# Group nilpotency from a graph point of view

Valentina Grazian, Andrea Lucchini and Carmine Monetta

###### Abstract.

Let $\Gamma_{G}$ denote a graph associated with a group $G$. A compelling
question about finite groups asks whether or not a finite group $H$ must be
nilpotent provided $\Gamma_{H}$ is isomorphic to $\Gamma_{G}$ for a finite
nilpotent group $G$. In the present work we analyze the problem for different
graphs that one can associate with a finite group, both reporting on existing
answers and contributing to new ones.

MSC(2020): Primary: 20D15; Secondary: 05C25.
Keywords: graph isomorphism; nilpotency.

##  1\. Introduction

Given a finite group $G$, one can consider a graph $\Gamma_{G}$ associated
with $G$ which encodes certain group properties of $G$. Such an approach has
been extensively studied in the last decades, mainly for two reasons. The
former is to determine structure description of $G$ investigating the
invariants of $\Gamma_{G}$, while the latter aims to produce graphs fitting
specific features (see for instance [7, 6, 12, 19, 21, 28]).

A natural question in this research line is to understand if a graph
isomorphism - which is clearly a weaker relation than a group isomorphism -
may or may not preserve specific properties of a group. More precisely, we are
interested in the following question.

######  Question 1.1.

If $G$ and $H$ are finite groups with isomorphic graphs
$\Gamma_{G}\cong\Gamma_{H}$ and $G$ is nilpotent, is it true that $H$ is
nilpotent as well?

Of course the hardness of the problem, as well as the answer, change depending
on the graph choice. For instance we may easily produce a negative answer to
Question 1.1 considering the soluble graph of a group $G$, that is the graph
whose set of vertices is $G$ and in which two vertices are adjacent if they
generate a soluble group. Indeed, both the soluble graph of a nilpotent group,
and the soluble graph of a soluble group are complete, thus in this case
Question 1.1 has negative answer whenever we pick a nilpotent group $G$ and a
non-nilpotent soluble group $H$ having the same order: the smallest example is
given by the cyclic group of order $6$ and the symmetric group of degree $3$.

However, our report on this problem will clarify that the situation is not so
easy in general and that in some cases Question 1.1 remains still open.

In the following we will analyze the distinct circumstances corresponding to
the following graph choices: the non-commuting graph, the power graph, the
prime graph (also known as Gruenberg-Kegel graph), the generating graph, the
Engel graph and the join graph; our study is summarized in Table 1.

| Answer to Q 1.1 | Positive answer if |  Examples with $H$ non-nilpotent
---|---|---|---
Power graph | Yes (Thm 2.2) |  |
Engel graph | Yes (Prop 3.1) |  |
Non-commuting graph | Open |  $|G|=|H|$ (Thm 4.1) or $G$ is an AC-group and $|Z(G)|\geq|Z(H)|$ (Thm 4.2) |
Generating graph | Open |  $H$ is supersoluble (Thm 5.1) or The subgraph of the non-generating graph obtained by removing all universal vertices is disconnected (Thm 5.13) |
Prime graph | No |  $|H|$ is square-free (Prop 6.2) |  $G\cong C_{6}\times C_{6}$ and $H\cong S_{3}\times C_{6}$
Join graph | No |  |  $G\cong C_{p}\times C_{p}$ and $H\cong D_{2p}$, for $p$ an odd prime
Table 1. Answers to Question 1.1 depending on the graphs.

We will mainly adopt standard terminology in graph theory and known notation
in group theory. If $G$ is a finite group, $|G|$ denotes the order of $G$,
$Z(G)$ denotes the center of $G$, ${\rm{Fit}}(G)$ stands for the Fitting
subgroup of $G$ and $\Phi(G)$ denotes the Frattini subgroup of $G$. Also, we
will write $C_{n}$, $S_{n}$, $D_{2n}$ and $Q_{n}$ for the cyclic group of
order $n$, the symmetric group of degree $n$, the dihiedral group of order
$2n$ and the quaternion group of order $n$, respectively.

##  2\. The power graph and the enhanced power graph

The power graph of a semigroup was first introduced by Chakrabarty et al.
[14], taking inspiration from the definition of directed power graph given by
Kelarev and Quinn [20]. The power graph of a finite group $G$ is a graph whose
vertices are the elements of $G$ and in which two distinct vertices $x$ and
$y$ are adjacent if there exists $k\geq 2$ such that $x^{k}=y$ or $y^{k}=x$.
The enhanced power graph of $G$, also known as cyclic graph of $G$, is the
graph with vertex set $G$ in which distinct elements $x$ and $y$ are joined by
an edge if and only if $\langle x,y\rangle$ is cyclic. Note that the power
graph is a subgraph of the enhanced power graph, and the two graphs coincide
if and only if any element of the group has prime power order (see Theorem 28
of [1]). However, Question 1.1 is equivalent for power graphs and enhanced
power graphs, thanks to the following result:

######  Theorem 2.1.

[34, Corollary 3.1] Two finite groups $G$ and $H$ have isomorphic power graph
if and only if they have isomorphic enhanced power graph.

In [13], Cameron proved that if $G$ and $H$ are finite abelian groups with
isomorphic power graph, then they are isomorphic (see [13, Theorem 1]). Then
in [11], the author proved that groups with isomorphic power graph have the
same numbers of elements of each order (see [11, Corollary 3]). Thanks to this
result, Mirzargar and Scapellato managed to give a positive answer to Question
1.1 for power graphs:

######  Theorem 2.2.

[29, Corollary 3.2] Let $G$ and $H$ be finite groups with isomorphic power
graphs. If $G$ is nilpotent, then $H$ is nilpotent.

###### Proof.

Let $p$ be a prime dividing $|G|$, let $P$ be the unique Sylow $p$-subgroup of
$G$ and set $|P|=p^{n}$. Since $G$ and $H$ have isomorphic power graph, we get
$|G|=|H|$ and so the Sylow $p$-subgroups of $H$ must have order $p^{n}$. Also,
$G$ has exactly $p^{n}$ elements of $p$-power order and by [11, Corollary 3],
the graph isomorphism implies that $H$ has exactly $p^{n}$ elements of
$p$-power order too. Thus we conclude that $H$ has a unique Sylow
$p$-subgroup. As this holds for every prime divisor of $|H|$, we deduce that
$H$ is nilpotent. ∎

Unfortunately, the proof of the previous theorem does not allow to obtain an
explicit criterion to detect the nilpotency of a finite group $G$ looking at
the associated power graph. The situation is much better if one considers the
directed power graph associated to $G$: in this case the vertex set is again
$G$ and there is an arc $y\mapsto x$ if $x$ is a power of $y$. In particular
the number of vertices $y\neq x$ such that $x\mapsto y$ is $|x|-1$, so $|x|$
is determined by the graph. Thus just looking at the direct power graph of $G$
we may compute, for any prime divisor $p$ of $|G|$, the number of $p$-elements
of $G$ and deduce immediately whether the Sylow $p$-subgroup is normal in $G$.

##  3\. The Engel graph

Following a suggestion given by Cameron (see [12, Section 11.1]), we may
define a graph $\Gamma_{\mathord{\mathrm{eng}}}(G)$, where the vertices are
the elements of $G$ and where two vertices are adjacent if they satisfy a
suitable Engel relation; more precisely if $x$ and $y$ are different elements
of $G,$ then there is an edge joining $x$ and $y$ if and only if either
$[x,{}_{r}y]=1$ or $[y,{}_{r}x]=1$ for some $r\in\mathbb{N}$. Cameron proposes
to call this graph the Engel graph of $G$, although, as he notices, the same
term was used by Abdollahi [2] to denote a related but different graph.

It is easy to see that the answer to Question 1.1 is affirmative for Engel
graphs:

######  Proposition 3.1.

A finite group $G$ is nilpotent if and only if its Engel graph is complete. In
particular, if $G$ and $H$ are finite groups with isomorphic Engel graphs and
$G$ is nilpotent, then $H$ is nilpotent.

###### Proof.

If $G$ is nilpotent, then every element of $G$ is a right Engel element. Hence
the graph $\Gamma_{\mathord{\mathrm{eng}}}(G)$ is complete. On the other hand,
if the graph $\Gamma_{\mathord{\mathrm{eng}}}(G)$ is complete, then by [32,
12.3.4], we conclude that $G$ is nilpotent. ∎

Cameron proposed to investigate the relation between the Engel graph and the
Nilpotent graph (where the Nilpotent graph
$\Gamma_{\mathord{\mathrm{nil}}}(G)$ has as vertices the elements of $G$ and
$x$ and $y$ are adjacent if and only if $\langle x,y\rangle$ is nilpotent). In
particular he asks (see [12, Question 24]) for which groups $G$ the two graphs
coincide. We answer to this question with the following result.

######  Theorem 3.2.

Let $G$ be a finite group. Then
$\Gamma_{\mathord{\mathrm{eng}}}(G)=\Gamma_{\mathord{\mathrm{nil}}}(G)$ if and
only if $G$ is nilpotent.

###### Proof.

We prove by induction on the order of $G$ that if
$\Gamma_{\mathord{\mathrm{eng}}}(G)=\Gamma_{\mathord{\mathrm{nil}}}(G)$, then
$G$ is nilpotent. The property that
$\Gamma_{\mathord{\mathrm{eng}}}(G)=\Gamma_{\mathord{\mathrm{nil}}}(G)$ is
inherited by all the subgroups of $G.$ So by induction all the proper
subgroups of $G$ are nilpotent, and this implies that $G$ is soluble. The set
of universal vertices of the graph $\Gamma_{\mathord{\mathrm{nil}}}(G)$
coincides with the hypercenter of $G$ (see [6, Proposition 2.1]), while the
set of universal vertices of $\Gamma_{\mathord{\mathrm{eng}}}(G)$ is the set
of elements that are either left or right Engel and coincides with the Fitting
subgroup of $G$ (see e.g. [32, 12.3.7]). Since
$\Gamma_{\mathord{\mathrm{eng}}}(G)=\Gamma_{\mathord{\mathrm{nil}}}(G)$, if
follows that $G$ is a finite soluble group, whose hypercenter and Fitting
subgroup ${\rm{Fit}}(G)$ coincide, and this is possible only if $G$ is
nilpotent. Indeed, arguing by contradiction, assume that $G$ is not nilpotent.
Since $G$ is soluble and ${\rm{Fit}}(G)<G$ by assumption, there exists a
normal subgroup $N$ of $G$ such that $1\neq
N/{\rm{Fit}}(G)={\rm{Fit}}(G/{\rm{Fit}}(G))$. In particular, there is an
integer $h\geq 2$ such that the $h$-term of the lower central series of $N$,
denoted $\gamma_{h}(N)$, is contained in ${\rm{Fit}}(G)$. Since
${\rm{Fit}}(G)$ coincides with the hypercenter of $G$, we can find an integer
$k\geq 2$ such that $\gamma_{h+k}(N)=1$, proving that $N$ is nilpotent and
contradicting the fact that $N\nleq{\rm{Fit}}(G)$. This proves that $G$ must
be nilpotent.

Conversely, if $G$ is nilpotent, then every element of $G$ is a right Engel
element, so
$\Gamma_{\mathord{\mathrm{eng}}}(G)=\Gamma_{\mathord{\mathrm{nil}}}(G)$ is the
complete graph on $|G|$ vertices. ∎

##  4\. The non-commuting graph

The non-commuting graph of a group was first considered by Paul Erdős in 1975,
while stating a problem solved by Neumann in [31]. If $G$ is a finite group,
the non-commuting graph of $G$ is the graph whose vertices are the non-central
elements of $G$ (i.e. $G\backslash Z(G)$) and in which two vertices $x$ and
$y$ are adjacent if and only if they do not commute, or equivalently, if the
group $\langle x,y\rangle$ is non-abelian. Note that the non-commuting graph
of $G$ is the complement of the commuting graph of $G$ (where two elements are
joined if they commute).

Question 1.1 in terms of the non-commuting graph was first posed by Abdollahi,
Akbari and Maimani in [4], and it is in fact still open. Nevertheless,
Question 1.1 is known to be true under certain extra conditions.

######  Theorem 4.1.

[4, Theorem 3.24] Let $G$ and $H$ be finite non-abelian groups with isomorphic
non-commuting graphs. If $G$ is nilpotent and $|G|=|H|$, then H is nilpotent.

###### Proof.

Note that by the main result in [15], it is enough to prove that $G$ and $H$
have the same number of conjugacy classes of the same size. For every integer
$i\geq 1$, let $m_{i}(G)$ and $m_{i}(H)$ denote the number of conjugacy
classes of size $i$ of $G$ and $H$, respectively. We will show that
$m_{i}(G)=m_{i}(H)$ for every $i\geq 1$. First note that since the non-
commuting graphs of $G$ and $H$ are isomorphic, we get $|G|-|Z(G)|=|H|-|Z(H)|$
(looking at the number of vertices). By assumption, $|G|=|H|$, so we deduce
that $m_{1}(G)=|Z(G)|=|Z(H)|=m_{1}(H)$. Now, for every $g\in G\backslash
Z(G)$, if $h$ is the image of $g$ under the graph isomorphism, then looking at
the number of vertices joined to $g$ and $h$, respectively, we deduce that
$|G|-|C_{G}(g)|=|H|-|C_{H}(h)|$. Using again the hypothesis $|G|=|H|$, we
obtain $|g^{G}|=[G\colon C_{G}(g)]=[H\colon C_{H}(h)]=|h^{H}|$. This implies
that for every $i>1$ we have $m_{i}(G)=m_{i}(H)$, concluding the proof. ∎

In the same work, Abdollahi, Akbari and Maimani conjectured that any two
finite groups having isomorphic non-commuting graph should have the same
order, but this was proven to be false in [30]. The counterexample built in
[30] involves two finite groups that are nilpotent, and so it does not affect
Question 1.1. Also, such groups are finite AC-groups, that is, finite groups
whose centralizers of non-central elements are all abelian. Taking inspiration
from such result, in a recent paper Grazian and Monetta proved the following:

######  Theorem 4.2.

[19, Corollary 1.4] Let $G$ and $H$ be finite non-abelian groups with
isomorphic non-commuting graphs. If $G$ is a nilpotent AC-group and
$|Z(G)|\geq|Z(H)|$, then $H$ is nilpotent.

###### Idea of the proof.

First notice that the isomorphism between the non-commuting graphs of the
finite groups $G$ and $H$ implies that $H$ is an AC-group. By [4, Proposition
3.14], if $H$ is non-solvable, then $|G|=|H|$ and we conclude by Theorem 4.1.
Therefore we can assume that $H$ is a solvable AC-group. Such groups have been
classified in [33, Satz 5.12], showing that if $H$ is non-abelian then $H$
must follow one of $4$ possible characterizations. The proof is completed
analyzing the different possibilities. ∎

Observe that a nilpotent AC-group has a unique Sylow subgroup that is non-
abelian. Question 1.1 in the case of a finite nilpotent group $G$ with at
least two distinct non-abelian Sylow subgroups has also been covered:

######  Theorem 4.3.

[5, Theorem 2.4] Let $G$ and $H$ be finite non-abelian groups with isomorphic
non-commuting graphs. If $G$ is nilpotent, $G$ has at least two non-abelian
Sylow subgroups and $|Z(G)|\geq|Z(H)|$, then $|G|=|H|$.

Note that if the assumptions of Theorem 4.3 are satisfied, then once again the
conclusion that $H$ is nilpotent is reached by Theorem 4.1.

Theorems 4.2 and 4.3 imply that in order to give a positive answer to Question
1.1 for non-commuting graphs with the additional assumption
$|Z(G)|\geq|Z(H)|$, it remains to consider the case in which $G$ is a finite
nilpotent group of the form $G=P\times A$, for a non-abelian $p$-group $P$ and
an abelian group $A$ with $(|P|,|A|)=1$, containing at least one element $x\in
G\backslash Z(G)$ such that $C_{G}(x)$ is not abelian. In this view, Grazian
and Monetta posed the following conjecture:

######  Conjecture 4.4.

[19, Conjecture 3] Let $p$ be a prime and suppose $G=P\times A$ is a finite
group where $P\in\rm{Syl}_{p}(G)$ is non-abelian and $A$ is an abelian
$p^{\prime}$-group. If $H$ is a finite group whose non-commuting graph is
isomorphic to the one of $G$ and $|Z(G)|\geq|Z(H)|$ then $H=Q\times B$, where
$q$ is a prime, $Q\in\rm{Syl}_{q}(H)$ is non-abelian and $B$ is an abelian
$q^{\prime}$-group. In particular, $H$ is nilpotent.

######  Remark 4.5.

Theorem 4.2 implies that Conjecture 4.4 is true when $G$ is an AC-group. Also,
if $G$ is a finite $p$-group (so $A=1$), then $|G|=|H|$ by [3, Theorem 1.2],
and so Conjecture 4.4 is true in this instance too.

##  5\. The generating graph

The generating graph of a finite $2$-generated group $G$ is the graph defined
on the elements of $G$ in such a way that two distinct vertices are connected
by an edge if and only if they generate $G$. It was defined by Liebeck and
Shalev in [22], and has been further investigated by many authors: see for
example [8, 9, 10, 16, 17, 23, 25, 26, 27] for some of the range of questions
that have been considered.

Question 1.1 is still open for generating graphs (even with the extra
assumption that $H$ is solvable) and appears to be a difficult problem.

In this section, we prove that the answer to Question 1.1 is affirmative at
least in the particular case when $H$ is a finite supersoluble group:

######  Theorem 5.1.

Let $G$ and $H$ be finite $2$-generated groups with isomorphic generating
graphs. If $G$ is nilpotent and $H$ is supersoluble, then $H$ is nilpotent.

First we need an easy numerical lemma.

######  Lemma 5.2.

Let $\alpha=(a_{1},\dots,a_{r})$ and $\beta=(b_{1},\dots,b_{s})$ be two
sequences of prime numbers, with $a_{1}\leq\dots\leq a_{r}$ and
$b_{1}\leq\dots\leq b_{s}.$ If

$\prod_{1\leq i\leq r}\left(1-\frac{1}{a_{i}}\right)=\prod_{1\leq j\leq
s}\left(1-\frac{1}{b_{j}}\right),$

then $\alpha=\beta.$

###### Proof.

By induction on $r+s.$ If $r+s=2$, then $r=s=1$ and the statement is trivial.
So suppose $r+s>2$. We have

(5.1) $\prod_{1\leq i\leq r}a_{i}\prod_{1\leq j\leq s}(b_{j}-1)=\prod_{1\leq
i\leq r}(a_{i}-1)\prod_{1\leq j\leq s}b_{j}.$

Let $p=\max\\{a_{1},\dots,a_{r},b_{1},\dots,b_{s}\\}$, $r^{*}=\max\\{i\mid
a_{i}\neq p\\}$, $s^{*}=\max\\{j\mid b_{j}\neq p\\}$. Note that we can assume
that both $r^{*}$ and $s^{*}$ exist. Indeed, if for example $a_{i}=p$ for
every $1\leq i\leq r$, then from (5.1) we get

$p^{r}\prod_{1\leq j\leq s}(b_{j}-1)=(p-1)^{r}\prod_{1\leq j\leq s}b_{j};$

thus $\beta$ contains exactly $r$ primes equal to $p$ and if $\alpha\neq\beta$
then $\prod_{b_{j}\neq p}(b_{j}-1)=\prod_{b_{j}\neq p}b_{j}$, a contradiction.

Now, since $p$ does not divide $a_{i}-1$ nor $b_{j}-1$, it divides $a_{i}$ if
and only if $i>r^{*}$ and divides $b_{j}$ if and only if $j>s^{*}.$ We deduce
that $r-r^{*}$ is the multiplicity of $p$ in the left term of (5.1) and
$s-s^{*}$ is the multiplicity of $p$ in the right term of (5.1). In particular
$r-r^{*}=s-s^{*}$ and $a_{r^{*}+1}=\dots=a_{r}=b_{s^{*}+1}=\dots=b_{s}=p.$ But
then

$\prod_{i\leq r^{*}}\left(1-\frac{1}{a_{i}}\right)=\prod_{j\leq
s^{*}}\left(1-\frac{1}{b_{j}}\right),$

and we conclude by induction. ∎

Now we need some information on the degrees of the vertices of the generating
graph of a finite nilpotent group. From now on, let $\Gamma(G)$ denote the
generating graph of $G$.

######  Lemma 5.3.

Let $G$ be a 2-generated, non cyclic, finite nilpotent group. Let $\pi(G)$ be
the set of prime divisors of $G,$ $\pi_{1}(G)=\\{p_{1},\dots,p_{r}\\}$ the set
of the primes $p\in\pi(G)$ such that the Sylow $p$-subgroup of $G$ is cyclic
and $\pi_{2}(G)=\\{q_{1},\dots,q_{s}\\}$ the set of the remaining primes. For
every subset $I$ of $\\{1,\dots,r\\}$, let

$\displaystyle\alpha_{I}$ $\displaystyle=|G|\prod_{1\leq j\leq
s}\left(1-\frac{1}{q_{j}^{2}}\right)\prod_{i\in
I}\left(1-\frac{1}{p_{i}}\right)\prod_{i\notin I}\frac{1}{p_{i}},$
$\displaystyle\beta_{I}$ $\displaystyle=|G|\prod_{1\leq j\leq
s}\left(1-\frac{1}{q_{j}}\right)\prod_{i\notin
I}\left(1-\frac{1}{p_{i}}\right).$

For $g\in G$, denote by $\delta_{G}(g)$ the degree of $g$ in the generating
graph of $G$. If $g$ is a non-isolated vertex, then $\delta_{G}(g)=\beta_{I}$
for some subset $I$ of $\\{1,\dots,r\\}$. Moreover for every
$I\subseteq\\{1,\dots,r\\}$, the generating graph of $G$ contains precisely
$\alpha_{I}$ vertices of degree $\beta_{I}.$

###### Proof.

An element $g\in G$ is not isolated in $\Gamma(G)$ if and only if $g\Phi(G)$
is not isolated in $\Gamma(G/\Phi(G))$, and this occurs if and only if
$q_{1}\cdots q_{s}$ divides $|g\Phi(G)|$. Given $I\subseteq\\{1,\dots,r\\},$
there are precisely $\alpha_{I}$ elements $g$ such that
$|g\Phi(G)|=q_{1}\cdots q_{s}\prod_{i\in I}p_{i}.$ All these elements have
degree $\beta_{I}$. ∎

Now we collect some information on the generating graph of a 2-generated
supersoluble group.

######  Lemma 5.4.

Assume that $X$ is a finite 2-generated supersoluble group and that
$\Phi(X)=1.$ Then

$X\cong(V_{1}\times\dots\times V_{t})\rtimes Y$

where $Y$ is abelian and $V_{1},\dots,V_{r}$ are pairwise non $Y$-isomorphic
nontrivial irreducible $Y$-modules.

###### Proof.

The Fitting subgroup ${\rm{Fit}}(X)$ of $X$ is a direct product of minimal
normal subgroups of $X$, and is complemented in $X$. Let $T$ be a complement
of ${\rm{Fit}}(X)$ in $X$. Since $X$ is supersoluble,
$X^{\prime}\leq{\rm{Fit}}(X),$ and consequently $T$ is abelian. Let $W$ be a
complement of $Z(X)$ in ${\rm{Fit}}(X).$ Then $X=W\rtimes Y,$ with $Y=\langle
T,Z(X)\rangle.$ We may decompose $W=V_{1}^{n_{1}}\times\dots V_{t}^{n_{t}},$
where $V_{1},\dots,V_{r}$ are pairwise non $Y$-isomorphic nontrivial
irreducible $Y$-modules. The condition that $X$ is 2-generated implies that
$n_{1}=\dots=n_{t}=1.$ ∎

######  Lemma 5.5.

Let $X=(V_{1}\times\dots\times V_{t})\rtimes Y$ be as in Lemma 5.4. For $1\leq
i\leq t,$ let $|V_{i}|=r_{i}$ (since $X$ is supersoluble, $r_{i}$ is a prime).
Assume that $x=(v_{1},\dots,v_{t})y$ is a non-isolated vertex of the
generating graph of $X$ and let $J_{y}=\\{j\in\\{1\dots,t\\}\mid[y,V_{j}]\neq
0\\}.$ Then

(5.2) $\delta_{X}(x)=\delta_{Y}(y)\prod_{j\notin J_{y}}r_{j}\prod_{j\in
J_{y}}(r_{j}-1).$

###### Proof.

For $1\leq i\leq t,$ we may identify $V_{i}$ with the additive group of the
field $F_{i}$ with $r_{i}$ elements. For every $z\in Y$ and $1\leq i\leq t,$
there exists $\alpha_{i}(z)\in F_{i}$ such that $w_{i}^{z}=\alpha_{i}(z)w_{i}$
for all $w_{i}\in V_{i}.$ Let
$\tilde{x}=(\tilde{v}_{1},\dots,\tilde{v}_{t})\tilde{y}\in X.$ It follows from
Propositions 2.1 and 2.2 in [25], that $\langle x,\tilde{x}\rangle=X$ if and
only if $\langle y,\tilde{y}\rangle=Y$ and

$\delta_{i}(x,\tilde{x}):=\det\begin{pmatrix}1-\alpha_{i}(y)&1-\alpha_{i}(\tilde{y})\\\
v_{i}&\tilde{v}_{i}\end{pmatrix}\neq 0\quad\text{ for all
$i\in\\{1,\dots,t\\}$}.$

If $i\notin J_{y},$ then $\alpha_{i}(y)=1$. Since $\langle
y,\tilde{y}\rangle=Y$ and $[Y,V_{i}]=V_{i},$ it must be that
$\alpha_{i}(\tilde{y})\neq 1$, and therefore $\delta_{i}(x,\tilde{x})\neq 0$
if and only if $v_{i}\neq 0,$ independently on the choice of $\tilde{v}_{i}.$
If $i\in J_{y},$ then, for every choice of $v_{i}$ and $y,$ the probability
that $\tilde{v}_{i}$ satisfies the condition
$\det\begin{pmatrix}1-\alpha_{i}(y)&1-\alpha_{i}(\tilde{y})\\\
v_{i}&\tilde{v}_{i}\end{pmatrix}\neq 0$ coincides with $1-1/r_{i}$.

We conclude that $x$ is not isolated in $\Gamma(X)$ if and only if $v_{i}\neq
0$ for every $i\notin J_{y}.$ Moreover if $x$ is not isolated, then (5.2)
holds. ∎

###### Proof of Theorem 5.1.

If $H$ is cyclic, then $H$ is nilpotent and there is nothing to prove. So
assume $H$ is not cyclic and let $n=|G|=|H|$. Moreover assume that
$\pi_{1}(G)$ and $\pi_{2}(G)$ are as described in Lemma 5.3 and that
$X=H/\Phi(H)$ is as described in Lemmas 5.4 and 5.5.

For every element $h\in H$, we write $h\Phi(H)=w_{h}y_{h}$ with $w_{h}\in
V_{1}\times\dots\times V_{t}$ and $y_{h}\in Y$. Since $\langle
h_{1},h_{2}\rangle=H$ if and only if $\langle h_{1},h_{2}\rangle\Phi(H)=H$, it
follows from Lemma 5.5 that

(5.3) $\delta_{H}(h)=\frac{n\cdot\delta_{Y}(y_{h})}{|Y|}\prod_{j\notin
J_{y_{h}}}r_{j}\prod_{j\in J_{y_{h}}}(r_{j}-1).$

To reach our conclusion, we will prove a series of consecutive claims.

Claim 1. Let $y\in Y$ be such that $Y/\langle y\rangle$ is cyclic and let
$\omega$ be the set of prime divisors of $|Y/\langle y\rangle|$. Then
$\\{r_{j}\mid j\in J_{y}\\}\cap\omega=\emptyset.$ Moreover if $j_{1},j_{2}\in
J_{y}$ and $j_{1}\neq j_{2},$ then $r_{j_{1}}\neq r_{j_{2}}.$

Indeed there exists a non isolated vertex $h$ in $\Gamma(H),$ with $y=y_{h},$
and by (5.3),

(5.4) $\delta_{H}(h)=n\prod_{j\in
J_{y}}\left(1-\frac{1}{r_{j}}\right)\prod_{u\in\omega}\left(1-\frac{1}{u}\right).$

Since $\Gamma(G)\cong\Gamma(H)$, there exists $g\in G$ with
$\delta_{H}(h)=\delta_{G}(g).$ It follows from Lemma 5.3, that there exists
$I\subseteq\\{1,\dots,r\\}$ such that

(5.5) $\prod_{j\in
J_{y}}\left(1-\frac{1}{r_{j}}\right)\prod_{u\in\omega}\left(1-\frac{1}{u}\right)=\prod_{1\leq
j\leq s}\left(1-\frac{1}{q_{j}}\right)\prod_{i\notin
I}\left(1-\frac{1}{p_{i}}\right).$

The factors in the right term of (5.5) are all distinct. By Lemma 5.2 the same
must be true for the left term, and this implies that Claim 1 is true.

Claim 2. The prime $r_{i}$ does not divide $|Y|,$ for every $1\leq i\leq t.$

Indeed, assume by contradiction that $r_{i}$ divides $|Y|.$ Since $Y$ is a
2-generated abelian group and $Y/C_{Y}(V_{i})\leq{\rm{Aut}}(V_{i})$ is cyclic
of order dividing $r_{i}-1,$ there exists $y\in Y$ such that $i\in J_{y}$,
$Y/\langle y\rangle$ is cyclic and $r_{i}$ divides $|Y/C_{Y}(V_{i})|,$ in
contradiction with Claim 1.

Claim 3. If $1\leq i<j\leq t,$ then $r_{i}\neq r_{j}$.

Assume $i\neq j.$ We can find $y\in Y$ such that $\\{i,j\\}\subseteq J_{y}$
and $Y/\langle y\rangle$ is cyclic (indeed let $Y=\langle y_{1},y_{2}\rangle$:
at least one of the three elements $y_{1},y_{2},y_{1}y_{2}$ does not
centralizes neither $V_{i}$ nor $V_{j}$ and this is the element we need). By
Claim 1 it must be $r_{i}\neq r_{j}.$

Now let $\pi_{1}=\\{r_{1},\dots,r_{t}\\},$ $\pi_{2}$ be the set of the prime
divisors $p$ of $|Y|$ such that the Sylow $p$-subgroup of $Y$ is cyclic,
$\pi_{3}$ the set of the remaining prime divisors of $|Y|.$ We have proved
that $\pi=\pi(G)=\pi(H)$ is the disjoint union of $\pi_{1},$ $\pi_{2}$ and
$\pi_{3}.$

In particular it follows from (5.3) and Claim 3 that every non isolated vertex
$h\in\Gamma(H)$ uniquely determines a subset $\pi_{h}$ of $\pi$ such that

$\delta_{H}(h)=n\prod_{p\in\pi_{h}}\left(1-\frac{1}{p}\right).$

Since the degrees in $\Gamma(G)$ and $\Gamma(H)$ are the same, denoting by
$\Lambda(H)$ the set of the non isolated vertices of $\Gamma(H),$ it follows
from Lemma 5.3 that

(5.6) $\pi_{2}(G)=\bigcap_{h\in\Lambda(H)}\pi_{h}.$

Let $r_{i}\in\pi_{1}.$ Since ${\rm{Aut}}(V_{i})$ is cyclic, there exists $y\in
Y$ such that $i\notin J_{y}$ and $Y/\langle y\rangle$ is cyclic. Moreover
there exists $h\in\Lambda(H)$ with $y_{h}=y.$ By (5.4) and Claim 2,
$r_{i}\notin\pi_{h},$ and therefore, by (5.6), $r_{i}\notin\pi_{2}(G).$ Hence

(5.7) $\pi_{1}\subseteq\pi_{1}(G).$

If $r\in\pi_{2},$ then there exists $y\in Y$ such that $Y/\langle y\rangle$ is
cyclic and has order coprime with $r.$ As before, take $h\in\Lambda(H)$ with
$y_{h}=y.$ By (5.4) and Claim 2, $r\notin\pi_{h},$ hence $r\notin\pi_{2}(H).$
It follows that

(5.8) $\pi_{2}\subseteq\pi_{1}(G).$

It follows easily from (5.4), that if $p\in\pi_{3},$ then $p\in\pi_{h}$ for
every $h\in\Lambda(H),$ hence

(5.9) $\pi_{3}\subseteq\pi_{2}(G).$

Since $\pi$ is the disjoint union of $\pi_{1},\pi_{2},\pi_{3},$ but also the
disjoint union of $\pi_{1}(G)$ and $\pi_{2}(G),$ combining $(\ref{11}),$
$(\ref{12})$ and $(\ref{13}),$ we conclude

$\pi_{1}(G)=\pi_{1}\cup\pi_{2}{\text{ and }}\pi_{2}(G)=\pi_{3}.$

Let $\eta_{1},\eta_{2}$ be, respectively, the number of edges of $\Gamma(G)$
and $\Gamma(H).$ Notice that $2\eta_{1}=|G|^{2}P_{G}(2)$ and
$2\eta_{2}=|H|^{2}P_{H}(2)$, where, given a 2-generated finite group $T$, we
denote by $P_{T}(2)$ the probability that a pair of uniformly randomly chosen
elements of $T$ generates $T$. Since $\Gamma(G)\cong\Gamma(H),$ we must have
that $\eta_{1}=\eta_{2}$ and consequently $P_{G}(2)=P_{H}(2).$ It follows from
[18, Satz 4] that

$\displaystyle P_{H}(2)$
$\displaystyle=\prod_{p\in\pi_{1}}\left(1-\frac{1}{p}\right)\prod_{p\in\pi_{2}}\left(1-\frac{1}{p^{2}}\right)\prod_{p\in\pi_{3}}\left(1-\frac{1}{p}\right)\left(1-\frac{1}{p^{2}}\right),$
$\displaystyle P_{G}(2)$
$\displaystyle=\prod_{p\in\pi_{1}(H)}\left(1-\frac{1}{p^{2}}\right)\prod_{p\in\pi_{2}(G)}\left(1-\frac{1}{p}\right)\left(1-\frac{1}{p^{2}}\right).$

Since $\pi_{1}\cup\pi_{2}=\pi_{1}(G)$ and $\pi_{3}=\pi_{2}(G),$ we deduce

$\prod_{p\in\pi_{1}}\left(1+\frac{1}{p}\right)=1$

and consequently $\pi_{1}=\emptyset.$ Therefore $H/\Phi(H)\cong Y$ is abelian
and we conclude that $H$ is nilpotent. ∎

###  5.1. An interesting subgraph

In this subsection we aim to obtain some indications that information on the
structure of a finite group $G$ can be obtained investigating the connectivity
properties of the complement of the generating graph of $G$. Notice that the
isolated vertices of the generating graph are joined to all the other vertices
of the complement graph, so the complement of the generating graph is always
connected. But an interesting graph is obtained if we consider the complement
$\Delta(G)$ of the subgraph of the generating graph induced by the subset of
its non-isolated vertices. In other word the vertices of $\Delta(G)$ are the
elements $x\in G$ with the property that $\langle x,y\rangle=G$ for some $y\in
G$ and there is an edge between $x_{1}$ and $x_{2}$ if and only if $\langle
x_{1},x_{2}\rangle\neq G.$ Note that the vertex set of $\Delta(G)$ is
contained in the set $G\backslash\Phi(G)$, but in general it can be smaller.
For example, in $\Delta(S_{4})$, the elements $(12)(34)$, $(13)(24)$ and
$(14)(23)$ do not belong to the vertex-set. We will show in particular that we
have a positive answer to Question 1.1 with respect to the graph $\Delta(G)$,
whenever it is disconnected.

We start analysing the graph $\Delta(G)$ when $G$ is either a cyclic group or
a $p$-group.

######  Lemma 5.6.

Let $G$ be a cyclic group of order $n$. Then $\Delta(G)$ contains $n$
vertices, and at least two of them are isolated.

###### Proof.

Suppose $G=\langle x\rangle$ is a cyclic group of order $n$. For every element
$g\in G$, we have $\langle g,x\rangle=G$. Thus none of the elements of $G$ is
a universal vertex, implying that $\Delta(G)$ contains $n$ vertices. Now, if
$n=2$, then the elements $1$ and $x$ are isolated vertices, while for $n\geq
3$, the elements $x$ and $x^{-1}$ are distinct isolated vertices. So in any
case there are at least two isolated vertices. ∎

######  Lemma 5.7.

Let $p$ be a prime and let $P$ be a $2$-generated $p$-group of order $p^{n}$
that is not cyclic. Then $\Delta(P)$ contains $p+1$ connected components, each
one complete and containing $p^{n-2}(p-1)$ vertices. In particular $\Delta(P)$
is disconnected and contains isolated vertices if and only if $P\cong
C_{2}\times C_{2}$.

###### Proof.

Note that $P$ contains $p+1$ maximal subgroups $M_{1},\dots M_{p+1}$ and the
following holds:

  * •

for every $x,y\in M_{i}$, $x\neq y$, we have $\langle x,y\rangle\leq M_{i}<P$,
so $x$ and $y$ are joined;

  * •

for every $x\in M_{i}$, $y\in M_{j}$ with $i\neq j$, we have $\langle
x,y\rangle=P$, so $x$ and $y$ are not joined.

Therefore $\Delta(P)$ contains $p+1$ connected components, each one complete
and containing $|M_{i}|-|\Phi(P)|=p^{n-2}(p-1)$ vertices. In particular
$\Delta(P)$ contains isolated vertices if and only if $p^{n-2}(p-1)=1$, that
is equivalent to $P\cong C_{2}\times C_{2}$. ∎

######  Lemma 5.8.

Suppose $G$ is a cyclic group and $P$ is a $2$-generated $p$-group that is not
cyclic, for some prime $p$. Then $\Delta(G)\cong\Delta(P)$ if and only if
$G\cong C_{3}$ and $P\cong C_{2}\times C_{2}$.

###### Proof.

By Lemma 5.6, $\Delta(G)$ contains isolated vertices. Hence by Lemma 5.7 we
deduce that $P\cong C_{2}\times C_{2}$. In particular
$\Delta(G)\cong\Delta(P)$ consists of $3$ vertices and no edges. Hence again
by Lemma 5.6 we conclude that $G\cong C_{3}$. ∎

Next, we consider the dihedral group $D_{2p}$, for $p$ an odd prime:

######  Lemma 5.9.

Let $p$ be an odd prime. Then $\Delta(D_{2p})$ has $D_{2p}\backslash\\{1\\}$
as vertex-set and consists of one connected component of size $p-1$ that is
complete and $p$ isolated vertices (corresponding to the $p$ involutions of
$D_{2p}$).

###### Proof.

Note that the trivial element is a universal vertex (as $D_{2p}$ is non-
cyclic) and it is the only one, so $\Delta(D_{2p})$ contains
$D_{2p}\backslash\\{1\\}$ as vertex-set. Now, all involutions of $D_{2p}$ are
isolated vertices, as each of them generates $D_{2p}$ together with any other
non-trivial element. Finally, if $P\in{\rm{Syl}}_{p}(D_{2p})$, then for every
pair of distinct non-trivial elements $x,y\in P$ we have $\langle
x,y\rangle=P<D_{2p}$, and so the non-trivial elements of $P$ form a connected
component of size $p-1$, that is complete. ∎

Lucchini and Nemmi characterized all finite $2$-generated groups $G$ for which
the graph $\Delta(G)$ contains isolated vertices:

######  Proposition 5.10.

[28, Proposition 2] Let G be a $2$-generated finite group. Then $\Delta(G)$
has an isolated vertex if and only if one of the following holds:

  1. (1)

$G$ is cyclic;

  2. (2)

$G\cong C_{2}\times C_{2}$; or

  3. (3)

$G\cong D_{2p}$ for an odd prime $p$.

As a consequence of Proposition 5.10 we obtain the following:

######  Corollary 5.11.

Let $G$ and $H$ be finite $2$-generated groups with $\Delta(G)\cong\Delta(H)$.
If $\Delta(G)$ contains isolated vertices and $G$ is nilpotent, then $H$ is
nilpotent.

###### Proof.

By Proposition 5.10, the group $G$ is either cyclic or isomorphic to
$C_{2}\times C_{2}$ and it is enough to prove that $H$ cannot be isomorphic to
the group $D_{2p}$. Aiming for a contradiction, suppose $H\cong D_{2p}$. If
$G\cong C_{2}\times C_{2}$, then $\Delta(H)$ should have $3$ vertices, and by
Lemma 5.9 we should have $2p-1=3$, that is impossible. Thus $G$ must be
cyclic, say $G=\langle x\rangle$, and by Lemma 5.6 we deduce that
$|G|=2p-1\geq 5$. Also, by Lemma 5.9 the graph $\Delta(G)$ has a complete
connected component of size $p-1$ and $p$ isolated vertices. In particular
$|G|$ is not prime (otherwise all vertices would be isolated). Suppose
$q_{1}\neq q_{2}$ are prime numbers dividing $|G|$. Then both $x^{q_{1}}$ and
$x^{q_{2}}$ are joined to the trivial element, but they are not joined among
each other (as they generate $G$ together). However by Lemma 5.9
$\Delta(D_{2p})$ cannot have this shape. Thus $G$ must be a cyclic $q$-group
for some prime $q$, of order $q^{k}=2p-1$ for $k\geq 2$. In particular the
number of isolated vertices is $q^{k}-q^{k-1}$ and this must equal $p$. Hence
$q$ divides $p$, implying $q=p$ a contradiction.

This proves that $H$ is either cyclic or isomorphic to $C_{2}\times C_{2}$,
and so in particular it is nilpotent. ∎

More generally, Lucchini and Nemmi described the finite $2$-generated groups
$G$ in which $\Delta(G)$ is disconnected:

######  Proposition 5.12.

[28, Theorem 1] Let G be a $2$-generated finite group. Then $\Delta(G)$ is
disconnected if and only if one of the following holds:

  1. (1)

$G$ is cyclic;

  2. (2)

$G$ is a $p$-group;

  3. (3)

$G$ is not a $p$-group, $G/\Phi(G)\cong(V_{1}\times\dots\times V_{t})\rtimes
Y$, where $Y\cong C_{p}$ for some prime $p$, and $V_{1},\dots,V_{t}$ are
pairwise non-$Y$-isomorphic non-trivial irreducible $Y$-modules; or

  4. (4)

$G$ is not a $p$-group, $G/\Phi(G)\cong(V_{1}\times\dots\times V_{t})\rtimes
Y$, where $Y\cong C_{p}\times C_{p}$ for some prime $p$, $V_{1},\dots,V_{t}$
are pairwise non-$Y$-isomorphic non-trivial irreducible $Y$-modules and we
have $C_{Y}(V_{1}\times\dots\times V_{t})\cong C_{p}$.

We are now ready to prove that Question 1.1 has a positive answer when
$\Delta(G)$ is disconnected:

######  Theorem 5.13.

Let $G$ and $H$ be finite $2$-generated groups with $\Delta(G)\cong\Delta(H)$.
If $\Delta(G)$ is disconnected and $G$ is nilpotent, then $H$ is nilpotent.

###### Proof.

Thanks to Corollary 5.11, we can assume that $\Delta(G)$ does not have
isolated vertices, so in particular $G$ is not cyclic by Lemma 5.6. By
Proposition 5.12, we can suppose that $G$ is a $p$-group of order $p^{n}$, for
some prime $p$ and integer $n\geq 2$, and we have to show that $\Delta(G)$ is
not isomorphic to $\Delta(X)$ whenever $X$ is a group of type (3) or (4).

By Lemma 5.7 $\Delta(G)$ contains $p^{n-2}(p^{2}-1)$ vertices and it is
regular of valency $p^{n-2}(p-1)-1$.

Aiming for a contradiction, suppose that $H$ is of type (3) or (4), with
$K\cong C_{q}$ or $K\cong C_{q}\times C_{q}$ for some prime $q$. Note that in
both cases $H$ contains a maximal subgroup $M$ that is normal in $H$ and such
that $|H|=|M|q$. Indeed, if $H$ is of type (3) then $M/\Phi(H)\cong
V_{1}\times\dots\times V_{t}$, while if $H$ is of type (4) then
$M/\Phi(H)\cong(V_{1}\times\dots\times V_{t})\times\langle k\rangle$ where
$\langle k\rangle=C_{K}(V_{1}\times\dots\times V_{t})$. Let $\Omega$ denote
the set of vertices of $\Delta(H)$ corresponding to elements of $M$. In [28,
Lemmas 24 and 25] the authors studied the structure of $\Delta(H)$, proving
that $\Omega$ is a proper connected component of $\Delta(H)$. In particular,
every vertex $x\in\Omega$ has degree $|\Omega|-1$ (as $x$ is not joined to
itself). Also, if $h\in H$ is such that $H=M\langle h\rangle$, then every
element of $H$ can be written as $mh^{j}$ for some $m\in M$ and $1\leq j\leq
q$ and $\langle h,mh^{j}\rangle=\langle h,m\rangle<H$ if and only if $m\in
M\backslash\Omega.$ Moreover, if $m\in M\backslash\Omega$, then $mh^{j}$ is a
vertex if and only if $h^{j}\neq 1$ (that is equivalent to $j\neq q$). This
implies that $h$ has degree $(|M|-|\Omega|)(q-1)-1$.

Since $\Delta(G)$ and $\Delta(H)$ are isomorphic, the graph $\Delta(H)$ must
be regular. Hence

$|\Omega|-1=(|M|-|\Omega|)(q-1)-1$

that gives

$|\Omega|=|M|\frac{q-1}{q}.$

Comparing the valencies of $\Delta(G)$ and $\Delta(H)$ we obtain

(5.10) $p^{n-2}(p-1)=|\Omega|=|M|\frac{q-1}{q}.$

Also, the total number of vertices of $\Delta(H)$ is

$|\Omega|+(|H|-|M|)=|\Omega|+|M|(q-1)=|M|\frac{q-1}{q}+|M|(q-1)=|M|(q-1)\left(\frac{1}{q}+1\right).$

Dividing the number of vertices of $\Delta(G)$ and $\Delta(H)$ by the values
obtained in equation 5.10 we get

$\frac{p^{n-2}(p^{2}-1)}{p^{n-2}(p-1)}=\frac{|M|(q-1)\left(\frac{1}{q}+1\right)}{|M|\frac{q-1}{q}}.$

Simplifying the above equation, we conclude that $p=q$. Finally, again by
equation 5.10 we obtain $|M|=p^{n-1}$ and so $|H|=p|M|=p^{n}$ and $H$ is a
$p$-group, a contradiction. This proves the statement. ∎

##  6\. The prime graph (or Gruenberg-Kegel graph)

The prime graph (also known as Gruenberg-Kegel graph) was introduced by
Gruenberg and Kegel in an unpublished paper in 1975. For a finite group $K$,
denote by $\pi(K)$ the set of prime divisors of the order of $K$. The prime
graph of a finite group $G$ is the graph having $\pi(G)$ as vertex set and
such that two distinct vertices $p$ and $q$ are adjacent if and only if $G$
contains an element of order $pq$. Note that if $p$ is a prime and $G$ is a
finite $p$-group, then the prime graph of $G$ contains only one vertex and no
edges. In particular Question 1.1 is trivially true for $p$-groups. However,
it is not hard to see that it is false in general:

######  Proposition 6.1.

Let $G$ be a finite nilpotent group and suppose there exists a finite non-
nilpotent group $K$ with $\pi(G)=\pi(K)$. Then $H:=K\times K$ is not nilpotent
and the prime graphs of $G$ and $H$ are isomorphic.

###### Proof.

Note that $H$ is not nilpotent, because $K$ is not nilpotent by assumptions,
$\pi(H)=\pi(G)$ and the prime graph of $H$ is complete, and so isomorphic to
the one of $G$. ∎

We can even find finite groups with the same order representing a negative
answer to Question 1.1. Take for example $G=C_{6}\times C_{6}$ and
$H=S_{3}\times C_{6}$, both having a complete prime graph on two vertices.
However, Question 1.1 has a positive answer if we assume that $H$ has square-
free order:

######  Proposition 6.2.

Let $G$ and $H$ be finite groups with isomorphic prime graphs. If $G$ is
nilpotent and $|H|$ is square-free, then $H$ is cyclic (hence nilpotent).

###### Proof.

Note that the assumption that $|H|$ is square-free implies that every Sylow
subgroup of $H$ is cyclic of prime order. If $H$ is a $p$-group for a prime
$p$, we are done. Assume that there exist distinct primes $p$ and $q$ dividing
the order of $H$. By hypothesis, the prime graph of $H$ is isomorphic to the
one of $G$, that is complete. Therefore there exists an element $x$ in $H$ of
order $pq$ and $P=\langle x^{q}\rangle$ is a Sylow $p$-subgroup of $H$. We
show that $P$ is contained in the center of $H$, by proving that for any prime
$r$ dividing the order of $H$ there exists a Sylow $r$-subgroup of $H$
commuting with $P$. If $r=q$, then $P$ commutes with $Q=\langle
x^{p}\rangle\in{\rm{Syl}}_{q}(H)$. Hence assume that $r$ is distinct from $p$
and $q$. Then there exists an element $y\in H$ of order $pr$ and the Sylow
$p$-subgroup $P_{1}=\langle y^{r}\rangle$ commutes with the Sylow $r$-subgroup
$R=\langle y^{p}\rangle$. Since $P$ and $P_{1}$ are conjugate, it follows that
$P$ commutes with a conjugate of $R$. The arbitrary choice of $p$, shows that
$H$ is isomorphic to the direct product of cyclic groups of coprime orders,
hence it is cyclic. ∎

##  7\. The join graph

We conclude our survey with a graph that focuses the attention on the subgroup
lattice of a finite group. The join graph of a finite group $G$ has been
introduced by Lucchini in [24] as follows: it is the graph having as vertex-
set the set of proper subgroups of $G$ and in which two subgroups $M$ and $N$
are joined if and only if $G=\langle M,N\rangle$. In general Question 1.1 has
a negative answer for the join graph: consider for example the groups
$C_{p}\times C_{p}$ and $D_{2p}$, for $p$ odd. However the following holds:

######  Theorem 7.1.

[24, Corollary 5] Let $G$ and $H$ be finite groups with isomorphic join
graphs. If $G$ is nilpotent, then $H$ is supersoluble.

We can say more if the Frattini subgroup of $H$ is trivial.

######  Lemma 7.2.

Let $G$ and $H$ be finite groups with isomorphic join graphs. If $\Phi(H)=1$
then $\Phi(G)=1$. In particular, if $\Phi(H)=1$ and $G$ is nilpotent then G is
a direct product of elementary abelian groups.

###### Proof.

Since $\Phi(H)=1$, the identity subgroup is the unique isolated vertex in the
join graph of $H$. Using the graph isomorphism, we deduce that the join graph
of $G$ has a unique isolated vertex that must correspond to the trivial
subgroup, and so $\Phi(G)$ is trivial too. Finally, if $G$ is nilpotent then
for every prime divisor $p$ of $|G|$, if $P\in\rm{Syl}_{p}(G)$ then
$\Phi(P)\leq\Phi(G)=1$ and so $P$ is elementary abelian. So all Sylow
subgroups of $G$ are elementary abelian and $G$ is a direct product of
elementary abelian groups. ∎

Recall that a finite group $K$ is a $P$-group if it is either non-cyclic
elementary abelian or a semidirect product of an elementary abelian $p$-group
$A$ by a group of prime order $q\neq p$ which induces a non-trivial power
automorphism on $A$. Using Lemma 7.2 and properties of the lattice of $G$,
Lucchini obtained the following result:

######  Theorem 7.3.

[24, Proposition 6] Let $G$ and $H$ be finite groups with isomorphic join
graphs. If $G$ is nilpotent and $\Phi(H)=1$ then $H$ is a direct product of
groups with pairwise coprime orders that are either $P$-groups or elementary
abelian groups.

Acknowledgments

The authors are partially supported by the “National Group for Algebraic and
Geometric Structures, and their Applications” (GNSAGA - INdAM).

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Valentina Grazian
Dipartimento di Matematica e Applicazioni, Università di Milano-Bicocca,
Milano, Italy
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Andrea Lucchini
Dipartimento di Matematica, Università di Padova, Padova, Italy
Email<EMAIL_ADDRESS>

Carmine Monetta
Dipartimento di Matematica, Università di Salerno, Salerno, Italy
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