Indian J. Pure Appl. Math., 48(3): 423-427, September 2017 °c Indian National Science Academy DOI: 10.1007/s13226-017-0240-9 ON ABSOLUTE CENTRAL AUTOMORPHISMS OF A GROUP FIXING THE CENTER ELEMENTWISE S. Hajizadeh and M. M. Nasrabadi Department of Mathematics, University of Birjand, Birjand, Iran e-mail: s [email protected], [email protected] (Received 16 August 2016; after final revision 25 January 2017; accepted 24 March 2017) Let G be a finite p-group. The automorphism ® of a group G is said to be an absolute central automorphism, if for all x 2 G, x¡1x® 2 L(G), where L(G) is the absolute center of G. In this paper, we obtain a necessary and sufficient condition that each absolute central automor- phism of G fixes the center element-wise. Key words : Autonilpotent group; absolute central automorphism; absolute center of group; purely non-abelian group; autocommutator subgroup. 1. INTRODUCTION AND RESULTS Throughout, p denotes a prime number. Let G be a finite group. We denote by G0, Z(G), Á(G), and Aut(G), respectively, the commutator subgroup, the center, the Frattini subgroup, and the au- tomorphism group of G. For a group H and abelian group K, Hom(H; K) denotes the group of all homomorphisms from H to K. If ® 2 Aut(G) and g 2 G then, [g; ®] = g¡1g® = g¡1®(g) is the autocommutator of g and ®. Clearly for x 2 G, by taking ® = 'x (an inner automorphism) we ¡1 'x ¡1 ¡1 have [g; 'x] = g g = g x gx, which is the ordinary commutator for the element g and x of G. The subgroup K(G) = [G; Aut(G)] = h[g; ®]jg 2 G; ® 2 Aut(G)i is called the autocommuta- tor subgroup of G. (see [3, 4]) We may define the autocommutator of higher weight inductively as follows: £ ¤ [g; ®1; :::; ®n] = [g; ®1; :::; ®n¡1]; ®n , for all ®1; ®2; :::; ®n 2 Aut(G); g 2 G and n ¸ 1. Assume K0(G) = G and K1(G) = K(G). Then for n ¸ 1 we may define: 424 S. HAJIZADEH AND M. M. NASRABADI ­ ® Kn(G) = [Kn¡1(G); Aut(G)] = [g; ®1; ®2; :::; ®n]jg 2 G; ®1; ®2; :::; ®n 2 Aut(G) . One can easily see that γn(G) · Kn(G), n ¸ 1 and Kn(G) is characteristic subgroup of G. Hence we obtain the following descending series of G : G ¶ K1(G) = K(G) ¶ K2(G) ¶ ::: ¶ Kn(G) ¶ ::: The absolute center of G is defined as follows: L(G) = fx 2 G j [x; ®] = 1; for all ® 2 Aut(G)g, which is contained in Z(G), the center th of G. Now assume L1(G) = L(G). Then n -absolute center of G is defined in the following ¡ ¢ way Ln(G)=Ln¡1(G) = L G=Ln¡1(G) for n ¸ 2. Now we recall (from [4]) a group G is an autonilpotent group if Ln(G) = G for some n ¸ 1. Since (from [3]) Ln(G) · Zn(G) so every autonilpotent group is nilpotent. One observe (from [4]) that if Ln(G) = G then, Kn(G) =< 1 >. An automorphism ® called absolute central if [g; ®] 2 L(G) for all g 2 G [3]. We define the subgroup V ar(G) = f® 2 Aut(G)j[g; ®] 2 L(G) for all g 2 Gg which is normal subgroup of Aut(G). For a group G we define [3]: CAut(G)(V ar(G)) = f® 2 Aut(G); ®¯ = ¯® for all ¯ 2 V ar(G)g the centralizer of V ar(G) in Aut(G). We denote by CV ar(G)(Z(G)) the group of all absolute central automorphisms of G fixing Z(G) element-wise and E(G) = [G; CAut(G)(V ar(G))]. One can easily see that E(G) is subgroup of K(G) which is contained in K(G). If G be a group then, E(G) is characteristic subgroup of G and containing G0.(G0 = [G; Inn(G)]) [3]. Lemma 1.1 — Let G be an autonilpotent group. Then for any nontrivial normal subgroup N of G, L(G) \ N 6= 1. PROOF : By induction we obtain Li(G) = fx 2 Gj[x; ®1; :::; ®i] = 1 for all ®1; :::; ®i 2 Aut(G)g. Since G is an autonilpotent group then, for some natural number n, Ln(G) = G. So there exist at least a positive integer i, such that N \ Li(G) 6=< 1 >. Now we have [N \ Li(G); Aut(G)] ⊆ N \ Li¡1(G) =< 1 > and N \ Li(G) ⊆ N \ L(G). Hence N \ L(G) = N \ Li(G) 6=< 1 >. Specially by taken N = L(G) we obtain L(G) 6= 1. 2 The following lemma gives the important property of E(G); while K(G) does not carry over such a property. Lemma 1.2 [3] — Let G be a group. Then V ar(G) acts trivially on the subgroup E(G) of G. ON ABSOLUTE CENTRAL AUTOMORPHISMS OF A GROUP 425 A non-abelian group G is called purely non-abelian if it has no non-trivial abelian direct factor. p For a finite p-group G we define Ω1(G) = hx 2 G j x = 1i. We recall that an automorphism ® is called central automorphism if [g; ®] = g¡1g® 2 Z(G) and ¯ define Autc(G) = f® 2 Aut(G)¯[g; ®] 2 Z(G)g which is a normal subgroup of Aut(G) [1]. Adney and Yen in [1] prove that if G is a purely non-abelian finite group then, there exist a bijection between ¡ ¢ 0 Autc(G) and Hom G=G ;Z(G) . Also, Jamali and mousavi in [2] prove that if G is a finite group ¡ ¢ 0 » 0 such that Z(G) · G then, Autc(G) = Hom G=G ;Z(G) . Similarly we have the following theorems about absolute central automorphisms [3] : Theorem 1.3 — [3]. Let G be a group such that L(G) is contained in E(G). Then: ¡ ¢ V ar(G) =» Hom G=E(G);L(G) : Theorem 1.4 — [3]. Let G be a purely non-abelian finite group. Then: ¡ ¢ V ar(G) =» Hom G; L(G) . The following result gives a description of the centralizer of the center of G in V ar(G). Corollary 1.5 — Let G be a purely non-abelian finite group. Then: ¡ ¢ V ar(G) =» Hom G=E(G);L(G) : PROOF : By Lemma 1.2 and Theorem 1.4. 2 Theorem 1.6 — [3]. Let G be a group. Then ¡ ¢ ¡ ¢ » CV ar(G) Z(G) = Hom G=E(G)Z(G);L(G) . Attar in [5] find necessary and sufficient condition that Autc(G) = CAutc(G)(Z(G)): Similarly in this paper we obtain the necessary and sufficient condition that we have V ar(G) = CV ar(G)(Z(G)). Let G be a non-abelian finite p-group . Then by assumption: G=E(G) = Cpa1 £ ¢ ¢ ¢ £ Cpak ; ai where Cpai is a cyclic group of order p , and a1 ¸ a2 ¸ ¢ ¢ ¢ ¸ ak ¸ 1. Let c c G=E(G)Z(G) = Cpb1 £ ¢ ¢ ¢ £ Cpbl and L(G) = Cp 1 £ ¢ ¢ ¢ £ Cp m 426 S. HAJIZADEH AND M. M. NASRABADI where b1 ¸ b2 ¸ ¢ ¢ ¢ ¸ bl ¸ 1 and c1 ¸ ¢ ¢ ¢ cm ¸ 1. Since G=E(G)Z(G) is a quotient of G=E(G), we have l · k and bi · ai for all 1 · i · l. 2. MAIN RESULT Theorem 2.1 — Let G be a non-abelian finite p-group which is autonilpotent. Then V ar(G) = CV ar(G)(Z(G)) if and only if Z(G) · E(G) or Z(G) · Φ(G), k = l and c1 · bt where t is the largest integer between 1 and k such that at > bt. PROOF : Suppose that V ar(G) = CV ar(G)(Z(G)) and Z(G) 6· E(G). We claim that Z(G) · Φ(G). Assume that Z(G) is not contained in Φ(G). Choose an element g in Z(G) such that g 62 M for some maximal subgroup M of G. Therefore G = Mhgi. k k k Let 1 6= z 2 Ω1(L(G)) \ M (by Lemma 1.1). Then the map ® define on G by ®(mg ) = mg z for every m 2 M and k 2 f0; 1; : : : ; p ¡ 1g, is an absolute central automorphism. By the given hypothesis g = ®(g) = gz, whence z = 1, which is a contradiction. Hence Z(G) · Φ(G). Since ¡ ¢ ¡ ¢ Z(G) · Φ(G), it follows that l = rank G=E(G)Z(G) = rank G=E(G) = k and G is purely ¯ ¯ ¯ ¡ ¢¯ non-abelian. Thus (by Corollary 1.5) we have ¯V ar(G)¯ = ¯Hom G=E(G);L(G) ¯. On the other hand (by Theorem 1.6) we have ¯ ¯ ¯ ¯ ¯ ¡ ¢¯ ¯ ¯ ¯ ¯ ¯ ¯ V ar(G) = CV ar(G)(Z(G)) = Hom G=E(G)Z(G);L(G) ; since V ar(G) = CV ar(G)(Z(G)), therefore ¯ ¡ ¢¯ ¯ ¡ ¢¯ ¯Hom G=E(G)Z(G);L(G) ¯ = ¯Hom G=E(G);L(G) ¯: Hence Y Y pminfai;cj g = pminfbi;cj g: 1·i·k 1·i·l 1·j·m 1·j·m Since ai ¸ bi, for all 1 · i < k, we have minfai; cjg ¸ minfbi; cjg for all 1 · i · k, 1 · j · m. Hence minfai; cjg = minfbi; cjg for all 1 · i · k, 1 · j · m. Since Z(G) 6· E(G), there exists some 1 · i · k such that ai > bi. Let t be the largest integer between 1 and k such that at > bt. We claim that c1 · bt. Suppose that c1 > bt. Thus bt = minfc1; btg = minfc1; atg, which is impossible.