MINIMAL NUMBER OF GENERATORS AND MINIMUM ORDER OF A NON-ABELIAN GROUP WHOSE ELEMENTS COMMUTE WITH THEIR ENDOMORPHIC IMAGES

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1 Communications in Algebra, 36: , 2008 Copyright Taylor & Francis roup, LLC ISSN: print/ online DOI: / MINIMAL NUMBER OF ENERATORS AND MINIMUM ORDER OF A NON-ABELIAN ROUP WHOSE ELEMENTS COMMUTE WITH THEIR ENDOMORPHIC IMAES A. Abdollahi, A. Faghihi, and A. Mohammadi Hassanabadi Department of Mathematics, University of Isfahan, Isfahan, Iran A group in which every element commutes with its endomorphic images is called an E-group. If p is a prime number, a p-group which is an E-group is called a pe-group. Every abelian group is obviously an E-group. We prove that every 2-generator E-group is abelian and that all 3-generator E-groups are nilpotent of class at most 2. It is also proved that every infinite 3-generator E-group is abelian. We conjecture that every finite 3-generator E-group should be abelian. Moreover, we show that the minimum order of a non-abelian pe-group is p 8 for any odd prime number p and this order is 2 7 for p = 2. Some of these results are proved for a class wider than the class of E-groups. Key Words: Endomorphisms of groups; 2-Engel roups; p-roups; Near-rings Mathematics Subject Classification: 20D45; 20E INTRODUCTION AND RESULTS A group in which each element commutes with its endomorphic images is called an E-group. It is well known (see, e.g., Malone, 1995) that a group is an E-group if and only if the near-ring generated by the endomorphisms of in the near-ring of maps on is a ring. Since in an E-group every element commutes with its image under inner automorphisms, every E-group is a 2-Engel group, and so they are nilpotent of class at most 3 (see Levi, 1942, or Robinson, 1995, Theorem ). Throughout the article, p denotes a prime number. We call an E-group which is also a p-group, a pe-group. Since a finite E-group can be written as a direct product of its Sylow subgroups, and any direct factor of an E-group is an E-group (Malone, 1977), so we need only consider pe-groups. The first examples of non-abelian pe-groups are due to Faudree (1971), which are defined as follows: = a 1 a 2 a 3 a 4 a p2 i = 1 a i a j a k = 1 i j k a 1 a 2 = a p 1 a 1 a 3 = a p 3 a 1 a 4 = a p 4 a 2 a 3 = a p 2 a 2 a 4 = 1 a 3 a 4 = a3 p Received May 8, 2006; Revised September 25, Communicated by M. R. Dixon. Address correspondence to A. Abdollahi, Department of Mathematics, University of Isfahan, Hezar Jerib St., Isfahan , Iran; Fax: ; a.abdollahi@math.ui.ac.ir 1976

2 MINIMAL NUMBER OF ENERATORS 1977 where p is any odd prime number. Note that the above example is false for p = 2. This is because when p = 2 then the map defined by a 1 = a 1 1 a 2a 4 a 2 = a 3 a 3 = a 4 a 4 = a 1a 4 can be extended to an endomorphism of. But a 3 a 3 = a 4 a 3 1, so that is not an E-group. All known examples of non-abelian E-groups have nilpotency class 2 (see Caranti, 1985; Caranti et al., 1987; Faudree, 1971; Malone, 1980). In this article, we see new examples of E-groups. Caranti posed the question (The Kourovka Notebook, 2002, Problem 11.46a) of whether there exists a finite 3E-group of nilpotency class 3. (Note that every E- group without elements of order 3 is nilpotent of class at most 2). Some partial (negative) answers to this question are as follows: finite 3E-groups of exponent dividing 9 are nilpotent of class at most 2 (Malone, 1977); every 2-generator E-group is nilpotent of class at most 2 (since they are 2-Engel). Here we concentrate on the following questions: (1) What is the least number of generators of a finitely generated non-abelian E-group? (2) What is the minimum order of a finite non-abelian pe-group? We prove the following theorem. Theorem 1.1. Every finite 3-generator E-group is nilpotent of class at most 2. Theorem 1.2. most 2. Let be a 3E-group. If 3 10, then is nilpotent of class at As we said, it is easily seen that every 2-generator E-group is nilpotent of class at most 2. In fact a stronger result holds, namely, the following theorem. Theorem 1.3. (i) Every 2-generator group is abelian if and only if it is an E-group. (ii) Every infinite 3-generator group is abelian if and only it is an E-group. Thus, by Theorem 1.3 and Faudree s examples of E-groups, the minimal number of generators of a non-abelian E-group is 3 or 4. We conjecture that this minimal number must be 4, that is every 3-generator E-group is abelian. In response to question (2), we prove the following theorem. Theorem 1.4. For any prime number p, every pe-group of order at most p 6 is abelian. Theorem 1.5. (i) For any odd prime number p, every pe-group of order at most p 7 is abelian. (ii) There exist non-abelian 2E-groups of order 2 7.

3 1978 ABDOLLAHI ET AL. As a result of Theorems 1.4 and 1.5 and Faudree s examples of E-groups, we conclude that the minimum order of a finite non-abelian pe-group is p 8, for any odd prime number p and this order is 2 7 for p = PRELIMINARY DEFINITIONS AND RESULTS Let be a group, H be an element or a subgroup of, and n be a positive integer. We denote by, n, 3, Z, Z 2, exp, Aut, End, C H, Q 8, and C n, respectively, the derived subgroup, the Frattini subgroup, the subgroup generated by n-powers of the elements, the third term of the lower central series, the center, the second center, the exponent, the automorphism group, the set of endomorphisms of, the centralizer of H in, the quaternion group of order 8, and the cyclic group of order n. For a finite p-group and a positive integer n, we denote by n the subgroup generated by elements x such that x pn = 1. If is a nilpotent group, we denote by cl the nilpotency class of. If is a finite group, d will denote the minimum number of generators of. Ifa b, and c are elements of a group, we denote by a b the commutator a 1 b 1 ab and we define a b c = a b c, as usual. An automorphism of a group is called central if x 1 x Z for all x. As we have found that some of our results are valid for a class of finite p-groups larger than pe-groups, we introduce the following class of p-groups for every prime number p. Definition 2.1. A finite p-group is called a pe-group if is a 2-Engel group and there exists a non-negative integer r such that r Z and exp ( ) = p r. Remark 2.2. We know that a finite pe-group is a pe-group (Malone, 1977); but the converse is false in general as one can see that Q 8 and the group = a 1 a 2 a 3 a i a j a k = 1 i j k a 4 1 = 1 a2 1 = a2 2 = a2 3 = a 1 a 2 a 2 a 3 2 = a 1 a 3 2 = 1 are 2E-groups which are not 2E-groups. The quaternion group Q 8 = a b a b = b 1 a 2 = b 2 a 4 = 1 is not an E-group, since a b 1 and the map which sends a to b and b to a, can be extended to an automorphism of Q 8. Lemma 2.3. If is a finite 2-Engel p-group, p>2, and m, then pm = g pm g and is regular. Proof. This follows easily from the following fact which is valid for all a b : a pm b pm = ab pm a b pm p m 1 2 = (ab a b pm 1 2 ) p m Lemma 2.4. Let be a pe-group such that exp ( ) = p r. (i) exp = exp /Z and exp = p r exp.

4 MINIMAL NUMBER OF ENERATORS 1979 (ii) If cl = 2, then exp p r. (iii) If cl = 3, then p = 3 and exp = 3 r+1. Proof. (i) We have exp /Z divides n a n b = 1 a b a b n = 1 a b exp divides n This shows that exp = exp /Z. For the second part of (i), note that if exp = p t, then pr+t pr pt pt = 1. Also if pr+t 1 = 1, then pt 1 r Z and so exp p t 1, which is impossible. It follows that exp = p r+t. (ii) This follows from (i) and the fact that Z. (iii) Note that in every 2-Engel group K, 3 K 3 = 1 (see Robinson (1995, Theorem )). Thus, since cl = 3, we have p = 3 and one can write ( ) 3 r+1 = [ 3r ] 3 [ ] 3 = 3 3 = 1 which gives exp 3 r+1. If exp 3 r, then r Z which is not possible, as cl = 3. Therefore, exp = 3 r+1. Theorem 2.5. Every 2-generator pe-group is either abelian or isomorphic to Q 8. Proof. Suppose that = a b is a pe-group and that exp / = n = p r. Then a n b n = a b and since is a cyclic p-group, we have a n b n = a n or b n. Without lose of generality, we can suppose that b n = a ns for some integer s. Then ba s n = b a sn n 1 /2 (1) which by Lemma 2.4 is trivial if p is odd or if p = 2 and either exp 2 r 1 or 2 s. In any of these cases we would have that ba s is in Z and is abelian. So one can suppose that p = 2, that s is odd and that exp = 2 r. In this case the equality 1 implies that ba s 2n = 1 and since isa2e-group, we have ba s 2 Z. Thus 1 = [ ba s 2 a ] = [ ba s a ] 2 = b a 2 It follows that r 1 and so = 8 and is either abelian or Q 8 or the dihedral group D 8 of order 8. But D 8 is not a 2E-group, since there are elements of order 2 in D 8 which are not central. On the other hand, Q 8 isa2e-group, since exp Q 8 /Q 8 = 2 and the only element of order 2 in Q 8 is central. This completes the proof.

5 1980 ABDOLLAHI ET AL. Lemma 2.6. Let be a pe-group and let exp ( ) = p r. Then Z 2 pr = Z pr. In particular if cl = 3, then exp Z 2 = Z 3r. Proof. By Robinson (1995, Theorem ), we may assume that p = 3. Let x Z 2 3r. By Lemma 2.3, x = y 3r for some y Z 2. Since 3r Z 2, we have x g = [ y 3r g ] = [ y g 3r ] = 1 for all g. This implies that x Z. Now assume that x 3r Z. Then x = y 3r 1 = x g = [ y 3r g ] = y g 3r for some y and so for all g. Then y g r Z which implies that y Z 2. Hence x Z 2 3r. This completes the proof of the first part. If cl = 3 and Z 2 3r 1 Z, then we have 3r = 3 3r 1 Z 2 3r 1 Z (note that since 3 3 = 1, 3 Z 2 ). Thus 3r Z which is impossible by Lemma 2.4(ii). Therefore exp ( Z 2 ) Z = 3 r. Remark 2.7. If is a 2-Engel group, then 3 Z 2. This is because, cl 3 and 3 3 = 1 (see Robinson, 1995, Theorem ). Thus if is a finite 2-Engel 3-group, then we always have that Z 2. Lemma 2.8. Let be a 2-Engel group. If is 2-generator, then cl 2. Z 2 Proof. If = az Z 2 2 bz 2, then = a b Z 2. Thus = x y 3 x y a b Z 2 Since is 2-Engel, we have Z. This completes the proof. Theorem 2.9. Every 3-generator pe-group is nilpotent of class at most 2. Proof. For a contradiction suppose that = x y z is a pe-group of class 3. Suppose that exp / = 3 r. Let H = 3 3. Notice that, by Lemma 2.4, H = H 3r = 1. Modulo H we have that x 3r = x y y z t 1 z x y 3r = x y y z z x t 2 z 3r = x y t 3 y z z x for some integers t 1 t 2 t Since x y z 1, it follows that t i s must all be zero. Now since x 3r y = x y 3r, one can see that =. Similarly one can deduce that = and =. Therefore, modulo H we have x 3r = x y z x y 3r = x y y z z 3r = y z z x

6 MINIMAL NUMBER OF ENERATORS 1981 It follows that x y 3r = x y z z x 3r = x y z y z 3r = x y z Then x 32r = y 32r = z 32r = 1. Now since is regular, it follows that 32r = 1 that contradicts Lemma 2.4. Theorem Let be a non-abelian 3-generator pe-group, exp ( ) = p r, exp = p t, and p>2. Then =p 3 r+t and has the following presentation x y z x p r+t = y pr+t = z pr+t = x pt y = x pt z = y pt x = y pt z = z pt x = z pt y = 1 x y = x pr t 11 y pr t 12 z pr t 13 x z = x pr t 21 y pr t 22 z pr t 23 y z = x pr t 31 y pr t 32 z pr t 33 where 1 t r and t ij L 3 p t. Moreover, every group with the above presentation is a pe-group. Proof. By Theorem 2.9, cl = 2. Since is 3-generator, there exist elements a b c such that = az bz cz. Thus, since has Z exponent p t, we have az = bz =p t and cz =p s for some integers t s 0 with t s. We also have = a b a c b c, since cl = 2 and = a b c Z. Since az =p t and cz =p s, we have a b p t, a c p s and b c p s. Therefore p t+2s. Also, since is regular, r = pr. Then r pr Z and so Z. Hence p 2t+s p t+2s and t s. It follows that s = t, = = p 3t and Z = a b a c b c. Since is not abelian, t 1. Thus C p Z p C p C p. This implies that = a b c. Now, since pr and = Z r = pr, we have = pr. Also we have pr = a pr b pr c pr (since t r). By Lemma 2.4 exp = p r+t and since = pr is an abelian group of order p 3t it follows that pr = a pr b pr c pr, t r and a = b = c =p r+t. Also since pt = a pt b pt c pt and pr = a pr b pr c pr pt, it is not hard to see that pt = a pt b pt c pt and so p 3r = pt r Z = p 3r It follows that pt = r = Z, t = pt = and so = t. Thus we have the following information about : =p 3 r+t exp = p r+t = a b c a = b = c =p r+t Z = r = pt = a pt b pt c pt = t = pr = a pr b pr c pr Hence there exists a 3 3 matrix T = t ij L 3 p t such that a b = a pr t 11 b pr t 12 c pr t 13

7 1982 ABDOLLAHI ET AL. a c = a pr t 21 b pr t 22 c pr t 23 b c = a pr t 31 b pr t 32 c pr t 33 and every element of can be written as a i b j c k for some i j k and ( a i b j c k)( a i b j c k ) = a i+i i jp r t 11 i kp r t 21 j kp r t 31 b j+j i jp r t 12 i kp r t 22 j kp r t 32 c k+k i jp r t 13 i kp r t 23 j kp r t 33 Now consider = p r+t p r+t p r+t on : and define the following binary operation i j k i j k = ( i + i i jp r t 11 i kp r t 21 j kp r t 31 j+ j i jp r t 12 i kp r t 22 j kp r t 32 k+ k i jp r t 13 i kp r t 23 j kp r t 33 ) It is easy to see that with this binary operation is a group and. Now one can easily see that the group has the required presentation. Theorem 2.11 (The Main Result of Morigi, 1995). For p an odd prime, there exists no finite non-abelian 3-generator p-group having an abelian automorphism group. Theorem Let be a non-abelian finite 3-generator pe-group and p > 2. Then exp <exp ( ). Proof. Suppose, for a contradiction, that exp exp. By Lemma 2.4, exp = exp ( ) = p r and by the proof of Theorem 2.10, we have = a b c and Z = r = pr = = a b a c b c a = b = c =p 2r a b = a c = b c =p r If we prove that Aut is abelian, then Theorem 2.11 completes the proof. Let Aut. There exist integers i n m, and an element w such that a = a i b n c m w. Since a a = 1, we have b a n c a m = 1 and so n m 0 (mod p r ). Therefore a = a i w a and similarly b = b j w b, c = c k w c, where 1 i j k p r 1 and w a w b w c = Z. From ab ab = 1 and ac ac = 1, it follows, respectively, that i = j and i = k. Also from equality pr =, we have a pr = a b t b c r a c s where t r, and s are integers. Then a pr = a b t b c r a c s and we obtain a pr i = x pr i 2. Therefore, i 2 i (mod p r ) and so i = 1. Therefore, all automorphisms of are central so that they fix the elements of = Z. If Aut, then x = x for every x a b c. Hence Aut is abelian which contradicts Theorem Corollary Let be a finite 3-generator pe-group and p>2. Ifexp p 2, then is abelian.

8 MINIMAL NUMBER OF ENERATORS 1983 Proof. If is non-abelian, then by Lemma 2.4, we have exp = exp = p which contradicts Theorem Lemma Let be a pe-group such that p 5. Then is abelian or is isomorphic to one of the following groups: Q 8 Q 8 C 2 Q 8 C 2 C 2 x y z x 4 = y 4 = y z = 1 x 2 = z 2 = x y xz 2 = y 2 or x y z x 4 = z 4 = y z = 1 x 2 = y 2 = x y x z = z 2 Proof. It can be easily checked by AP (2005) that there are exactly five non-abelian 2E-groups, which are the same as listed in the lemma. Thus we may assume that p>2 and is non-abelian. If d 4, then Z 1 = p p 4 (since is regular) and so Z p, which is impossible. Hence d = 3 and so p 4 which contradicts Theorem Remark Suppose that is a finite p-group such that 1 Z. If has no nontrivial abelian direct factor, then 1. To see this, let x be of order p and x. Then there exists a maximal subgroup M such that x M. Since x Z, we have x, so that = M x, a contradiction. Theorem Let be a pe-group having no abelian direct factor and p>2. If =p 7, then is abelian. Proof. Suppose, for a contradiction, that is not abelian. If d 4, by Remark 2.15, we have 1 = p p 4 and so p 8 which is impossible. Therefore, by Lemma 2.5, we have d = 3 which is a contradiction by Theorem Lemma Let be an E-group and a be such that a is an infinite direct summand of. Then a Z. Proof. By assumption, we have = a X for some X. Let be the natural epimorphism and a X a the projection map on the first component. Now for each x let x a x be the map defined by a i x i for all i. Since a, x is a group homomorphism mapping a to x. Thus x is an endomorphism of mapping a to x. Since is an E-group, we have that a x = 1. This completes the proof. Theorem Let be an infinite finitely generated E-group. Then = K H, where K is a central torsion-free subgroup of and H is a finite subgroup of. In particular, if is infinite and indecomposable, then is infinite cyclic.

9 1984 ABDOLLAHI ET AL. Proof. Thus Since is infinite and nilpotent, is an infinite finitely generated group. = a 1 a n b 1 b m (*) for some a 1 a n b 1 b m such that a i is infinite, b i is finite and = a 1 a n b 1 b m. By Lemma 2.17, K = a 1 a n Z. It follows that = H, where H = b 1 b m. Thus H is finite and since H is a nilpotent H group, H is finite. Since K Z and we have the decomposition, K is a torsion-free group. It follows that = K H. Now if is indecomposable, we must have H = 1 and n = 1. This completes the proof. Lemma Let be a finite nilpotent p-group of class 3. If Z =p, then Z 2 =p 2. Proof. Suppose = a b c 1 c r where Z = a b Z. By replacing c i by a suitable c i a i b i one can assume that c i a c i b Z for i = 1 r. We claim that c 1 c r Z 2. For this it suffices to show that c i c j Z for 1 i<j r. Suppose with z Z. As c i c j = a b r z 1 = a b c k b c k a c k a b = a b c k this is clear. Hence /Z 2 = a b Z 2 /Z 2 is of order p PROOFS OF THE MAIN RESULTS Proof of Theorem 1.1. Let be a 3-generator E-group. If is finite, since is nilpotent, is finite and is a direct product of its Sylow subgroups. Every Sylow subgroup of is endomorphic image of and so by Malone (1977) they are at most 3-generator E-groups. In this case, Theorem 2.9 completes the proof. If is infinite, then by the fundamental theorem of finitely generated abelian groups, we have = a b c for some a b c such that a has infinite order. Thus by Lemma 2.17, a Z and since is 2-Engel, it follows easily that = b c and since is 2-Engel, 3 = 3 b c = 1. This completes the proof. Proof of Theorem 1.2. Suppose, for a contradiction, that is a finite 3E-group of the least order subject to the properties cl = 3 and Then is indecomposable and so 1, by Remark Thus 1 Z. Ifd 5, then 35. Since is regular and = 3, Z 1 = and since cl = 3, Z. It follows that 3 11, a contradiction. Thus Theorem 1.1 implies that d = 4. If Z =3, then by Lemma 2.19, we have Z 2 =9. Therefore Z 2

10 MINIMAL NUMBER OF ENERATORS 1985 is a 2-generator group and by Lemma 2.8, cl 2, a contradiction. Hence Z 9. Since and we have = Z Z Z = 3 3 Z = Z 3 Z = Z 3 Z 3 Z = Z 3 Z 3 Z 3 10 Thus =3 10, 1 = Z =3 4, Z =9 and 3 Z. Since 3 = 1, 3 = Z 2, is an abelian group of order 3 6 and d = 4. Hence C 27 C 3 C 3 C 3 or C 9 C 9 C 3 C 3 Also we have 9 = = 1 and Lemma 2.4(ii) yields that exp ( ) = 3. Hence by Lemma 2.6, we have Z2 3 = Z. Now Lemma 2.8 implies that d ( Z 2 ) = 3 or 4. Then Z2 =3 6 or 3 7, which implies that Z 2 = or Z 2 =3. If Z 2 =, then Z 2 3 = Z =9, a contradiction. Thus Z 2 =3. Since Z 2 = 1, we have Z Z 2, Z 2 is an abelian group and d Z 2 = 4. Hence Z 2 C 81 C 3 C 3 C 3 or C 27 C 9 C 3 C 3 or C 9 C 9 C 9 C 3 Thus Z 2 3 = Z =27. This contradiction completes the proof. Proof of Theorem 1.3. i) Let be a 2-generator E-group. Suppose first that is finite. Since is nilpotent, is finite and so it is the direct product of its Sylow subgroups. Every Sylow subgroup of is also at most 2-generator and an E-group. Now Theorem 2.5 and Remark 2.2 imply that every Sylow subgroup of is abelian and so is abelian. Therefore, we may assume that is infinite. It follows from the fundamental theorem of finitely generated abelian groups, that = a b for some a b such that a is infinite. Since, = a b and Lemma 2.17 completes the proof of (i). ii) Let be an infinite 3-generator E-group. Since is infinite and nilpotent, is an infinite finitely generated 3-generator group. Thus = a b c for some a b c such that a is infinite and = a b c. By Lemma 2.17, a Z. If either b or c is infinite, then Lemma 2.17 implies that is abelian. Thus we may assume that b and c are both finite. It

11 1986 ABDOLLAHI ET AL. follows that = b c H = b c is finite. Thus = a H, since a Z H is of infinite order and H is a finite group. Hence H is a 2-generator E-group, as it is a direct factor of. Now part (i) completes the proof. Proof of Theorem 1.4. Suppose, for a contradiction, that is a non-abelian pe-group of order p 6. We see that non-abelian groups in Lemma 2.14 are not E-groups and so =p 6.Ifp = 2, then one can see (e.g., by AP, 2005) that there exist ten pe-groups T. We have checked by the package AutPrp in AP (2005), that for each such a group T, there are Aut T and x T such that x x 1 (the automorphism is in a set of generators given by the package for Aut T ); thus they are not E-groups. Therefore, p is odd. Similarly by the proof of Theorem 2.16, we have d = 3 (since has no nontrivial abelian direct factor). Then by Theorem 2.10, we have exp = p 2. Hence by Corollary 2.13, the proof is complete. Proof of Theorem 1.5. i) Suppose, for a contradiction, that is a non-abelian pe-group of least order subject to the property p 7. Then by Theorem 1.4, =p 7. By the choice of and that every direct factor of an E-group is again an E-group, has no abelian direct factor. Now Theorem 2.16 completes the proof. ii) Let = x y z t y 2 = z 2 x y = x t = y 2 x z = z 2 t 2 y z = x 2 y t = z t = y z t = 1 We have = Z = 2 = 1 = g 2 g exp = 4 and =2 7. By Package AutPrp of AP (2005), Aut is abelian and so every automorphism of is central. Then for all Aut and all a, a a = 1. Now we prove that every endomorphism of which is not an automorphism, maps into Z. We denote by Ker and Im the kernel and image of, respectively. We first show that if y 2 Ker, then Im Z. Since y 2 = z 2 = 1, we have y and z Z. Then x 2 = y z = 1 and so x Z. Also t 2 = z 2 t 2 = x z = 1 which implies that t Z. Therefore in this case, Im Z. Thus we can assume y 2 Ker. Since Aut, Ker Z 1. Now it follows from the equality = 1 = Z = g 2 g that there exists an element g such that 1 g 2 Ker. Then g and g. Thus g = x i y j z k t l w, where 0 i j k l 1 w and at least one of the integers i j k l is nonzero. For all a, a g = a g = 1 and so a g Ker. Since t g Ker, y 2i Ker and so i = 0. Also y g Ker, implies that x 2k Ker. Ifk = 1 then x 2 = 1 and so x Z. Therefore, y 2 = x t = 1 which is impossible. Therefore k = 0. Since z g Ker and x g Ker, we have j = l = 0 which is a contradiction and the proof is complete. By a similar proof one can see that the following groups are non-abelian 2E-groups of order 2 7. x y z t y 2 = z 2 t 2 x y = x t = z 2 x z = t 2 y z = x 2

12 MINIMAL NUMBER OF ENERATORS 1987 y t = z t = y z t = 1 x y z t y 2 = z 2 x z = y 2 x t = t 2 y 2 x y = t 2 y z = x 2 t 2 y t = t 2 z t = y z t = 1 These groups are not isomorphic and their automorphism groups are abelian and every endomorphism which is not an automorphism maps the group into its center. ACKNOWLEDMENT The authors are grateful to the referee for his/her valuable comments which caused the article to be shorter. This work was supported partially by the Center of Excellence for Mathematics, University of Isfahan. REFERENCES Caranti, A. (1985). Finite p-groups of exponent p 2 in which each element commutes with its endomorphic images. J. Algebra 97:1 13. Caranti, A., Franciosi, S., de iovanni, F. (1987). Some examples of infinite groups in which each element commutes with its endomorphic images, roup Theoy, Proc. Conf., Brixen/Italy (1986), Lect. Notes Math. Vol. 1281, pp Faudree, R. (1971). roups in which each element commutes with its endomorphic images. Proc. Amer. Math. Soc. 27: Levi, F. W. (1942). roups in which the commutator operation satisfies certain algebraic condition. J. Indian Math. Soc. 6: Malone, J. J. (1977). More on groups in which each element commutes with its endomorphic images. Proc. Amer. Math. Soc. 65: Malone, J. J. (1980). A non-abelian 2-group whose endomorphisms generate a ring, and other examples of E-groups. Proc. Edinburgh Math. Soc. 23: Malone, J. J. (1995). Endomorphism near-rings through the ages, Kluwer Academic Publishers. Math. Appl., Dordr. 336: Morigi, M. (1995). On the minimal number of generators of finite non-abelian p-groups having an abelian automorphism group. Comm. Algebra 23(6): Robinson, D. J. S. (1995). A Course in the Theory of roups. 2nd ed. New York: Springer-Verlag. The AP roup, AP-roups, Algorithms, and Programming. (2005). Version 4.4; ( The Kourovka Notebook, Unsolved Problems in roup Theory. (2002). 15th augm. ed., Novosibirisk. Rossiuiskaya Akademiya Nauk Sibirskoe Otdelenie, Institut Matematiki im. S. L. Soboleva.