The Absolute Center of Finite Groups

J. Group Theory 18 (2015), 887–904 DOI 10.1515/jgth-2015-0015 © de Gruyter 2015

The absolute center of ﬁnite groups

Hangyang Meng and Xiuyun Guo Communicated by Evgenii I. Khukhro

Abstract. In this paper we first investigate the relationship between the absolute center L.G/ and the Frattini subgroup ˆ.G/ for a finite group G, and then we describe the structure of finite groups G satisfying L.G/ ˆ.G/. For example, we prove that a finite group G is an F -group if and only if G=L.G/Ä is an F -group, where F is a saturated formation containing Z2. Next we determine all finite groups X such that X=L.X/ is isomorphic to Zp! Zp, where ! is a positive integer and Zp! is a cyclic group of order p! .

1 Introduction

A group G is said to be capable if there exists a group H such that H=Z.H / is isomorphic to G. The study of capable groups plays a central role in various group- theoretical problems. Many mathematicians, such as Baer, Hall, Senior, Schur, Isaacs and so on, have investigated capable groups [1, 4, 8] and many interesting results have been given. For example, a classical result due to Schur states that if the central quotient G=Z.G/ of a group G is ﬁnite, then the commutator subgroup G0 is also ﬁnite, see [7]. In 1994, Hegarty [5] introduced the autocommutator subgroup K.G/ g 1g˛ g G; ˛ Aut.G/ D h W 2 2 i of a group G and its absolute center

L.G/ g G g˛ g for all ˛ Aut.G/ : D ¹ 2 W D 2 º Furthermore, Hegarty proved an analogue of Schur’s theorem for the absolute center and the autocommutator subgroup of a group, that is, if G is a group such that G=L.G/ is finite, then K.G/ is also finite. Following Chaboksavar, Ghouchan and Saeedi [3], we say that a group G is autocapable if there exists a group M such that M=L.M / is isomorphic to G. In 1997, Hegarty [6] proved that for any finite autocapable group G there are finitely

The research of the work was partially supported by the National Natural Science Foundation of China (11371237) and a grant of “The First-Class Discipline of Universities in Shanghai”. 888 H. Meng and X. Guo many finite groups M such that M=L.M / is isomorphic to G. However, it is not easy to see if a group G is autocapable. In 2014, Chaboksavar, Ghouchan and Saeedi proved that there is no group G such that G=L.G/ is isomorphic to Q8, and there are infinitely many finite and infinite non-autocapable groups. Now the following questions naturally arise.

Can we describe the structure of a finite group G if G=L.G/ is known? Which finite groups X satisfy the equation X=L.X/ G for a given finite Š autocapable group G? In this paper we first investigate the relationship between L.G/ and the Frattini subgroup ˆ.G/ for a finite group G, and then we describe the structure of finite groups G satisfying L.G/ ˆ.G/. For example, we prove that a finite group G Ä is an F -group if and only if G=L.G/ is an F -group, where F is a saturated formation containing Z2 (Corollary 3.8). Next we determine the structure of the absolute center of all finite minimal non-abelian p-groups, and then we determine all finite groups X such that X=L.X/ is isomorphic to Zp! Zp, where ! is ! a positive integer and Zp! is a cyclic group of order p .

2 Preliminary results

In this section we list some basic results, which are frequently used in this paper.

Lemma 2.1. Let H1;:::;Hs be subgroups of a ﬁnite group G such that

G H1 Hs: D Then L.G/ L.H1/ L.Hs/: Ä Moreover, if Hi is characteristic in G for i 1; : : : ; s, then D L.G/ L.H1/ L.Hs/: D Proof. Let ˛i be an automorphism of Hi for i 1; : : : ; s. Then we deﬁne a map D ˛ G G; x .h1; : : : ; hs/ ˛.x/ .h1; : : : ; hi 1; ˛i .hi /; hi 1; : : : ; hs/: W ! D 7! D C It is easy to see that ˛ is an automorphism of G. If x .h1; : : : ; hs/ L.G/, then D 2 it follows from

.h1; : : : ; hs/ x ˛.x/ .h1; : : : ; ˛i .hi /; : : : ; hs/ D D D that ˛i .hi / hi and so hi L.Hi /. Thus L.G/ L.H1/ L.Hs/. D 2 Ä If Hi is characteristic in G for each i, then Aut.G/ Aut.H1/ Aut.Hs/, D which implies that L.H1/ L.Hs/ L.G/. Whence the result follows. Ä The absolute center of ﬁnite groups 889

Lemma 2.2 ([2, Theorem 3.2.]). Let H and K be subgroups of a ﬁnite group G such that G H K. If H and K have no common direct factor, then D Aut.G/ ® ˛ ˇ ˛ Aut.H /; ˇ hom.K; Z.H //; Š ı W 2 2 hom.H; Z.K//; ı Aut.K/¯: 2 2 Lemma 2.3. Let G be a ﬁnite group. Then L.G/ G if and only if G 1 or Z2. D Š Proof. This lemma is a special case of [3, Lemma 2.2].

3 The absolute center and the Frattini subgroup

Recall that a group H is said to be a direct factor of a group G if there exist subgroups K and T of G such that K H and G K T . In this section, we Š D investigate the relationship between L.G/ and the Frattini subgroup ˆ.G/ for a ﬁnite group G. The following lemma is the key lemma in this paper.

Lemma 3.1. Let G be a ﬁnite group. If Z2 is not a direct factor of G, then L.G/ ˆ.G/: Ä Proof. Suppose that the lemma is false. Let G be a group such that Z2 is not a direct factor of G and L.G/ is not contained in ˆ.G/. Then there exists a maximal subgroup M of G such that L.G/ M . Since L.G/ Z.G/, the maximality — Ä of M implies that G ML.G/ and M C G. Hence we assume that G=M p, D j j D where p is a prime number. Now we consider the following two cases.

Case 1: p L.G/ M . Choose z L.G/ T M such that o.z/ p and ﬁx j j \ j 2 D g L.G/ M . It is clear that 2 n p 1 G G Mgi : D i 0 D where “ ” means disjoint union of sets. For any x G, there exist unique m M t 2 2 and i 0; 1; : : : ; p 1 such that x mgi . Deﬁne a map 2 ¹ º D ˛ G G; x mgi mgi zi : W ! D 7! We claim that ˛ is an automorphism of G. In fact, for any two elements x mgi D and y ngj of G, where i; j 0; 1; : : : ; p 1 and m; n M , we see D 2 ¹ º 2 xy mgi ngj mngi j : D D C Since 0 i j < 2p 1 and gp M , we consider two possibilities. Ä C 2 890 H. Meng and X. Guo

If i j p, then C ˛.xy/ ˛..mngp/gi j p/ D C .mngp/gi j pzi j p D C C .mgi zi /.ngj zj / D ˛.x/˛.y/: D If i j < p, then C ˛.xy/ ˛..mn/gi j / D C .mn/gi j zi j D C C .mgi zi /.ngj zj / D ˛.x/˛.y/: D In a word, ˛.xy/ ˛.x/˛.y/. In addition, for any element x. mgi / Ker.˛/, D D 2 in which m M and i 0; 1; : : : ; p 1 , 2 2 ¹ º 1 ˛.x/ ˛.mgi / mgi zi ; D D D which implies i 0 and m 1. Thus Ker.˛/ 1, and so the finiteness of G D D D implies that ˛ is an automorphism of G. Since g L.G/, it follows from the definition of L.G/ and ˛ that 2 g ˛.g/ ˛.1 g1/ gz: D D D Hence z 1, which is a contradiction. D Case 2: p − L.G/ T M . In this case, since j j \ p G=M ML.G/=M L.G/=L.G/ M ; D j j D j j D j j we see that p divides L.G/ and we may choose z L.G/ such that o.z/ p j j 2 D and z M . Hence G M; z M z : … D h i D h i The hypothesis of the lemma implies p > 2. It is clear that for any x G, there 2 exist unique m M and i 0; 1; : : : ; p 1 such that x mzi . Consider the 2 2 ¹ º D map ˇ G G; x. mzi / mz2i : W ! D 7! We claim that the map ˇ is an automorphism of G; the proof is similar to Case 1. Since z L.G/, it is clear that z ˛.z/ ˛.1 z1/ z2, a contradiction. 2 D D D The proof of the lemma is complete. The absolute center of finite groups 891

Remark 3.2. The hypothesis that Z2 is not a direct factor of G in Lemma 3.1 can not be removed. In fact, let a be an element of order 2, let H be a group of odd order and let G a H: D h i Then L.G/ L. a / L.H / a L.H / D h i D h i by Lemmas 2.1 and 2.2, and ˆ.G/ ˆ. a / ˆ.H / ˆ.H /: D h i D Hence L.G/ ˆ.G/. —

Lemma 3.3. Let G be a ﬁnite group such that G En H , where H is a group Š such that Z2 is not a direct factor of H and

En Z2 Z2 Z2: Š „ ƒ‚ … n If n > 1, then L.G/ ˆ.G/. Ä Proof. Lemma 3.1 implies that L.H / ˆ.H / and [3, Lemma 2.1] implies that Ä L.En/ 1. Hence, by Lemma 2.1, we see that D L.G/ L.En/ L.H / 1 L.H / ˆ.En/ ˆ.H / ˆ.G/: Ä D Ä D The lemma follows.

Lemma 3.4. Let G a H be a ﬁnite group, where a is an element of order D h i two and H has no normal subgroup of order two. Then L.G/ a L.H /. D h i Proof. By Lemma 2.1, L.G/ L. a / L.H / a L.H /. So we only need Ä h i D h i to prove a L.H / L.G/. Since H has no normal subgroup of order two, h i Ä a and H have no common direct factor. Using Lemma 2.2, we see that every h i automorphism of G can be expressed as ! ˛ ˇ ; ı where ˛ Aut. a / 1, ˇ hom.H; a /, hom. a ; Z.H //, ı Aut.H /, 2 h i D 2 h i 2 h i 2 which is deﬁned by the rule ! ! ! ! ai ˛ ˇ ai ˛.ai /ˇ.h/ , h ı h D .ai /ı.h/ for all i 0; 1 and h H . D 2 892 H. Meng and X. Guo

By hypothesis, Z.H / has no subgroup of order two. Thus . a / 1 and h i D 0 (where 0 denotes the zero homomorphism). Hence, any Aut.G/ can D 2 be expressed as ! 1 ˇ ; 0 ı where ˇ hom.H; a /; ı Aut.H /. Therefore, for any . ai / a L.H /, 2 h i 2 h 2 h i ! ! ! ! ! ai 1 ˇ ai ai ˇ.h/ ai ˇ.h/ : h D 0 ı h D ı.h/ D h

Noticing that H Ker.ˇ/ or Ker.ˇ/ is a maximal subgroup of H for any element D ˇ hom.H; a /, we see ˆ.H / Ker.ˇ/. By Lemma 3.1, we have 2 h i Ä L.H / ˆ.H / Ker.ˇ/: Ä Ä Thus ˇ.h/ 1 for all ˇ hom.H; a / and h L.H / and therefore D 2 h i 2 a L.H / L.G/: h i Ä The lemma is proved.

From the above lemmas, we obtain the key theorem in this paper.

Theorem 3.5. Let G be a ﬁnite group. The following conclusions are equivalent:

(1) G Z2 H, where H is a group such that Z.H / is odd. Š j j (2) G Z2 H, where H has no normal subgroup of order two. Š (3) L.G/ ˆ.G/. — Proof. It is clear that (1) implies (2) and that (3) follows from (2) by Lemma 3.4. Next we prove that (3) implies (1). If L.G/ ˆ.G/, we may assume G a H — D h i by Lemma 3.1 and Lemma 3.3, where o.a/ 2 and Z2 is not a direct factor of H. D If 2 Z.H / , then we may choose b Z.H / such that o.b/ 2. Deﬁne a map j j j 2 D 0 hom. a ; Z.H //, 2 h i i i 0 a Z.H /; a b ; i 0; 1: W h i ! 7! D

According to Lemma 2.2, there is an automorphism 0 of G such that it can be expressed as ! 1 0 : 0 1 The absolute center of ﬁnite groups 893

We claim that L.G/ H. In fact, for any x . ai / L.G/ with i 0; 1 and Ä D h 2 D h H, we have 2 ! ! ! ! ai ai ai ai 0 i i : h D h D 0.a /h D b h

Thus bi 1 and i 0. It follows that x H and therefore L.G/ H. On the D D 2 Ä other hand, by Lemma 2.1, we have L.G/ a L.H /. Thus Ä h i L.G/ H . a L.H // L.H / .H a / L.H /: Ä \ h i D \ h i D Since H has no direct factor of order two, L.H / ˆ.H / by Lemma 3.1. Hence Ä L.G/ L.H / ˆ.H / ˆ.G/, in contradiction to L.G/ ˆ.G/ . Ä Ä D — Now we give some applications of Theorem 3.5.

Corollary 3.6. If G is a group of odd order, then L.G/ ˆ.G/. Ä Proof. It is clear that the corollary is true by Lemma 3.1.

Corollary 3.7. Let p be a prime number and let G be a ﬁnite p-group except Z2. Then L.G/ ˆ.G/. Ä Proof. If p is an odd prime, then it is clear that L.G/ ˆ.G/ by Lemma 3.1. Ä Now we assume that G is a 2-group. If L.G/ ˆ.G/, then, by Lemma 3.5, we — may assume that G a H such that o.a/ 2 and Z.H / is odd. It follows D h i D j j that Z.H / 1 and therefore H 1. Hence G Z2, a contradiction. D D D Recall that a formation is a class of groups closed under taking homomorphic images and such that if G=M and G=N are in the formation, then so is G=M N . \ Let F be a formation; a group G is called an F -group if G F . Then F is said 2 to be saturated provided the following condition is satisﬁed: If G is a group but not an F -group and M is a minimal normal subgroup of G such that G=M is an F -group, then M has a complement and all such complements are conjugate in G. A famous result states that if F is a saturated formation, then G is an F -group if and only if G=ˆ.G/ is an F -group. The following corollary is an analogue of that result.

Corollary 3.8. Let F be a saturated formation containing Z2. Then a ﬁnite group G is an F -group if and only if G=L.G/ is an F -group.

Proof. It is clear that we only need to prove the sufﬁciency part. Suppose that the sufﬁciency is false and let G be a counterexample of minimal order. 894 H. Meng and X. Guo

If L.G/ ˆ.G/, then it is clear that G=L.G/ F implies that G=ˆ.G/ F . Ä 2 2 Since F is a saturated formation, it follows that G F . Now we assume that 2 L.G/ ˆ.G/. By Theorem 3.5, we may assume that G a H such that — D h i o.a/ 2 and H has no normal subgroup of order two. By Lemma 3.4, we have D L.G/ a L.H /. Thus we have D h i G=L.G/ a L.H /H= a L.H / H=.L.H / a H/ H=L.H /; D h i h i Š h i \ D and so H=L.H / F . By induction, H F . It follows from G=H Z2 F 2 2 Š 2 and G= a H F that G G=.H a / F , a contradiction. h i Š 2 Š \ h i 2

Remark 3.9. The condition “Z2 F ” in Corollary 3.8 cannot be removed. In fact, 2 let F be the class of all groups of odd order. It is clear that F is a saturated formation and Z2 F . We take a group G a H such that o.a/ 2 and H … D h i D is a group of odd order. It is clear that G F but G=L.G/ is a group of odd order. … Recall that a group G is said to be p-solvable group if its upper p-series reaches G:

1 P0.G/ E M0.G/ E P1.G/ E M1.G/ E E P .G/ E M .G/ G; D l l D where

Mi .G/=Pi .G/ Op .G=Pi .G//; Pi .G/=Mi 1.G/ Op.G=Mi 1.G//: D 0 D The minimum number l is called the p-length of G and denoted by lp.G/. The p-length of p-solvable groups is an interesting topic in ﬁnite group theory. The following corollary shows that for p 2 a p-solvable group G and its quotient ¤ group G=L.G/ have the same p-length.

Corollary 3.10. Let G be a p-solvable group, where p is an odd prime. Then lp.G/ lp.G=L.G//. D Proof. It is clear that lp.G/ lp.G=L.G//. If L.G/ ˆ.G/, then G=ˆ.G/ is Ä a factor group of G=L.G/. Thus lp.G=L.G// lp.G=ˆ.G// lp.G/ and so we D get lp.G/ lp.G=L.G//. Now we assume that L.G/ ˆ.G/. By Theorem 3.5, D — we may assume that G a H such that o.a/ 2 and Z2 is not a direct factor D h i D of the group H. By Lemma 3.4, we have L.G/ a L.H /, which implies that D h i G=L.G/ H=L.H / and hence lp.G=L.G// lp.H=L.H //. By Lemma 3.1, Š D L.H / ˆ.H / and thus lp.H / lp.H=L.H //. Furthermore, it follows from Ä D lp.G/ max lp. a /; lp.H / lp.H / (as lp. a / 0) that D ¹ h i º D h i D lp.G=L.G// lp.H=L.H // lp.H / lp.G/: D D D The proof is now complete. The absolute center of ﬁnite groups 895

4 The absolute center of ﬁnite minimal non-abelian p-groups

Recall that a p-group is called a minimal non-abelian p-group if it is a non-abelian group and all its maximal subgroups are abelian. It is well-known that finite minimal non-abelian p-groups play an important role in finite group theory. In this section, we determine the absolute center of finite minimal non-abelian p-groups. First we list the following result due to Redei [9].

Lemma 4.1 ([9]). Let G be a ﬁnite minimal non-abelian p-group. Then G is one of the following groups:

(1) Q8, pn pm pn 1 (2) Mp.n; m/ a; b a b 1; Œa; b a , where n 2, m 1 (metacyclic),D h W D D D i pn pm p (3) Mp.n; m; 1/ a; b; c a b c 1; Œa; b c; Œa; c 1; D h W D D D D D Œb; c 1 , where n m 1, m n > 2 (non-metacyclic). D i C The following equivalent conditions about ﬁnite minimal non-abelian p-groups are always used.

Lemma 4.2 ([9]). Let G be a ﬁnite p-group. Then the following conditions are equivalent: (1) G is a minimal non-abelian p-group. (2) d.G/ 2 and G p. D j 0j D (3) d.G/ 2 and Z.G/ ˆ.G/. D D The following lemma plays an important role in the theory of automorphism groups of ﬁnite groups.

Lemma 4.3. Let G a1; : : : ; an fi .a1; : : : ; an/ 1; i I be a ﬁnite group, D h W D 2 i where fi i I is a set of generated relations of G. Let Gen.G/ a1; : : : ; an ¹ W 2 º D ¹ º be a set of generators. If ' is a map from Gen.G/ to G such that

G '.a1/; : : : ; '.an/ fi .'.a1/; : : : ; '.an// 1; i I ; (4.1) D h W D 2 i then there exists a unique automorphism ' Aut.G/ such that ' '. e 2 ejGen.G/ D For convenience, we use

Gen.G/ '.Gen.G// .a1; '.a1//; : : : ; .an; '.an// D ¹ º to denote the automorphism e' of G determined by the set Gen.G/ and the map ' from Gen.G/ to G with (4.1). 896 H. Meng and X. Guo

Lemma 4.4. Let G a1 an be a ﬁnite abelian p-group, in which D h i h i ni o.ai / p and n1 n : D k Then ´ 1 if p is odd, or p 2 and n1 n2, L.G/ n1 1 D D p D a if p 2, and k 1 or n1 > n2. h 1 i D D Proof. If p is odd or p 2 with n1 n2, then L.G/ 1 by [3, Lemma 2.1]. D D D If p 2 and k 1, then L.G/ Z2 by [3, Lemma 2.1]. Now we assume that D D Š p 2 and n1 > n2. In this case we assume that D s11 s12 s1k sk1 sk2 skk ˛ .a1; a a a /; : : : ; .a ; a a a / D ¹ 1 2 k k 1 2 k º n is an automorphism of G. Lemma 4.3 implies o.˛.a1// 2 1 so that D s s s n1 1 2n1 1s .a 11 a 12 a 1k /2 a 11 1: 1 2 k D 1 ¤ Thus 2 − s11 and so

n 1 n 1 n 1 n1 1 n 1 2 1 2 1 s11 s12 s1k 2 1 2 s11 2 1 ˛.a / .˛.a1// .a a a / a a : 1 D D 1 2 k D 1 D 1 n1 1 Hence a2 L.G/. By [3, Lemma 2.1], the lemma follows. h 1 i Ä n m n 1 Lemma 4.5. Let G a; b ap bp 1; Œa; b ap be a finite group, D h W D D D i where m > n 2 or p 2; n m 2. If D D D ˇ ˇ.i; j; s; t/ .a; ai bj /; .b; asbt / D D ¹ º is an automorphism of G with 1 i; s pn and 1 j; t pm, then Ä Ä Ä Ä .i; p/ 1; t 1 .mod p/; pm n 1 j: (4.2) D Á C j Proof. By Lemma 4.3, ˇ satisfies relation (4.1). If m > n 2, then o.ˇ.a// pn implies that D ´ i j pn ipn jpn j i 1 pn.pn 1/ jpn 1 ijp2n 1.pn 1/ 1 .a b / a b Œb ; a 2 b a 2 ; D i j pn D1 ipn 1 jpn 1 j i 1 pn D1.pn 1 1/ 1 .a b / a b Œb ; a 2 : ¤ D We see p j from m > n and thus Œbj ; ai Œb; aij 1. It follows that j D D ´ pm n j; j .i; p/ 1 or pm n 1 − j: D C n 1 In addition, Œˇ.a/; ˇ.b/ ˇ.a/p implies that D n 1 Œai bj ; asbt .ai bj /p : D The absolute center of finite groups 897

Considering p j , we see j n 1 n 1 n 1 .ai bj /p .a/ip .b/jp D and j n 1 Œai bj ; asbt Œai ; asbt b Œbj ; asbt Œa; bit aitp ; nD1 n 1 D D which implies that aip .t 1/bjp 1. Clearly, pm n 1 divides j and thus D C .i; p/ 1, which implies that t 1 .mod p/. Hence ˇ satisﬁes relation (4.2). D Á If p 2 and m n 2, then o.ˇ.a// 4 implies that D D D D i j 2 2i 2j j i 2.2 1/ 2i.1 j / 2j 1 .a b / a b Œb ; a 2 a b ; ¤ D D and therefore 2 − i or 2 − j . Also o.ˇ.b// 4 implies that 2 − s or 2 − t. It fol- D lows from Œˇ.a/; ˇ.b/ ˇ.a/2 that Œai bj ; asbt .ai bj /2. Notice that D D .ai bj /2 .a/2i .b/2j Œbj ; ai a2i.1 j /b2j D D and j Œai bj ; asbt Œai ; asbt b Œbj ; asbt Œa; bit a2it ; D D D we see a2i.t j 1/b2j 1. Clearly, 2 j and thus 2 − i. Furthermore, 2 t 1. C D j j Hence ˇ satisﬁes relation (4.2).

Lemma 4.6. Let

n m G a; b; c a2 b2 c2 1; Œa; b c; Œa; c 1; Œb; c 1 D h W D D D D D D i be a ﬁnite group, where n > m 1; n 3. If .i; j; k; r; s; t/ .a; ai bj ck/; .b; ar bsct /; .c; c/ D D ¹ º is an automorphism of G, where 1 i; r 2n; 1 j; s 2m; 1 k; t 2, then Ä Ä Ä Ä Ä Ä 2 − i; 2 − s; 2n m r: (4.3) j Proof. By Lemma 4.3, satisﬁes relation (4.1). Since n > m 1, o. .b// 2m D implies that

m m m m m 1 m m m 1 m .ar bsct /2 a2 r b2 sc2 t Œbs; ar 2 .2 1/ a2 r c2 .2 1/rs 1: D D D Then we have 2n m r. Considering Œai bj ck; ar bsct c, we see j D s Œai bj ck; ar bsct Œai bj ; ar bs Œai bj ; bsŒai bj ; ar b D D Œai bj ; bs Œai ; bs cis; D D D which implies that 2 is 1. It is clear that 2 − i and 2 − s. Thus satisﬁes rela- j tion (4.3). 898 H. Meng and X. Guo

If G is a finite group and A is a subset of Aut.G/, then we use CG.A/ to denote the centralizer of A in G, which consists of all elements in G fixed by A. The following theorem shows the structure of the absolute center of finite minimal non-abelian p-groups.

Theorem 4.7. Let G be a ﬁnite minimal non-abelian p-group. (1) If p > 2, then the following hold.

(a) If G Mp.n; m/, n 2, m 1, then D ´ m 1 bp if n < m, L.G/ h i D 1 if n m.

(b) If G Mp.n; m; 1/, n m 1, m n > 2, then D C L.G/ 1: D (2) If p 2, then the following hold. D (a) If G Q8, then Š

L.G/ Z.Q8/ .Q8/ ˆ.Q8/ Z2: D D 0 D Š

(b) If G M2.n; m/, n 2, m 1, then D ´ n 1 m 1 a2 b2 if n < m or n m 2; L.G/ h n 1 i h i D D D a2 if n > m or n m 3: h i D

(c) If G M2.n; m; 1/, n m 1, m n > 2, then D C ´ c if n m or n 2, m 1; L.G/ h i n 1 D D D D a2 c otherwise: h i h i Proof. We prove the results by considering the following cases.

Case 1: p > 2. Since G is a ﬁnite minimal non-abelian p-group and p > 2, the map ˛ G G; x x1 p W ! 7! C is an automorphism of G. For any element x of L.G/, x ˛.x/ x1 p and so D D C xp 1. Hence exp.L.G// p. D D The absolute center of ﬁnite groups 899

2 Case (a): G Mp.n; m/, n 2; m 1. In this case ˛1 .a; a /; .b; b/ D D ¹ º is an automorphism of G and therefore L.G/ CG.˛1/ b . It follows from m 1 Ä D h i exp.G/ p that L.G/ bp . If n m, then we consider the automorphism D pn mÄ h i ˛2 .a; a/; .b; a b/ and so D ¹ º pm 1 pm 1 L.G/ CG.˛2/ b a b 1: Ä \ h i D h i \ h i D If n < m, for any automorphism ˇ .a; ai bj /; .b; asbt / of G, ˇ satisﬁes rela- D ¹ º tion (4.2) by Lemma 4.5. We see

m 1 m 1 m 1 ˇ.bp / .ˇ.b//p .asbt /p D D spm 1 tpm 1 1 stpm 1.pm 1 1/ a b Œb; a 2 : D It follows from m 1 n 2 and t 1 .mod p/ that Á m 1 m 1 ˇ.bp / bp : D m 1 Therefore L.G/ bp and (a) follows. D h i

Case (b): G Mp.n; m; 1/, n m 1, m n > 2. In this case, consider- D C ing the following two automorphisms and their centralizers in G, we have 2 2 ˛3 .a; a /; .b; b/; .c; c / ;CG.˛3/ b ; D ¹ º D h i 2 2 ˛4 .a; a/; .b; b /; .c; c / ;CG.˛4/ a : D ¹ º D h i Then L.G/ CG.˛3/ CG.˛4/ 1 and (b) follows. Ä \ D Case 2: p 2. Noticing that G 2 and G char G, we see G L.G/. D j 0j D 0 0 Ä Case (a): G Q . It is easy to verify (a). D 8

3 3 Case (b): G M2.n; m/, n 2, m 1. In this case ˛5 .a; a /; .b; b / D D ¹ º is an automorphism of G and therefore

2n 1 2m 1 L.G/ CG.˛5/ a b : Ä D h i h i Thus n 1 n 1 m 1 G a2 L.G/ a2 b2 : 0 D h i Ä Ä h i h i 2n m If n > m or n m 3, then ˛6 .a; a/; .b; a b/ is an automorphism D D ¹ º of G and CG.˛6/ a . We see D h i 2n 1 2m 1 2n 1 L.G/ CG.˛6/ . a b / a : Ä \ h i h i D h i n 1 Therefore L.G/ a2 . D h i 900 H. Meng and X. Guo

Now we assume n < m or m n 2. As in Case (1.a), we may have D D m 1 b2 L.G/ h i Ä n 1 m 1 by Lemma 4.5. Therefore, L.G/ a2 b2 and (b) follows. D h i h i Case (c): G M .n; m; 1/, n m 1, m n > 2. In this case it is clear D 2 C that c G L.G/. Consider the following automorphisms and their central- h i D 0 Ä izers in G:

2n m ˛7 .a; a/; .b; a b/; .c; c/ ;CG.˛7/ a c ; D ¹ º D h i h i 3 2n 1 ˛8 .a; a /; .b; b/; .c; c/ ;CG.˛8/ a b c : D ¹ º D h i h i h i 2n 1 Thus c L.G/ CG.˛7/ CG.˛8/ a c . h i Ä Ä \ D h i h i If m n, then ˛9 .a; b/; .b; a/; .c; c/ is an automorphism of G. It follows D D ¹ º that

2n 1 c L.G/ . a c / CG.˛9/ h i Ä Ä h i h i \ n 1 . a2 c / . a2b2 c / c ; D h i h i \ h i h i D h i so L.G/ c . D h i If n > m 1 and n 2, then n 2; m 1 and ˛10 .a; ab/; .b; b/; .c; c/ D D D D ¹ º is an automorphism of G. Thus

2n 1 2n 1 L.G/ . a c / CG.˛10/ . a c / . b c / c Ä h i h i \ D h i h i \ h i h i D h i and therefore L.G/ c . D h i If n > m 1 and n 3, then .i; j; k; r; s; t/ .a; ai bj ck/; .b; ar bsct /; .c; c/ D D ¹ º as an automorphism of G. By Lemma 4.5, satisﬁes relation (4.3). Notice that

n 1 n 1 n 1 .a2 / . .a//2 .ai bj ck/2 D D 2n 1.2n 1 1/ij 2n 1i 2n 1j 2n 1k 2n 1i a b Œb; a 2 c a : D D n 1 n 1 It follows from i 1 .mod 2// that .a2 / a2 . Since is arbitrary, we Á D see n 1 a2 L.G/: h i Ä n 1 Thus L.G/ a2 c and (c) follows. D h i h i The absolute center of ﬁnite groups 901

Remark 4.8. We can ﬁnd many interesting examples using Theorem 4.7. For example,

there exist ﬁnite groups of odd order whose absolute centers are non-trivial, such as Mp.m; n/ with p > 2 and m > n 2, there exists a ﬁnite group G such that ˆ.G/ L.G/ 1, such as M2.2; 2/ D ¤ or Q8,

there exists a ﬁnite group G such that 1 < L.G/ < ˆ.G/, such as M2.n; m; 1/ with n m 1 and m n > 2. C

5 Applications

In this section we hope to extend the result of Chaboksavar, Farrokhi Derakhshan- deh Ghouchan and Saeedi [3]. In fact, we determine all solutions of the equation X=L.X/ G whenever G Zp! Zp, where ! is a positive integer and Zp! Š Š is a cyclic group of order p!.

Theorem 5.1. Let p be a prime number and let ! be a positive integer. Then G is a ﬁnite group such that G=L.G/ Zp! Zp if and only if G is isomorphic to one of the following groups: Š (I) When p is odd,

(1) Zp! Zp, (2) Zp! Zp Z2. (II) When p 2, D (1) Q8 .! 1/, D (2) Z2 Z2 .! 1/, D (3) Z2! 1 Z2, C (4) M2.! 1; 1/, C (5) M2.2; ! 1/, C (6) M2.! 1; 1; 1/ .! 2/, C (7) M2.2; 1; 1/ .! 2/. D Proof. (I) Let p be odd. Since L.G/ Z.G/ and G=L.G/ is abelian, G is nilpo- Ä tent and c.G/ 2. Suppose that G P Q1 Qs, where P is a Sylow Ä D p-subgroup of G and Qi is a Sylow qi -subgroup of G, for i 1; : : : ; s. By D Lemma 2.1, L.G/ L.P / L.Q1/ L.Qs/: D 902 H. Meng and X. Guo

Therefore,

G=L.G/ P=L.P / Q1=L.Q1/ Qs=L.Qs/ Š is a p-group. It follows that L.Qi / Qi for i 1; : : : ; s. By Lemma 2.3, we see D D Qi 1 or Qi Z2. Hence we may assume that G P or G P Z2, where D Š D Š P is a p-group such that P=L.P / Zp! Zp. Š Now we only discuss the structure of the p-group P . By Corollary 3.6, we see L.P / ˆ.P /. It is clear that Ä P=ˆ.P / P=L.P /=ˆ.P /=L.P / P=L.P /=ˆ.P=L.P // Zp Zp: Š D Š Therefore d.P /, the rank of P , is two. Now suppose that P=L.P / a b , D h i h i where a aL.P /; b bL.P / and o.a/ p!; o.b/ p. Clearly, D D D D P a; b; L.P / a; b D h i D h i and thus P Œa; b; P3 , where P3 ŒŒP; P ; P . It follows from c.P / 2 that 0 D h i D Ä P Œa; b . Since bp L.P / Z.P /, we have Œa; bp Œa; bp 1, which 0 D h i 2 Ä D D implies that P 1 or p. j 0j D If P p, then P is a ﬁnite minimal non-abelian p-group by Lemma 4.2. j 0j D Since P=L.P / is abelian, we have P L.P /, which is contradictory to 0 Ä Theorem 4.7 (1). If P 1, then P is an abelian p-group. It follows from Lemma 4.7 that j 0j D L.P / 1. Hence P Zp! Zp. D Š In a word, G is isomorphic to the group (1) or (2).

(II) Let p 2. As in (I), we have that G is a 2-group. The hypotheses of the D theorem imply G Z2. It follows from Corollary 3.7 that L.G/ ˆ.G/. Thus © Ä we have d.G/ 2 and G 1 or 2 as in (I). D j 0j D If G 1, it is obvious that G is of type (4) or (5) .! 1/ by Lemma 4.4. j 0j D D If G 2, then G is a ﬁnite minimal non-abelian 2-group. Using the results j 0j D of Theorem 4.7 (2), we consider the following cases. (a) If G Q8, then G=L.G/ Z2 Z2, which satisﬁes the condition when Š Š ! 1. In this case, G is isomorphic to the group (3). D (b) If G M2.n; m/; n 2; m 1, then we discuss the following cases. D n 1 m 1 If m > n or m n 2, then L.G/ a2 b2 and hence D D D h i h i

G=L.G/ Z2n 1 Z2m 1 : Š We see n 2, m 1 ! .! 1/ and G is isomorphic to the group (7). D D C The absolute center of ﬁnite groups 903

n 1 If n > m or n m 3, then L.G/ a2 and hence D D h i

G=L.G/ Z2n 1 Z2m : Š It is clear that n 1 !, m 1 .! 1/. In this case, G is isomorphic to the D C D group (6).

(c) If G M2.n; m; 1/, n m 1, n m > 2, then we discuss the following D C two cases. If m n or n 2, m 1, at this time, L.G/ c . Therefore D D D D h i G=L.G/ Z2n Z2m : Š This implies that n 1, m 1 .! 1/ or n 2, m 1 .! 2/, which is D D D D D D impossible because of n m > 2. C n 1 If n > m 1 and n 3, then L.G/ a2 c . Therefore D h i h i

G=L.G/ Z2n 1 Z2m : Š Clearly, n 1 !, m 1 .! 2/. Thus G is of type (8) or (9). D C D Conversely, it is easy to verify that a group isomorphic to one of the groups (1)–(9) satisﬁes G=L.G/ Zp! Zp. Š

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Received November 20, 2014; revised February 18, 2015.

Author information Hangyang Meng, Department of Mathematics, Shanghai University, Shanghai 200444, P.R. China. Xiuyun Guo, Department of Mathematics, Shanghai University, Shanghai 200444, P.R. China. E-mail: [email protected]