Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60 https://doi.org/10.1007/s12044-018-0441-0

Nilpotent groups related to an automorphism

AHMAD ERFANIAN∗ and MASOUMEH GANJALI

Department of Mathematics, Ferdowsi University of Mashhad, P.O. Box 1159-91775, Mashhad, Iran *Corresponding author. E-mail: [email protected]; [email protected]

MS received 16 March 2017; revised 23 May 2017; accepted 18 June 2017; published online 25 October 2018

Abstract. The aim of this paper is to state some results on an α-nilpotent group, which was recently introduced by Barzegar and Erfanian (Caspian J. Math. Sci. 4(2) (2015) 271–283), for any fixed automorphism α of a group G. We define an identity nilpotent group and classify all finitely generated identity nilpotent groups. Moreover, we prove a theorem on a generalization of the converse of the known Schur’s theorem. In the last section of the paper, we study absolute normal subgroups of a finite group.

Keywords. Nilpotent group; identity nilpotent group; absolute normal subgroup.

2010 Mathematics Subject Classification. Primary: 20F12; Secondary: 20D45.

1. Introduction The extension of a nilpotent group has been studied by different authors. For instance an autonilpotent group has been investigated in [13]. Recently, Barzegar and Erfanian [2] defined a new extension of nilpotent and solvable groups. Actually, this extension displays nilpotency and solvability of a group with respect to an automorphism. Similar to the definition of a nilpotent group, this extension needs an introduction of a normal series of group G. Assume that α is an automorphism of group G. Define the subgroup α − − α Z (G) ={y ∈ G :[x, y]α = x 1 y 1xy = 1 ∀x ∈ G} which is called the α- center of G. A central α-series is a normal series {1}=G0  G1  ···  Gn = G α = / ≤ α¯ ( / ) α¯ ∈ ( / ) such that Gi Gi and Gi+1 Gi Z G Gi , where Aut G Gi by the rule α¯ = α ∈ / ≤ ≤ − α gGi g Gi for all gGi G Gi ,0 i n 1. An -nilpotent group is a group which has at least a central α-series. One can easily see that Z α(G) = Z(G) ∩ Fix(α), where Fix(α) ={g ∈ G : gα = g}. As a result of Theorem 3.10 of [2], Z α(G) ={1} whenever G is α-nilpotent, so G is not α-nilpotent for a fixed point free automorphism α. Now, put α( ) α α α¯ G Zi G Z (G) = Z (G) and define Z ( α ( ) ) = α ( ) for i ∈ N. Then the normal series 1 Zi−1 G Zi−1 G

{ }= α( )  α( )  α( )  ··· 1 Z0 G Z1 G Z2 G is said to be an upper central α-series. One of the interesting results in [2] is a theorem that provides a sufficient and necessary condition for a given group to be α-nilpotent. It

© Indian Academy of Sciences 1 60 Page 2 of 12 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60 states that a group G is α-nilpotent if and only if there exists a positive integer n such α( ) = α that Zn G G. Also, Barzegar and Erfanian defined, in an analogous manner, the - ( ) =[ , ] : , ∈ α( ) = α( ) = ( ) commutator subgroup Dα G x y α x y G . Put 1 G G, 2 G Dα G α ( ) =[ ,α( )] =[ , ] : ∈ , ∈ α( ) and define inductively n+1 G G n G α x y α x G y n G for all n ≥ 1. It has been proved that a group G is α-nilpotent if and only if there exists an integer α( ) ={ } α  r such that r G 1 . It is easy to see that if G is an -nilpotent group, H G and H α = H, then G/H is α¯ -nilpotent. Also we can prove that for every normal subgroup N contained in Z α(G),ifG/N is α¯ -nilpotent, then G is α-nilpotent. One can check that an α- nilpotent group is nilpotent, but the converse is not true in general. Because consider a group with a fixed point free automorphism α of prime order which is nilpotent (Theorem 10.5.4 α α α( ) = ( ) of [14]), but is not -nilpotent. For an inner automorphism of group G, Zn G Zn G for all n ∈ N, so nilpotency and α-nilpotency are equivalent. Therefore, one question that may arise is: can we find a non-inner automorphism α of a nilpotent group G such that G is α-nilpotent? In section 3, we answer this question for abelian groups, indeed we classify all finite groups that are α-nilpotent for at least one non-identity automorphism α. Suppose that α ∈ Aut(G) and β ∈ Aut(H), then we define α × β as an automorphism of G × H by α × β : (g, h) → (gα, hβ ) for all g ∈ G and h ∈ H. By Theorem 3.11 of [2], if G is α-nilpotent and H is β-nilpotent, then G × H is α × β-nilpotent. Also we know that a finite nilpotent group is the direct product of its Sylow subgroups, so it is enough to answer the above question for finite p-groups. We remind an old result proved by Gaschütz [7], asserting that finite non-abelian p- groups possess non-inner automorphisms of p-power order. Also, some have investigated finite non-abelian p-groups with at least one non-inner automorphism of order p (for example, see [1] and [4]). Here, we prove a theorem that states the relation between α- nilpotency of a group and the order of α. Although, the above question has not been answered as yet for non-abelian groups, but we do hope to give a certain answer by using results of this paper and papers similar to [1], [4] and [7] in near future. In section 3, we will give an example of a group that is α-nilpotent if and only if α is the identity automorphism of G. We call a group with this property an identity nilpotent group and we classify all finitely generated identity nilpotent groups. At the end of this paper, we recall the definition of absolute normal subgroups [5], and study absolute normal subgroups of some finite groups. All groups in this paper are considered finite unless we notify. We show the identity element of G and Aut(G) by 1 and I respectively.

2. Basic results In this section, we investigate some conditions that a nilpotent group G is α-nilpotent for a non-identity automorphism α. Also, we show that an α-solvable group is αn-solvable for an arbitrary positive integer n. Moreover, we are going to give a generalization of the converse of Schur’s theorem.

Theorem 2.1. If G is a finite group and I = α ∈ Aut(G) such that (|G|, |α|) = 1, then G is not α-nilpotent.

Proof. Assume that (|G|, |α|) = 1 and G is α-nilpotent. Then there exists a normal series, { }= ≤ ≤ ··· ≤ = α = / ≤ α¯ ( / ) 1 G0 G1 Gn G such that Gi Gi and Gi+1 Gi Z G Gi for α¯ 0 ≤ i ≤ n − 1. So for an arbitrary g ∈ Gi+1,wehavegGi ∈ Gi+1/Gi ≤ Z (G/Gi ) = Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60 Page 3 of 12 60

(α)¯ ∩ ( / ) ∈ (α)¯ α¯ = . . α Fix Z G Gi . Thus, gGi Fix and gGi gGi and by Theorem 1 6 6of[9], is identity, which is a contradiction. Hence, G is not α-nilpotent. 

The structure of automorphisms of an extra-special p-groupmaybeusedinthefollowing example, verified in [15].

Example 2.2. If P is an extra-special p-group of order p2n+1, then Aut(P) =θ H, where θ ∩H ={I }, |θ|=p − 1 and H is a normal subgroup of Aut(P) consisting of all members of Aut(P) which act trivially on Z(P) and p is an odd prime number. Since (|P|, |θ|) = 1, then by Theorem 2.1, P is not θ-nilpotent.

α( ) [ , ] ∈ α ( ) The fact that an element x is a member of Zn G if and only if y x α Zn−1 G for all y ∈ G, may used in the rest of this paper.

DEFINITION 2.3

An automorphism α of a group G is central if α commutes with every inner automorphism or equivalently if g−1gα ∈ Z(G), for all g ∈ G.

Lemma 2.4.Ifα is a central automorphism that fixes Z(G) element wise, then G is α- nilpotent if and only if G is nilpotent.

∈ N ( ) = α( ) Proof. We are going to prove that for every n , Zn G Zn G . It is not difficult α( ) ≤ ( ) ( ) ≤ (α) to see that Zn G Zn G in general. By assumption in lemma, Z G Fix , then α( ) = ( ) α ( ) = ( ) ∈ ( ) ∈ Z G Z G . Now, suppose that Zn−1 G Zn−1 G and x Zn G and y G are [ , ] =[ , ] −1 α [ , ]∈ ( ) = α ( ) two arbitrary elements. Then y x α y x x x and y x Zn−1 G Zn−1 G , −1 α ∈ ( ) = α( ) ≤ α ( ) [ , ] −1 α =[ , ] ∈ α ( ) also x x Z G Z G Zn−1 G . Hence, y x x x y x α Zn−1 G ∈ ∈ α( ) ( ) ≤ α( ) α for all y G and so, x Zn G . Therefore, Zn G Zn G and -nilpotency and nilpotency are equivalent. 

A power automorphism of a group G is an automorphism which maps every subgroup of G into itself.

Lemma 2.5.Ifα, β ∈ Aut(G), β is a power automorphism and G is α-nilpotent. Then it is αβ -nilpotent, where αβ = β ◦ α ◦ β−1.

[ , ]β =[β , β ] , ∈ Proof. We can see that y x αβ y x α for all x y G. We will prove that α( ) ≤ αβ ( ) ∈ N ∈ α( ) β Zn G Zn G for every n .Ifx Z G , then since is a power automorphism, β ∈ α( ) [ β , β ] = [ , ]β = ∈ x Z G . Therefore y x α 1 and so y x αβ 1 for all y G, which means ∈ αβ ( ) α ( ) ≤ αβ ( ) ∈ α( ) that x Z G . Now, assume that Zn−1 G Zn−1 G and x Zn G . Then for every ∈ [ , ] ∈ α ( ) β ∈ α( ) [ , ]β =[ β , β ] ∈ α ( ) y G, y x α Zn−1 G and x Zn G , then, y x αβ y x α Zn−1 G . α ( ) β [ , ] ∈ α ( ) ≤ αβ ( ) ∈ Since Zn−1 G is invariant under , then y x αβ Zn−1 G Zn−1 G for all y G, ∈ αβ ( ) α( ) ≤ αβ ( ) ∈ N which means that x Zn G . Hence, Zn G Zn G for all n and the proof is completed.  60 Page 4 of 12 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60

There are some open problems at the end of the paper [2]. One of them has been answered here by Theorems 2.6 and 2.8. Actually, we prove that if G is α-nilpotent (α-solvable), then it is αn-nilpotent (αn-solvable).

Theorem 2.6. If G is α-nilpotent, then it is αn-nilpotent for all positive integers n.

2 Proof. Assume that x ∈ Fix(α), then x ∈ Fix(α2) and we conclude that Z α(G) ⊆ Z α (G). α( ) ⊆ α2 ( ) ∈ α ( ) [ , ] ∈ α( ) Now suppose that Zi G Zi G and x Zi+1 G , then g x α Zi G for all g ∈ G. Now for an arbitrary g ∈ G,wehave

[ , ] =[ , ] ( −1 α)α ∈ α( ) ⊆ α2 ( ). g x α2 g x α x x Zi G Zi G

α ( ) ⊆ α2 ( ) (α) ⊆ (α3) α( ) ⊆ α3 ( ) Therefore Zi+1 G Zi+1 G . Now since Fix Fix , then Z G Z G . α( ) ⊆ α3 ( ) ∈ α ( ) ∈ [ , ] ∈ Let Zi G Zi G and x Zi+1 G . Then for all g G, we can see that g x α2 α( ) −1 α −1 α2 ∈ α( ) ∈ Zi G and x x , x x Zi G . So for all g G,wehave

[ , ] =[ , ] ( −1 α2 )α ∈ α( ) ⊆ α3 ( ), g x α3 g x α x x Zi G Zi G

∈ α3 ( ) α ( ) ⊆ αn ( ) thus x Zi+1 G . By a similar method, we can see that Zm G Zm G for arbitrary positive integers m and n. Hence, if G is an α-nilpotent group, then so is αn-nilpotent. The proof is completed. 

1 Here, we may state some results about α-solvable groups. Put Dα(G) = Dα(G), i i−1 Dα(G) = Dα(Dα (G)). Then the derived α-series is defined as follows:

2 1 0 ··· Dα(G)  Dα(G) = Dα(G)  Dα(G) = G.

A group G is called α-solvable if there is a subnormal series

1 = G0  G1  ··· Gn = G,

α = ( / ) = α such that Gi Gi and Dα¯ Gi+1 Gi 1. It has been shown that G is -solvable if and n only if Dα(G) = 1forsomen ∈ N (see [2] for more details). It has been shown that an α-nilpotent group is α-solvable. The converse is not true, for instance, if we consider group α α S3 ={e,(12), (13), (23), (123), (132)} and α ∈ Aut(S3) as (12) = (13), (23) = (12), α α α (13) = (23), (123) = (123), (132) = (132), then it is easy to see that S3 is α-solvable which is not α-nilpotent.

n n n Lemma 2.7.G ⊆ Dα(G) for an arbitrary positive integer n, where G ’s are terms of the ordinary derived series.

 1 Proof. The proof is based on induction on n. Assume that [x, y]∈G , then [x, y]∈Dα(G) −1 i i because [x, y]=[x, y]α[y, y]α . Now, suppose that G ⊆ Dα(G) for all i  n and let n n−1 n−1 G =[x, y]:x, y ∈ G . Then by induction hypothesis, x, y ∈ Dα (G).Nowbythe n n n definition of Dα(G), we can see that [x, y]α and [y, y]α ∈ Dα(G), thus [x, y]∈Dα(G). Hence we are done.  Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60 Page 5 of 12 60

As an immediate result of the previous lemma, every α-solvable group is solvable. The group Z2 ×Z2 shows that the converse is not valid, which means that this group is solvable but it is not α-solvable for a fixed point free automorphism α.

Theorem 2.8. If a group G is α-solvable, then it is αn-solvable for an arbitrary positive integer n.

( ) ⊆ ( ) r ( ) ⊆ r ( ) Proof. It is easy to see that Dα2 G Dα G . Now if we assume that Dα2 G Dα G [ , ] ∈ r+1( ) [ , ] =[ , ] [ , ]α [ , ] ∈ and x y α2 Dα2 G , then by x y α2 x y α y y α, we conclude that x y α2 r+1 r r Dα (G). We can continue this process and prove that Dαn (G) ⊆ Dα(G) for all positive integers r and so if G is α-solvable then it should be αn-solvable. 

Assume that G is an infinite group, then we are going to give a generalization of the converse of Schur’s theorem. In [2], the following theorem has been proved that is a generalization of Schur’s theorem.

/ α( ) α( ) Theorem 2.9. If G Z G is finite, then so is 2 G . α ( ) =∼ / α( ) Also, it has been investigated that if G is an abelian group, then n+1 G G Zn G . / α( ) α ( ) Hence G Zn G is finite if and only if n+1 G is a finite group. So, we have a general- ization of Baer’s theorem for abelian groups (Theorem 3.14. of [2]). The validation of the converse of Schur’s theorem has been studied by some authors. For instance, Niroomand [12] proved the following theorem.

Theorem 2.10. Let G be an arbitrary group such that d(G/Z(G)) and G are finite. Then |G/Z(G)|≤|G|d(G/Z(G)), in which d(X) is the minimal number of generators of a group X.

Now, we prove a theorem on the converse of Theorem 2.9.

α Theorem 2.11. If Dα(G) is a finite subgroup and G/Z (G) is finitely generated, then G/Z α(G) is finite.

 α  Proof. Since G ≤ Dα(G) and Z (G) ≤ Z(G), then G is finite and G/Z(G) is finitely generated. Thus by Theorem 2.10, G/Z(G) is finite and so the index of Z(G)/Z α(G) in G/Z α(G) is finite. Also G/Z α(G) is finitely generated, then Z(G)/Z α(G) is finitely generated. The proof will be completed if we show that Z(G)/Z α(G) is finite. α α φ − α Define function φ : Z(G)/Z (G) → Dα(G) by the rule (yZ (G)) = y 1 y . Then φ α α α is well-defined because if y1 Z (G) = y2 Z (G) then there is an element z ∈ Z (G) such α that y2 = y1z and since Z (G) = Z(G) ∩ Fix(α), then

−1 α = −1( −1)α = −1 α ( −1)α = −1 α. y1 y1 zy2 y2z y2 y2 z z y2 y2

φ −1 α = −1 α −1 = ( −1 )α Also, is one-to-one because if y1 y2 y2 y2 , then y1 y2 y1 y2 and since −1 ∈ ( ) −1 ∈ α( ) ( )/ α( ) y1 y2 Z G ,wehavey1 y2 Z G . Hence Z G Z G is finite and the proof is completed.  60 Page 6 of 12 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60

3. Identity nilpotent groups We know there are some groups that are α-nilpotent for some non-identity automorphisms α. But, we have the situation that G is α-nilpotent only for the identity automorphism α. The following definition will state such groups.

DEFINITION 3.1

A group G is called identity nilpotent whenever it is α-nilpotent if and only if α is the identity automorphism.

α α( ) = ( ) We remind that if is an inner automorphism, then Zn G Zn G for all positive integer n.

Lemma 3.2. Let G be an identity nilpotent group. Then G is abelian.

Proof. Since G is nilpotent, Zn(G) = G for some positive integer n. Assume that α ∈ ( ) α( ) = ( ) α α Inn G . Then Zn G Zn G and G is -nilpotent. By our hypothesis, is the identity automorphism and Inn(G) ={I } and hence G is an abelian group. 

We know that every automorphism α of a cyclic group Zn can be determined as the form α m = um, where m ∈ Zn and (u, n) = 1. The following theorem also has been proved in [2].

c n Theorem 3.3. A cyclic group Zn is α-nilpotent of class c if and only if (u − 1) ≡ 0.

In [2], there is a theorem that says an abelian group G is α-nilpotent if and only if there is a positive integer n such that (α − I )n = 0 and (α − I )n−1 = 0, where I is the identity automorphism. The converse of Lemma 3.2 is not true. Consider the cyclic group Z9. By Theorem 3.3, Zp, for prime number p,isα-nilpotent if and only if α is the identity automorphism. We can prove the same result for the cyclic group Zn when n = p1 p2 ...pt for different primes pi ’s, i = 1, 2,...,t. We remind that if G = G1 × G2 ×···×Gt is a finite group and (|Gi |, |G j |) = 1, ∼ i = j, then Aut(G) = Aut(G1) × Aut(G2) ×···×Aut(Gt ) ([3], Lemma 2.1). Thus G is identity nilpotent if and only if Gi is identity nilpotent, for all i = 1, 2,...,t.The next theorem classifies all finitely generated groups that are identity nilpotent. Indeed, it answers the question that we ask in the Introduction of this paper for an abelian group.

Theorem 3.4. Let G be finitely generated. Then G is identity nilpotent if and only if one of the following holds: =∼ Z ,..., ≥ (i) G p1 p2...pt for distinct primes p1 pt ; t 1. (ii) G =∼ Z.

n1 n2 ··· nt Proof. Assume first that G is a finite identity nilpotent group of order p1 p2 pt . Then by Lemma 3.2, G is abelian and so is a direct product of cyclic groups. If there exists an ∼ integer 1 ≤ i ≤ t such that G = Z ri × A and ri  1, then G is not identity nilpotent. pi Because if we consider non-identity automorphism α = αi × I , then by Theorem 3.3, G Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60 Page 7 of 12 60

α α i = ( + ) ∈ Z r ≤ ≤ is -nilpotent where x pi 1 x for all x p i . Also, if there exists 1 i t such ∼ i that G = Zp × Zp × B, then G again is not an identity nilpotent. Indeed, if we define i i α α : Z × Z → Z × Z ( , ) i = ( + , ) ( , ) ∈ Z × Z function i pi pi pi pi by a b a b b for all a b pi pi , α ∈ (Z × Z ) (α − )2 = (α − ) = Z × Z α then i Aut pi pi , i I 0 and i I 0. Therefore, pi pi is i - α × =∼ Z = , ,..., nilpotent and G is i I . Hence, G p1 p2...pt for different primes pi ’s, i 1 2 t. ( = Z ) =∼ (Z ) × (Z ) ×···× (Z ) Conversely, since Aut G p1 p2...pt Aut p1 Aut p2 Aut pt and Z ≤ ≤ pi is identity nilpotent for all 1 i t, then G is identity nilpotent. Now, we may study infinite finitely generated identity nilpotent groups. Let us start with infinite cyclic group Z. We know that Z has only two automorphisms, the identity automorphism I and α that corresponds every element to its converse. It is easy to see that Z α(Z) ={x ∈ Z : 2x = 0}={0} and Z is not α-nilpotent. Hence Z is an identity nilpotent group. Now, we consider Z × Z. Then it is α-nilpotent whenever α is (a, b)α = (a + b, b), because (α − I )2 = 0 and α − I = 0. Thus, Z × Z is not an identity nilpotent. Similarly, Z × Zp is not identity nilpotent. Using structure theorem of finitely generated abelian groups and the above ideas, one can prove that an infinite finitely generated group G is identity nilpotent if and only if G =∼ Z. 

4. On absolute normal subgroups In [5], we defined an α-normal subgroup of a finite group G. Here, we recall this definition and state a theorem about α-normality and α-nilpotency. Indeed, we prove a generalization of the famous Fitting’s theorem. Also, we study absolute normal subgroups of some finite groups.

DEFINITION 4.1

α A subgroup H is called an α-normal subgroup of G, and we denote it by H  G, whenever g−1hgα ∈ H for every h ∈ H and g ∈ G.

It is clear that every α-normal subgroup is normal but the converse is not true, in general. For instance, consider the cyclic group of order 2n, Z2n ={0, 1,...,2n − 1} and that n is an odd number. Then the subgroup H ={0, n} is α-normal if and only if α is the identity automorphism.

Lemma 4.2. If G is the cyclic group of prime power order, then for every subgroup H there exists a non-identity automorphism α such that H is α-normal.

Proof. It is an immediate result of Theorem 2.11 of [5]. 

α Lemma 4.3. If G is a group, H, N  G and K  G, then

(i) [K, HN]α ≤[K, H]α[K, N]α, (ii) [HN, K ]α ≤[N, K ]α[H, K ]α.

Proof. − − α − α − − − α α (i) Since [k, hn]α = k 1(h 1kh )(h 1kh ) 1n 1(h 1kh )n , then we have [k, hn]α =   − α [k, h]α[k , n]α, where k = h 1kh ∈ K because K is an α-normal subgroup of G. 60 Page 8 of 12 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60

− − − − − − − − α (ii) We have [hn, k]α = n 1(h 1k 1h)n(h 1k 1h) 1h 1k 1hk , so we can conclude   −  α −  − that [hn, k]α =[n, k ]α((k ) 1(k ) ) 1[h, k]α, where k = h 1kh ∈ K because K  G. 

One can see that if H is an α-normal subgroup of a given group G, then it is invariant under α. Therefore, the restriction of α on H, denoted by α|H , is an automorphism of H and we write H is α-nilpotent instead of α|H -nilpotent.

Theorem 4.4. Assume that M and N are two α-normal subgroups of G for α ∈ Aut(G). Then if both M and N are α-nilpotent, then M N is also α-nilpotent.

α ( ) ={} α ( ) ={} Proof. There exists integers r and s such that r+1 N 1 and s+1 M 1 . α ( ) ={} α =[ ,..., ] : We will prove that r+s+1 MN 1 . Define Kr+s+1 L1 Lr+s+1 α Li = M or N, i = 1,...,r + s + 1 . Then we may prove by induction on (r + s) α ( ) ≤ α + = that r+s+1 MN Kr+s+1.Ifr s 1, then by the previous lemma, we can α( ) =[ , ] ≤[ , ] , [ , ] , [ , ] = α conclude that 2 MN MN MN α M M α M N α N N α K2 . α ( ) ≤ α α ( ) ≤[ ,α ( )] ≤ Now, suppose that r+s MN Kr+s. Then r+s+1 MN MN r+s MN α [ ,α ( )] [ ,α ( )] ≤[ , α ( )] [ , α ( )] ≤ α N r+s MN α M r+s MN α N Kr+s MN α M Kr+s MN α Kr+s+1. Now, we show that [L1, L2,...,Lr+s+1]α ={1}, where Li = M or N, i = 1,...,r +s + 1. If there exist at most s copies of M and at most r copies of N in the above α-commutator subgroup, then we have at most r + sLi ’s which is a contradiction. Hence, there exist at least s + 1 copies of M or at least r + 1 copies of N. Suppose, without loss of generality, + [ , ,..., ] ≤ α ( ) ={} there exist s 1 copies of M, then L1 L2 Lr+s+1 α s+1 M 1 . Hence, α ={ } α ( ) ={ } α  Kr+s+1 1 and so r+s+1 MN 1 and MN is -nilpotent. DEFINITION 4.5

α A subgroup H of group G is called absolute normal if H  G for all automorphisms α ∈ Aut(G).

We recall the definition of the autocommutator subgroup that has been introduced by Hegarty [8]. The autocommutator subgroup of group G is K (G) =g−1gα : g ∈ G,α ∈ Aut(G) . The next theorem shows that K (G) is an absolute normal subgroup of G.

Theorem 4.6. A subgroup H of group G is absolute normal if and only if K (G) ≤ H.

Proof. Assume first that H is absolute normal, then g−1hgα ∈ H for all g ∈ G, h ∈ H and α ∈ Aut(G).Sog−1gα ∈ H and hence K (G) ≤ H. Conversely, if K (G) ≤ H, then g−1hgα = hh−1g−1hgg−1gα = h[h, g]g−1gα. Since [h, g]∈G ≤ K (G) and g−1gα ∈ K (G), then h[h, g]g−1gα ∈ H for all g ∈ G, h ∈ H and α ∈ Aut(G), and thus H is absolute normal. 

If a finite group G possesses a fixed point free automorphism α, then x−1xα = y−1 yα for different elements x and y. Therefore K (G) = G and G does not have a proper absolute normal subgroup. Moreover if G is an abelian group of odd order, then it does not have a proper absolute normal subgroup. One can find a proof in [6], Proposition 2.1, page 139. Now, we deduce absolute normal subgroups of some finite groups. Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60 Page 9 of 12 60

Lemma 4.7. n 2 −1 (i) If D2n =x, y : x = y = 1, yxy = x , then x is the only proper absolute normal subgroup of D2n. =, : 2n = 2 = , = 2n−1−1 , ≥ , (ii) If SD2n+1 x y x y 1 xy yx n 3 then subgroup H  ,  2 , is absolute normal in SD2n+1 if and only if H is one of the four subgroups x x x2, y , x2, yx while n ≥ 3.

Proof. (i) It has been proved in [5]. α ∈ ( ) αs,t = t αs,t = s (ii) Suppose that s,t Aut SD2n+1 such that x x , y yx , where t is an − α − − odd number and s is even, 0 ≤ s, t ≤ 2n − 1. Then x 1x s,t = x 1xt = xt 1 and −1 αs,t = −1 s = s ( ) = 2 y y y yx x and so, K SD2n+1 x . Hence, by Theorem 4.6, subgroups :=  2 :=  H1 x and H2 x are proper absolute normal subgroups of SD2n+1 .Thetwo 2 2 2n−1 subgroups H3 := x , y and H4 := x , yx are isomorphic to D2n =x, y : x = 2 −1 2n−1 2n−2 2 −1 −1 y = 1, yxy = x and Q2n =x, y : x = 1, x = y , yxy = x , respectively, which follows from the relations

n−1 yx2 y−1 = (yxy)2 = (x2 −1)2 = (x2)−1 and

n−2 (yx)x2(yx)−1 = (yxy)2 = (x2)−1,(yx)2 = (x2)2 .

j 2k n−1 Every element of H3 is expressed as y x , where 0 ≤ k ≤ 2 − 1, j = 0, 1. To prove H3 is absolute normal, we should investigate the following cases. − α ( − )+ Case I: g = x j , h = x2k;wehaveg 1hg s,t = x j t 1 2k ∈x2, y , since t is odd. − α ( n−1− )(− )+ + Case II: g = x j , h = yx2k;wehaveg 1hg s,t = yx 2 1 j 2k jt ∈x2, y , since (2n−1 − 1)(− j) + 2k + jt is even as t is odd. − α − + + ( − ) Case III: g = yx j , h = x2k;wehaveg 1hg s,t = x 2k s j t 1 ∈x2, y , since s is even and t is odd. − α (− + )( n−1− )+ + Case IV: g = yx j , h = yx2k;wehaveg 1hg s,t = yx j 2k 2 1 s jt ∈x2, y as above. j 2k H4 is absolute normal, because every element of H4 is expressed as (yx) x , where 0 ≤ k ≤ 2n−1 − 1, j = 0, 1. − α ( − )+ Case I: g = x j , h = x2k;wehaveg 1hg s,t = x j t 1 2k ∈x2, yx . − α ( n−1− )(− )+ + Case II: g = x j , h = (yx)x2k;wehaveg 1hg s,t = (yx)x 2 1 j 2k jt ∈x2, yx . − α − + + ( − ) Case III: g = yx j , h = x2k;wehaveg 1hg s,t = x 2k s j t 1 ∈x2, yx . Case IV: g = yx j , h = (yx)x2k;wehave

− α (− + )( n−1− )+ + +( n−1− ) g 1hg s,t = (yx)x j 2k 2 1 s jt 2 2 ∈x2, yx . 

Lemma 4.7(ii) can be improved towards a classification of proper absolute normal subgroups of all finite 2-groups with maximal cyclic subgroup of index 2. For more details about these groups, see [10] and [11]. 60 Page 10 of 12 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60

Theorem 4.8. Let G be a finite 2-group which contain a maximal cyclic subgroup of index 2. Then these groups have the proper absolute normal subgroups as given by (i) Dihedral 2-group = , : 2n = 2 = , −1 = −1 , ≥ D2n+1 x y x y 1 yxy x n 1 has the only proper absolute normal subgroup x if n ≥ 2. While n = 1, G has no proper absolute normal subgroup. (ii) Semi-dihedral 2-group = , : 2n = 2 = , −1 = 2n−1−1 , ≥ SD2n+1 x y x y 1 yxy x n 3 has the only proper absolute normal subgroups given by x , x2 , x2, y , x2, yx . (iii) Generalized quaternion 2-group = , : 2n = , 2n−1 = 2, −1 = −1 , ≥ Q2n+1 x y x 1 x y yxy x n 3 has the only proper absolute normal subgroup x . (iv) Twisted dihedral 2-group = , : 2n = 2 = , −1 = 2n−1+1 , ≥ S∗ D2n+1 x y x y 1 yxy x n 3 has the only proper absolute normal subgroup x2, y .

Proof. (i) Already covered in Lemma 4.7(i), except for n = 1, which represents the Klein 4-group, which has no proper absolute normal subgroup. (ii) Already covered in Lemma 4.7(ii). i j ≤ ≤ n − = , (iii) Every element of Q2n+1 is expressed as x y , where 0 i 2 1, j 0 1. All n−1 elements of the form xi y are of order 4. There is a unique element y2 = x2 of order α α i, j = i 2 which is central. Any automorphism of Q2n+1 is of the form i, j , where x x , α y i, j = x j y, and where i odd and 0 ≤ i, j ≤ 2n − 1. − α 1 i,2n −1 =  ⊆ ( )  = ( ) Now y y x. Hence, x K Q2n+1 .Tosee x K Q2n+1 , notice that while g = xs,wehave − α ( − ) g 1g i, j = xs i 1 and while g = xs y,wehave − α − ( − )− g 1g i, j = x s i 1 j . [ : ] =  Since Q2n+1 x 2, the only proper absolute normal subgroup of Q2n+1 is x . n+1 (iv) The group S∗ D2n+1 has 2 elements. Every element of S∗ D2n+1 is expressed as xi y j , where 0 ≤ i ≤ 2n − 1, j = 0, 1. The order of xi y is the same as the order of xi .An automorphism α needs to look like

n−1 xα = xi y j , yα = x2 k yl , where i is odd, 0 ≤ i ≤ 2n −1, j, k, l ∈{0, 1} and (k, l) = (0, 0).Now,ifl = 0, then k = 1 α ∈ ( ) ( i j )2n−1 = 2n−1 = α  α, α = α and y Z S∗ D2n+1 .Also, x y x y , which implies x y x is n a subgroup of order 2 , a contradiction. Hence, l = 1 and we denote α = αijk. Verifying

α α α n−1+ α y ijkx ijk = (x ijk)2 1 y ijk, Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60 Page 11 of 12 60 one can see that each such αijk is a legitimate automorphism. Next, we notice that − α − α 1 1,1,1 = 1 3,0,1 = 2  2, ⊆ ( ) x x y, x x x which means x y K S∗ D2n+1 . To prove − α ( ) = 2, 1 ijk K S∗ D2n+1 x y , the various values of g g are computed: Case I: While g = xt , t even,

− α ( − )+ n−1 g 1g ijk = xt i 1 2 j,t ∈x2, y , where  1ifj = 1, t/2 odd,  j,t = 0 otherwise.

Case II: While g = xt , t odd,

− α ( − )+ n−2(( − ) − ) g 1g ijk = xt i 1 2 t 1 i 1 j,t y j ∈x2, y , where  1ift ≥ 3, j = 1,  j,t = 0 otherwise.

Case III: While g = xt y,

− α − − n−1 g 1g ijk = y 1[x t (xi y j )t ](x2 k y) ∈x2, y , since the term in the middle x−t (xi y j )t ∈x2, y is as seen in Cases I and II. ( ) = 2,  2, These prove K S∗ D2n+1 x y . Finally, since x y is an index 2 subgroup of  S∗ D2n+1 , it follows that it would be the only proper absolute normal subgroup.

pn−1 p Theorem 4.9. If p is an odd prime number and Mn(p) =x, y : x = y = 1, xy = pn−2+1 yx , n ≥ 2, then Mn(p) does not possess a proper absolute normal subgroup.

i j n−1 Proof. We can see that Mn(p) ={x y : 0 ≤ i ≤ p − 1, 0 ≤ j ≤ p − 1} and n |Mn(p)|=p . The automorphism group of Mn(p) consists of elements αijk, where α α n−2 − x ijk = xi y j , y ijk = xkp y,0≤ i ≤ pn 1 − 1, i ≡ 0(mod p),0≤ j, k ≤ p − 1. If n = 2, then Mn(p) is abelian and since p is an odd number, it does not have a proper absolute normal subgroup. Assume that n ≥ 3 and put i = 2 and j = 0. Then −1 α , , −1 2 0 −1 2 x x 2 0 k = x x y = x x = x,sox ∈ K (Mn(p)). Now, put i = 1 and j = 1. Then −1 α , , −1 x x 1 1 0 = x xy = y ∈ K (Mn(p)). Hence, K (Mn(p)) = Mn(p) and Mn(p) does not possess a proper absolute normal subgroup. 

Acknowledgements The authors would like to thank the referee for helpful suggestions. This research was supported by a Grant from Ferdowsi University of Mashhad-Graduate Studies (No. 29078). 60 Page 12 of 12 Proc. Indian Acad. Sci. (Math. Sci.) (2018) 128:60

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