# Finite Groups Which Are the Union of Autocentralizers of Some Automorphisms

Authors: Esmat Alamshahi; Mohammad Reza R. Moghaddam; Mohammad Amin Rostamyari
Journal: Journal of Mathematics
 (2022)

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## Abstract

Letα be an automorphism of a given group G; then, CGα=g∈G|αg=g is called the autocentralizer of α in G. In this work, we study finite groups G, which can be written as the union of autocentralizers of some automorphisms of G. In particular, if a group G is 5-A-centralizers, then we determine the absolute central quotient G/LG of G, where A=AutG. Finally, it is shown that autocommutative property of a group is equivalent to the one, in which every A-centralizer of its nontrivial elements is abelian.


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## Body

## 1. Introduction

LetA=AutG be the group of all automorphisms of a given group G and α∈A. The autocentralizer or A-centralizer of α in G is the subgroup of G consisting of all elements of G which are fixed by α and denoted by CGα, i.e.,(1)CGα=g∈G|gα=αg=g.Also, we denote the set of all such autocentralizes inG by CGA. Note that, if CGα=e, where e is the identity element of G, then α is called fixed-point-free automorphism. We remark that, for the identity automorphism id. of A, we have G=CGid, and hence, G∈CGA.For any elementg∈G and α∈A, the element,(2)g,α=g−1gα=g−1αg,is the autocommutator of g and α. Clearly, if α=ϕx is the inner automorphism induced by the element x of G, then the autocommutator g,ϕx coincides with the ordinary commutator g,x. The set LG, consisting of all elements g in G which are fixed by every automorphism of G, is a central characteristic subgroup of G, which is called absolute centre or the autocentre of G (see [1–5], for more information). Hence, we may write(3)LG=g∈G:g,α=1,∀α∈A=∩α∈ACGα.We denote the number of distinct autocentralizers of automorphisms inG by CGA, and now a question arises that “how many distinct autocentralizers can a group have?”This article is motivated by providing some answers to the above question.


## 2. Preliminary Results

It is clear that, in a given groupG, there may exist an element x, which is not fixed by none of nontrivial automorphisms of G. For example, consider cyclic groups of order p, for odd prime p.In the present article, we consider all finite groups which do not have the above property, so the following result is obvious.Lemma 1.
A groupG is the union of autocentralizers of all nontrivial automorphisms of A in G,LG, i.e., G=∪id≠α∈ACG,LGα.Proof.
Clearly,∪id≠α∈ACG,LGα⊆G. Now, if x∈G,LG, then, by the above assumption, there exists α∈A, such that x∈CG,LGα. Therefore, G⊆∪id≠α∈ACG,LGα, which completes the proof.Remark 1.
Note that ifCGA=1, then (3) implies that LG=G, so x,α=1, for all x∈G and α∈A. Hence, A=id, and consequently, G=1 or ℤ2. Clearly, the converse is also true. Therefore, CGA=1 if and only if G=1 or ℤ2.Lemma 2.
LetG be a nontrivial group except ℤ2; then, CGA≥4.Proof.
By Remark1, CGA≠1. If CGA=2, then G is the proper subgroup of itself, which is impossible. Suppose that CGA=3; then, CGA=G,CGα1,CGα2, for some automorphisms α1 and α2∈A, where CGα1 and CGα2 are proper subgroups of G. Therefore, G=CGα1∪CGα2, which is impossible, since a group cannot be written as the union of two proper subgroups. Thus, CGA≥4.A groupG is called n-A-centralizer if CGA=n (see also [6]).Example 1.
(i)
ConsiderD8=x,y|x4=y2=1,xy=x−1, the Dihedral group of order 8; then, clearly, the set of all automorphisms of D8 is(4)αij:x↦xi,i=1,3,y↦xjy,j=1,2,3,4.Thus, one can calculate that(5)CD8α11=CD8α12=CD8α13=x,CD8α31=CD8α33=1,x2,CD8α14=D8,CD8α32=1,x2,xy,x3y,CD8α34=1,x2,y,x2y.Therefore,D8 is 4-A-centralizer, as CD8α31 and CD8α32 are not distinct.Similarly, one can easily see thatCQ8A=4.(ii)
ConsiderS3=x,y|x3=y2=1,xy=x−1, the symmetric group of order 6; then, one can calculate that S3 has 5 A-centralizers as follows:(6)S3,1,y,1,x,x2,1,x2y,1,xy.


## 3. CountingA-Centalizers in Groups

For an arbitrary elementx of a given group G, the set CGx=y∈G:y,x=y,ϕx=1, where ϕx is the inner automorphism of G, is called the centralizer of x in G. The set of all such centralizers in G is denoted by CentG. Clearly, G is abelian if and only if CentG=1.Many authors have studied the influence of the number of centralizers of a finite groupG on the structure of the group. In 1994, Belcastro and Sherman [7] proved that CentG≥4, for any nonabelian finite group G. They also showed that G has 4 centralizers if and only if G/ZG≅C2×C2 and G has 5 centralizers if and only if G/ZG≅C3×C3 or S3. In [8], Ashrafi proved that if G has 6 centralizers, then G/ZG≅D8,A4,C2×C2×C2 or C2×C2×C2×C2, where A4 is the alternating group of degree 4.In this section, we computeCGA under certain conditions on the group G.Proposition 1.
LetD2n be the Dihedral group of order 2n; then,(7)CD2nA|=n+2,nis odd,n2+2,nis even.Proof.
Clearly, forn≥3, the group D2n has the following presentation:(8)D2n=x,y|xn=y2=1,xy=x−1=1,x,…,xn−1,y,xy,…,xn−1y.
Then,A=αij:x↦xiy↦xjy|gcdi,n=1,0≤j≤n−1, so CD2nα10=D2n. Assume that n is odd; then, we have the following A-centralizers:(i)
CD2nα1j=x, as xk,α1j=x−kxk=1, y,α1j=y−1xjy=x−j≠1, and xky,α1j=y−1x−kxkxjy=x−j≠1, for every 0<k≤n−1.(ii)
CD2nαi0=1,y, as xk,αi0=x−kxik≠1, y,αi0=y−1y=1, and xky,αi0=y−1x−kxiky=xk−ik≠1, for every 0<k≤n−1 and even i.(iii)
CD2nα21=1,xn−1y, as xk,α21=x−kx2k≠1, for every 0<k≤n−1, y,α21=y−1xy=x−1≠1, and xsy,α21=y−1x−sx2sxy=xs+1=1, when s=n−1.By similar argument, one hasCD2nα22=1,xn−2y,CD2nα23=1,xn−3y,…,CD2nα2n−1=1,xy, so CD2nA=n+2 if n is odd.Now, assume thatn is even; then, we have the following cases(iv)
CD2nα1j=x, as xk,α1j=x−kxk=1, y,α1j=y−1xjy=x−j≠1, and xky,α1j=y−1x−kxkxjy=x−j≠1, for every 0<k≤n−1.(v)
CD2nαn−10=1,y,xn/2,xn/2y, as xk,αn−10=x−kxnk−k=1, when k=n/2, y,αn−10=y−1y=1, and xky,αn−10=y−1x−kxnk−ky=x2k−nk=1, when k=n/2.
Using similar argument forCD2nαn−1j, when j is even, we have the following A-centralizers CD2nαn−12=1,xy,xn/2,xn/2+1y,CD2nαn−14=1,x2y,xn/2,xn/2+2y,…,CD2nαn−1n−2=1,xn/2−1y,xn/2,xn−1y, so CD2nA=n/2+2 if n is even.The following result of [9] is useful in our further investigations.Theorem 1 (see Theorem 3.1 in [9]).
LetG be a group with G/LG≅Cp×Cp, for any prime number p. Then, G is isomorphic to one of the following groups:(1)
Cp×Cp(2)
Cp×Cp×C2pis odd(3)
C4×C2(4)
D8(5)
Q8(6)
x,y:x4=y4=1,xy=x−1Using the above theorem, we have the following.Theorem 2.
LetG be a group such that G/LG≅Cp×Cp, for any prime number p. Then, CGA=p+2.Proof.
Letp=2 and G/LG≅C2×C2; then, Proposition 1 implies that G≅C2×C2=x,y:x2=1,y2=1,x,y=1. Clearly, A=AutC2×C2≅S3, and it is easily checked that CC2×C2A=4.
AssumeG=C4×C2=x,y:x4=y2=1,x,y=1; then, clearly,(9)AutC4×C2=αij:x↦xorx3orxyorx3yy↦yorx2y.
Hence, one can calculate that the groupC4×C2 has the following A-centralizers:(10)C4×C2,x,1,x2,xy,x3y,1,x2,y,x2y.
In the casesG≅D8 and Q8, example 1 and Proposition 1 show that G is 4-A-centralizer.
IfG=x,y:x4=y4=1,xy=x−1, then with relatively simple and long calculations, we obtain that G has the following A-centralizers:(11)G,1,x2,y2,x2y2,x,x3,xy2,x3y2,1,x2,y2,x2y2,y,x2y,y3,x2y3,1,x2,y2,x2y2,xy,x3y,xy3,x3y3.
Now, letG≅Cp×Cp×C2 and p be an odd prime number. Then, it is clear that CGα=2,2p or 2p2, for every α∈A. Thus, we have the following A-centralizers:(12)G=x,y,z:xp=yp=z2=1,x,y=x,z=y,z=1,1,z,1,z,x,x2,…,xp−1,zx,zx2,…,zxp−1,1,z,y,y2,…,yp−1,zy,zy2,…,zyp−1,1,z,xyi,xyi2,…,xyip−1,zxyi,…,zxyip−1,where 1≤i≤p−1, so G has p+2 distinct A-centralizers.The following theorem of [10] is needed to prove our next result.Theorem 3 (see Theorem 1 in [10]).
A groupG is the nontrivial union of three subgroups if and only if it is homomorphic to the Klein four-group.Proposition 2.
LetG be a group; then, CGA=4 if and only if G/LG≅C2×C2.Proof.
Using Theorem1, it is enough to show that CGA=4, which implies G/LG≅C2×C2.
SupposeCGA=4; then, CGA=G,CGα1,CGα2,CGα3, for some α1,α2,α3∈A. Hence, by Lemma 1, G=∪i=13CGαi. Consider CGα1α2, which will be one of the A-centralizers G,CGα1,CGα2, or CGα3. If CGα1α2=G, then α2x=α−1x, for all x∈G, so CGα1=CGα2, which is a contradiction.
AssumeCGα1α2=CGα1, then for any x∈CGα1α2=CGα1, we have α1α2x=α1x=x. Therefore α2x=α1−1x=x, and hence, CGα1α2=CGα1⊆CGα2, which is again a contradiction. Similarly, if CGα1α2=CGα2, then we have a contradiction, so CGα1α2=CGα2α1 must be equal with CGα3.
Now, Theorem3 implies that G/∩i=13CGαi=G/LG is isomorphic with Klein four-group.Theorem 4.
LetG be a finite group with CGA=5; then, G/LG≅S3,C3×C3,D12,C2×C6,C3⋊C4 or A4.Proof.
AssumeCGA=5 and CGαi is A-centralizer, for some automorphisms α1,…,α4 of G. Hence, Lemma 1 implies that G=CGα1∪CGα2∪CGα3∪CGα4.
Now, considerCGα1α2, which should be one of G,CGα1,CGα2,CGα3, or CGα4. Thus, we have the following cases:(a)
IfCGα1α2=G, then α2x=α−1x, for all x∈G, so CGα1=CGα2, which is a contradiction.(b)
IfCGα1α2=CGα1, then, for all x∈CGα1α2=CGα1, we have α1α2x=α1x=x. Therefore, α2x=α1−1x=x, and hence, CGα1α2=CGα1⊆CGα2 and gives a contradiction.
Similarly, ifCGα1α2=CGα2, then we obtain a contradiction, so CGα1α2 can be equal to either CGα3 or CGα4.(i)
Assume thatCGα1,CGα2,CGα1α2=CGα1α2, and CGα4 are the A-centralizers of G. Then, using similar argument as in parts (a) and (b), we have CGα1α4=CGα2 and CGα2α4=CGα1. On the contrary, LG⊆CGα1∩CGα2. Thus, for all x∈CGα1∩CGα2, we have(13)α1x=α2α4x=x⇒α2−1x=α4x,α2x=α1α4x=x⇒α2−1x=x,and hence,x∈CGα4. Also, α1α2x=α1x=x and α2α1x=α2x=x imply that x∈CGα1α2=CGα2α1. Therefore, x∈LG and LG=CGα1∩CGα2.(ii)
Assume thatCGα1,CGα2,CGα1α2, and CGα2α1 are the A-centralizers of G. Then, similar argument as in part (i) implies that LG=CGα1∩CGα2.
Hence,∩i=14CGαi=LG=CGαi∩CGαj=CGαi∩CGαj∩CGαk, for 1≤i,j,k≤4, when CGA=5.
Now, for computing the value ofLG, we show that if CGαi and CGαj are arbitrary distinct proper A-centralizers of G, for 1≤i≠j≤4; then,(14)CGαiCGαjG≤LG≤G6⋅∗.
Clearly,CGαiCGαj/CGαiCGαj=CGαi∩CGαj. As CGαiCGαj⊆G, we have 1/CGαiCGαj≥1/G. Therefore, CGαi∩CGαj≥CGαiCGαj/G implies that LG≥CGαiCGαj/G. On the contrary, one observes that(15)G=CGα1+CGα2+CGα3+CGα4−3LG≥2LG+2LG+2LG+2LG−3LG=5LG,and hence, G/|LG|≥5. Assume G/|LG=5; then, G/|LG| is cyclic and Theorem 2.2 of [9] implies that G≅ℤ5 or ℤ10, which both give contradictions as ℤ5 has fixed-point-free automorphism, and ℤ10 does not conform to the conditions given at the beginning of the second section. Therefore, G/|LG|≥6.
Now, without loss of generality, we may assume thatCGα1≥CGα2≥CGα3≥CGα4. Suppose CGα1≤G/4; then, we have(16)G=CGα1+CGα2+CGα3+CGα4−3LG≤G4+G4+G4+G4−3LG=G−3LG,which is a contradiction. Hence, CGα1=G/2 or G/3. If CGα1=G/2, we obtain(17)G=CGα1+CGα2+CGα3+CGα4−3LG=G2+CGα2+CGα3+CGα4−3LG.
One can easily calculate that(18)G2<CGα2+CGα3+CGα4≤3CGα2,so G/6<CGα2.
Now, applying (3) to CGα1 and CGα2, we have CGα1CGα2/G≤G/6, and hence, CGα2≤2G/6. That is, G/6<CGα2≤G/3, so CGα2=G/5,G/4 or G/3. The property CGα1CGα2/G≤LG≤G/6 implies that G/10≤LG≤G/6. Therefore, the value of LG must be one of G/6,G/7,G/8,G/9, or G/10.
Now, ifLG=G/7, then LG divides CGα1, and hence, 2|7 is impossible. Similarly, LG≠G/9. Hence, we have the following cases:(19)LG=G6⇒GLG=6⇒GLG≅C6orS3.
IfG/LG≅S3, then Theorem 3.5 of [9] and example 1 (iii) imply that G≅S3 and CS3A=5. Again Theorem 2.2 of [9] implies that if G/LG≅C6, then G≅C12, which does not conform to the conditions given at the beginning of Section 2.
LetLG=G/8; then, as LG divides CGα2 if CGα2=G/3, then 3|8, and if CGα2=G/5, then 5|8, which both give contradictions. Therefore, CGα2=G/4. On the contrary, the property G=CGα1+CGα2+CGα3+CGα4−3LG implies that G/4=CGα3+CGα4−3G/8, and hence, 5G/8=CGα3+CGα4. As CGα3,CGα4≤G/4, we obtain 5G/8=CGα3+CGα4≤G/2, which is again a contradiction. So, LG cannot be equal to G/8.
Finally, assume thatLG=G/10 and LG divides CGα2. If CGα2=G/3, then 3|10, and if CGα2=G/4, then 4|10, which are both impossible. Therefore, CGα2=G/5. Now, again G=CGα1+CGα2+CGα3+CGα4−3LG implies that CGα3+CGα4=6G/10. Also, CGα2≥CGα3≥CGα4 implies that 6G/10=CGα3+CGα4=2G/5, which is a contradiction. Hence, LG≠G/10.
Now, assume thatCGα1=G/3. In this case, using(20)G=CGα1+CGα2+CGα3+CGα4−3LG,we have 2G/3<CGα2+CGα3+CGα4≤3CGα2. Thus, CGα2>2G/9. On the contrary, CGα1≥CGα2, so 2G/9<CGα2≤G/3. Therefore, CGα2=G/3 or G/4. Again applying ∗ on CGα1 and CGα2, we obtain(21)CGα1CGα2G≤LG≤G6.
Thus,G/12≤LG≤G/6, and hence, LG=G/6,G/7,G/8,G/9,G/10,G/11, or G/12.
Assume thatLG=G/7, and as LG divides CGα1, we must have 3|7, which is impossible. Similarly, LG≠G/8,G/10, and G/11. Also, assume that LG=G/6, CGα1=G/3, and CGα2=G/4 or G/3; then, again(22)G=CGα1+CGα2+CGα3+CGα4−3LGimplies that 11G/12=CGα3+CGα4≤G/2 or 5G/6=CGα3+CGα4≤2G/3, respectively, which are both impossible. Hence, LG≠G/6, so we have one of the following cases:(23)LG=G12⇒GLG=12⇒GLG≅C12,A4,D12,C3⋊C4,C2×C6,or(24)LG=G9⇒GLG=9⇒GLG≅C9,C3×C3.


## 4. Groups with AbelianA-Centralizers

The concept of commutative transitive groups was first introduced and studied by Weisner [11] in 1925.In this section, we introduce the new concept ofautocommutative transitive groups, which is a generalized version of commutative transitive groups. Also, we study a group G, in which every A-centralizer of a nontrivial element of G is abelian. We show that such groups are equivalent to autocommutative transitive groups.Definition 1.
A groupG is autocommutative transitive henceforthA−CT if x,α=1 and α,y=1 imply that x,y=1, for any nontrivial elements x,y in G and every α∈A.Ifα runs over the inner automorphisms of G, then one has the usual commutative transitive groups.Lemma 3.
For any groupG, the following statements are equivalent:(i)
G is A-CT group(ii)
TheA-centralizers of nontrivial automorphisms of G are abelianProof.
i⇒ii Let G be A-CT group. For any nonidentity automorphism element α∈A, if x,y∈CGα, we have x,α=1 and α,y=1. The definition of A-CT implies that x,y=1. Hence, CGα is abelian.ii⇒i Assume x,y are nontrivial elements of G, with x,α=1 and α,y=1, for every α∈A. Obviously, x,y∈CGα, by the assumption CGα is abelian, and hence, x,y=1. Thus, G is autocommutative transitive.Using the above lemma, we have the following.Corollary 1.
LetG be a finite A-CT group and x1,…,xr be a set of pairwise noncommuting elements of G with maximal size. Then, CGA≥r+1.



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