\documentstyle{SibMatJ}
% \TestXML
\topmatter

  \Author           Lyngdoh
    \Initial        D.
    \Sign           Daphimosha Lyngdoh
    \Email          daphisalyngdoh\@gmail.com
    \AffilRef       1
  \endAuthor

  \Author           Nongsiang
    \Initial        D.
    \Sign           Deiborlang Nongsiang
    \Email          ndeiborlang\@yahoo.in
    \AffilRef       1
    \Corresponding
  \endAuthor

  \Affil 1
    \Division       Department of Mathematics
    \Organization   North-Eastern Hill University
    \City           Meghalaya
    \Country        India
  \endAffil

  \datesubmitted    May 1, 2025\enddatesubmitted
  \dateaccepted     November 7, 2025\enddateaccepted

  \UDclass
    ???  %\?05C10, 05C25
  \endUDclass

  \thanks  %\?сюда это из \acknowledgment
    The first author wishes to express her sincere thanks to CSIR (India) for its financial assistance.
  \endthanks

  \title
    On the Genus of the Nonnilpotent Graphs of Nonweakly Nilpotent Groups %\?Non- всюду слитно сделала
  \endtitle

  \abstract
    In this paper, we focus on a topological aspect, namely, the genus of the nonnilpotent graphs associated with nonweakly nilpotent groups. We determine the genus of the  nonnilpotent graphs of some classes of finite nonnilpotent groups. We also classify all nonweakly nilpotent groups whose nonnilpotent graphs are planar, toroidal, double-toroidal, triple-toroidal, quadruple-toroidal, or pentuple-toroidal.
  \endabstract

  \keywords
    nonweakly nilpotent group,
    nonnilpotent graph,
    genus
  \endkeywords

\endtopmatter

\head
  1. Introduction
\endhead	
	
		Let $G$ be a group and $\roman{nil}(G)=\{x \in G \mid \langle x,y\rangle$ is nilpotent for all $y\in G$\}. It is still unknown whether the subset $\roman{nil}(G)$ of $G$ is a subgroup of $G$ or not. In the case when $G$ is finite, $\roman{nil}(G)$ equals the hypercenter $Z^*(G)$ of $G$
(see [1, Lemma 3.1]).
		
		Associate with a group $G$, a (simple) graph $\goth{R}_G$ as follows: the vertex set $V(\goth{R}_G)$ is $G \setminus \roman{nil}(G)$ and two distinct vertices $x$ and $y$ are adjacent if and only if $\langle x,y\rangle$ is not nilpotent. We call $\goth{R}_G$, the nonnilpotent graph of $G$. The nonnilpotent graph can be seen as a generalization of the noncommuting graph, 
considered in [2] and [3]. Recently, in [4] and [5], the group and graph properties of the nonnilpotent graphs of groups were examined. 
For a finite nonnilpotent group $G$, in [4], Abdollahi and Zarrin, show that $\goth{R}_G$ is planar if and only if $G\cong S_3$ and in [5],
Nongsiang and Saikia show that $\goth{R}_G$ is not toroidal.
		
		In this paper, we examine a topological feature, specifically the genus of nonnilpotent graphs associated with nonweakly nilpotent groups. In \Sec*{Section 2}, %\?\Sec*{Section 3}, как раз про это в Section 3 вроде
we obtained some properties of the nonnilpotent graphs. We show that, if $G$ is a nonweakly nilpotent group
(a~group $G$ is said to be weakly nilpotent if every two generated subgroups of $G$ are nilpotent)
and if $\gamma(\goth{R}_G)=n$, where $n$ is a nonnegative integer, then $G$ is finite. We also show that, for any nonnegative integer $n$, there can be only finitely many nonweakly nilpotent groups whose nonnilpotent graphs are of genus $n$.

In \Sec*{Section 4},  %\?опять про Section 3 ничего не написано
we determine the genus of the  nonnilpotent graphs of nonnilpotent groups of some well-known classes of finite groups and in
\Sec*{Section 5}, we classify all nonweakly nilpotent groups whose nonnilpotent graphs are planar, toroidal, double-toroidal, triple-toroidal, quadruple-toroidal, or pentuple-toroidal.
		
\head
2. Some Prerequisites
\endhead

In this section, we review specific group-theoretic and graph-theoretic terminology [6--8]
along with some well-established results that will be referenced in the coming sections.

		The {\it nilpotent graph\/} of a group $G$ [1], denoted by $\Gamma_{\roman{nil}}(G)$, is a~simple undirected graph whose vertex set is $G\setminus \roman{nil}(G)$ and any two distinct vertices $x$ and $y$ are adjacent if and only if $\langle x,y\rangle$ is nilpotent.
Let $\operatorname{Sol}(G) = \{x\in G : \langle x,y\rangle$ is solvable for all $y\in G\}$. The {\it nonsolvable graph\/} of $G$ [9], denoted by $\Cal{S}_G$, is a~simple undirected graph whose vertex set is $G\setminus \operatorname{Sol}(G)$ and any two distinct vertices $x$ and $y$ are adjacent if and only if $\langle x,y\rangle$ is not solvable. The {\it noncommuting graph\/} of $G$ [2], denoted by~$\Gamma_G$, is
a~simple undirected graph whose vertex set is $G\setminus Z(G)$ and any two distinct vertices $x$ and $y$ are adjacent if and only if $x$ and $y$ do not commute.
		
		 A graph is said to be {\it complete\/} if there exists an edge between every pair of distinct vertices in $\Gamma$.
We denote the complete graph with $n$ vertices by $K_n$. The {\it complete bipartite graph\/} and the {\it complete tripartite graph\/} are the one whose vertex set can be partitioned into two disjoint parts and three disjoint parts, respectively, and two vertices are adjacent if and only if they lie in different parts. Similarly, a {\it complete multipartite graph\/} or a {\it complete $k$-partite graph\/} is a simple graph whose vertices can be partitioned into $k$ sets so that two distinct vertices $u$ and $v$ are adjacent if and only if $u$ and $v$ belong to different sets of the partition. We write $K_{n_1,\dots,n_k}$ for the complete $k$-partite graph with partite sets of sizes $n_1,\dots, n_k$. If $n_1=\dots= n_k=m$, then we get the regular complete $k$-partite graph $K_{m,m,\dots,m}$ and it is denoted
by~$K_{k(m)}$. If $n_1=\dots= n_{l_1}=m_1,\dots,n_{l+1}=\dots= n_{l+l_r}=m_r$ $(l=l_1+l_2+\dots+l_{r-1})$ then we denote the complete $k$-partite graph $K_{n_1,\dots,n_k}$ by $K_{l_1(m_1),\dots,l_r(m_r)}$. Given a graph $\Gamma$, let $U$ be a nonempty subset of $V(\Gamma) $. Then the {\it induced subgraph\/} of $\Gamma$ on $U$ is defined to be the graph $\Gamma[U]$ in which the vertex set is $U$ and the edge set consists precisely of those edges in $\Gamma$ whose endpoints lie in $U$. The graph obtained by taking the union of graphs $\Gamma_1$ and $\Gamma_2$ with disjoint vertex sets is the disjoint union or sum, written %\?as
$\Gamma_1 + \Gamma_2$. In general, $m\Gamma$ is the graph consisting of $m$ pairwise disjoint copies of $\Gamma$. The join of two graphs $\Gamma_1$ and $\Gamma_2$, denoted by $\Gamma_1 \vee \Gamma_2$, is the graph obtained from $\Gamma_1 +\Gamma_2$ by joining each vertex of $\Gamma_1$ to each vertex of $\Gamma_2$. Further, given a graph $\Gamma$, its {\it complement} is defined to be the graph in which the vertex set is the same as the one in $\Gamma$ and two distinct vertices are adjacent if and only if they are not adjacent vertices in $\Gamma$.
		
		 The {\it genus\/} of a graph $\Gamma$, denoted by $\gamma(\Gamma)$, is the smallest nonnegative integer $n$ such that the graph can be embedded on the surface obtained by attaching $n$ handles to a sphere. Clearly, if $\widetilde{\Gamma}$ is a subgraph of~$\Gamma$, then $\gamma(\widetilde{\Gamma}) \leq \gamma(\Gamma)$. The sphere with one, two, three, four, and five handles are the {\it torus, double-torus, triple-torus, quadruple-torus,  and pentuple-torus}, respectively. The graphs embeddable on the surfaces of genus 0, 1, 2, 3, 4, and 5 are the {\it planar, toroidal, double-toroidal, triple-toroidal, quadruple-toroidal, and pentuple-toroidal graphs}, respectively. A $block$ of a graph $\Gamma$ is a connected subgraph $B$ of $\Gamma$ that is maximal with respect to the property that removal of a single vertex (and the incident edges) from $B$ does not make it disconnected, that is, the graph $B\setminus\{v\}$ is connected for all $v\in V(B)$.
Given a~graph~$\Gamma$, there is a unique finite collection $\goth {B}$ of blocks of $\Gamma$ such that $\Gamma = \bigcup\nolimits_{B\in \goth {B}}B$. The collection $\goth {B}$ is called the {\it block decomposition\/} of $\Gamma$. In [4, Theorem 1], it has been proved that the genus of a graph is the sum of the genera of its blocks. Some of the important helpful results are listed below:
		
		\proclaim{Lemma 2.1 \rm [8, Theorem 6-38]}
If $n\geq 3$, then
			$$
				\gamma(K_n)=\Bigl \lceil \frac{(n-3)(n-4)}{12}\Bigr \rceil.
			$$			
		\endproclaim

		\proclaim{Lemma 2.2 \rm	[8, Theorem 6-37]}
If $m,n\geq 2$, then
			$$
				\gamma(K_{m,n})=\Bigl \lceil  \frac{(m-2)(n-2)}{4}\Bigr \rceil.
			$$	
		\endproclaim
		
		\proclaim{Lemma 2.3 \rm [10, Theorem 6.1]}
For all $n\neq 2$,
			$$
				\gamma(K_{2n,n,n,n})=\Bigl \lceil \frac{(n-1)(3n-2)}{2}\Bigr \rceil.
			$$	
		\endproclaim		
		
		\proclaim{Lemma 2.4 \rm [8, Corollary 6-14]}
If $G$ is connected, with $p$ the number of vertices, $p\geq 3$, and $q$ the number  of edges,
then $\gamma(G) \geq \frac{q}{6}-\frac{p}{2} + 1$. Furthermore, equality %\?the
holds if and only if a triangular embedding can be found for $G$.
\endproclaim		
		
\head
3. Some Properties of Nonnilpotent Graphs
\endhead

			Given a group $G$ with $x \in G$, the nilpotentizer of $x$ is defined as $\roman{nil}_G(x)=\{y \in G \mid \langle x,y\rangle$ is nilpotent\}. As in [4], a group $G$ is said to be an $\goth{n}$-group if $\roman{nil}_G(x)$ is a subgroup of $G$ for all $x \in G$ and a~group $G$ is said to be an $\goth{nn}$-group (see [1]), if $\roman{nil}_G(x)$ is a nilpotent subgroup of $G$ for all $x \in G\setminus \roman{nil}(G)$. We call $G$ an $\goth{n}_p$-group [4] if $\roman{nil}_G(x)$ is a subgroup of $G$ for all $p$-elements $x$ of $G$. A group $G$ is said to be an $AC$-group if the centralizer $C_G(x)$ of every noncentral element $x$ of $G$ is an abelian subgroup of $G$.
			
			For any group $G$, the upper central series of a group $G$ is given by $\{e\} = Z_0(G) \leq Z_1(G) \leq \cdots$, where $Z_{n+1}(G)/Z_n(G)$ is the center of $G/Z_n(G)$ for all $n \geq 0$. The subgroup $Z_n(G)$ is called the
$n^{th}$-center  %\?\it $n$th-center
of~$G$. Clearly, $Z_1(G) = Z(G)$, the center of $G$.
This series need not reach $G$, but if $G$ is finite then this %\?2 this
series terminates at a subgroup called the hypercenter of $G$, denoted by $Z^*(G)$.

 We start this section with the following result, which allows us to use $Z^*(G)$ and $\roman{nil}(G)$ interchangeably, whenever the group $G$ is finite.

			\proclaim{Lemma 3.1 \rm [1, Lemma 3.1]}
				  Let $G$ be a finite group. Then, the following assertions hold:
	
\Item (1)
$\langle x,Z^*(G)\rangle$ is nilpotent for every $x\in G$.
\Item (2)
$Z^*(G) =\roman{nil}(G)$.
			\endproclaim
			
	\proclaim{Lemma 3.2}
		A periodic finitely generated abelian group is finite.
	\endproclaim

	\demo{Proof}
		Let $G$ be a periodic finitely generated abelian group. Let $g_i$, $1\leq i\leq k$, be the generators of $G$ of order $n_i$.
Every %\?Each
element of $G$ can be written as a product of powers of these generators, i.e., if $x\in G$, then
$x=g_1^{a_1}g_2^{a_2}....g_k^{a_k}$,  %\?.... \dots или \cdots
where $0\leq a_i\leq n_i$ for $1\leq i\leq k$.
The number of such possible products is finite as $G$ is abelian and each $a_i$
can only take %\?take only
finitely many values. Thus, $G$ is finite.
	\qed\enddemo	
	
	\proclaim{Lemma 3.3}
		Let $G$ be a periodic finitely generated nilpotent group. Then, $G$ is finite.
	\endproclaim

	\demo{Proof}
		Let $m$ be the nilpotency class of $G$. We will prove the result by induction on the nilpotency class.
If nilpotency class %\?the
of $G$ is 1, then $G$ is abelian. By \Par*{Lemma 3.2}, $G$ is finite. Suppose the result holds for all groups of nilpotency class at most $m-1$. Let the upper central series of $G$ be $Z_0(G)\leq Z_1(G)\leq \dots \leq Z_m(G)=G$. Let $G_i=Z_{i+1}(G)/Z_1(G)$, $i=0,1,2,\dots,m-1$. Then $G_0\leq G_1\leq \dots \leq G_{m-1}$ forms a central series of $\overline{G}=G/Z_1(G)$ as $[\overline{G},G_{i+1}]\leq G_i$ for all $i=0,1,2,\dots,m-2$. It follows that $\overline{G}$ is nilpotent and its nilpotency class is at most $m-1$. By hypothesis, $\overline{G}$ is finite. Since $G$ is periodic finitely generated, we have %\?that
$Z(G)$ is finitely generated, periodic, and abelian. By \Par*{Lemma 3.2}, $Z(G)$ is finite. Thus, $G$ is finite.
	\qed\enddemo	
	
	\proclaim{Lemma 3.4}
		Let $G$ be a nonweakly nilpotent group and $x\in G\setminus \roman{nil}(G)$. If $\deg(x)$ in $\goth{R}_G$ is finite, then the order of $x$ is finite.
	\endproclaim

	\demo{Proof}
		Suppose $x$ has infinite order, that is, the order of the subgroup $\langle x\rangle$ is infinite. Since $x\in G\setminus\roman{nil}(G)$, there exist %\?exists
$y\in G\setminus\roman{nil}(G)$ such that $\langle x,y\rangle$ is nonnilpotent. We will show that $\langle x,x^iy\rangle$ is nonnilpotent for all $i\in \Bbb{Z}$. Clearly, $x\in \langle x,x^iy\rangle$ and so $y\in \langle x,x^iy\rangle$. Thus, $\langle x,y\rangle= \langle x,x^iy\rangle$ and so $\langle x,x^iy\rangle$ is nonnilpotent for all $i\in \Bbb{Z}$. Hence, $\deg(x)$ is infinite, which is a contradiction.
	\qed\enddemo
	
	\proclaim{Lemma 3.5}
		Let $G$ be a nonweakly nilpotent group. Then $\goth{R}_G$ has at most one planar connected component.
	\endproclaim

	\demo{Proof}
If $\goth{R}_G$ has no planar connected component, then there is nothing to prove. So suppose $\goth{R}_G$ has a planar connected component. Let $\Gamma$ be the subgraph of $\goth{R}_G$ consisting of all the planar connected components of $\goth{R}_G$.
	
		 Firstly, we will show that $\circ(x)\leq 4$ for all $x\in V(\Gamma)$. Suppose that $x\in V(\Gamma)$ such that $\circ(x)>4$. Since $x\notin\roman{nil}(G)$, there exist %\?exists
$y\in G\setminus\roman{nil}(G)$ such that $\langle x,y\rangle$ is nonnilpotent. Note that $y \in V(\Gamma)$.
		
		 Suppose $\circ(y)>2$. Clearly, $x^{-1},y^{-1},xy,yx,xy^{-1}\notin\roman{nil}(G)$. Suppose $|xy|=2$. Then $yx=x^{-1}y^{-1}$ and so the element $xy^{-1}$ is distinct from $x,x^{-1},y,y^{-1},xy,yx$. %\?$x$, $x^{-1}$, $y$, $y^{-1}$, $xy$, $yx$.
Let $H=\{x,x^{-1},y,y^{-1},xy,yx,xy^{-1}\}$ and $\Gamma'=\goth{R}_G[H]$. Then, $\Gamma'$ contains $K_{3,2,2}$ as a subgraph with partite sets $\{x,x^{-1}\}$, $\{y,y^{-1}\}$, and $\{xy,yx,xy^{-1}\}$. Since $x\in V(\Gamma')$ and $\Gamma'$ is connected, we have %\?that
$\Gamma'$ is a subgraph of $\Gamma$, which is a contradiction. So, suppose $|xy|>2$. Now $(xy)^{-1}\notin\roman{nil}(G)$. If $(xy)^{-1}\neq yx$, then $\Gamma$ contains $K_{3,2,2}$ as a subgraph with partite sets $\{x,x^{-1}\}$, $\{y,y^{-1}\}$, and $\{xy,yx,(xy)^{-1}\}$, which is not planar. If $(xy)^{-1}=yx$, then $y^2=x^{-2}$ and if $x^{-1}y=yx$, then $y = xyx$ and so $y^2 = xyx^2yx =xy(y^{-1})^2yx = x^2$, that is %\?,
$x^4 = e$; a contradiction. It follows that $x^{-1}y\neq yx$. Since $x^{-1}y \notin\roman{nil}(G)$ is distinct
from $x,x^{-1},y, y^{-1},xy$, %\?$x$, $x^{-1}$, $y$, $y^{-1}$, $xy$,
we have %\?that
$\Gamma$ contains $K_{3,2,2}$ as a subgraph with partite sets $\{x,x^{-1}\}$, $\{y,y^{-1}\}$, and $\{xy,yx,x^{-1}y\}$, which is not planar. Thus $\circ(y)>2$ is not possible.
		
		Suppose $\circ(y)=2$. If $|xy|>2$, then $\Gamma$ contains $K_{3,2,1}$ as a subgraph  with partite sets $\{x,x^{-1}\}$, $\{y\}$, and $\{xy,yx,(xy)^{-1}\}$, which is not planar. If $|xy|=2$, then $\langle x,y\rangle\cong D_{2|x|}$ and either $D_{2|x|}$ is nilpotent or $\Gamma$ contains $\goth{R}_{D_{2|x|}}$ which is a connected nonplanar graph; a contradiction. Hence, $|x|\leq 4$ for all $x\in V(\Gamma)$.
		
		Secondly, we will prove the existence of $2$-elements and $3$-elements in $V(\Gamma)$. Let $x,y\in V(\Gamma)$ such that $\langle x,y\rangle$ is nonnilpotent. Suppose $V(\Gamma))$ consists of only $2$-elements. If $|x|=|y|=|xy|=2$, then $xy=yx$; a contradiction. If $|x|=|y|=2$ and $|xy|=4$, then $\langle x,y\rangle\cong D_8$, which is nilpotent. If $|x|=2$, $|y|=4$ and $|xy|=2$, then again $\langle x,y\rangle\cong D_8$, which is nilpotent. If $|x|=2$, $|y|=4$, $|xy|=4$ and if $yx=xy^{-1}$, then $\langle x,y\rangle\cong D_8$. If $|x|=2$, $|y|=4$, $|xy|=4$ and if $yx\neq xy^{-1}$, then  $\Gamma$ contains $K_{3,2,1}$ as a subgraph with partite sets $\{x\}$, $\{y,y^{-1}\}$, and $\{xy,yx,xy^{-1}\}$, which is not planar. If $|x|=|y|=4$ and $|xy|=2$, then $yx=x^{-1}y^{-1}$ and so $yx\neq xy^{-1}$. Thus $\Gamma$ contains $K_{3,2,2}$ as a subgraph with partite sets $\{x,x^{-1}\}$, $\{y,y^{-1}\}$, and $\{xy,yx,xy^{-1}\}$, which is not planar. If $|x|=|y|=|xy|=4$ and $(xy)^{-1}=x^{-1}y=xy^{-1}=yx$, then $\langle x,y\rangle\cong Q_8$, which is a contradiction. Thus $\Gamma$ contains $K_{3,2,2}$ as a subgraph with partite sets $\{x,x^{-1}\}$, $\{y,y^{-1}\}$, and $\{xy,yx, u\}$, where $u\in \{(xy)^{-1},x^{-1}y,xy^{-1}\}, u\neq yx$, which is a contradiction. Thus, there exists a $3$-element in $V(\Gamma)$. If $V(\Gamma))$ consists of only $3$-elements, i.e., if $|x|=|y|=|xy|=3$, then $\Gamma$ contains $K_{3,2,2}$,
as a subgraph with partite sets $\{x,x^{-1}\}$, $\{y,y^{-1}\}$, and $\{xy,yx,y^{-1}x^{-1}\}$, which is not planar.
Thus, there exists a $2$-element in $V(\Gamma)$.
		
		Now, we will show that $\Gamma$ is connected. Suppose $x,y\in V(\Gamma)$ such that $\langle x,y\rangle$ is nilpotent.
Then, by \Par*{Lemma 3.3}, $\langle x,y\rangle$ is finite. If $x$ is a $2$-element and $y$ is a $3$-element,
then since $\operatorname{gcd}(|x|,|y|)=1$, we have $\langle x,y\rangle=\langle xy\rangle$ (by [11, Lemma 2.2]). Since $x,y\notin\roman{nil}(G)$, there exist $u,v\in G\setminus\roman{nil}(G)$ such that $\langle x,u\rangle$ and $\langle y,v\rangle$ are nonnilpotent and thus $u,v\in V(\Gamma)$. Now, $\langle x,u\rangle\subseteq \langle xy,u\rangle$ and $\langle y,v\rangle\subseteq \langle xy,v\rangle$. Thus, $xy$ is adjacent to both $u$ and $v$ and $x-u-xy-v-y$ is the path connecting $x$ and $y$. If $x$ and $y$ are both $2$-elements then
by the previous argument $\exists$ %\?there exists
a~$3$-element $z$ which is connected to $x$ and $y$ and thus there exists a path connecting $x$ and $y$. If $x$ and $y$ and both $3$-elements, then by the same argument, $\exists$ %\?there exists
a~$2$-element $z$ which is connected to $x$ and $y$ and thus there exists a path connecting $x$ and $y$. Thus, $\Gamma$ is connected. Hence $\goth{R}_G$ has at most one planar connected component.
	\qed\enddemo
	
	\proclaim{Theorem 3.6}
		Let $G$ be a nonweakly nilpotent group and $n\in \Bbb{N}\cup \{0\}$. If $\gamma(\goth{R}_G)=n$, then $G$ is finite.
	\endproclaim

	\demo{Proof}
		Since  $\gamma(\goth{R}_G)=n$; therefore, $\goth{R}_G$ has an embedding in $S_n$. By [8, Definition 6-10], the embedding is a minimal embedding. If $\goth{R}_G$ is connected, then by [8, Theorem 6-11], the embedding of $\goth{R}_G$ is a 2-cell embedding or a cellular embedding. By [12, Proposition 3.1], $\goth{R}_G$ must be finite and hence  by [5, Proposition 4.2] $G$ is finite. If $\goth{R}_G$ is disconnected, then the number of nonplanar connected components of $\goth{R}_G$ is finite (since $\gamma(\goth{R}_G)=n$) and by
\Par*{Lemma 3.5}, there is at most one planar connected component. Each of these connected components of $\goth{R}_G$ is also finite, otherwise it would contradict [12, Proposition 3.1]. Hence, $\goth{R}_G$ is finite which implies that $G$ is finite by [5, Proposition 4.2].
	\qed\enddemo
		
			\proclaim{Proposition 3.7}
		Let $n$ be a nonnegative integer. Then, there are finitely many nonweakly nilpotent groups whose nonnilpotent graphs are of genus $n$.
	\endproclaim

	\demo{Proof}
Let $G$ be a nonweakly nilpotent group such that $\gamma(\goth{R}_G)=n$.
Then by \Par*{Theorem 3.6}, $G$ is a~finite group.
	
		If $n=0$, then by [4, Theorem 6.1], there is only one finite nonnilpotent   group whose nonnilpotent graph is planar, namely $S_3$.
		
		So suppose $n\geq 1$. Let $h=\lfloor\frac{7+
\sqrt{1+48n}}{2} %\?
\rfloor$ and $\omega(\goth{R}_G)=m$. By Heawood's formula
[7, Theorem~6.3.25], we have $m\leq \chi(\goth{R}_G)\leq h$. If $G$ is a solvable group, then by [4, Theorem 4.4], $|G/Z^*(G)|<m^{m^4}\leq h^{h^4}$ and thus by [5, Lemma 5.1], we have $|G|<h^{h^4}
(\sqrt{n}+1)$.  %\?
If $G$ is a nonsolvable group, then $\Cal{S}_G$ is a~subgraph of $\goth{R}_G$. This implies that $\gamma(\Cal{S}_G)\leq n$. By [13, Corollary 2.4], $|G/\operatorname{Sol}(G)|< c^{2m^2[\log_{21}m]}[\log_{21}m]!$, where $c$ is a~constant. By [9, Proposition 38], $|\operatorname{Sol}(G)|\leq
\sqrt{2\gamma(\Cal{S}_G)}+2$. Thus, we have $|G|<c^{2h^2[\log_{21}h]}[\log_{21}h]!(\sqrt{2n}+2)$.
Hence, the result follows.		
	\qed\enddemo	

\proclaim{Proposition 3.8}
Let $G$ be a nonweakly nilpotent group such that $\gamma(\goth{R}_G)=n$. Then
$$
|E(\goth{R}_G)|\leq 3(2n+|G|-3).
$$	
\endproclaim

\demo{Proof}
By \Par*{Theorem 3.6}, $G$ is a finite group. From \Par*{Lemma 2.4}, we have $\gamma(\goth{R}_G)\geq \frac{q}{6}-\frac{p}{2}+1$, where $p$ and $q$ denote the number %\?numbers таких случаев много
of vertices and edges of $\goth{R}_G$, respectively. Thus it follows that $q\leq 6n+3p-6$. Since $p= |G|-|Z^*(G)|\leq |G|-1$, the result follows.
\qed\enddemo	

\proclaim{Proposition 3.9}
				Let $G$ be a finite nonnilpotent group and let $k$ be the number of conjugacy classes of $G$. Then,
				$$
					|E(\goth{R}_G)|\leq (|G|^2-k|G|)/2.
				$$
			\endproclaim

\demo{Proof}
By [2, Lemma 3.27], $2|E(\Gamma_G)|=|G|^2-k|G|$, where $\Gamma_G$ is the noncommuting graph of a~group $G$. Since $\goth{R}_G$ is a subgraph of $\Gamma_G$, the result follows.
\qed\enddemo	

\proclaim{Proposition 3.10}
Let $G$ be a nonweakly nilpotent group. Then $\goth{R}_G$ is not a complete graph. If $G$ is finite, then $\goth{R}_G$ is neither a bipartite graph nor a tripartite graph.
\endproclaim

\demo{Proof}
The first part follows from [4, Proposition 4.8] and the second part follows from [4, Theorem~4.2].
\qed\enddemo	
			
\head
4. Genus of Nonnilpotent Graphs of Some Groups
\endhead

			In this section, we compute the genus of nonnilpotent graphs of some nonnilpotent finite groups.
We need the following result in this section. %\?in this section может, убрать, наверху уже написано
		
			\proclaim{Proposition 4.1 \rm [5, Proposition 3.12]}
 Let $G$ be a finite nonnilpotent group. Then $\goth{R}_G$ is a complete multipartite
 graph if and only if $G$ is an $nn$-group. In particular,
 $\goth{R}_G \cong K_{\mid X_1\mid,\dots,\mid X_n\mid}$,  %\?\mid заменить просто на |
 where $\goth{P} = \{\roman{nil}_G(u) \setminus\roman{nil}(G) \mid u \in G \setminus\roman{nil}(G)\} = \{X_1,X_2,\dots,X_n\}$. %\?,..... заменила на \dots
			\endproclaim	
		
			\proclaim{Proposition 4.2}
				If $t\geq 1$, $m\geq 3$ and $m$ is odd then $\goth{R}_{D_{2^{t+1}m}}\cong\goth{R}_{Q_{2^{t+1}m}}$ and they are isomorphic to the $(m+1)$-partite graph $K_{2^{t}m-2^t,m(2^t)}$. In particular,
				$$
					\gamma(\goth{R}_{D_{2^{t+1}m}})=\gamma(\goth{R}_{Q_{2^{t+1}m}})=\gamma(K_{2^{t}m-2^t,m(2^t)}).
				$$
				
			\endproclaim

\demo{Proof}
If $G$ denotes any of the groups $D_{2^{t+1}m}=\langle x,y : y^{2^tm}=x^2=e,x^{-1}yx=y^{-1}\rangle$ or
$Q_{2^{t+1}m}=\langle x,y : y^{2^tm}=x^4=e,y^{2^{t-1}m}=x^2,x^{-1}yx=y^{-1}\rangle$,
then $G$ is a nonnilpotent $\goth{nn}$-group [1] with $Z^*(G)=\langle y^m \rangle$, for $m\nmid i$ we have $\roman{nil}_G(y^i)=\langle y\rangle$ and for $1\leq i\leq 2^tm$ we have $\roman{nil}_G(xy^i)=Z^*(G) \cup xy^{i}Z^*(G)$. Thus, the distinct nilpotentizers of the nonhypercenter elements of $G$ are $\langle y\rangle$ and $Z^*(G) \cup xy^{i}Z^*(G)$, $1\leq i\leq m$. Hence, by \Par*{Proposition 4.1}, $\goth{R}_{D_{2^{t+1}m}}\cong \goth{R}_{Q_{2^{t+1}m}}\cong K_{2^tm-2^t,m(2^t)}$ and the result follows.
\qed\enddemo
			
			\proclaim{Corollary 4.3}
				If $m=3$, then $\goth{R}_{D_{2^{t+1}3}}\cong \goth{R}_{Q_{2^{t+1}3}}\cong K_{2^{t+1},2^t,2^t,2^t}$ and $\gamma(\goth{R}_{D_{2^{t+1}3}})=\gamma(\goth{R}_{Q_{2^{t+1}3}})= \lceil (2^t-1)(3.2^{t-1}-1)\rceil$.
			\endproclaim

\demo{Proof}
  Follows from \Par*{Lemma 2.3}.
\qed\enddemo
			
				\proclaim{Proposition 4.4}
				If $m$ is odd, then $\goth{R}_{D_{2m}}$ is isomorphic to the $(m+1)$-partite graph $K_{m-1,m(1)}$. In particular,
				$$
					\gamma(\goth{R}_{D_{2m}})=\gamma(K_{m-1,m(1)}).
				$$
			\endproclaim
\demo{Proof} The group $G\cong D_{2m}=\langle x,y :y^m=x^2=e, x^{-1}yx=y^{-1}\rangle$ is a nonnilpotent $\goth{nn}$-group with $Z^*(G)=\{e\}$, for $1\leq i<m$, we have $\roman{nil}_G(y^i)=\langle y\rangle$ and for $1\leq i\leq m$, we have $\roman{nil}_G(xy^i)=\{e,xy^i\}$. Thus, the distinct nilpotentizers of the nonhypercenter elements are $\langle y\rangle$ and $\{e,xy^i\}$, $1\leq i\leq m$. So, in view of \Par*{Proposition 4.1}, $\goth{R}_{D_{2m}}\cong K_{m-1,m(1)}$ and the result follows.
\qed\enddemo			
			
			\proclaim{Corollary 4.5}
				Let $m=3^q(2^p+1/2)+3/2$, where $p\geq 3$,  $q\geq 1$, and $q$ is odd. Then,		
				$$
\gamma(\goth{R}_{D_{2m}})=\gamma(K_{m-1,m})=\Bigl \lceil  \frac{(m-3)(m-2)}{4}\Bigr \rceil.
$$
			\endproclaim

\demo{Proof}
Here $m$ is odd and so $\goth{R}_{D_{2m}}\cong K_{m-1,m(1)} \cong \overline{K_{m-1}}\vee K_m$. By [10, Theorem 5.6], the result follows.
\qed\enddemo
				
\proclaim{Proposition 4.6}
If $N$ is a nilpotent group of order $n$, $t\geq 1$, $m\geq 3$, and $m$ is odd, then $\goth{R}_{N\times D_{2^{t+1}m}}\cong \goth{R}_{N\times Q_{2^{t+1}m}}$ and they are isomorphic to the $(m+1)$-partite graph $K_{2^tmn-2^tn,m(2^tn)}$. In particular,
$$
\gamma(\goth{R}_{N\times D_{2^{t+1}m}})=\gamma(\goth{R}_{N\times Q_{2^{t+1}m}})=\gamma(K_{2^tmn-2^tn,m(2^tn)}).
$$
\endproclaim
	
\demo{Proof}
If $G$ denotes the group $N\times K$, where $K$ is any of the groups $D_{2^{t+1}m}=\langle x,y : y^{2^tm}=x^2=e,x^{-1}yx=y^{-1}\rangle$ or $Q_{2^{t+1}m}=\langle x,y : y^{2^tm}=x^4=e,y^{2^{t-1}m}=x^2,xyx^{-1}=y^{-1}\rangle$, then $G$ is a nonnilpotent $\goth{nn}$-group with $Z^*(G)=N\times \langle y^m\rangle$, for $m\nmid i$ we have $\roman{nil}_G((a,y^i))=N\times \langle y\rangle$ and for $1\leq j\leq 2^tm$ we have $\roman{nil}_G((a,xy^j))=Z^*(G) \cup (a,xy^j)Z^*(G)$. Thus, the distinct nilpotentizers of the nonhypercenter elements of $G$ are $N\times \langle y\rangle$ and $Z^*(G) \cup (e',xy^i)Z^*(G)$, $1\leq i\leq m$, where $e'$ is the identity element of $N$. Hence, $\goth{R}_G\cong K_{2^tmn-2^tn,m(2^tn)}$ and the result follows.
\qed\enddemo
		
		\proclaim{Corollary 4.7}
			If $m=3$, %\?then
$\gamma(\goth{R}_{N\times D_{2^{t+1}3}})=\gamma(\goth{R}_{N\times Q_{2^{t+1}3}})=\lceil (2^tn-1)(3.2^{t-1}n-1)\rceil$.
		\endproclaim

		\demo{Proof}
		Follows from \Par*{Lemma 2.3}.
		\qed\enddemo		
		
		\proclaim{Proposition 4.8}
			If $N$ is a nilpotent group of order $n$, $m\geq 3$, and $m$ is odd, then $\goth{R}_{N\times D_{2m}}$ is isomorphic to the $(m+1)$-partite graph $K_{mn-n,m(n)}$. In particular,
			$$
				\gamma(\goth{R}_{N\times D_{2m}})=\gamma(K_{mn-n,m(n)}).
			$$
		\endproclaim

\demo{Proof}
If $G= N\times \langle x,y : y^m=x^2=e, x^{-1}yx=y^{-1}\rangle$, then $G$ is a nonnilpotent $\goth{nn}$-group with $Z^*(G)=N\times \{e\}$. Also, for each $i$, $1\leq i<m$, $\roman{nil}_G((a,y^i))=N\times\langle y\rangle$  and  $\roman{nil}_G((a,xy^i))=N\times \{e,xy^i\}$. Thus, the distinct nilpotentizers of the nonhypercenter elements are $N\times\langle y\rangle$ and $N\times \{e,xy^i\}$, $1\leq i\leq m$. Hence, $\goth{R}_G\cong K_{mn-n,m(n)}$ and the result follows.
\qed\enddemo
		
		\proclaim{Corollary 4.9}
			If $m=3$ and $n\neq 2$, %\?then
$\gamma(\goth{R}_{N\times D_6})=\lceil \frac{(n-1)(3n-2)}{2}\rceil$.
		\endproclaim
	
		\proclaim{Proposition 4.10}
Let $G$ be a nonnilpotent group of order $pq$, where $p$ and $q$ are primes with  $p \mid q-1$. Then, $\goth{R}_G$ is isomorphic to the $(q+1)$-partite graph $K_{q-1,q(p-1)}$. In particular, $\gamma(\goth{R}_G)=\gamma(K_{q-1,q(p-1)})$.
		\endproclaim

\demo{Proof}
			Note that if $x \in G\setminus\{e\}$ then $|\langle x \rangle|= p$ or $q$. In either case we have $\roman{nil}_G (x) = \langle x \rangle$. It is now not difficult to see that the nilpotent graph of $G$ is isomorphic to the complete $q+1$-partite graph $K_{q-1,p-1,\dots,p-1}$. Thus the result follows.
\qed\enddemo

\proclaim{Lemma 4.11}
			If the order of the centralizer of a nontrivial element of a group is  $pq$, where $p$ and $q$ are primes, then the centralizer is abelian.
		\endproclaim

\demo{Proof}
If $p=q$, then we are done. So suppose $p\neq q$. Let $C_G(x)$ be the centralizer of a nontrivial element $x$ of a group $G$. If $x$ is of order $pq$, then we are done. So, without any lost of generality, we can assume that $x$ is of order $p$. Since $q$ is a prime dividing $|C_G(x)|$, $C_G(x)$ has an element $y$ of order~$q$. Then, $x$ and $y$ commute, which implies that $xy$ is an element of order $pq$ and thus $C_G(x)$ is abelian. Hence, the result follows.
\qed\enddemo

\demo{Remark 4.12}
Let $G$ be a finite group and let $p$ be a prime such that $p\mid |G|$. If $x$ is a $p$-element of~$G$ then $x\in P$, where $P$ is a Sylow $p$-subgroup of $G$.
\enddemo
		
		\proclaim{Proposition 4.13}
			Let $G$ be a nonnilpotent group of order $p^2q$, where $p$ and $q$ are primes and $p<q$. Then,			
			$$\goth{R}_G \cong \cases
					K_{3,2,2,2,2}, & \text{if $p=2$, $q=3$, $|Z^*(G)|=1$,}
\\
					K_{q-1,q(p^2-1)}, & \text{if $p\neq 2$ or $q\neq 3$, $|Z^*(G)|=1$,}
\\
					K_{pq-p,q(p^2-p)}, & \text{if $|Z^*(G)|=p$.}
				\endcases
$$
				
In particular,
				$$\gamma(\goth{R}_G)= \cases
					\gamma(K_{3,2,2,2,2}), & \text{if $p=2$, $q=3$, $|Z^*(G)|=1$,}
\\
					\gamma(K_{q-1,q(p^2-1)}), & \text{if $p\neq 2$ or $q\neq 3$, $|Z^*(G)|=1$,}
\\
					\gamma(K_{pq-p,q(p^2-p)}), & \text{if $|Z^*(G)|=p$.}
				\endcases
$$
		\endproclaim

\demo{Proof}
Here, every Sylow subgroup of $G$ is abelian. So by [4, Lemma 3.5], $G$ is an $\goth{n}_p$-group, an $\goth{n}_q$-group, and $C_G(x)=\roman{nil}_G(x)$ for every %\?all
$p$-elements and $q$-elements $x$ of $G$. Thus, by [4, Lemma 3.4], $G$ is an $\goth{n}$-group.  Now, for any $p$-element or $q$-element $x$ of $G\setminus Z^*(G)$, if $|C_G(x)|=p$ or $p^2$ or $q$, then $C_G(x)$ is abelian. If $|C_G(x)|=pq$, then by \Par*{Lemma 4.11},  $C_G(x)$ is abelian. Also, for any element $x$ of $G\setminus Z^*(G)$ of order $pq$, $|C_G(x)|=|\roman{nil}_G(x)|=pq$ and so $\roman{nil}_G(x)$ is cyclic and hence nilpotent. Thus, $G$ is an $\goth{nn}$-group.
Let $n_p$ and $n_q$ be the number of Sylow $p$-subgroups and Sylow $q$-subgroups of $G$ respectively. Then, $n_p=1$ or $q$ and $n_q=1$ or $p^2$. If $n_q=p^2$, then $p=2,q=3$ and thus $|G|=12$. Hence, $G\cong A_4$ and $\goth{R}_G\cong K_{3,2,2,2,2}$. So suppose that $n_q=1$.

		{\sc Case I}:
$|Z^*(G)|=1$. Since $n_q=1$, we have $n_p=q$ (as $G$ is nonnilpotent).
Let $Q$ be the Sylow $q$-subgroup of $G$ and let $P_i$, $1\leq i\leq q$, be the Sylow $p$-subgroups of $G$. Suppose $x\in G\setminus Z^*(G)$ and $\circ(x)=pq$. Note that $|\roman{nil}_G(x)|\neq p^2q$; otherwise $x\in Z^*(G)$, which is a contradiction. Thus $|\roman{nil}_G(x)|=pq$. Let $y\in\roman{nil}_G(x), \circ(y)=p$. %\?
Then $y\in P_i$ for some $i=1,2,\dots,q$ and so $|\roman{nil}_G(y)|=p^2q$. Thus $y\in Z^*(G)$; a contradiction as $|Z^*(G)|=1$. Thus it follows that $G\setminus Z^*(G)=Q'\sqcup P_1'\sqcup P_2'\sqcup \dots \sqcup P_q'$, where $Q'=Q\setminus Z^*(G)$, $P_i'=P_i\setminus Z^*(G)$, $i=1,2,\dots,q$. Let $x\in P_i\setminus Z^*(G)$. Then $P_i\subseteq\roman{nil}_G(x)$. Note that $|\roman{nil}_G(x)|\neq p^2q$, otherwise $x\in Z^*(G)$. Thus $|\roman{nil}_G(x)|=p^2$. Let $x\in Q\setminus Z^*(G)$. With the same argument as previous, we see that $|\roman{nil}_G(x)|\neq pq$. Thus $|\roman{nil}_G(x)|=q$. It follows that the distinct nilpotentizers of $G$ are
$Q$ and $P_i$,  %\?
$1\leq i\leq q$. Thus in view of \Par*{Proposition 4.1}, $\goth{R}_G\cong K_{q-1,q(p^2-1)}$.

		{\sc Case II}:
$|Z^*(G)|=p$. Since $n_q=1$, we have $n_p=q$ (as $G$ is nonnilpotent).
Let $Q$ be the Sylow $q$-subgroup of $G$ and let $P_i$, $1\leq i\leq q$, be the Sylow $p$-subgroups of $G$.
Note that $P_i\cap P_j=Z^*(G)$ for $i\neq j$. Also $P_i\cap QZ^*(G)=Z^*(G)$ for all $i=1,2,\dots,q$.
Thus $G\setminus Z^*(G)=Q'\sqcup P_1'\sqcup P_2'\sqcup \dots \sqcup P_q'$, where $Q'=QZ^*(G)\setminus Z^*(G)$, $P_i'=P_i\setminus Z^*(G)$, $i=1,2,\dots,q$. Now, for $x\in QZ^*(G)\setminus Z^*(G)$, we have $\roman{nil}_G(x)=QZ^*(G)$ and for $x\in P_i$, $\roman{nil}_G(x)=P_i$.  Therefore the distinct nilpotenttizers are $QZ^*(G)$ and~$P_i$, $1\leq i\leq q$.
Thus in view of \Par*{Proposition 4.1}, $\goth{R}_G\cong K_{pq-p,q(p^2-p)}$.

		 {\sc Case III}:
$|Z^*(G)|=q$. For any $p$-element $x$ of $G\setminus Z^*(G)$, $|\roman{nil}_G(x)|=p^2q$ and so $ x\in Z^*(G)$; a~contradiction.

		 {\sc Case IV}:
$|Z^*(G)|=pq$. Again, for any $p$-element $x$ of $G\setminus Z^*(G)$, $|\roman{nil}_G(x)|=p^2q$ and so $ x\in Z^*(G)$;
a contradiction.
		
		Hence, the result follows.
\qed\enddemo
		
		\proclaim{Proposition 4.14}
			Let $G$ be a nonnilpotent group of order $pq^2$, where  $p$ and $q$ are primes and $p<q$. Then,
$$
\goth{R}_G\cong \cases
				K_{q^2-1,q^2(p-1)}, & \text{if $|Z^*(G)|=1$,}
\\
				K_{q^2-q,q(pq-q)}, & \text{if $|Z^*(G)|=q$.}
			\endcases
$$
		In particular,	
			$$\gamma(\goth{R}_G)= \cases
				\gamma(K_{q^2-1,q^2(p-1)}), & \text{if $|Z^*(G)|=1$,}
\\
				\gamma(K_{q^2-q,q(pq-q)}), & \text{if $|Z^*(G)|=q$.}
			\endcases$$
		\endproclaim

\demo{Proof}
Following the same argument as in the previous proposition, we get that $G$ is an $\goth{nn}$-group.
If $n_p$ and $n_q$ are the number %\,
of Sylow $p$-subgroups and Sylow $q$-subgroups of $G$ respectively, then $n_q=1$ or $p$. Since $p<q$, we have $n_q=1$ and thus $n_p=q$ or $q^2$.

		{\sc Case I}:
$|Z^*(G)|=1$. Let $Q$ be the Sylow $q$-subgroup of $G$ and let $P_i$, $1\leq i\leq n$, $n=q$, or $q^2$, be the Sylow $p$-subgroups of $G$. Suppose $x\in G\setminus Z^*(G)$ with $\circ(x)=pq$. Note that $|\roman{nil}_G(x)|\neq pq^2$; otherwise $x\in Z^*(G)$, which is a contradiction. Thus $|\roman{nil}_G(x)|=pq$. Let $y\in\roman{nil}_G(x), \circ(y)=q$. %\? and
Then $y\in Q$ and so $|\roman{nil}_G(y)|=pq^2$. Thus $y\in Z^*(G)$; a contradiction as $|Z^*(G)|=1$. Thus it follows that $n=q^2$ and $G\setminus Z^*(G)=Q'\sqcup P_1'\sqcup P_2'\sqcup \dots \sqcup P_{q^2}'$, where $Q'=Q\setminus Z^*(G)$, $P_i'=P_i\setminus Z^*(G)$, $i=1,2,\dots,q^2$. Let $x\in P_i\setminus Z^*(G)$. Then $P_i\subseteq\roman{nil}_G(x)$. With the same argument as above, we see that $|\roman{nil}_G(x)|\neq pq$. Note that $|\roman{nil}_G(x)|\neq pq^2$, otherwise $x\in Z^*(G)$. Thus $\roman{nil}_G(x)=P_i$. Let $x\in Q\setminus Z^*(G)$. Then $\roman{nil}_G(x)=Q$. It follows that the distinct nilpotentizers of $G$ are $Q$ and $P_i$, $1\leq i\leq q^2$. Thus in view of \Par*{Proposition 4.1}, $\goth{R}_G\cong K_{q^2-1,q^2(p-1)}$.
		
		{\sc Case II}:
$|Z^*(G)|=p,pq$. If $x\in G\setminus Z^*(G)$ is any $q$-element, then $|\roman{nil}_G(x)|=pq^2$ and so $x\in Z^*(G)$; a contradiction.

		{\sc Case III}:
$|Z^*(G)|=q$. Let $Q$ be the Sylow $q$-subgroup of $G$ and let $P_i$, $1\leq i\leq n$, $n=q$, or $q^2$ be the Sylow $p$-subgroups of $G$. Note that $P_iZ^*(G)\cap P_jZ^*(G)=Z^*(G)$ for $i\neq j$. Also $P_iZ^*(G)\cap Q=Z^*(G)$ for all $i=1,2,\dots,n$. Thus $n=q$ and $G\setminus Z^*(G)=Q'\sqcup P_1'\sqcup P_2'\sqcup \dots \sqcup P_q'$, where $Q'=Q\setminus Z^*(G)$, $P_i'=P_iZ^*(G)\setminus Z^*(G)$, $i=1,2,\dots,q$. Now, for $x\in Q\setminus Z^*(G)$, we have $\roman{nil}_G(x)=Q$ and for $x\in P_iZ^*(G)\setminus Z^*(G)$, $\roman{nil}_G(x)=P_iZ^*(G)$.  Therefore the distinct nilpotenttizers are $Q$ and $P_iZ^*(G)$, $1\leq i\leq q$. Thus in view of \Par*{Proposition 4.1}, $\goth{R}_G\cong K_{q^2-q,q(pq-q)}$.

		{\sc Case IV}:
$|Z^*(G)|=q^2$. If $x\in G\setminus Z^*(G)$ is any $p$-element, then $|\roman{nil}_G(x)|=pq^2$ and so $x\in Z^*(G)$; a~contradiction.
		
		Hence, the result follows.
\qed\enddemo

\proclaim{Lemma 4.15}
			A group of order $p^2q$, where $p$ and $q$ are distinct primes, $p<q$, $p>2$, and $p\nmid q-1$ is nilpotent.
		\endproclaim

		\demo{Proof}
		 Let $G$ be a group of order $p^2q$. Let $n_p$ and $n_q$ be the number
of Sylow $p$-subgroups and Sylow $q$-subgroups of $G$ respectively. Then, $n_q=1$ or $p^2$. If $n_q=p^2$, then
$p^2\equiv 1(\text{mod q})$. %\?
Thus $q\mid (p+1)$ and so $p=2,q=3$, which is a contradiction. Thus, $n_q=1$. Since $p\nmid q-1$, we have $n_p=1$.
Thus, every Sylow subgroup of $G$ is unique which implies that $G$ is the direct product of its Sylow subgroups.
Hence, the result follows.
		 \qed\enddemo	

\proclaim{Lemma 4.16}
		Let $x$ be a $q$-element or an element of order $pq$ of a group $G$, where $p$ and $q$ are primes, $p<q$, and $p\nmid (q-1)$. If $|C_G(x)|=pq^2$, then $C_G(x)$ is nilpotent.
	\endproclaim

	\demo{Proof}
		We have $\circ(x)=q,q^2,pq$. Note that $\langle x\rangle\subseteq Z(C_G(x))$. Thus $|\frac{C_G(x)}{Z(C_G(x))}|=1,p,q,pq$. It follows that, since $p\nmid q-1$, we have $\frac{C_G(x)}{Z(C_G(x))}$ is cyclic and so $C_G(x)=Z(C_G(x))$, that is %\?,
$C_G(x)$ is abelian.
	\qed\enddemo
		
		\proclaim{Proposition 4.17}
			Let $G$ be a nonnilpotent group of order $p^2q^2$, where $p$ and $q$ are distinct primes, $p<q$, $p>2$, $p\nmid q-1$, and $p\mid q+1$. Then,
$$
\goth{R}_G\cong\cases
				K_{q^2-1,q^2(p^2-1)}, & \text{if }|Z^*(G)|=1,
\\
				K_{pq^2-p,q^2(p^2-p)}, & \text{if }|Z^*(G)|=p.			
			\endcases
$$
			
			In particular,			
$$
\gamma(\goth{R}_G)=
\cases
				\gamma(K_{q^2-1,q^2(p^2-1)}), & \text{if }|Z^*(G)|=1,
\\
				\gamma(K_{pq^2-p,q^2(p^2-p)}), & \text{if }|Z^*(G)|=p.			
			\endcases
$$
		\endproclaim

\demo{Proof}
Here, every Sylow subgroup of $G$ is abelian and so by [4, Lemma 3.5], $G$ is
an $\goth{n}_p$-group, an $\goth{n}_q$-group,
and $C_G(x)=\roman{nil}_G(x)$ for every %\?
$p$-element and $q$-element $x$ of $G$. Thus, by [4, Lemma 3.4], $G$ is an $\goth{n}$-group. Let $n_p$ and $n_q$ be the number of Sylow $p$-subgroups and Sylow $q$-subgroups of $G$ respectively. Then, $n_q=1$ or $p^2$. If $n_q=p^2$, then $p=2$ and $q=3$, which is a contradiction. Thus, $n_q=1$ and since $G$ is nonnilpotent, therefore $n_p=q^2$. Now, for any $p$-element or $q$-element $x$ of $G\setminus Z^*(G)$, if $|C_G(x)|=p^2$ or $q^2$, then $C_G(x)$ is abelian and thus nilpotent. If $|C_G(x)|=p^2q$ or $pq^2$, then by
\Par{Lemma 4.15}{Lemmas 4.15} and \Par{Lemma 4.16}{4.16}, $C_G(x)$ is nilpotent. If $x\in G\setminus Z^*(G)$ is an element of order $pq$, then $|\roman{nil}_G(x)|=pq$ or $p^2q$ or $pq^2$. Using \Par{Lemma 4.15}{Lemmas 4.15} and \Par{Lemma 4.16}{4.16}, $\roman{nil}_G(x)$ is nilpotent. If $x\in G\setminus Z^*(G)$ is an element of order $p^2q$, then $|\roman{nil}_G(x)|=p^2q$ and so $\roman{nil}_G(x)$ is cyclic and thus nilpotent. Similar argument holds if $x\in G\setminus Z^*(G)$ is an element of order $pq^2$. Hence, $G$ is an $\goth{nn}$-group. Let $P_1,P_2,\dots,P_{q^2}$ be the Sylow $p$-subgroups of $G$. Let $Q$ be the Sylow $q$-subgroup of $G$.

		{\sc Case I}: \Label{CI}
$|Z^*(G)|=1$. Let $x\in Q\setminus Z^*(G)$. Suppose $|\roman{nil}_G(x)|=pq^2$. Let $y\in\roman{nil}_G(x)$ such that $\circ(y)=p$. Then $y\in P_i$ for some $i$. Therefore $\roman{nil}_G(x),P_i\subset\roman{nil}_G(y)$ and so $|\roman{nil}_G(y)|=p^2q^2$, that is %\?,
$y\in Z^*(G)$; a contradiction. Thus $|\roman{nil}_G(x)|=q^2$ and so $\roman{nil}_G(x)=Q$. Similarly, if $x\in P_i\setminus Z^*(G)$, then $\roman{nil}_G(x)=P_i$. If $i\neq j$ and $x\in P_i\cap P_j$, then $P_i,P_j\subset\roman{nil}_G(x)$ and so $\roman{nil}_G(x)=G$. Thus $x=e$. It follows that $G=P_1\cup P_2\cup\dots\cup P_{q^2}\cup Q$ and $P_1, P_2,\dots, P_{q^2}, Q$ are the distinct nilpotentizers of $G$. Thus, in view of \Par*{Proposition 4.1}, we have $\goth{R}_G\cong K_{q^2-1,q^2(p^2-1)}$.

		{\sc Case II}:
$|Z^*(G)|=p$. As in \Par{CI}{Case I}, we see that if $x\in P_i\setminus Z^*(G)$, then $\roman{nil}_G(x)=P_i$
and if $x\in QZ^*(G)\setminus Z^*(G)$, then $\roman{nil}_G(x)=QZ^*(G)$. Note that $P_i\cap P_j=Z^*(G)$ for $i\neq j$ and $P_j\cap QZ^*(G)=Z^*(G)$. Thus $G=P_1\cup P_2\cup\dots\cup P_{q^2}\cup QZ^*(G)$ and $P_1, P_2,\dots, P_{q^2}, QZ^*(G)$ are the distinct nilpotentizers of $G$. Thus, in view of \Par*{Proposition 4.1}, we have $\goth{R}_G\cong K_{pq^2-p,q^2(p^2-p)}$.

		{\sc Case III}:
If $|Z^*(G)|=q$, then $|\frac{G}{Z^*(G)}|=p^2q$ and so by \Par*{Lemma 4.15}, $\frac{G}{Z^*(G)}$ is nilpotent. Since $Z^*(G)$ is nilpotent, we have $G$ is nilpotent; a contradiction. If $|Z^*(G)|=pq$, then $|\frac{G}{Z^*(G)}|=pq$ and so, since $p\nmid q-1$, we have $\frac{G}{Z^*(G)}$ is cyclic and hence nilpotent. Since $Z^*(G)$ is nilpotent, we have $G$ is nilpotent; a contradiction.
Similarly $|Z^*(G)|\neq p^2q,pq^2$.
		
		This completes the proof.
\qed\enddemo	

\proclaim{Lemma 4.18}\Label{L4.18}
			Let $G$ be a group of order $pqr$, where $p$, $q$, and $r$ are distinct primes and $p<q<r$. Then, the following hold:

				\Item (a)
The Sylow $r$-subgroup is normal in $G$.

				\Item (b)
$G$ has a normal subgroup of order $qr$.

				\Item (c)
If $q\nmid r-1$, then the Sylow $q$-subgroup is normal in $G$.
		\endproclaim

\demo{Proof}
\Par{L4.18}{(a)}:
Let $n_q$ and $n_r$ be the number of Sylow $q$-subgroups and Sylow $r$-subgroups of $G$ respectively. Then, $n_r=1$ or $pq$. Suppose $n_r=pq$. Also, $n_q=1$ or $r$ or $pr$. If $n_q\geq r$, then $G$ contains at least $r(q-1)$ elements
of order $q$.
Thus, the number of $r$-elements and $q$-elements is $pq(r-1)+r(q-1)=pqr+r(q-1)-pq>pqr$; a contradiction. Thus, $n_q=1$. Let $Q$ be the normal Sylow $q$-subgroup of $G$. The quotient group $G/Q$ has order $pr$ which implies that $G/Q$ has a normal Sylow $r$-subgroup $N/Q$. Then, by the correspondence theorem,  %\?
$N$ is a normal subgroup of $G$ of order $qr$. Again, $N$ has a normal Sylow $r$-subgroup, say $M$. Since $M$ is a normal Sylow $r$-subgroup of $N$ and $N$ is normal in $G$, it follows that $M$ is normal in $G$; a contradiction to the fact that $n_r=pq$. Thus, \Par{L4.18}{(a)} holds.

\Par{L4.18}{(b)}:
Let $H$ and $K$ be Sylow $q$-subgroup and Sylow $r$-subgroup of $G$, respectively. Since $K$ is normal in $G$, therefore $HK$ is a subgroup of $G$. Also, $H\cap K$ is trivial. Thus, $|HK|=qr$. Now $I_G(HK)=p$ and since $p$ is the smallest prime dividing $|G|$, therefore $HK$ is normal in $G$. Hence, the result follows.

\Par{L4.18}{(c)}:
 Let $H$ and $K$ be as in \Par{L4.18}{(b)}. Since $q\nmid r-1$, the subgroup $HK$ is cyclic, which implies that $HK$ is abelian. Since $H$ is a subgroup of $HK$, we have $H$ is a normal subgroup of $G$. Thus, \Par{L4.18}{(c)} holds.
\qed\enddemo		
		
		\proclaim{Proposition 4.19}
			Let $G$ be a nonnilpotent group of order $pqr$, where $p$, $q$, and $r$ are distinct primes, $p<q<r$, and $q\nmid r-1$. Then,
			$$ \goth{R}_G\cong\cases
				K_{qr-1,qr(p-1)}, & \text{if }|Z^*(G)|=1,
\\
				K_{qr-q,r(pq-q)}, & \text{if }|Z^*(G)|=q,
\\
				K_{qr-r,q(pr-r)}, & \text{if }|Z^*(G)|=r.
			\endcases
$$
In particular,
			$$ \gamma(\goth{R}_G)=\cases
				\gamma(K_{qr-1,qr(p-1)}), & \text{if }|Z^*(G)|=1,
\\
				\gamma(K_{qr-q,r(pq-q)}), & \text{if }|Z^*(G)|=q,
\\
				\gamma(K_{qr-r,q(pr-r)}), & \text{if }|Z^*(G)|=r.
			\endcases
$$
		\endproclaim

\demo{Proof}
Here, every Sylow subgroup of $G$ is abelian. So, following the same argument as in \Par*{Proposition 4.13}, we get that $G$ is an $\goth{nn}$-group. By \Par*{Lemma 4.18}, $G$ has a~normal Sylow $q$-subgroup and a~normal Sylow $r$-subgroup and thus a normal subgroup of order $qr$. Since $q\nmid r-1$, this normal subgroup is abelian. Let $n_p$ be the number of Sylow $p$-subgroups of $G$. Then, $n_p=r$ or $q$ or $qr$.

		{\sc Case I}:
$|Z^*(G)|=1$. Let $P_1,P_2,\dots,P_n$ be the Sylow $p$-subgroups of $G$. Let $Q$ and $R$ be the Sylow $q$-subgroup and Sylow $r$-subgroup of $G$, respectively. Then $QR$ is the unique abelian subgroup of $G$ of order $qr$. Let $x\in P_i\setminus Z^*(G)$. Then $|\roman{nil}_G(x)|=p,pq,pr$. Suppose $|\roman{nil}_G(x)|=pq$. Then there exist %\?exists
$y\in\roman{nil}_G(x)$ such that $\circ(y)=q$. Thus $y\in Q$ and so $QR\subset\roman{nil}_G(y)$. Also $\langle x\rangle\subset\roman{nil}_G(y)$. Thus $|\roman{nil}_G(y)|=pqr$ and so $y\in Z^*(G)$; a contradiction. Therefore, $|\roman{nil}_G(x)|\neq pq$. Similarly, $|\roman{nil}_G(x)|\neq pr$ and so $|\roman{nil}_G(x)|=p$. It is obvious now that if $x\in QR\setminus Z^*(G)$, then $\roman{nil}_G(x)=QR$ and there are no elements of order $pq$ or $pr$. Thus if $x\in G$, then $x\in H=P_1\cup P_2\cup\cdots P_n\cup QR$  %\?\cup пропущено
and so $G=H$, it follows that $n=qr$. Also we have shown that the distinct nilpotentizers are $P_1, P_2,\dots, P_{qr}$ and $QR$. Hence, in view of \Par*{Proposition 4.1}, we have $\goth{R}_G\cong K_{qr-1,qr(p-1)}$.

		{\sc Case II}:
$|Z^*(G)|=p$. For any $r$-element or $q$-element $x\in G\setminus Z^*(G)$, $|\roman{nil}_G(x)|=pqr$ and so $x\in Z^*(G)$; a contradiction.

		{\sc Case III}: \Label{CIII}
$|Z^*(G)|=q$. Let $P_1,P_2,\dots,P_n$, be the Sylow $p$-subgroups of $G$. Let $Q$ and $R$ be the Sylow $q$-subgroup and Sylow $r$-subgroup of $G$, respectively. Then $Q=Z^*(G)$. If $x\in QR\setminus Z^*(G)$, then $\roman{nil}_G(x)=QR$. If $x\in P_iQ\setminus Z^*(G)$, then $\roman{nil}_G(x)=P_iQ$. Let $x\in G\setminus Z^*(G)$. If $\circ(x)=p,q,pq$, then $x\in P_iQ$ for some $i\in \{1,2,\dots,n\}$. If $\circ(x)=q,r,qr$, then $x\in QR$. Suppose $\circ(x)=pr$. Then $\langle x\rangle,Z^*(G) \subset\roman{nil}_G(x)$ and so $|\roman{nil}_G(x)|=pqr$, that is %\?,
$x\in Z^*(G)$; a contradiction. For $i\neq j$, $P_iQ\cap P_jQ=Q$ and $P_iQ\cap QR=Q$. Thus $n=r$ and $G=P_1Q\cup P_2Q\cup\dots\cup P_rQ\cup QR$. Also, we have shown that the distinct nilpotentizers are $P_1Q, P_2Q,\dots, P_rQ$ and $QR$. Hence, in view of \Par*{Proposition 4.1}, we have $\goth{R}_G\cong K_{qr-q,r(pq-q)}$.

		{\sc Case IV}:
$|Z^*(G)|=r$. With a similar argument as in \Par{CIII}{Case III}, we see that $\goth{R}_G\cong K_{qr-r,q(pr-r)}$.

		{\sc Case V}:
$|Z^*(G)|=pq$ or $pr$ or $qr$. In this case, for any $p$-element or $q$-element or %\?запятые перед or
$r$-element $x\in G\setminus Z^*(G)$, we have $|\roman{nil}_G(x)|=pqr$ and so $x\in Z^*(G)$; a contradiction.

		Hence, the result follows.
\qed\enddemo		
		
		\proclaim{Proposition 4.20}
			Let $G$ be a finite Frobenius group with complement $H$ and kernel $N$. If $H$ is nilpotent, then $\goth{R}_G$ is isomorphic to the complete $(n+1)$-partite graph $K_{n-1,n(m-1)}$, where $n=|N|$ and $m=|H|$. In particular, $\gamma(\goth{R}_G)=\gamma(K_{n-1,n(m-1)})$.
		\endproclaim

\demo{Proof}
By [1, Proposition 4.6], $G$ is a nonnilpotent $\goth{nn}$-group with $Z^*(G)=\{e\}$, $\roman{nil}_G(x)=N$ for all $x\in N\setminus \{e\}$, and $\roman{nil}_G(y)=gHg^{-1}$ for all $y\in gHg^{-1}\setminus \{e\}$. Moreover, the collection $\{gHg^{-1}\setminus\{e\}\mid g \in G\}\cup \{N\setminus \{e\}\}$ forms a partition of $G\setminus\{e\}$ and $H$ has exactly $n$ conjugates in $G$. Thus, by \Par*{Proposition~4.1}, $\goth{R}_G\cong K_{n-1,n(m-1)}$ and the result follows.
\qed\enddemo
		
		\proclaim{Proposition 4.21}
Let $k\geq 2$ and $G=PSL(2,2^k)$. Then, $\goth{R}_{PSL(2,2^k)}$ is isomorphic to the complete $(2^k+1)+(2^{2k-1}+2^{k-1})+(2^{2k-1}-2^{k-1})$-partite graph
$$
K_{(2^k+1)(2^k-1),(2^{2k-1}+2^{k-1})(2^k-2),(2^{2k-1}-2^{k-1})(2^k)}.
$$
		 In particular,
			$$
				\gamma(\goth{R}_{PSL(2,2^k)})=\gamma(K_{(2^k+1)(2^k-1),(2^{2k-1}+2^{k-1})(2^k-2),(2^{2k-1}-2^{k-1})(2^k)}).
			$$	
		\endproclaim

		\demo{Proof}
It is well known that $PSL(2,2^k)$ is a centerless group of order $2^k(2^{2k} -1)$. Moreover, in view of [2, Proposition 3.21], the following assertions hold for $PSL(2,2^k)$:		
				
\Item (1)
$PSL(2,2^k)$ has an elementary abelian $2$-subgroup $P$ of order   $2^k$ such that the number of conjugates of $P$ in
$PSL(2,2^k)$ is $2^k +1$.

\Item (2)
$PSL(2,2^k)$ has a cyclic subgroup $A$ of order $2^k -1$ such that the number of conjugates of $A$ in $PSL(2,2^k)$ is $2^{k-1}(2^k +1)$.

\Item (3)
$PSL(2,2^k)$ has a cyclic subgroup $B$ of order $2^k +1$ such that the number of conjugates of $B$ in $PSL(2,2^k)$ is $2^{k-1}(2^k -1)$.

\Item (4)
The centralizers of the nontrivial elements of $PSL(2,2^k)$  constitute  precisely the family
$\{xPx^{-1},xAx^{-1},xBx^{-1} \mid x \in G\}$; in particular, $PSL(2,2^k)$ is an $AC$-group.
			
			Thus, using [4, Lemma 3.6], we see that $PSL(2,2^k)$ is an $\goth{nn}$-group and so $\goth{R}_{PSL(2,q)}\cong K_{(2^k+1)(2^k-1),(2^{2k-1}+2^{k-1})(2^k-2),(2^{2k-1}-2^{k-1})(2^k)}$. Hence, the result follows.
		\qed\enddemo	
		
		\proclaim{Proposition 4.22}
			Let $G\cong Sz(2^{2m+1})$ be the Suzuki groups over the field with $2^{2m+1}$ elements,
$m>0$. Let $q=2^{2m+1}$, $r=2^m$, and $s=|Z^*(G)|$. Then, $\goth{R}_G$ is isomorphic to the complete $(q^2+1)+\frac{q^2(q^2+1)}{2}+\frac{q^2(q^2+1)(q-1)}{4(q+2r+1)}+\frac{q^2(q^2+1)(q-1)}{4(q-2r+1)}$-partite graph			 $$K_{(q^2+1)(q^2-s),\frac{q^2(q^2+1)}{2}(q-1-s),\frac{q^2(q^2+1)(q-1)}{4(q+2r+1)}(q+2r+1-s),\frac{q^2(q^2+1)(q-1)}{4(q-2r+1)}(q-2r+1-s)}.
$$
			
			In particular,
			$$				 \gamma(\goth{R}_G)=\gamma
\bigl(K_{(q^2+1)(q^2-s),\frac{q^2(q^2+1)}{2}(q-1-s),\frac{q^2(q^2+1)(q-1)}{4(q+2r+1)}(q+2r+1-s),\frac{q^2(q^2+1)(q-1)}{4(q-2r+1)}(q-2r+1-s)}
\bigr).
			$$
		\endproclaim

\demo{Proof}
Here, $G$ is a simple group of order $q^2(q^2+1)(q-1)$. By [5, Proposition 3.7],

			\Item (a)
$G$ has a Sylow $2$-subgroup $F$ of order $q^2$ and $|\{F^x$ : $x\in G\}|=q^2+1$.

			\Item (b)
$G$ contains a cyclic subgroup $A$ of order $q-1$ and $|\{A^x$ : $x\in G\}|=q^2(q^2+1)/2$.

			\Item (c)
$G$ contains a cyclic subgroup $B$ of order $q+2r+1$ and $|\{B^x$ : $x\in G\}|=q^2(q^2+1)(q-1)/4(q+2r+1)$.

			\Item (d)
$G$ contains a cyclic subgroup $C$ of order $q-2r+1$ and $|\{C^x$ : $x\in G\}|=q^2(q^2+1)(q-1)/4(q-2r+1)$.

			\Item (e)
Suppose $x$ is a nontrivial element of $G$. Then,
			$$
\roman{nil}_G(x)=
\cases
				F^x, & \text{if }x\in F^x,\\
				A^x, & \text{if }x\in A^x,\\
				B^x, & \text{if }x\in B^x,\\
				C^x, & \text{if }x\in C^x.
			\endcases
$$

		In particular, $G$ is an $\goth{nn}$-group and thus by \Par*{Proposition 4.1},
		$$
\goth{R}_G\cong K_{(q^2+1)(q^2-s),\frac{q^2(q^2+1)}{2}(q-1-s),\frac{q^2(q^2+1)(q-1)}{4(q+2r+1)}(q+2r+1-s),\frac{q^2(q^2+1)(q-1)}{4(q-2r+1)}(q-2r+1-s)}$$ and the result follows.
\qed\enddemo		
			
		\proclaim{Proposition 4.23}
			Let $G\cong \Bbb{Z}_m\rtimes \Bbb{Z}_{2n}=\langle a,x : a^m=x^{2n}=1, xax^{-1}=a^{-1}\rangle$, where $m$ is odd, $n=2^t$, and $t\geq 2$. Then, $\goth{R}_G$ is isomorphic to the complete $(m+1)$-partite graph $K_{mn-n,m(n)}$ and $\gamma(\goth{R}_G)=\gamma(K_{mn-n,m(n)})$. In particular, if $m=3$, then $\gamma(\goth{R}_G)=   \lceil \frac{(n-1)(3n-2)}{2} \rceil$.
		\endproclaim

\demo{Proof}
The group $G$ is a nonnilpotent $\goth{nn}$-group with $Z^*(G)=\langle x^2\rangle$. For $1\leq i<m$ and $j$ even, we have $\roman{nil}_G(a^ix^j)=\langle a,Z^*(G)\rangle$. For $1\leq i\leq m$, $1\leq j<2n$ and $j$ odd, we have $\roman{nil}_G(a^ix^j)=Z^*(G)\cup a^ix^jZ^*(G)$. Thus, the distinct nilpotentizers of the nonhypercenter elements are $\langle a,Z^*(G)\rangle$ and $Z^*(G)\cup a^ix^jZ^*(G)$, where $1\leq i< m$ and $1\leq j<2n$, $j$ odd. By \Par*{Proposition 4.1}, $\goth{R}_G\cong K_{mn-n,m(n)}$ and hence by \Par*{Lemma 2.3}, the result follows.
\qed\enddemo			
			
		\proclaim{Proposition 4.24}
			If $N$ is a nilpotent group of order $s$ and $H=\Bbb{Z}_m\rtimes \Bbb{Z}_{2n}=\langle a,x$ : $a^m=x^{2n}=1, xax^{-1}=a^{-1}\rangle$, where $m$ is odd, $n=2^t$, $t\geq 2$, then $\goth{R}_{N\times (\Bbb{Z}_m\rtimes \Bbb{Z}_{2n})}\cong K_{mns-ns,m(ns)}$ and $\gamma(\goth{R}_{N\times (\Bbb{Z}_m\rtimes \Bbb{Z}_{2n})})=\gamma(K_{mns-ns,m(ns)})$. In particular, if $m=3$, then $\gamma(\goth{R}_G)=\lceil \frac{(ns-1)(3ns-2)}{2}\rceil$.
		\endproclaim

\demo{Proof}
Let $G=N\times H$. Then the group $G$ is a nonnilpotent $\goth{nn}$-group with $Z^*(G)=N\times Z^*(H)$. For $1\leq i<m$ and $j$ even, we have $\roman{nil}_G(n',a^ix^j)=N\times\langle a,Z^*(H)\rangle$. For $1\leq i\leq m$, $1\leq j<2n$ and $j$ odd, we have $\roman{nil}_G(n',a^ix^j)=N\times (Z^*(H)\cup a^ix^jZ^*(H))$. Thus, the distinct nilpotentizers of the nonhypercenter elements are $N\times\langle a,Z^*(H)\rangle$ and $N\times (Z^*(H)\cup a^ix^jZ^*(H))$, where $1\leq i< m$ and $1\leq j<2n$, $j$ odd. Thus, by \Par*{Proposition 4.1}, $\goth{R}_G\cong K_{mns-ns,m(ns)}$ and hence by \Par*{Lemma 2.3}, the result follows.
\qed\enddemo

			\proclaim{Lemma 4.25}
		Let $G$ be a finite nonnilpotent group such that every %\?
Sylow subgroups of $G$ is abelian. Then,
		$$
			|E(\goth{R}_G)|\geq \frac{p-1}{2p}|G|^2,
		$$
		where p is the smallest prime dividing the order of $G$.
	\endproclaim

	\demo{Proof}
		As in [14], we have $v_0=n_0/|G|^2$, where
$n_0=|\{(x,y)\in G^2 : \langle x,y\rangle \text{ is nonnilpotent}\}|$. Clearly, $n_0=\sum\nolimits_{v\in V(\goth{R}_G)}\deg(v)=2|E(\goth{R}_G)|$ and thus $v_0=\frac{2|E(\goth{R}_G)|}{|G|^2}$. Since $G$ is nonabelian, by [14, Lemma 5], we have $v_0\geq (p-1)/p$. Hence, the result follows.
	\qed\enddemo

		\proclaim{Proposition 4.26}
			Let $G$ be a nonnilpotent group of order $n=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$, %\?\cdots
where $p_i's$ are distinct primes and $0\leq a_i\leq 2$, $1\leq i\leq k$. Then,
			$$
				\frac{p-1}{12p}n^2-\frac{(n-3)}{2}\leq \gamma(\goth{R}_G)\leq \gamma(K_{n-|Z^*(G)|}),
			$$
where $p$ is the smallest prime dividing order of $G$.
		\endproclaim

\demo{Proof}
Here, every Sylow subgroup of $G$ is abelian. Thus, the result follows easily from
\Par{Lemma 2.4}{Lemmas~2.4}, \Par{Lemma 4.25}{4.25}, and the fact that $|Z^*(G)|\geq 1$.
\qed\enddemo
		
		\proclaim{Proposition 4.27}
			Let $G$ be a nonnilpotent group of order $n$, where $n$ is odd. Then,
			$$
				\frac{p-1}{12p}n^2-\frac{(n-3)}{2}\leq \gamma(\goth{R}_G)\leq \gamma(K_{n-|Z^*(G)|}),
			$$
			where $p$ is the smallest prime dividing $|G|$.
		\endproclaim

\demo{Proof}
By the main theorem of [15], $G$ is a solvable group. Thus, the result follows easily from \Par*{Lemma 2.4}, [4, Proposition 4.11], and the fact that $|Z^*(G)|\geq 1$.
\qed\enddemo
		
		\proclaim{Proposition 4.28}
			Let $G$ be a finite nonnilpotent group such that $\goth{R}_G\cong \Gamma_G$, the noncommuting graph of $G$. Then,
			$$
				\frac{|G|^2-k|G|-6|G|}{12}+\frac{3}{2}\leq \gamma(\goth{R}_G)\leq \gamma(K_{|G|-|Z^*(G)|}),
			$$
			where $k$ is the number of conjugacy classes of $G$.
		\endproclaim
\demo{Proof} By [2, Lemma 3.27], $2|E(\Gamma_G)|=|G|^2-k|G|$, where $\Gamma_G$ is the noncommuting graph of a~group $G$. Since  $\goth{R}_G\cong \Gamma_G$, therefore $|E(\goth{R}_G)|= (|G|^2-k|G|)/2$. Thus, by \Par*{Lemma 2.4} and the fact that $|Z^*(G)|\geq 1$, the result follows.
\qed\enddemo
		
\head
5. Groups Whose Nonnilpotent Graph is Planar, Toroidal, Double-Toroidal, Triple-Toroidal, Quadruple-Toroidal, or Pentuple-Toroidal
\endhead

			In this section, we will classify all nonweakly nilpotent groups whose nonnilpotent graphs are planar, toroidal, double-toroidal, triple-toroidal, quadruple-toroidal, or pentuple-toroidal.
	
	\proclaim{Theorem 5.1}
 Let $G$ be a nonweakly nilpotent group. Then $\goth{R}_G$ is planar if and only if $G \cong S_3$.	
\endproclaim

\demo{Proof}
Follows from \Par*{Theorem 3.6} and [4, Theorem 6.1].
\qed\enddemo
	
			\proclaim{Theorem 5.2}
 Let $G$ be a nonweakly nilpotent group. Then $\goth{R}_G$ is not toroidal.
			\endproclaim

			\demo{Proof}
			Follows from \Par*{Theorem 3.6} and [5, Proposition 5.3].
			\qed\enddemo
			
We need the following result in the sequel.

\proclaim{Lemma 5.3}
Let $G$ be a nonnilpotent solvable group such that $\gamma(\goth{R}_G)=n$. Then,
$$
		|G|(|G|(p-1)-6p)\leq 6p(2n-3)
$$
where $p$ is the smallest prime dividing the order of $G$.
\endproclaim

\demo{Proof} It follows from \Par*{Lemma 2.4}, [4, Proposition 4.11], and the fact that $|Z^*(G)|\geq 1$.
\qed\enddemo

			\proclaim{Theorem 5.4}
			Let $G$ be a nonweakly nilpotent group. Then, the nonnilpotent graph of G is double-toroidal if and only if $G$ is isomorphic to $D_{10}$.
			\endproclaim
			
			\proclaim{Theorem 5.5}
				Let $G$ be a nonweakly nilpotent group. Then, the nonnilpotent graph of $G$ is triple-toroidal if and only if $G$ is isomorphic to $D_{12}$ or $Q_{12}$.
			\endproclaim
			
			\proclaim{Theorem 5.6}
				Let $G$ be a nonweakly nilpotent group. Then, $\goth{R}_G$ is quadruple-toroidal or pentuple-toroidal if and only if $G\cong A_4$.
			\endproclaim
	
			\demo{Proof of \Par{Theorem 5.4}{Theorems 5.4}--\Par{Theorem 5.6}{5.6}}
%\?			\demo{Proof of \Par{Theorem 5.4}{Theorems 5.4}, \Par{Theorem 5.5}{5.5}, and \Par{Theorem 5.6}{5.6}}
By \Par*{Theorem 3.6}, $G$ is a finite group.
Suppose  %\?that
the nonnilpotent graph of $G$ is double-toroidal, triple-toroidal, quadruple-toroidal, or pentuple-toroidal. Since $\gamma(\goth{R}_G)\leq 5$, we have $\omega (\goth{R}_G)\leq 11$ and by [5, Lemma 5.1], we have $|Z^*(G)|\leq \sqrt{5}+1=3.23$ and so $|Z^*(G)|\leq 3$. Since $G$ is a finite nonnilpotent group, by [4, Theorem 4.2], $\omega(\goth{R}_G) \geq 4$ and $G$ is solvable. Suppose $p \geq 11$ be a prime dividing the order of $G$. Suppose $g\in G\setminus Z^*(G)$ such that $\circ(g)=p$.
Then there exist %\?exists
$x \in G$ such that $\langle g,x\rangle$ is not nilpotent. Let $H=\{g,g^2,\dots,g^{p-1}\}$ %\?
and $L=xH\cup \{x\}$. Then the induced subgraph $\goth{R}_G[H\cup L]$ has a subgraph isomorphic to $K_{10,11}$, a contradiction as $\gamma ( K_{10,11}) =18>5$. Thus $|G|=2^r3^s5^t7^u$. If $2$ is the smallest prime dividing the order of $G$, then by \Par*{Lemma 5.3}, $|G|< 17$. If $3$ is the smallest prime dividing the order of $G$, then by \Par*{Lemma 5.3}, $|G|< 14$. If $5$ is the smallest prime dividing the order of $G$, then by \Par*{Lemma 5.3}, $|G|< 12$. Thus $G$ is isomorphic to $S_3$, $D_{10}$, $A_4$, $D_{14}$, $Q_{12}$, or $D_{12}$. By [4, Theorem 6.1], $\goth{R}_{S_{3}}$ is planar. The group $D_{10}$ is a {\it centerless AC-group\/} and so by [5, Lemma 4.5], $\goth{R}_G \cong \Gamma (G)$, where $\Gamma (G)$ is the noncommuting graph of G. It follows from [5, Proposition~5.5] that $\goth{R}_{D_{10}}$ is double-toroidal. By [5, Remark 5.2], $\goth{R}_{D_{12}}\cong \goth{R}_{Q_{12}}\cong K_{4,2,2,2}$. Thus, by [10, Corollary~6.2], $\gamma (\goth{R}_{D_{12}}) = \gamma (\goth{R}_{Q_{12}})=3$. By [5, Remark 5.2], $4\leq \gamma(\goth{R}_{A_4})\leq 5$. The nonnilpotent graph of the group $D_{14}$ is isomorphic to $K_{6,1,1,1,1,1,1,1}$. The graph $K_{6,1,1,1,1,1,1,1}$ has $13$ vertices and $63$ edges and so, by \Par*{Lemma 4.25}, we have $\gamma(\goth{R}_{D_{14}})\geq 5$. Note that $\goth{R}_{D_{14}}\cong K_{13}-K_6$ and by [16], $K_{13}-K_6$ has no triangular embedding. Thus, $\goth{R}_{D_{14}}$ has no triangular embedding. It follows from \Par*{Lemma 4.25} that $\gamma(\goth{R}_{D_{14}})> 5$. This completes the proof.
\qed\enddemo
	
\acknowledgment
The authors would like to thank the referee for his/her valuable comments and suggestions.
The first author wishes to express her sincere thanks to CSIR (India)
for its financial assistance. %\?это в \thanks надо, пока поместила, может, номер гранта укажут
	
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