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

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).


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 Υ be a simple graph.The degree of a vertex x in Υ, denoted by deg(x), is the number of vertices adjacent with x.Υ is said to be regular if all of its vertices have the same degree.If deg(x) = n, we say that Υ 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 ∈ G | yx = xy}, and the center of G is Z(G) = {x ∈ G | xy = yx, ∀y ∈ G}.For an x in G, the coset xZ(G) is denoted by x.The set Cent(G) = {C G (x) | x ∈ G} is the set of distinct centralizers in G.
The notion of non-centralizer graph was introduced by Tolue in [14]: The noncentralizer graph of a finite group G, denoted by Υ G , is the simple graph whose vertices are the elements of G with two vertices x and y are adjacent if C G (x) = C G (y).The induced subgroup of Υ G associated with the vertex set G \ Z(G) is called the induced non-centralizer graph and denoted by Υ G\Z(G) .It is clear that Υ G is a complete |Cent(G)|−partite graph and Υ G\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 Υ G is 6-regular if and only if G ∼ = D 8 or Q 8 , and Υ G is not n-regular, for n = 4, 5, 7, 8, 11, 13, leading to the following conjecture.
Conjecture 1.1.[14] Υ G is not p−regular graph, where p is a prime integer.
In this paper we prove this conjecture by showing that Υ 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 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 Υ G is regular (resp.Υ G\Z(G) is regular).In section 2, we study the regularity of the noncentralizer 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 ≤ |G| ≤ 4n/3.We end this section with a table listing reduced n−regular 2−groups for all possible values of n ≤ 60.
Section 3 is devoted to the regularity of Υ G\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.

Regularity of Υ G
We start this section by listing some known results that will be used later in the sequel.
The following proposition can be directly obtained from [6, Proposition 2.2] and its proof.
Proposition 2.4.Let G be a non-abelian group.Then G is regular if and only if Proof.Suppose G is regular, and let x ∈ G.By Proposition 2.4, we have β G (x −1 ) = x −1 .Since x ∈ β G (x −1 ), we get x = x −1 z for some z ∈ Z(G), and so x 2 = z ∈ 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 In fact, using GAP we find that for each one of the groups with ID's [32, 27], ..., [32,35] yet none of them is regular.The next two theorems show two cases in which the covers of Theorem 2.1 is true.
Proof.Suppose G is regular, and let x ∈ G. Since G/Z(G) is abelian, we can associate with every g ∈ G a unique element The following corollary proves Conjecture 1.1.
Corollary 2.2.Let G be a non-abelian group.Then Υ G is not n−regular, where n is a prime power integer.
Then Proposition 2.4 yields regularity of H. Now we prove the converse.Assume that G = H × A where A is an abelian and H is regular 2−group.We have Therefore, by Proposition 2.4 G, is regular.
Example 2.2.Let A be an abelian group of order k ≥ 1 and let 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.
Let G be an induced regular group, and let C G (x) be a maximal centralizer such that Hence we obtain that yB ⊆ β G (w 0 y) for some w 0 ∈ B.
But also, since y p ∈ Z(G) and p = 2, we get that y ∈ β G (x), a contradiction.So y(H x \ T ) and y −1 (H x \ T ) are disjoint.This implies that Proof.Let G be a non-abelian induced regular group.If C G (x) = β G (x)∪Z(G), for all x ∈ G, then G is an AC-group.Thus by [14,Theorem 2.11] and [1, Proposition 2.6], we have G = P × A, where P is a p−group and A is an abelian group.Hence, , 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 . By Proposition 3.2, there exists an element y in C G (x 0 )\H x0 such that o(y) is prime p.The next goal is to show the following claim. Claim: If there is an element y in C G (x 0 ) \ H x0 such that o(y) = p = 2, then by Proposition 3.3, H x0 /Z(G) is an elementary p−group.Now suppose that for each element y in C G (x 0 ) \ H x , o(y) is either 2 or not prime.In this case we show that H x0 /Z(G) is a 2−group.This is achieved in three steps .
Step 1: We show that C G (x 0 )/H x0 is a 2−group.For simplicity, we will denote the coset yH x0 in C G (x 0 ) by ỹ.Assume there is y ∈ C G (x 0 ) such that o(ỹ) is divisible by a prime p = 2, and let y 0 = y o( ỹ) p .Clearly o(ỹ 0 ) = p and x 0 y 0 / ∈ H x0 .Now y p 0 and (x 0 y 0 ) p belong to H x0 .But since o(y 0 ) and o(x 0 y 0 ) are either 2 or not prime, we have y p 0 and (x 0 y 0 ) p belong to . 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 ) ∩ C G (y 0 ) = C G (y 0 ) for all w ∈ β G (x). Thus we obtain that y 0 β G (x 0 ) ⊆ β(y 0 ).From the fact that Step 2: We show that for all y / ∈ H x0 , o(y) is a power of 2. Let y / ∈ H x0 , and suppose that o(y) = 2 k (2t − 1).If y 2t−1 belongs to β G (x 0 ), then (ỹ) 2t−1 = 1 in C G (x 0 )/H x0 , and so by Step 1, we get that 2 divides 2t − 1, a contradiction.Thus y 2t−1 / ∈ β G (x). Now suppose that p is a prime divisor of 2t − 1 and let y p = y 2t−1 p .Then o(y p ) = p, and so y p ∈ β G (x 0 ).This implies that 2 divides (2t − 1)/p, a contradiction.Thus 2t − 1 = 1, and hence o(y) is a power of 2.
Step 3: We show that Proof.Suppose [G : Z(G)] = p q , for some q ≥ 2. Consider a maximal centralizer , 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.
On the other hand, o(x i ) = p for all i.Then for each , for all i.If we assume that |βG(xi)| |Z(G)| > (p − 1), for some i, then we obtain that , for all i, and hence Υ G\Z(G) is regular.Theorem 3.2.A group G is induced regular if and only if G ∼ = H × A where A is an abelian group, and H is an induced regular p−group for some prime p.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.
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 Υ G is n−regular, then n is even, |G| ≡ 0 (mod 8), and n + 2 ≤ |G| ≤ 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| = [G:Z(G)]n [G:Z(G)]−1 , form Proposition 2.1 we get [G : Z(G)] ≥ 4, and so we obtain that |G| ≤ 4n/3.In addition, 2 ≤ |Z(G)| = |G| − n.Therefor n + 2 ≤ |G| ≤ 4n/3.Theorem 2.6.Let k ≥ 3 and t ≥ 1 be integers and let G be a group of order 2 k (2t − 1).Then G is regular if and only if G ∼ = H × 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 ≤ s ≤ 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 ≤ G. Also, we have H ∩ 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 ∈ A and h ∈ H, we have HA ∼ = H × A. It remains to show that H is regular.Since A ≤ Z(G), we obtain that for any two elements x, y ∈ H, C G , and so ghy = gyh, which yields h ∈ C G (y).This proves our claim.Since G is induced regular group, we get that |yβ G (x)| = |β G (y)|, and so yβ G (x) = β G (y).This implies that 1 ∈ β G (x), a contradiction.Therefore, there must be y ∈ C G (x) \ β G (x) such that o(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

Theorem 3 . 1 .
and hence H x /Z(G) is an elementary p−group.If G be a non-abelian induced regular group then G/Z(G) is a p-group.

Hx 0 Z
(G) is a 2−group.Assume that there is an element w ∈ H x0 such that o(w) = p = 2.By Proposition 3.2, there is y ∈ C G (x 0 ) \ H x0 such that o(y) = 2. Then we have wy / ∈ H x0 and o(wy) = 2p, which contradicts the result in Step 2. Hence Hx 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 = β G (x) ∪ Z(G).Then from the regularity of Υ G\Z(G) we obtain that |H x | = |H x0 |.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 = p.Let y ∈ G such that o(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 yβ G (x) ⊆ β G (y).But since y / ∈ yβ G (x) we obtain that |β G (x)| = |yβ G (x)| < |β 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 ∈ G, |β G (x)| = (p − 1)|Z(G)|.

Corollary 3 . 1 .
Let G be an induced regular group of odd order.IfC G (x) = β G (x) ∪ Z(G), for all x ∈ 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 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 ≤ s ≤ 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 ≤ G. Also, we have H ∩ A = {e} (because (|H|, |A|) = 1), and hence |HA| = |H||A| = |G|.Therefor G = HA.Moreover, since ha = ah for all a ∈ A and h ∈ H, we have HA ∼ = H × A. It remains to show that H is induced regular.Fix x in H \ Z(H) and let s = |β G (x)|/|Z(G)|.Since A ≤ Z(G), we obtain that y belongs to β G (x) if and onlyif C H (x) = C H (y).So β H (x) = β G (x) ∩ H. Also since Z(G) = Z(H)A, there are y 1 , • • • , y s in H such that β G (x) = s i=1 y i Z(G).So β H (x) = β G (x) ∩ H = s i=1 y i Z(G) ∩ H = s i=1 y i (Z(G) ∩ H) = s i=1 y i Z(H).Thus |β H (x)| = |β G (x)| |Z(G)| |Z(H)|.Therefore H is induced regular.Now we prove the converse.Assume that G = H × A where A is an abelian group, H is an induced regular group of order pk .Now Z(G) = Z(H) × Z(A) = Z(H) × A. Also, for each (h, a) ∈ G, C G (h, a) = C H (h) × C A (a) = C H (h) × A. So β G (h, a) = β H (h) × A,for all (h, a) ∈ G. Therefore G is induced regular.
Proof.From Proposition 2.4, we see that G is regular if and only if the number of parts of Υ G is equal to the number of distinct cosets of Z(G).Hence the result.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 .
A group G is called p−group (where p is a prime) if the order of G is a power of 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.