G)$.

b) $2$ does not divide $|F(G)|$. In this case $F(G)$ has a complement $L$ in
$P$ and any element of $L$ has centralizer of even order, since $P/F(G)$ is a
central product of a Sylow $2$-subgroup of $E(G/F(G))$ and $J/F(G)$. As $G$ is
neither Frobenius nor almost simple, from Corollary 3.6 it follows that
$J\setminus\\{1\\}$ is contained in a strong component of $\Gamma(G)$, as
claimed. ∎

Indeed a stronger statement can be claimed.

######  Theorem 3.12.

Suppose $Z_{\infty}(G)=1.$ If $\Gamma_{2}(G)$ is not strongly connected, then
$G$ is either almost simple or Frobenius.

###### Proof.

The proof of Theorem 1.2 relies on the following observations:

  * •

If $[x,y]=1,$ then $x\mapsto y.$

  * •

If $x\in N_{G}(\langle y\rangle),$ then $[x,_{2}y]=1,$ hence $x\mapsto y.$

  * •

If $y\in F(G),$ then $x\mapsto y.$

Clearly if either $[x,y]=1$ or $x\in N_{G}(\langle y\rangle),$ then
$x\mapsto_{2}y.$ If $z\in Z(F(G))$, then $[x,_{2}z]=1,$ so $x,z,y$ is a path
in $\Gamma_{2}(G)$ for any $y\in F(G).$ ∎

######  Remark 3.13.

The previous theorem is no more true if the assumption $Z_{\infty}(G)=1$ is
removed. Consider for example $G=\operatorname{GL}(2,3).$ We have
$Z_{\infty}(G)=Z(G)\cong C_{2}$ and $G/Z(G)\cong S_{4}.$ By Lemma 3.11,
$\Gamma(G)$ is strongly connected. However, if $x$ has order 3 or 6 and
$x\mapsto_{2}y$, then $y$ has order $3$ or $6$. Indeed there are $9$ (non-
central) elements $g\in G$ with $x\mapsto g$: the three elements of order 3 or
6 different from $x$ in the unique cyclic subgroup of $G$ of order $6$
containing $x$, and $6$ elements of order $4$. If $g$ is one of these $6$
elements of order $4$, then $[x,_{3}g]=1$ but $[x,_{2}g]\neq 1.$ So
$\Gamma_{3}(G)$ is connected, but $\Gamma_{2}(G)$ is not.

##  4\. An example

The aim of this section is to complete the proof of Theorem 1.3: we will
present a family of soluble groups $G$ such that the diameter of $\Gamma(G)$
is $4$.

We select $q$ a power of an odd prime such that the prime $r,$ with $r\geq 3$,
divides $q-1$ exactly. Let $\mathbb{F}=\operatorname{GF}(q^{r})$ and let
$\beta$ the Frobenius automorphism of $\mathbb{F}$ of order $r$. Note that
$r^{2}$ divides exactly $q^{r}-1$ (see [6, Lemma 8.1 (e)]). Let $t$ be a prime
that divides $(q^{r}-1)/(q-1)$ but not $q-1$. As an example we may take
$q=7,r=3,t=19.$ Let

$z:=\begin{pmatrix}e&0\\\ 0&e\end{pmatrix}\in\operatorname{GL}(2,\mathbb{F}),$

with $|e|=r^{2}.$ and let

$c:=\begin{pmatrix}1&0\\\ 0&f\end{pmatrix}\in\operatorname{GL}(2,\mathbb{F}),$

with $|f|=t.$ Recall that $\beta$ induces an automorphism of the affine group
$\operatorname{AGL}(2,q^{r})\cong F\rtimes\operatorname{GL}(2,q^{r})$, with
$F\cong\mathbb{F}^{2}.$ We also denote this induced automorphism by $\beta.$
Note that $\beta$ normalizes $F,\langle z\rangle$ and $\langle c\rangle.$
Consider $x=z\beta$ in the semidirect product
$\operatorname{AGL}(2,q^{r})\rtimes\langle\beta\rangle$ and let

$G=F\rtimes D,\text{ with }D=\langle x,c\rangle.$

######  Lemma 4.1.

The element $x^{r}$ has order $r$, commutes with $c$ and acts fixed-point-
freely on $F$. Moreover $c^{x}=c^{q}\neq c.$

###### Proof.

We have

$x^{r}=(z\beta)^{r}=z^{1+q+\dots+q^{r-1}}=\begin{pmatrix}e^{1+q+\dots+q^{r-1}}&0\\\
0&e^{1+q+\dots+q^{r-1}}\end{pmatrix}.$

Since $|e|=r^{2}$ and $r^{2}$ does not divides $(q^{r}-1)/(q-1),$ it follows
that $x^{r}$ has order $r$ and acts fixed-point-freely on $F$. The other
information in the statement can be easily verified. ∎

By the previous lemma $|G|=q^{2r}\cdot t\cdot r^{2}$ (so in particular
$|G|=7^{6}\cdot 19\cdot 3^{2}$ if $q=7,r=3,t=19).$

######  Lemma 4.2.

The following statements hold.

  1. (1)

$C_{G}(x)=\langle x\rangle.$

  2. (2)

$C_{G}(x^{r})=D.$

  3. (3)

Let $C=\langle c\rangle.$ Then $[F,C]$ is a non-trivial subgroup of $F$
normalized by $D$ and $[F,C]D$ is a Frobenius group.

###### Proof.

  1. (1)

Let $g=fd\in C_{G}(x)$ for some $f\in F$ and $d\in D$. So,
$1=[fd,x]=[f,x]^{d}[d,x]$, and therefore $[f,x]=1$ and $[d,x]=1$. Since $x$
acts fixed-point-freely on $F$, we deduce $f=1$. Moreover $c^{x}=c^{q}\neq c$,
hence $C_{G}(x)=C_{D}(x)=\langle x\rangle$.

  2. (2)

Arguing as before, since $x^{r}$ acts fixed-point-freely on $F$, we deduce
$C_{G}(x^{r})=C_{D}(x^{r})=D.$

  3. (3)

Since $C\unlhd D,$ $[F,C]$ is normalized by $D$. Notice that $[F,C]$ is
precisely the eigenspace of $c$ corresponding to the eigenvalue $f.$ In
particular $[F,C]$ is a 1-dimensional $\mathbb{F}$-subspace of $F.$ Let $1\neq
d\in D.$ If $r$ divides $|d|$, then $x^{r}\in\langle d\rangle,$ so
$C_{[F,C]}(d)\leq C_{[F,C]}(x^{r})\leq C_{F}(x^{r})=1.$ If $r$ does not divide
$|d|,$ then $\langle d\rangle=\langle c\rangle,$ so
$C_{[F,C]}(d)=C_{[F,C]}(c)=1.$∎

We denote by $\operatorname{B}_{i}(\\{x\\})$ the set of vertices of
$\Gamma(G)$ having distance at most $i$ from $x$. By Corollary 3.11 the
diameter of $\Gamma(G)$ is at most $4$. We will show that diam$(\Gamma(G))=4$,
by proving, in Lemma 4.5, that $\operatorname{B}_{3}(\\{x\\})\neq
G\setminus{\\{1\\}}$.

######  Lemma 4.3.

$\operatorname{B}_{1}(\\{x^{i}\\})=\langle x\rangle\setminus\\{1\\},$ for
every $i$ prime to $r$.

###### Proof.

Let $g\in\operatorname{B}_{1}(\\{x\\}).$ Then $[g,_{n}x]=1$ for some positive
integer $n.$ Choose $n$ as small as possible. If $n>1,$ then let
$s=[g,_{n-1}x].$ It follows $s\in G^{\prime}\cap C_{G}(x)=FC\cap\langle
x\rangle=1,$ a contradiction with the minimality of $n.$ So $n=1,$ and $g\in
C_{G}(x)=\langle x\rangle.$ The same arguments apply whenever $x^{i}$ is a
generator of $\langle x\rangle$. ∎

######  Lemma 4.4.

$\operatorname{B}_{2}(\\{x\\})=D\setminus\\{1\\}.$

###### Proof.

By the previous lemma, $\operatorname{B}_{2}(\\{x\\})\subseteq\langle
x\rangle\cup\operatorname{B}_{1}(\\{x^{r}\\}).$ Assume, by contradiction, that
there exists an element $g\in\operatorname{B}_{1}(\\{x^{r}\\})$ that does not
belong to $D=C_{G}(x^{r})$. Then $[g,_{n}x^{r}]=1$ for some $n\geq 2$ and
$[g,_{n-1}x^{r}]\in C_{G}(x^{r})=D$. Choose the smallest $m$ with the property
that $[g,_{m}x^{r}]\in D$ and note that $m\geq 1$. Let us set
$w=[g,_{m-1}x^{r}].$ So, $w\notin D$ while $[w,x^{r}]\in D$. Write $w=fd\in
G=FD$, for some $f\in F$ and $d\in D$. Then
$[w,x^{r}]=[f,x^{r}]^{d}[d,x^{r}]\in D$ implies $[f,x^{r}]\in D\cap F=1$. So
$f\in C_{G}(x^{r})\cap F=1$ and hence $w\in D$, a contradiction. ∎

######  Lemma 4.5.

$[F,C]\setminus\\{1\\}\not\subseteq\operatorname{B}_{3}(\\{x\\}).$

###### Proof.

By the previous lemmas, if $y\in\operatorname{B}_{3}(\\{x\\})\cap[F,C]$, then
$y\mapsto d,$ for some non-trivial element $d\in D.$ But, by Lemma 3.1, this
is impossible since $[F,C]D$ is a Frobenius group. ∎

##  5\. Almost simple groups

###  5.1. Alternating group

In this subsection we investigate the strong connectivity of the alternating
groups. One of the tool that we can use is the knowledge of the connected
components of the prime graph of a finite simple group [15]. Recall that the
prime graph of a finite group $G$ is the graph whose vertices are the primes
dividing the order of $G$ and where two vertices $p$ and $q$ are joined by an
edge if and only if $G$ contains an element of order $p\cdot q.$

######  Lemma 5.1.

If $n\geq 6,$ then

$(1,3,5)(1,\dots,n)(1,3,5)^{-1}(1,\dots,n)^{-1}=(1,3,5)(2,n,4).$

######  Theorem 5.2.

If $n\geq 5,$ then $\Gamma(A_{n})$ is strongly connected if and only if $n\neq
5.$ Moreover the following holds:

  * •

if $n\geq 7$, then $\Gamma_{2}(A_{n})$ is strongly connected;

  * •

$\Gamma_{3}(A_{6})$ is strongly connected, but $\Gamma_{2}(A_{6})$ is not.

###### Proof.

Let $G=A_{n}.$ All the elements of order $2$ belong to the same strong
component of $\Gamma_{1}(G)$ (see [5] or the first lines of the proof of [10,
Theorem 7.1]). Let $\Omega$ be the strong component of $\Gamma_{2}(G)$
containing all the elements of order $2$ and let $\pi_{1}(G)$ be the connected
component of the prime graph of $G$ containing 2. Clearly $\Omega$ contains
also all the elements whose order is divisible by a prime $q\in\pi_{1}(G),$ as
every element commutes with its own powers.

Suppose $n\geq 8.$ By [15, Table I] either the prime graph of $G$ is connected
or there exists a prime $p$ such that $n\in\\{p,p+1,p+2\\}$ and
$\pi(G)=\pi_{1}(G)\cup\\{p\\}.$ In the first case we can immediately conclude
that $\Gamma_{2}(G)$ is connected. In the second case, we need to prove that
if $|x|=p,$ then $x\in\Omega.$ By Lemma 5.1, there exists a 3-cycle $y$ with
$[x,y,y]=1$, so $x\twoheadrightarrow_{2}y,$ with $y\in\Omega$. Moreover there
exists an element $z$ of order $(p-1)/2$ which normalizes $\langle x\rangle$.
In particular $z\in\Omega$ and $z\mapsto_{2}x.$

Assume $n=7$. Again all the elements of even order of $G$ belong to the same
strong component $\Omega$ of $\Gamma_{2}(G).$ A $3$-cycle commutes with a
double-transposition, so $\Omega$ contains all the $3$-cycles, as well as all
the products of two disjoint $3$-cycles. Moreover
$(2,5)(3,4)\mapsto_{2}(1,2,3,4,5)$ and $(1,2,3,4,5)\mapsto_{2}(3,5)(6,7)$ so
$\Omega$ contains all the $5$-cycles. Finally
$(1,2,3,4,5,6,7)\mapsto_{2}(1,4)(3,7)$ and
$(2,3,5)(4,7,6)\mapsto_{2}(1,2,3,4,5,6,7).$ So, $\Gamma_{2}(A_{7})$ is
strongly connected

Now, let $n=6.$ If $x=(1,2,3,4,5)$ and $x\mapsto_{2}y$, then $y\in\langle
x\rangle$ and therefore $\Gamma_{2}(G)$ is not strongly connected. But
$x\mapsto_{3}(2,4)(5,6)$ and $(2,5)(3,4)\mapsto_{3}x.$ So there is a strong
component $\Omega$ of $\Gamma_{3}(G)$ containing all the elements of $G$ of
even order and all the $5$-cycles. The other elements of $G$ are conjugated in
$\operatorname{Aut}(G)$ to $(1,2,3)$ so to conclude it suffices to notice that
$(2,3)(5,6)\mapsto_{2}(1,2,3)$ and $(1,2,3)\mapsto_{2}(1,2)(3,4).$

Finally assume $n=5.$ If $x=(1,2,3,4,5)$ and $x\mapsto y$, then $y\in\langle
x\rangle$ and therefore $\Gamma(G)$ is not strongly connected. ∎

###  5.2. Symmetric groups

The aim of this subsection is to prove that if $n\geq 5,$ then
$\Gamma_{2}(S_{n})$ is strongly connected. When $n\geq 7,$ this could be
immediately deduced combining Lemma 3.2 with Theorem 5.2. We prefer to give an
easy and self-contained proof.

######  Lemma 5.3.

If $x=(a,b)$ and $y=(c,d)$ are two transpositions in $S_{n},$ then
$x\twoheadrightarrow_{1}y.$

###### Proof.

It suffices to notice that either $[x,y]=1$ or there exists a transposition
$z=(e,f)$ with $[x,z]=[y,z]=1.$ ∎

######  Lemma 5.4.

Let $1\neq x\in S_{n}.$ Write $x=\sigma_{1}\cdots\sigma_{r}$ as product of
disjoint cycles (including possibly cycles of length 1), in such a way that
$|\sigma_{1}|\leq\dots\leq|\sigma_{r}|.$ If $|\sigma_{r}|\leq n-2,$ then there
exists a transposition $u$ such that $x\twoheadrightarrow_{1}u$ and
$u\twoheadrightarrow_{1}x.$

###### Proof.

Let $y=\sigma_{1}\cdots\sigma_{r-1}$. If $y\neq 1$, then set $u=(a,b)$ with
$\\{a,b\\}\subseteq{\rm{supp}}(\sigma_{r})$, so that $[x,y]=1$ and $[y,u]=1$.
Otherwise, take $u=(a,b)$ with
$\\{a,b\\}\cap{\rm{supp}}(\sigma_{r})=\emptyset.$ ∎

######  Lemma 5.5.

If $1\neq x\in S_{n},$ then there exists an involution $u$ such that
$u\twoheadrightarrow_{2}x.$

###### Proof.

Note that $x$ is conjugate to $x^{-1}$ and consequently $N_{G}(\langle
x\rangle)$ contains an involution. ∎

######  Lemma 5.6.

If $m\geq 4,$ then

$(1,3)\cdot(1,2,3,\dots,m)\cdot(1,3)\cdot(1,2,3,\dots,m)^{-1}=(1,3)(2,m)$

and consequently $[(1,2,3,\dots,m),(1,3),(1,3)]=1.$

######  Theorem 5.7.

If $n\geq 5,$ then $\Gamma_{2}(S_{n})$ is strongly connected.

###### Proof.

By Lemmas 5.3 and 5.4, all the elements of $G$ that are neither a
$(n-1)$-cycle nor a $n$-cycle belong to the same strong component of
$\Gamma_{1}(G)$, and consequently, to the same strong component, say $\Omega,$
of $\Gamma_{2}(G).$ Suppose that $x$ is an $r$-cycle, with $r\in\\{n-1,n\\}.$
By Lemma 5.6, $x\twoheadrightarrow_{2}u$ for a suitable involution $u.$
Combining this with Lemma 5.5, we conclude $x\in\Omega.$ ∎

###  5.3. Simple groups whose Engel graph is not strongly connected

It follows from the previous subsections that there are infinitely many simple
and almost simple groups with a strongly connected Engel graph. Now our aim is
to prove that there exist also infinitely many simple groups whose Engel graph
is not strongly connected. We first need a preliminary lemma.

######  Lemma 5.8.

Let $y$ be contained in a subgroup $K\leq G$ with the property that

  1. (1)

$N_{G}(K)=K$;

  2. (2)

$y\in K^{g}$ if and only if $K^{g}=K$.

Then $x\notin K$ implies $x\not\mapsto y$.

###### Proof.

Note that conditions (1) and (2) imply that $y^{-g}\in K$ if and only if $g\in
K$. Assume $x\mapsto y$, so that $[x,_{n}y]=1$ for some $n\geq 1$, and let $m$
be the minimal integer such that $[x,_{m}y]\in K$. Assume $m>0$. We set
$w=[x,_{m-1}y]$. Then from $[x,_{m}y]=[w,y]=y^{-w}y\in K$ and $y\in K$, we
deduce that $y^{-w}\in K$. So, $w\in K$, against the minimality of $m$. It
follows that $m=0$, that is $x\in K$. ∎

######  Theorem 5.9.

Let $G=\operatorname{PSL}(2,q)$ with $q=2^{f}.$ Then $\Gamma(G)$ is not
strongly connected.

###### Proof.

Note that the order of an element of $G$ can be:

  * •

2;

  * •

a divisor of $q-1;$

  * •

a divisor of $q+1.$

Now let $x\neq 1$ be an element whose order divides $q+1.$ We are going to
prove that if $x\mapsto y$, then $|y|$ divides $q+1.$ Let $y$ be a nontrivial
element of $G$.

(a) Assume $|y|=2$ and let $P$ be a Sylow 2-subgroup of $G$, with $y\in P.$
Then $P\cong C_{2}^{f}$ and $K=N_{G}(P)\cong P\rtimes C_{q-1}$ is a maximal
subgroup of $G.$ We may apply Lemma 5.8 to conclude that $x\not\mapsto y.$

(b) Assume that $|y|=t$ divides $q-1$ and set $K=N_{G}(\langle y\rangle).$
Then $K$ is a maximal subgroup of $G$ and $K\cong D_{2(q-1)}.$ Again we apply
Lemma 5.8 to conclude that $x\not\mapsto y.$ ∎

######  Theorem 5.10.

Let $G=\operatorname{Sz}(q)$ with $q=2^{2t+1}$ and $t\geq 1$. Then $\Gamma(G)$
is not strongly connected.

###### Proof.

If $g\in G$ then the order of $G$ can be:

  * •

$2$ or $4$;

  * •

a divisor of $q-1;$

  * •

a divisor of $q+\sqrt{2q}+1;$

  * •

a divisor of $q-\sqrt{2q}+1.$

Now let $x\neq 1$ be an element whose order divides $q+\sqrt{2q}+1.$ We are
going to prove that if $x\mapsto y$, for a nontrivial element $y\in G$, then
$|y|$ divides $q+\sqrt{2q}+1.$

(a) If $|y|$ is $2$ or $4,$ then $y$ is contained in a unique Borel maximal
subgroup, since the intersection of two distinct such subgroups does not
contain $2$-elements. Therefore by Lemma 5.8 we conclude $x\not\mapsto y$.

(b) If $|y|$ divides $q-1$, let $K=N_{G}(C_{G}(y))$. This is a maximal
subgroup of dihedral type, so it is self-normalizing. We have $\langle
y\rangle\mathrel{\unlhd}K$. If $y\in K^{g}$ with $K^{g}\neq K$, then
$\left\langle y\right\rangle\mathrel{\unlhd}K^{g}$ since it is the unique
subgroup of $K^{g}$ of order $|y|$. Therefore $\left\langle
y\right\rangle\mathrel{\unlhd}\left\langle K,K^{g}\right\rangle=G$, which is
impossible, so $K$ respects the requirements of Lemma 5.8 and $x\not\mapsto
y$.

(c) If $|y|$ divides $q-\sqrt{2q}+1$, let $K=N_{G}(C_{G}(y))$. $K$ is a
maximal subgroup of type $C_{q-\sqrt{2q}+1}\rtimes C_{4}$ and so it is self-
normalizing. Since $\left\langle y\right\rangle$ is the unique subgroup of $K$
of order $|y|$, we can argue as in (b) and conclude $x\not\mapsto y$. ∎

##  6\. Some computations

Combining the available information about the connected components of the
prime graph of $G$ (see [15]) with some direct computations with GAP [7], we
checked that $\Gamma(G)$ is strongly connected if $G$ is a sporadic simple
group, although its commuting graph $\Gamma_{1}(G)$ is not (notice that by
Lemma 3.2 this implies that $\Gamma(\operatorname{Aut}(G))$ is also strongly
connected). Indeed, let $\pi_{1}(G)$ be the connected components of the prime
graph of $G$ which contains 2. By [10, Lemma 6.1], there is a unique connected
component of the commuting graph $\Gamma_{1}(G)$ containing all the elements
of even order, and consequently there is a strong component of $\Gamma(G)$
containing all the elements whose order is divisible by a prime
$p\in\pi_{1}(G).$ In particular, $\Gamma(G)$ is strongly connected if for any
prime divisor $q$ of $|G|$ with $q\not\in\pi_{1}(G),$ there exists a subgroup
$H$ of $G$ whose order is divisible by $q$ and some prime in $\pi_{1}(G)$ and
such that $\Gamma(H)$ is strongly connected.

For example, the prime graph of $\rm{J}_{2}$ has two connected components,
$\pi_{1}(\rm{J_{2}})=\\{2,3,5\\}$ and $\\{7\\},$ so it suffices to notice that
$\rm{J}_{2}$ contains a subgroup $H\cong\operatorname{PSL}(2,7)$ and to check
with GAP that $\Gamma(\operatorname{PSL}(2,7))$ is strongly connected.

Similarly if $G=\rm{M},$ then 41, 59 and 71 are the only prime divisors of $G$
not contained in $\pi_{1}(G)$; since $\rm{M}$ contains subgroups isomorphic to
$\operatorname{PSL}(2,41),\operatorname{PSL}(2,59)$ and
$\operatorname{PSL}(2,71)$ (see [3] and [11]) we reduce to check that the
Engel graphs of these subgroups are strongly connected.

Another case worth mentioning is $G=\rm{J_{1}}$. In that case the prime
divisors of $|G|$ not contained in $\pi_{1}(G)$ are 7, 11, 19. However $G$
contains a maximal subgroup $M_{1}\cong 2^{3}:7:3$ and a maximal subgroup
$M_{2}\cong\operatorname{PSL}(2,11)$ and $\Gamma(M_{1}),\Gamma(M_{2})$ are
strongly connected. There is also a maximal subgroup isomorphic to 19:6 and
consequently, by Lemma 2.2, $\Gamma(G)$ contains and edge from an element of
order 6 to an element of order 19. Hence to prove that $\Gamma(G)$ is strongly
connected it suffices to prove that there is an edge $x\mapsto y$, with
$|x|=19$ and $|y|\neq 19.$ When we checked whether this is true, we found an
unexpected phenomenon: there exists $g$ of order $19$ such that $g\mapsto y$
only if $y\in\langle g\rangle,$ but $g^{2}$ has a different behavior, since
there are 38 involutions $z$ with $g\mapsto z.$

The four sporadic groups $\\{\rm{J_{4},Ly,F_{24}^{\prime},B}\\}$ are more
difficult to handle. In these cases the methods described above do not allow
to deduce that there is an edge $x\mapsto y$ with $|x|=p$ and $|y|$ divisible
by a prime in $\pi_{1}(G),$ when
$(G,p)\in\\{(\rm{J_{4}},29),(J_{4},43),(Ly,37),$
$(Ly,67),(F_{24}^{\prime},29),(B,47)\\},$ since $\Gamma(H)$ turns out to be
disconnected for every subgroup $H$ of $G$ whose order is divisible by $p.$ In
that case the following remark is of great help. As it can be easily verified,
if $g_{1},g_{2}$ are elements of an arbitrary group and $|g_{2}|=2,$ then
$[g_{1},_{n}g_{2}]=[g_{1},g_{2}]^{(-2)^{n-1}}$. In our case, testing the
commutators between a randomly chosen element of order $p$ and a random
involution, we produce pair $(x,y)$ with $|x|=p,|y|=2$ and
$[x,_{3}y]=[x,y]^{4}=1.$

We also used GAP to check that if $p$ is and odd prime with $p\leq 109,$ then
the graph $\Gamma(\operatorname{PSL}(2,p))$ is strongly connected if and only
if $p\notin\\{13,29,37,53,61,101,109\\},$ i.e. if and only if $p\neq 5\mod 8.$
We aim to investigate whether this is the general behavior in a subsequent
paper. It is interesting to notice that $G=\operatorname{PSL}(2,47)$ is the
smallest simple group we know, with the property that $\Gamma_{4}(G)$ is
strongly connected, but $\Gamma_{3}(G)$ is not. If
$G=\operatorname{PSL}(2,127),$ then $\Gamma_{7}(G)$ is strongly connected, but
$\Gamma_{6}(G)$ is not.

To perform these computations we wrote a function “$\rm{eng}$” defined on the
elements of $G$. This function, given two nontrivial elements $x,y\in G$,
recursively evaluates the commutators $[x,y],[x,_{2}y]\dots$ and stops when
either $[x,_{n}y]=1$ or $[x,_{n_{1}}y]=[x,_{n_{2}}y]\neq 1$ with $n_{1}\neq
n_{2}.$ The first component $\rm{eng}(x,y)[1]$ of the output of
$\rm{eng}(x,y)$ is either the trivial element, if $[x,_{n}y]=1$ for some
positive integer, or a nontrivial element $z$ with the property that
$z=[x,_{n_{1}}y]=[x,_{n_{2}}y]$ with $n_{1}\neq n_{2}.$ The second component
$\rm{eng}(x,y)[2]$ keeps track of the commutators that have been evaluated
during the procedure. From Lemma 2.2 (1) it follows that if
$1\neq[x,_{n_{1}}y]=[x,_{n_{2}}y]$ with $n_{1}\neq n_{2}$, then $x$ is not
adjacent to $y$. Therefore, $x\mapsto y$ if and only if
$\rm{eng}(x,y)[1]=Identity(Group(x,y))$. In particular $x\mapsto_{n}y$ if and
only if $\rm{eng}[1]=Identity(Group(x,y))$ and $\rm{Size(eng}(x,y)[2])\leq n.$

⬇

eng:=function(x,y)

local z,s;

s:=[Identity(Group(x,y))];z:=Comm(x,y);

while (z in s)=false do

Add(s,z);

z:=Comm(z,y);

od;

return[z, s];

end;

## References

  * [1] A. Abdollahi and M. Zarrin, Non-nilpotent graph of a group, Comm. Algebra 38 (2010), 4390–4403.
  * [2] T. Burness, A. Lucchini, D. Nemmi, On the soluble graph of a finite group, arXiv:2111.05697.
  * [3] J.H. Conway, R.T. Curtis, S.P. Norton, R.A. Parker, R.A. Wilson, Atlas of Finite Groups, Oxford University Press, 1985.
  * [4] D. Gorenstein, _Finite simple groups: An introduction to their classification_ , University Series in Mathematics. Plenum Publishing Corp., New York, 1982.
  * [5] A. Iranmanesh, A. Jafarzadeh, On the commuting graph associated with the symmetric and alternating groups. J. Algebra Appl. 7 (2008), no. 1, 129–146.
  * [6] B. Huppert, N. Blackburn, Finite groups II, Springer-Verlag, Berlin, 1982.
  * [7] The GAP Group, _GAP – Groups, Algorithms, and Programming, Version 4.11.1_ ; 2021, https://www.gap-system.org.
  * [8] M. Giudici and C. Parker, There is no upper bound for the diameter of the commuting graph of a finite group, J. Combin. Theory Ser. A 120 (2013), 1600–1603.
  * [9] R. Guralnick, B. Kunyavskiĭ, E. Plotkin and A. Shalev, Thompson-like characterizations of the solvable radical, J. Algebra 300 (2006), 363–375.
  * [10] G. L. Morgan and C. W. Parker, The diameter of the commuting graph of a finite group with trivial centre, J. Algebra 393 (2013), 41–59.
  * [11] S. Norton and R. Wilson, A correction to the 41-structure of the Monster, a construction of a new maximal subgroup $L_{2}(41)$ and a new Moonshine phenomenon, J. Lond. Math. Soc. (2) 87 (2013), no. 3, 943–962.
  * [12] C. Parker, The commuting graph of a soluble group. Bull. Lond. Math. Soc. 45 (2013), no. 4, 839–848.
  * [13] D. J. S. Robinson. A course in the theory of groups. Second edition. Graduate Texts in Mathematics, 80. Springer-Verlag, New York, 1996.
  * [14] Y. Segev and G. M. Seitz, Anisotropic groups of type $A_{n}$ and the commuting graph of finite simple groups, Pacific J. Math. 202 (2002), 125–225.
  * [15] J. S. Williams, Prime graph components of finite groups, J. Algebra 69 (1981), no. 2, 487–513.


\overline{A\Gamma}^{H,\mathcal{N}}(\beta_{n})\in\mathbb{L}^{H}.$ (9.21)
${\large{\Huge{\Gamma^{R}}}}(\beta_{n}):=\underset{\longleftarrow}{\mathrm{lim}}\
\overline{A\Gamma}^{R,\mathcal{N}}(\beta_{n})\in\mathbb{L}^{R}.$ (9.22)

Using that the three sequences of ideals behave well with respect to
inclusions, as shown in relation (9.12), we have well defined maps between the
limit rings, which we denote as below:

$\displaystyle\psi^{H,R}\ \ \ \ \ \ \ \ \psi^{R,U}$ (9.23)
$\displaystyle\mathbb{L}^{H}\ \ \rightarrow\ \ \mathbb{L}^{R}\ \ \rightarrow\
\ \hat{\mathbb{L}}.$

For the next part, we consider the specialisations of coefficients associated
to these limit rings, as follows.

######  Definition 9.9 (Specialisations)

Let $\psi^{R}_{\mathcal{N}}$ be the ring homomorphism induced by
$\psi^{u}_{\mathcal{N}}$ and the projection maps from
$\mathbb{L}/J^{R}_{\mathcal{N}}$ to $\mathbb{L}/\tilde{I}_{\mathcal{N}}$ (via
Definition 2.19, Definition 2.20):

$\displaystyle\psi^{R}_{\mathcal{N}}:\mathbb{L}^{R}\rightarrow\Lambda_{\mathcal{N}}$
(9.24)
$\displaystyle\psi^{R}_{\mathcal{N}}=\psi^{u}_{\mathcal{N}}\circ\psi_{\mathcal{N}}^{R,U}.$

Also, let $\psi^{H}_{\mathcal{N}}$ be the specialisation induced by
$\psi^{u}_{\mathcal{N}}$ and the projection maps from
$\mathbb{L}/J^{H}_{\mathcal{N}}$ to $\mathbb{L}/\tilde{I}_{\mathcal{N}}$:

$\displaystyle\psi^{H}_{\mathcal{N}}:\mathbb{L}^{H}\rightarrow\Lambda_{\mathcal{N}}$
(9.25)
$\displaystyle\psi^{H}_{\mathcal{N}}=\psi^{R}_{\mathcal{N}}\circ\psi_{\mathcal{N}}^{H,R}.$

Once we have defined these limit rings and the limits of graded intersections,
we can prove that they recover all ADO invariants through changes of
coefficients, as follows.

######  Theorem 9.10 (Universal extended intersection)

The extended intersection in the Habiro ring
${\large{\Huge{\Gamma^{H}}}}(\beta_{n})\in\mathbb{L}^{H}$ recovers all
coloured Alexander invariants:

$\Phi^{\mathcal{N}}(L)={\large{\Huge{\Gamma^{H}}}}(\beta_{n})|_{\psi^{H}_{\mathcal{N}}}.$
(9.26)

######  Theorem 9.11 (Universal refined ADO intersection)

The refined intersection
${\large{\Huge{\Gamma^{R}}}}(\beta_{n})\in\mathbb{L}^{R}$ constructed as a
limit of the graded intersections via configuration spaces on ovals, is a
braid invariant recovering all coloured Alexander invariants:

$\Phi^{\mathcal{N}}(L)={\large{\Huge{\Gamma^{R}}}}(\beta_{n})|_{\psi^{R}_{\mathcal{N}}}.$
(9.27)

This braid invariant belongs to $\mathbb{L}^{R}$, which is a module over the
extended Habiro ring.

###### Proof.

These two unification results follow from the fact that our universal
invariant ${\large\hat{{\Huge{\Gamma}}}}(L)$ recovers all ADO polynomials, as
in Theorem 1.14 (relation (9.27)) and the property that the ideals that all
projection maps between $\mathbb{L}/J^{H}_{\mathcal{N}}$,
$\mathbb{L}/J^{R}_{\mathcal{N}}$ and $\mathbb{L}/\tilde{I}_{\mathcal{N}}$
behave well with respect to the change of the level. ∎

######  Conjecture 9.12 (Universal refined ADO invariant)

The universal graded refined intersection

${\large{\Huge{\Gamma^{R}}}}\in\mathbb{L}^{R}$

is a well-defined link invariant in the refined universal ring
$\mathbb{L}^{R}$ (which is a module over the extended Habiro ring
$\mathbb{L}^{H}$).

Comment  In order to prove that ${\large{\Huge{\Gamma^{R}}}}$ is a link
invariant in the limit ring, we need to show that all the refined graded
intersections

$\overline{A\Gamma}^{R,\mathcal{N}}(\beta_{n})\in\mathbb{L}/J^{R}_{\mathcal{N}}$

are link invariants. These intersections are quotients of
$A\Gamma^{\mathcal{N}}(\beta_{n})$. So, one needs to show that
$A\Gamma^{\mathcal{N}}(\beta_{n})$ becomes a link invariant once we quotient
by the ideal $J^{R}_{\mathcal{N}}$. This means that we would need to check the
invariance of $A\Gamma^{\mathcal{N}}(\beta_{n})$ modulo the ideal
$J^{R}_{\mathcal{N}}$ under the Markov moves.

Invariance under the Markov I move This concerns the invariance of the
intersection under braid conjugation. It will follow from the fact that we
quotient by these cyclotomic polynomials in $d$, and the formulas for the
braid action on the configuration space.

Markov II move This is what remains to be shown for the proof of Conjecture
9.12: the invariance of the graded intersection once we perform a
stabilisation of the braid.

We have evidence that for the first level, associated to $\mathcal{N}=2$, this
holds.

######  Proposition 9.13 (Invariant at level $\mathcal{N}=2$)

The level $2$ refined intersection is a well-defined link invariant

$\overline{A\Gamma}^{R,2}(L)\in\mathbb{L}/J^{R}_{2}.$

###### Proof.

This follows from an analog argument as the proof from [7]. In that article we
proved by direct homological computations the invariance of a graded
intersection pairing under the Markov moves, once we quotient the ring of
coefficients. This ring was a Laurent polynomial ring where we quotiented by a
quadratic relation. Based on the same homological tools, we can show that we
have a link invariant also in this situation, where we consider the quotient
by $J^{R}_{2}$. ∎

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(9.28) Figure 9.1: Universal ADO invariant and extensions over the Habiro ring

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University of Geneva, Section de mathématiques, Rue du Conseil-Général 7-9, CH
1205 Geneva, Switzerland; Institute of Mathematics “Simion Stoilow” of the
Romanian Academy, 21 Calea Grivitei Street, 010702 Bucharest, Romania.

<EMAIL_ADDRESS><EMAIL_ADDRESS>
www.cristinaanghel.ro