On the Structure of Finite Groups Associated to Regular Non-Centralizer Graph

TL;DR

This paper investigates the structure of finite groups whose non-centralizer or induced non-centralizer graphs are regular, revealing that such groups have quotients over their centers that are elementary 2-groups or p-groups.

Contribution

It characterizes the structure of regular and induced regular finite groups based on their non-centralizer graph properties, a novel approach in group theory.

Findings

Regular groups have quotients over their centers that are elementary 2-groups.

Induced regular groups have quotients over their centers that are p-groups.

The study provides structural insights linking graph regularity to group properties.

Abstract

The non-centralizer graph of a finite group $G$ is the simple graph $\Upsilon_G$ whose vertices are the elements of $G$ with two vertices $x$ and $y$ are adjacent if their centralizers are distinct. The induced subgroup of $\Upsilon_G$ associated with the vertex set $G\setminus Z(G)$ is called the induced non-centralizer graph of $G$. The notions of non-centralizer and induced non-centralizer graphs were introduced by Tolue in \cite{to15}. A finite group is called regular (resp. induced regular) if its non-centralizer graph (resp. induced non-centralizer graph) is regular. In this paper we study the structure of regular groups as well as induced regular groups. Among the many obtained results, we prove that if a group $G$ is regular (resp. induced regular) then $G/Z(G)$ as an elementary $2-$group (resp. $p-$group).

Tables

Table 1. Table 1.

$n$	Reduced $n-$Regular $2-$Groups
6	$Q_8$, $D_8$
12	[16,3], [16,4], [16,6], [16,13]
24	[32,2], [32,4], [32,5], [32,12], [32,17], [32,24], [32,38]
30	[32,49], [32,50]
48	[64,3], [64,17], [64,27], [64,29], [64,44],
	[64,51], [64,57], [64, 86], [64, 112], [64, 185]
56	[64, 73], [64, 74], [64, 75], [64, 76], [64, 77]
	[64, 78], [64, 79], [64, 80], [64, 81],[64, 82]
60	[64, 199], [64, 200], [64, 201], [64, 226], [64, 227],
	[64, 228], [64, 229], [64, 230], [64, 231], [64, 232],
	[64, 233], [64, 234], [64, 235], [64, 236], [64, 237],
	[64, 238], [64, 239], [64, 240], [64, 249], [64, 266]

Equations

$$G\text{ is regular}$$

$$⟺|G|-|{\beta }_G(x)|=|G|-|Z(G)|$$

$$⟺|{\beta }_G(x)|=|Z(G)|=|\overline{x}|$$

$$⟺\overline{x}={\beta }_G(x)$$

$$M(2^k)=⟨a,b\mid a^{2^{k-1}}=b^2=1,bab=a^{2^{k-2}+1}⟩.$$

$$g^{-1}{C}_G(x)g=C_G(g^{-1}xg)=C_G(g^{-1}gx{z}_{xg})=C_G(x{z}_{xg})=C_G(x).$$

$${\beta }_H(x)={\beta }_G(x)\cap H=\overline{x}\cap H=xZ(H)A\cap H=xZ(H).$$

$${\beta }_G(h,a)={\beta }_H(h)\times A=hZ(H)\times aA=(h,a)(Z(H)\times A)=(h,a)Z(G).$$

$$2|y(H_x\setminus T)|=|y(H_x\setminus T)|+|y^{-1}(H_x\setminus T)|\le |{\beta }_G(w_{0}y)|<|H_x|.$$

$$\frac{1}{|Z(G)|}\sum _{i=1}^{p+1}|{\beta }_G(x_i)|=[G:Z(G)]-1=p^2-1.$$

$$p^2-1=\frac{1}{|Z(G)|}\sum _{i=1}^{p+1}|{\beta }_G(x_i)|>(p+1)(p-1)=p^2-1,$$

$${\beta }_H(x)={\beta }_G(x)\cap H=\left(\underset{i=1}{\overset{s}{⨆}}{y}_{i}Z(G)\right)\cap H=\underset{i=1}{\overset{s}{⨆}}{y}_i(Z(G)\cap H)=\underset{i=1}{\overset{s}{⨆}}{y}_{i}Z(H).$$

Full text

1. Introduction

For standard terminology and notion in graph theory and group theory, we refer the reader to the text-books of [8] and [13] respectively. Let $\Upsilon$ be a simple graph. The degree of a vertex $x$ in $\Upsilon$, denoted by $deg(x)$, is the number of vertices adjacent with $x$. $\Upsilon$ is said to be regular if all of its vertices have the same degree. If $deg(x)=n$, we say that $\Upsilon$ is $n-$regular (or regular of degree $n$).

Throughout this paper, $G$ denotes a finite group. The order of a group $G$ (respectively the order of $x$ in $G$) is denoted by $|G|$ (respectively $o(x)$. The centralizer of $x$ is $C_{G}(x)=\{y\in G\mid yx=xy\}$, and the center of $G$ is $Z(G)=\{x\in G\mid xy=yx,\forall y\in G\}$. For an $x$ in $G$, the coset $xZ(G)$ is denoted by $\overline{x}$. The set $Cent(G)=\{C_{G}(x)\mid x\in G\}$ is the set of distinct centralizers in $G$.

The notion of non-centralizer graph was introduced by Tolue in [14]: The non-centralizer graph of a finite group $G$, denoted by $\Upsilon_{G}$, is the simple graph whose vertices are the elements of $G$ with two vertices $x$ and $y$ are adjacent if $C_{G}(x)\neq C_{G}(y)$. The induced subgroup of $\Upsilon_{G}$ associated with the vertex set $G\setminus Z(G)$ is called the induced non-centralizer graph and denoted by $\Upsilon_{G\setminus Z(G)}$. It is clear that $\Upsilon_{G}$ is a complete $|Cent(G)|-$partite graph and $\Upsilon_{G\setminus Z(G)}$ is complete $(|Cent(G)|-1)-$partite graph. Classical properties of these graphs such as diameter, girth, domination and chromatic numbers, and independent set were studied in [14]. Also the author showed that the induced non-centralizer graph and the non-commuting graph associated to an $AC-$group are isomorphic (for information on non-commuting graph see [1, 11]). Interestingly, Tolue proved that $\Upsilon_{G}$ is $6$-regular if and only if $G\cong D_{8}$ or $Q_{8}$, and $\Upsilon_{G}$ is not $n$-regular, for $n=4,5,7,8,11,13$, leading to the following conjecture.

Conjecture 1.1**.**

[14]** $\Upsilon_{G}$ is not $p-$regular graph, where $p$ is a prime integer.

In this paper we prove this conjecture by showing that $\Upsilon_{G}$ in not $n$-regular, if $n$ is a prime power integer. This is direct consequence of deeper results giving the structure of regular groups such as Theorem 2.1 and Theorem 2.4.

In the last few decades, many researches showed interest in characterizing groups by properties of their centralizers such as commutativity and number of centralizer see for instant [2, 3, 4, 5, 6, 7, 10, 12, 15]. Our work is no exception of this trend. Indeed, the sets $\beta_{G}(x)=\{y\in G\mid C_{G}(y)=C_{G}(x)\}$, $x\in G$, as well as the notion of maximal centralizers play key rolls in this paper.

In the sequel, we say that $G$ is regular (resp. induced regular) if $\Upsilon_{G}$ is regular (resp. $\Upsilon_{G\setminus Z(G)}$ is regular). In section 2, we study the regularity of the non-centralizer graph. We show that if $G$ is regular then $G/Z(G)$ is elementary $2-$group. We also prove that a group is regular if and only if it is the direct product of a regular $2-$group and an abelian group. Following this result, the notion of reduced regular groups is introduced. Moreover, We observe that if $G$ is $n-$regular then every centralizer in $G$ is normal and $n+2\leq|G|\leq 4n/3$. We end this section with a table listing reduced $n-$regular $2-$groups for all possible values of $n\leq 60$.

Section 3 is devoted to the regularity of $\Upsilon_{G\setminus Z(G)}$. Using the concept of maximal centralizers, we obtain many results on the structure of induced regular groups. We prove that if $G$ is induced regular then $G/Z(G)$ is a $p-$group. We also show that a group $G$ is induced regular if and only if it is the direct product of an induced regular $p-$group and an abelian group.

2. Regularity of $\Upsilon_{G}$

We start this section by listing some known results that will be used later in the sequel.

Proposition 2.1**.**

[7, Fact 2]** If $G$ is non-abelian group then $|Cent(G)|\geq 4$.

Proposition 2.2**.**

[7, Fact 6]** Let $p$ be a prime. If $G/Z(G)\cong C_{p}\times C_{p}$ then $|Cent(G)|=p+2$.

The following proposition can be directly obtained from [6, Proposition 2.2] and its proof.

Proposition 2.3**.**

Let $G$ be a non-abelian group, such that $[G:Z(G)]=p^{3}$, where $p$ is the smallest prime dividing $|G|$. If $[G:C_{G}(x)]=p^{2}$ for all $x\in G\setminus Z(G)$, then $|Cent(G)|=p^{2}+p+2$. Otherwise $|Cent(G)|=p^{2}+2$.

For every element $x$ in $G$, define the set $\beta_{G}(x)=\{y\in G\mid C_{G}(y)=C_{G}(x)\}$. These sets form the parts of $\Upsilon_{G}$. Clearly $deg(x)=|G|-|\beta_{G}(x)|$ and $\beta_{G}(e)=Z(G)$. Also $\overline{x}\subseteq\beta_{G}(x)$ for all $x\in G$. So $\beta_{G}(x)$ is a disjoint union of cosets of $Z(G)$. Also we conclude that $|Z(G)|$ divides $deg(x)$ for every $x\in G$, and $|Cent(G)|\leq[G:Z(G)]$. The next proposition shows that $\Upsilon_{G}$ is regular if and only if for each $x$ in $G$, $\beta_{G}(x)$ is a coset of $Z(G)$.

Proposition 2.4**.**

Let $G$ be a non-abelian group. Then $G$ is regular if and only if $\beta_{G}(x)=\overline{x}$ for all $x\in G$.

Proof.

We know that $xZ(G)\subseteq\beta_{G}(x)$. So

[TABLE]

∎

Corollary 2.1**.**

Let $G$ be a non-abelian group. Then $G$ is regular if and only if $|Cent(G)|=[G:Z(G)]$.

Proof.

From Proposition 2.4, we see that $G$ is regular if and only if the number of parts of $\Upsilon_{G}$ is equal to the number of distinct cosets of $Z(G)$. Hence the result. ∎

A group $G$ is called $p-$group (where $p$ is a prime) if the order of $G$ is a power of $p$. A $p-$group that is the direct product of copies of $C_{p}$ is called elementary abelian $p-$group. Is it well known in group theory that a group is elementary abelian $2-$group if and only if the order of each element in $G$ is exactly $2$. For convenience, we call a group $G$ elementary $p-$group if the order of each element in $G$ is exactly $p$. With this convention, an elementary $p-$group is abelian if and only if it is the direct product of copies of $C_{p}$.

Theorem 2.1**.**

Let $G$ be a non-abelian group. If $G$ is regular then $G/Z(G)$ is an elementary abelian 2-group.

Proof.

Suppose $G$ is regular, and let $x\in G$. By Proposition 2.4, we have $\beta_{G}(x^{-1})=\overline{x}^{-1}$. Since $x\in\beta_{G}(x^{-1})$, we get $x=x^{-1}z$ for some $z\in Z(G)$, and so $x^{2}=z\in Z(G)$. Therefore, $G/Z(G)$ is an elementary abelian 2-group. ∎

The converse of Theorem 2.1 is not true in general. For example, the group $G=\langle a,b,c\mid a^{4}=b^{4}=c^{2}=e,ab=ba,cac^{-1}=a^{-1},cbc^{-1}=b^{-1}\rangle$ is not regular even though $G/Z(G)\cong C_{2}\times C_{2}\times C_{2}$. In fact, using GAP we find that for each one of the groups with ID’s $[32,27],...,[32,35]$, $G/Z(G)$ is isomorphic to $C_{2}\times C_{2}\times C_{2}$, yet none of them is regular. The next two theorems show two cases in which the covers of Theorem 2.1 is true.

Theorem 2.2**.**

Let $G$ be a non-abelian group. If $G/Z(G)\cong C_{2}\times C_{2}$ then $G$ is regular.

Proof.

Suppose that $G/Z(G)\cong C_{2}\times C_{2}$. Then by Proposition 2.2, $|Cent(G)|=4=[G:Z(G)]$. Hence by Corollary 2.1, $G$ is regular. ∎

Example 2.1**.**

For any positive integer $k\geq 3$, let $M(2^{k})$ be the group defined by

[TABLE]

We have $Z(M(2^{k}))=\langle a^{2}\rangle$. So $M(2^{k})/Z(M(2^{k}))\cong C_{2}\times C_{2}$. Then $M(2^{k})$ is regular of degree $3\cdot 2^{k-2}$.

Theorem 2.3**.**

Let $G$ be a non-abelian group such that $G/Z(G)\cong C_{2}\times C_{2}\times C_{2}$. Then $G$ is regular if and only if $[G:C_{G}(x)]=4$ for all non central elements $x$ of $G$.

Proof.

Suppose $G$ is regular. Then $|Cent(G)|=[G:Z(G)]=8$. Now for each $x$ in $G\setminus Z(G)$, $[G:Z(G)]=[G:C_{G}(x)][C_{G}(x):Z(G)]$, and so $[G:C_{G}(x)]\neq 1,8$. If $[G:C_{G}(x)]=2$, for some $x\in G$, then Proposition 2.3 implies that $|Cent(G)|=6$, a contradiction. So $[G:C_{G}(x)]=4$ for all $x\in G\setminus Z(G)$. Now for the converse, assume that $[G:C_{G}(x)]=4$ for all $x\in G\setminus Z(G)$. Then Proposition 2.3 implies that $|Cent(G)|=8=[G:Z(G)]$. Hence, by Corollary 2.1, $G$ is regular. ∎

Theorem 2.4**.**

Let $G$ be a regular group. Then for each $x\in G$, $C_{G}(x)$ is normal in $G$ and $G/C_{G}(x)$ is isomorphic to a subgroup of $Z(G)$.

Proof.

Suppose $G$ is regular, and let $x\in G$. Since $G/Z(G)$ is abelian, we can associate with every $g\in G$ a unique element $z_{xg}$ in $Z(G)$ such that $xg=gxz_{xg}$. By Proposition 2.4, we have that $C_{G}(xz_{xg})=C_{G}(x)$. So we obtain that

[TABLE]

Hence $C_{G}(x)$ is normal in $G$. Also the mapping $\phi_{x}:G\longrightarrow Z(G)$ defined by $\phi_{x}(g)=z_{xg}$ is a homomorphism with $Ker(\phi_{x})=C_{G}(x)$. Thus $G/C_{G}(x)\cong Img(\phi_{x})$. ∎

The following corollary proves Conjecture 1.1.

Corollary 2.2**.**

Let $G$ be a non-abelian group. Then $\Upsilon_{G}$ is not $n-$regular, where $n$ is a prime power integer.

Proof.

We know that $n=|G|-|Z(G)|=|Z(G)|([G:Z(G)]-1)$. By Theorem 2.1 and Theorem 2.4 we get that $[G:Z(G)]-1$ is odd and $|Z(G)|$ is even. Hence $n$ can not be a power of prime. ∎

Theorem 2.5**.**

If $G$ is a non-abelian group such that $\Upsilon_{G}$ is $n-$regular, then $n$ is even, $|G|\equiv 0\pmod{8}$, and $n+2\leq|G|\leq 4n/3$.

Proof.

Corollary 2.1 yields $[G:Z(G)]$ is dividable by $4$, and Theorem 2.4 implies that $|Z(G)|$ is even. Hence $|G|$ is divisible by $8$. Since $|G|=\frac{[G:Z(G)]n}{[G:Z(G)]-1}$, form Proposition 2.1 we get $[G:Z(G)]\geq 4$, and so we obtain that $|G|\leq 4n/3$. In addition, $2\leq|Z(G)|=|G|-n$. Therefor $n+2\leq|G|\leq 4n/3$. ∎

Theorem 2.6**.**

Let $k\geq 3$ and $t\geq 1$ be integers and let $G$ be a group of order $2^{k}(2t-1)$. Then $G$ is regular if and only if $G\cong H\times A$ where $A$ is an abelian group of order $2k-1$, $H$ is a regular $2-$group of order $2^{k}$.

Proof.

Suppose $G$ is regular. Then $G/Z(G)$ is a $2-$group, which yields $|Z(G)|=2^{s}(2t-1)$ for some $1\leq s\leq k$. Let $A$ be the subgroup of $Z(G)$ of order $2t-1$ and let $H$ be a $2-$sylow subgroup of $G$. Now $HA=AH$. So $HA\leq G$. Also, we have $H\cap A=\{e\}$ (this is because $H$ is a $2-$ group and $|A|$ is odd). Thus $|HA|=|H||A|=|G|$, and so $G=HA$. Moreover, since $ha=ah$ for all $a\in A$ and $h\in H$, we have $HA\cong H\times A$. It remains to show that $H$ is regular. Since $A\leq Z(G)$, we obtain that for any two elements $x,y\in H$, $C_{G}(x)=C_{G}(y)$ if and only if $C_{H}(x)=C_{H}(y)$. So, for each $x\in H$, $\beta_{H}(x)=\beta_{G}(x)\cap H$. Now $Z(G)=Z(H)A$, also by Proposition 2.4, we have $\beta_{G}(x)=\overline{x}$. Therefore

[TABLE]

Then Proposition 2.4 yields regularity of $H$.

Now we prove the converse. Assume that $G=H\times A$ where $A$ is an abelian and $H$ is regular $2-$group. We have $Z(G)=Z(H)\times Z(A)=Z(H)\times A$. Also, for each $(h,a)\in G$, $C_{G}(h,a)=C_{H}(h)\times C_{A}(a)=C_{H}(h)\times A$. Combining this we obtain

[TABLE]

Therefore, by Proposition 2.4 $G$, is regular. ∎

Example 2.2**.**

Let $A$ be an abelian group of order $k\geq 1$ and let $G=D_{8}\times A$ or $G=Q_{8}\times A$. Then $\Upsilon_{G}$ is $K_{2k,2k,2k,2k}$ which is $6k-$regular.

We get the following two remarks from Theorem 2.6 and its proof.

Remark 2.1**.**

The sylow subgroup $H$ is normal in $G$, and the abelian subgroup $A$ is the direct product of all other sylow subgroups. So all sylow subgroups of $G$ are normal, and $G$ is isomorphic to the direct product of its sylow subgroups.

Remark 2.2**.**

Characterizing regular groups reduces to studying regular $2-$groups that are not the direct product of a regular group and an abelian group.

Definition 2.1**.**

A regular $2-$group is called reduced regular if it is not the direct product of a regular group and an abelian group.

Using Theorem 2.5, we can find, for a fixed value of $n$, all $n-$regular and reduced $n-$regular groups. For example, for $n=6$, Theorem 2.5 implies that $|G|$ must be $8$. Among the groups of order $8$ we find that $Q_{8}$, $D_{8}$ are the only $6-$regular groups and both of them are reduced regular. For the case $n=10$, we obtain that $|G|=12$. But also $|G|$ must be divisible by $8$. So there no $10-$regular groups. For $n=12$ we find that $|G|=16$. After investigating the groups of order $16$, we find that six of them are $10-$regular, and four out of these six are reduced $10-$regular. Table 1 lists all reduced $n-$regular $2-$groups with $n\leq 60$.

3. Regularity of $\Upsilon_{G\setminus Z(G)}$

Definition 3.1**.**

Let $G$ be a non-abelian group. A centralizer $C_{G}(x)$ is called maximal if it is not contained in any other proper centralizer.

Proposition 3.1**.**

Let $G$ be a group. If $C_{G}(x)$ is a maximal centralizer in $G$, then $\beta_{G}(x)\cup Z(G)$ is a subgroup of $G$.

Proof.

Since $C_{G}(y)=C_{G}(y^{-1})$, we get that $y\in B_{G}(x)\cup Z(G)$ if and only if $y^{-1}\in\beta_{G}(x)\cup Z(G)$. Now, let $y,w\in\beta_{G}(x)$. Since $C_{G}(y)=C_{G}(x)=C_{G}(w)$, we get $C_{G}(x)\subseteq C_{G}(yw)$. So by maximality of $C_{G}(x)$, we obtain that $C_{G}(x)=C_{G}(yw)$ or $C_{G}(yw)=G$. In both cases $yw$ belongs to $\beta_{G}(x)\cup Z(G)$. Therefore $\beta_{G}(x)\cup Z(G)$ is a subgroup. ∎

Proposition 3.2**.**

Let $G$ be an induced regular group, and let $C_{G}(x)$ be a maximal centralizer such that $C_{G}(x)\neq\beta_{G}(x)\cup Z(G)$. Then there is an element $y$ in $C_{G}(x)\setminus\beta_{G}(x)$ such that $o(\overline{y})$ is prime.

Proof.

Assume that for each $y\in C_{G}(x)\setminus\beta_{G}(x)$, $o(\overline{y})$ is not prime. Then for each $w\in C_{G}(x)\setminus(\beta_{G}(x)\cup Z(G))$, $C_{G}(w)\subseteq C_{G}(x)$. Indeed, take a prime divisor $p$ of $o(\overline{w})$, then $w^{\frac{o(\overline{w})}{p}}\in\beta_{G}(x)$. Hence $C_{G}(w)\subseteq C_{G}(w^{\frac{o(\overline{w})}{p}})=C_{G}(x)$. Now let $y\in C_{G}(x)\setminus\beta_{G}(x)\cup Z(G)$. We claim that $y\beta_{G}(x)\subseteq\beta_{G}(y)$. By Proposition 3.1, $\beta_{G}(x)\cup Z(G)$ is a group. This implies that for all $g\in\beta_{G}(x)$, $yg\notin\beta_{G}(x)\cup Z(G)$. Now let $g\in\beta_{G}(x)$, we want to show that $C_{G}(gy)=C_{G}(y)$. Clearly $C_{G}(g)\cap C_{G}(y)\subseteq C_{G}(gy)$, and so $C_{G}(y)=C_{G}(x)\cap C_{G}(y)=C_{G}(g)\cap C_{G}(y)\subseteq C_{G}(gy)$. For the other inclusion, let $h\in C_{G}(gy)$, then $hgy=gyh$. But also $h$ is in $C_{G}(x)=C_{G}(g)$, and so $ghy=gyh$, which yields $h\in C_{G}(y)$. This proves our claim. Since $G$ is induced regular group, we get that $|y\beta_{G}(x)|=|\beta_{G}(y)|$, and so $y\beta_{G}(x)=\beta_{G}(y)$. This implies that $1\in\beta_{G}(x)$, a contradiction. Therefore, there must be $y\in C_{G}(x)\setminus\beta_{G}(x)$ such that $o(\overline{y})$ is prime. ∎

Proposition 3.3**.**

Let $G$ be an induced regular group, and let $C_{G}(x)$ be a maximal centralizer in $G$. If there exists an element $y$ in $C_{G}(x)\setminus\beta_{G}(x)$ such that $o(\overline{y})$ is a prime $p\neq 2$, then $\dfrac{\beta_{G}(x)\cup Z(G)}{Z(G)}$ is an elementary $p-$ group.

Proof.

Let $H_{x}=\beta_{G}(x)\cup Z(G)$ and let $B=\{w\in\beta_{G}(x)\mid o(\overline{w})\neq p\}$. For $w\in B$, $C_{G}(w)\cap C_{G}(y)\subseteq C_{G}(wy)$. Also , $C_{G}(wy)\subseteq C_{G}((wy)^{p})=C_{G}(w^{p})=C_{G}(w)$. Then $C_{G}(w)\cap C_{G}(y)=C_{G}(wy)$. Hence we obtain that $yB\subseteq\beta_{G}(w_{0}y)$ for some $w_{0}\in B$. Let $T=\left\langle\{w\in H_{x}\mid o(\overline{w})=p\}\right\rangle$. Then $T/Z(G)$ is an elementary $p-$subgroup of $H_{x}/Z(G)$. We have $y(H_{x}\setminus T)\subseteq\beta_{G}(w_{0}y)$ and $y^{-1}(H_{x}\setminus T)\subseteq\beta_{G}(w_{0}y)$. If $y(H_{x}\setminus T)$ and $y^{-1}(H_{x}\setminus T)$ are not disjoint then $\overline{y}\overline{u}=\overline{y}^{-1}\overline{v}$ for some $u,v\in H_{x}\setminus T$. This implies that $y^{2}\in\overline{v}\,\overline{u}^{-1}\subseteq\beta_{G}(x)\cup Z(G)$. But also, since $y^{p}\in Z(G)$ and $p\neq 2$, we get that $y\in\beta_{G}(x)$, a contradiction. So $y(H_{x}\setminus T)$ and $y^{-1}(H_{x}\setminus T)$ are disjoint. This implies that

[TABLE]

So $2(|H_{x}|-|T|)<|H_{x}|$, which implies that $|H_{x}|<2|T|$. Thus $H_{x}=T$, and hence $H_{x}/Z(G)$ is an elementary $p-$group. ∎

Theorem 3.1**.**

If $G$ be a non-abelian induced regular group then $G/Z(G)$ is a $p$-group.

Proof.

Let $G$ be a non-abelian induced regular group. If $C_{G}(x)=\beta_{G}(x)\cup Z(G)$, for all $x\in G$, then $G$ is an AC-group. Thus by [14, Theorem 2.11] and [1, Proposition 2.6], we have $G=P\times A$, where $P$ is a $p-$group and $A$ is an abelian group. Hence, $\dfrac{G}{Z(G)}=\dfrac{P\times A}{Z(P)\times A}\cong\dfrac{P}{Z(P)}$, and so it is a $p-$group. Now suppose that $G$ is not an AC-group. Let $C_{G}(x_{0})$ be a non-abelian maximal centralizer of $G$ and let $H_{x_{0}}=\beta_{G}(x_{0})\cup Z(G)$. By Proposition 3.2, there exists an element $y$ in $C_{G}(x_{0})\setminus H_{x_{0}}$ such that $o(\overline{y})$ is prime $p$. The next goal is to show the following claim.

Claim: $H_{x_{0}}/Z(G)$ is a $p-$group.

If there is an element $y$ in $C_{G}(x_{0})\setminus H_{x_{0}}$ such that $o(\overline{y})=p\neq 2$, then by Proposition 3.3, $H_{x_{0}}/Z(G)$ is an elementary $p-$group. Now suppose that for each element $y$ in $C_{G}(x_{0})\setminus H_{x}$, $o(\overline{y})$ is either $2$ or not prime. In this case we show that $H_{x_{0}}/Z(G)$ is a $2-$group. This is achieved in three steps .

Step 1: We show that $C_{G}(x_{0})/H_{x_{0}}$ is a $2-$group. For simplicity, we will denote the coset $yH_{x_{0}}$ in $C_{G}(x_{0})$ by $\tilde{y}$. Assume there is $y\in C_{G}(x_{0})$ such that $o(\tilde{y})$ is divisible by a prime $p\neq 2$, and let $y_{0}=y^{\frac{o(\tilde{y})}{p}}$. Clearly $o(\tilde{y}_{0})=p$ and $x_{0}y_{0}\notin H_{x_{0}}$. Now $y^{p}_{0}$ and $(x_{0}y_{0})^{p}$ belong to $H_{x_{0}}$. But since $o(\overline{y}_{0})$ and $o(\overline{x_{0}}\overline{y}_{0})$ are either $2$ or not prime, we have $y^{p}_{0}$ and $(x_{0}y_{0})^{p}$ belong to $\beta_{G}(x_{0})$. Therefor $C_{G}(x_{0}y_{0})\cup C_{G}(y_{0})\subseteq C_{G}((x_{0}y_{0})^{p})\cup C_{G}((y_{0})^{p})=C_{G}(x_{0})$. Using the same method as in the proof of Proposition 3.2 one can easily show that $C_{G}(wy_{0})=C_{G}(x_{0})\cap C_{G}(y_{0})=C_{G}(y_{0})$ for all $w\in\beta_{G}(x)$. Thus we obtain that $y_{0}\beta_{G}(x_{0})\subseteq\beta(y_{0})$. From the fact that $y_{0}\notin y_{0}\beta_{G}(x_{0})$, we get that $|\beta_{G}(y_{0})|>|y_{0}\beta_{G}(x_{0})|=|\beta_{G}(x_{0})|$, a contradiction. Therefore $C_{G}(x_{0})/H_{x_{0}}$ is a $2-$group.

Step 2: We show that for all $y\notin H_{x_{0}}$, $o(\overline{y})$ is a power of $2$. Let $y\notin H_{x_{0}}$, and suppose that $o(\overline{y})=2^{k}(2t-1)$. If $y^{2t-1}$ belongs to $\beta_{G}(x_{0})$, then $(\tilde{y})^{2t-1}=\tilde{1}$ in $C_{G}(x_{0})/H_{x_{0}}$, and so by Step 1, we get that $2$ divides $2t-1$, a contradiction. Thus $y^{2t-1}\notin\beta_{G}(x)$. Now suppose that $p$ is a prime divisor of $2t-1$ and let $y_{p}=y^{\frac{2t-1}{p}}$. Then $o(\overline{y}_{p})=p$, and so $y_{p}\in\beta_{G}(x_{0})$. This implies that $2$ divides $(2t-1)/p$, a contradiction. Thus $2t-1=1$, and hence $o(\overline{y})$ is a power of $2$.

Step 3: We show that $\frac{H_{x_{0}}}{Z(G)}$ is a $2-$group. Assume that there is an element $w\in H_{x_{0}}$ such that $o(\overline{w})=p\neq 2$. By Proposition 3.2, there is $y\in C_{G}(x_{0})\setminus H_{x_{0}}$ such that $o(\overline{y})=2$. Then we have $wy\notin H_{x_{0}}$ and $o(\overline{wy})=2p$, which contradicts the result in Step 2. Hence $\frac{H_{x_{0}}}{Z(G)}$ is a $2-$group. This completes the proof of the claim.

Now let $C_{G}(x)$ be an arbitrary maximal centralizer in $G$, and let $H_{x}=\beta_{G}(x)\cup Z(G)$. Then from the regularity of $\Upsilon_{G\setminus Z(G)}$ we obtain that $|H_{x}|=|H_{x_{0}}|$. So $|H_{x}|$ is a $p-$group. Now we come to the last step in the proof which is showing that $G/Z(G)$ is a $p-$group. Assume that $[G:Z(G)]$ is divisible by a prime $q\neq p$. Let $y\in G$ such that $o(\overline{y})=q$, and let $C_{G}(x)$ be maximal centralizer that contains $C_{G}(y)$. Again following the same argument as in the proof of Proposition 3.2 one can show that $y\beta_{G}(x)\subseteq\beta_{G}(y)$. But since $y\notin y\beta_{G}(x)$ we obtain that $|\beta_{G}(x)|=|y\beta_{G}(x)|<|\beta_{G}(y)|$, which contradicts the fact that $G$ is induced regular. Therefore $G/Z(G)$ is a $p-$group. ∎

Proposition 3.4**.**

Let $G$ be an induced regular group. If $[G:Z(G)]=p^{q}$ where $p$ and $q$ are primes then $G/Z(G)$ is an elementary $p-$group. Moreover, for each $x\in G$, $|\beta_{G}(x)|=(p-1)|Z(G)|$.

Proof.

Suppose $[G:Z(G)]=p^{q}$, for some $q\geq 2$. Consider a maximal centralizer $C_{G}(x)$ in $G$. Then $|\beta_{G}(x)|/|Z(G)|=p^{s}-1$ for some $1\leq s\leq q-1$ and $(|Cent(G)|-1)(p^{s}-1)=p^{q}-1$, hence $p^{s}-1$ divides $p^{q}-1$. This implies that $s$ divides $q$. Thus, if $q$ is prime, then $s=1$, which completes the proof. ∎

Corollary 3.1**.**

Let $G$ be an induced regular group of odd order. If $C_{G}(x)\neq\beta_{G}(x)\cup Z(G)$, for all $x\in G$ then $G/Z(G)$ is an elementary $p$-group.

Proof.

This result follows directly from Proposition 3.3. ∎

Proposition 3.5**.**

Let $G$ be a non-abelian group. If there is a prime $p$ such that $G/Z(G)\cong C_{p}\times C_{p}$, then $G$ is induced regular.

Proof.

Suppose that $G/Z(G)\cong C_{p}\times C_{p}$. Then by Proposition 2.2, $\Upsilon_{G\setminus Z(G)}$ has $p+1$ parts. Let $\beta_{G}(x_{1}),\cdots,\beta_{G}(x_{p+1})$ be the parts $\Upsilon_{G\setminus Z(G)}$ . Then

[TABLE]

On the other hand, $o(\overline{x}_{i})=p$ for all $i$. Then for each $i\in\{1,\cdots,p+1\}$, $\bigcup\limits_{j=1}^{p-1}\overline{x}_{i}^{j}\subseteq\beta_{G}(x_{i})$, which implies that $\frac{|\beta_{G}(x_{i})|}{|Z(G)|}\geq(p-1)$, for all $i$. If we assume that $\frac{|\beta_{G}(x_{i})|}{|Z(G)|}>(p-1)$, for some $i$, then we obtain that

[TABLE]

a contradiction. Thus $\frac{|\beta_{G}(x_{i})|}{|Z(G)|}=(p-1)$, for all $i$, and hence $\Upsilon_{G\setminus Z(G)}$ is regular. ∎

Theorem 3.2**.**

A group $G$ is induced regular if and only if $G\cong H\times A$ where $A$ is an abelian group, and $H$ is an induced regular $p-$group for some prime $p$.

Proof.

Suppose $G$ is induced regular. Then $G/Z(G)$ is a $p-$group. Let $|G|=p^{k}m$, where $(m,p)=1$. Then $|Z(G)|=p^{s}m$ for some $1\leq s\leq k$. Let $A$ be the subgroup of $Z(G)$ of order $m$ and let $H$ be a $p-$sylow subgroup of $G$. Now $HA=AH$. So $HA\leq G$. Also, we have $H\cap A=\{e\}$ (because $(|H|,|A|)=1$), and hence $|HA|=|H||A|=|G|$. Therefor $G=HA$. Moreover, since $ha=ah$ for all $a\in A$ and $h\in H$, we have $HA\cong H\times A$. It remains to show that $H$ is induced regular. Fix $x$ in $H\setminus Z(H)$ and let $s=|\beta_{G}(x)|/|Z(G)|$. Since $A\leq Z(G)$, we obtain that $y$ belongs to $\beta_{G}(x)$ if and only if $C_{H}(x)=C_{H}(y)$. So $\beta_{H}(x)=\beta_{G}(x)\cap H$. Also since $Z(G)=Z(H)A$, there are $y_{1},\cdots,y_{s}$ in $H$ such that $\beta_{G}(x)=\bigsqcup\limits_{i=1}^{s}y_{i}Z(G)$. So

[TABLE]

Thus $|\beta_{H}(x)|=\dfrac{|\beta_{G}(x)|}{|Z(G)|}|Z(H)|$. Therefore $H$ is induced regular.

Now we prove the converse. Assume that $G=H\times A$ where $A$ is an abelian group, $H$ is an induced regular group of order $p^{k}$. Now $Z(G)=Z(H)\times Z(A)=Z(H)\times A$. Also, for each $(h,a)\in G$, $C_{G}(h,a)=C_{H}(h)\times C_{A}(a)=C_{H}(h)\times A$. So $\beta_{G}(h,a)=\beta_{H}(h)\times A$, for all $(h,a)\in G$. Therefore $G$ is induced regular. ∎

It is worth mentioning that there are several induced regular groups $G$ for which $G/Z(G)$ is not abelian such as the groups with GAP ID’s $[243,2]$ to $[243,9]$. On the other hand, we checked many groups and we could not find one induced regular group $G$ for which $G/Z(G)$ is not elementary. Based on these remarks and the results obtained in this section we make the following conjecture.

Conjecture 3.1**.**

If $G$ is induced regular group then $G/Z(G)$ is an elementary $p-$group.

The following problem state the remaining part needed to completely prove conjecture 3.1 .

Problem 3.1**.**

Let $G$ is induced regular group, and let $C_{G}(x)$ be a maximal centralizer such that $C_{G}(x)=\beta_{G}(x)\cup Z(G)$ or $(\beta_{G}(x)\cup Z(G))/Z(G)$ is $2-$group. Then $\beta_{G}(x)\cup Z(G)$ is elementary $p-$group.