Houston Journal of Mathematics. c 2007 University of Houston Volume 33, No. 1, 2007

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1 Houston Journal of Mathematics c 2007 University of Houston Volume 33, No. 1, 2007 ROUPS WITH SPECIFIC NUMBER OF CENTRALIZERS A. ABDOLLAHI, S.M. JAFARIAN AMIRI AND A. MOHAMMADI HASSANABADI Communicated by Jutta Hausen Abstract. Let be a group and let cent() denote the set of centralizers of single elements of. A group is called n-centralizer if cent() = n. In this paper, for a finite group, we give some interesting relations between cent() and the maximum number of the pairwise non-commuting elements in. Also we characterize all n-centralizer finite groups for n = 7 and 8. Using these results we prove that there is no finite group with the property that cent() = cent( ) = 8, where Z() denotes the centre of. This Z() latter result answers positively a conjecture posed by A. R. Ashrafi. 1. Introduction and results Throughout we will use the usual notation, for example C n denotes the cyclic group of order n; S n and A n denote respectively symmetric and alternating groups on n letters and D 2n stands for dihedral group of order 2n. Let be a group. We denote by cent() = {C (g) g }, where C (g) is the centralizer of the element g in. Let n > 0 be an integer. A group is called n-centralizer if cent() = n and is called primitive n-centralizer if cent( Z() ) = cent() = n, where Z() denotes the centre of. A subgroup H of is called a proper centralizer of if H = C (x) for some x \Z(). It is clear that a group is 1-centralizer if and only if it is abelian. Belcastro and Sherman have the following results in [5]: (i) There is no 2-centralizer and no 3-centralizer group. (ii) A finite group is 4-centralizer if and only if Z() = C 2 C Mathematics Subject Classification. 20F99; 20D60. Key words and phrases. Covering groups by subgroups; n-centralizer groups. This research was in part supported by a grant from IPM (No ). This work was in part supported by the Center of Excellence of University of Isfahan for Mathematics. 43

2 44 A. ABDOLLAHI, S.M. JAFARIAN AMIRI, AND A.M. HASSANABADI (iii) A finite group is 5-centralizer if and only if Z() = C 3 C 3 or S 3. Using these results, it is easy to see that there is no finite primitive 4-centralizer group and a finite group is primitive 5-centralizer if and only if Z() = S 3. Ashrafi in [2, 3] has shown that if is a finite 6-centralizer group, then Z() = D 8, A 4, C 2 C 2 C 2 or C 2 C 2 C 2 C 2. Our main results are: Theorem A. Let be a finite group. Then is a 7-centralizer group if and only if Z() = C 5 C 5, D 10 or x, y x 5 = y 4 = 1, y 1 xy = x 3. Theorem B. Let be a finite 8-centralizer group. Then Z() = C 2 C 2 C 2, A 4 or D 12. Using Theorem B, we have confirmed Conjecture 2.3 of [2], namely: Theorem C. There is no finite primitive 8-centralizer group centralizer groups In this section, we give the proof of Theorem A. A cover Γ for a group is a collection of proper subgroups whose union is the whole. We use the term n-cover for a cover with n members. A cover is called irredundant if no proper subcollection is also a cover. A cover is called a partition with kernel K if the intersection of pairwise members of the cover is K. B. H. Neumann in [11] obtained a uniform bound for the index of the intersection of an irredundant n-cover in terms of n and Tomkinson [13] improved this bound. For a natural number n, let f(n) denote the largest index : D, where is a group with an irredundant n-cover whose intersection is D. We know that f(3) = 4, f(4) = 9, f(5) = 16 and f(6) = 36 (see [12], [9], [6] and [1], respectively). We will use these results in the sequel. We shall use the following lemma of Tomkinson repeatedly. Lemma 2.1. (Lemma 3.3 of [13]) Let M be a proper subgroup of the (finite) group and let H 1,..., H k be subgroups of with : H i = β i and β 1 β k. If = M H 1 H k, then β 1 k. Furthermore, if β 1 = k, then β 1 = β 2 = = β k = k and H i H j M for any two distinct i and j. Now we have

3 ROUPS WITH SPECIFIC NUMBER OF CENTRALIZERS 45 Proposition 2.2. Let n 2 be an integer and let be a finite group such that Z() = D 2n. Then cent() = n + 2. Proof. For an element x in, we write x for xz(). By the hypothesis there exist elements r, s such that = Z() = r, s r n = s 2 = 1, r s = s r n 1. Since r C (r)\z() and r is a maximal subgroup of, C (r) Z() = C (r) = r. If n is even, then C (r i s) = r i s r n 2. If C (r i s) Z() = C (r i s), then r n 2 C (r i s) and so r i s C (r n 2 ) = C (r), a contradiction. Therefore C (r i s) Z() = r i s = ri s Z() Z(). Thus cent() = {, C (r), C (r i s) : 1 i n} and so cent() = n + 2. If n is odd, then C (r i s) = r i s = ri s Z() Z(). Hence cent() = {, C (r), C (r i s) : 1 i n} and the proof is complete. Corollary 2.3. Let be a finite group and Z() = D 2n. Then is primitive n-centralizer if and only if n > 1 is an odd integer. Proof. Note that if n > 1 is an integer, then cent(d 2n ) = { n if n is even n + 2 if n is odd, (see for example; [2, Lemma 2.4] or [3, Lemma 2.4]). Now the proof follows from Proposition 2.2. Definition 2.1. A non-empty subset X = {x 1,..., x r } of a finite group is called a set of pairwise non-commuting elements if x i x j x j x i for all distinct i, j {1,..., r}. A set of pairwise non-commuting elements of is said to have maximal size if its cardinality is the largest one among all such sets. Remark 2.1. Let be a finite group and {x 1,..., x r } be a set of pairwise noncommuting elements of having maximal size. Then (1) {C (x i ) i = 1,..., r} is an irredundant r-cover with the intersection Z() = r i=1 C (x i ) (see Theorem 5.1 of [13]). (2) Z() f(r) (see Corollary 5.2 of [13]). (3) f(3) = 4, f(4) = 9, f(5) = 16 and f(6) = 36 (see [12], [9], [6] and [1], respectively).

4 46 A. ABDOLLAHI, S.M. JAFARIAN AMIRI, AND A.M. HASSANABADI (4) Let be a group such that every proper centralizer in is abelian. Then for all a, b \Z() either C (a) = C (b) or C (a) C (b) = Z(). For, if z ( C (a) C (b) ) \Z(), then C (z) contains both C (a) and C (b), since C (a) and C (b) are abelian. Since z is not in Z(), C (z) C (a) and C (z) C (b). Thus C (z) = C (a) = C (b). Hence, in such a group, {C (x) x \Z()} forms a partition with kernel Z(). It follows that { C (x) Z() x \Z()} forms a partition whose kernel is the trivial subgroup (see also Proposition 1.2 of [10]). Lemma 2.4. Let be a finite non-abelian group and {x 1,..., x r } be a set of pairwise non-commuting elements of with maximal size. Then (1) r 3. (2) r + 1 cent(). (3) r = 3 if and only if cent() = 4. (4) r = 4 if and only if cent() = 5. Proof. (1) Since is not abelian, there exist elements x, y such that xy yx. Thus {x, y, xy} is a set of pairwise non-commuting elements of, and so r is at least 3, as required. (2) It is clear, since C (x i )s are distinct for 1 i r. (3) Suppose that r = 3. Then Z() f(3) = 4 by Remark 2.1(1), (2) and (3). Since is not abelian, Z() = C 2 C 2. It now follows from Theorem 2 of [5] that cent() = 4. Conversely, suppose that cent() = 4. Then by part (2), r 3 and part (1) forces r to be 3. (4) Suppose that r = 4. Then Z() f(4) = 9, by Remark 2.1 (2) and (3). Assume that : C (x i ) = α i for 1 i r and α 1 α r. We claim that is not a 2-group. Assume, on the contrary, that is a 2-group. Z() Z() Then Z() = 4 or 8 and so C (x) is abelian for each (x) Z(C (x)) x \Z(), this is because C (x) C Z() = 2 or 4 and is cyclic. By Remark 2.1(4), we have Z() = C (x) C (y) for distinct proper centralizers C (x) and C (y). We have α 2 3 by Lemma 2.1 and so α 2 = 2. It follows that Z() = C 2 C 2. Thus cent() = 4 by Theorem 2 of [5], which is a contradiction. Hence = 6 or 9 from which it Z() follows that Z() = S 3 or C 3 C 3. Now Theorem 5 of [5] implies that cent() = 5.

5 ROUPS WITH SPECIFIC NUMBER OF CENTRALIZERS 47 Conversely, if cent() = 5, then r 4 by part (1). On the other hand r 3 by part (3). Thus r = 4, and the proof is complete. Proposition 2.5. Let be a finite group and let X = {x 1,..., x r } be a set of pairwise non-commuting elements of having maximal size. (a) If cent() < r + 4, then (1) For each element x, C (x) is abelian if and only if C (x) = C (x i ) for some i {1,..., r}. (2) If C (x i ) is a maximal subgroup of for some i {1,..., r}, then Z() = C (x i ) C (x j ) for all j {1,..., r}\{i}. In particular, if : C (x 1 ) : C (x 2 ) 2, then cent() = 4, and if : C (x 1 ) : C (x 2 ) = 3, then cent() = 5. (b) If cent() = r + 2, then there exists a proper non-abelian centralizer C (x) which contains C (x i1 ), C (x i2 ) and C (x i3 ) for three distinct i 1, i 2, i 3 {1,..., r}. (c) If cent() = r+3, then there exists a proper non-abelian centralizer C (x) which contains C (x i1 ) and C (x i2 ) for two distinct i 1, i 2 {1,..., r}. Proof. (a)(1) Suppose, for a contradiction, that there exists an index i such that K := C (x i ) is non-abelian. Then cent(c (x i )) 4 and so C (x i ) contains at least three proper centralizers, say C K (y i ), i = 1, 2, 3. By the hypothesis C (x t ) C (y j ) for every t {1,..., r} and j {1, 2, 3}. Therefore cent() r + 4, a contradiction. Conversely, suppose that C (x) is abelian. Since {x 1,..., x r } is a set of pairwise non-commuting elements of of maximal size, there exists an index j such that x C (x j ). By the previous part, C (x j ) is abelian which implies that C (x j ) C (x). Since x j is in C (x), we have C (x) C (x j ), which completes the proof. (a)(2) Let x C (x i ) C (x j ). By part (a)(1), C (x i ) is abelian and so C (x i ) C (x). Since x j C (x) and x i x j x j x i, C (x i ) C (x). Now since C (x i ) is a maximal subgroup of, C (x) =, hence x Z(). If : C (x 1 ) : C (x 2 ) 2, then by the hypothesis : C (x 1 ) = : C (x 2 ) = 2. Now by the previous part we have Z() = C 2 C 2, and so by Theorem 2 of [5], cent() = 4. If : C (x 1 ) : C (x 2 ) = 3, then C (x 2 ) is a maximal subgroup of and Z() = C (x 1 ) C (x 2 ), by the first part of (a)(2). Thus Z() 9. Now since 3 divides Z(), we have Z() = C 3 C 3 or S 3 and the result holds by Theorem 4 of [5]. (b) By the hypothesis and part (a)(1), there exists a proper centralizer K := C (x) which is not abelian. Therefore cent(k) 4. Assume that {C K (y i ) i =

6 48 A. ABDOLLAHI, S.M. JAFARIAN AMIRI, AND A.M. HASSANABADI 1, 2, 3} is a set of three proper centralizers of K. Then {C (y i ) i = 1, 2, 3} is a set of three proper centralizers of. Since cent() = r + 2, C (y i ) = C (x ji ) for distinct j 1, j 2, j 3 {1,..., r} and so for i = 1, 2, 3; C (x ji ) is contained in C (x) by part (a)(1). (c) By the hypothesis and part (a)(1), there exists a proper centralizer H := C (x) which is not abelian. Therefore cent(h) 4. Assume that {C H (w i ) i = 1, 2, 3} is a set of three proper centralizers of H. Then {C (w i ) i = 1, 2, 3} is a set of three proper centralizers of. Since cent() = r + 3, we may assume that C (w i ) = C (x ji ) for distinct j 1, j 2 {1,..., r}. Then C (w i ) is abelian for i = 1, 2 by part (a)(1). But x C (w i ) for i = 1, 2 and so C (x ji ) = C (w i ) is contained in C (x) for i = 1, 2. This completes the proof. Lemma 2.6. Let be a finite non-abelian group. Then every proper centralizer of is abelian if and only if cent() = r + 1, where r is the maximal size of a set of pairwise non-commuting elements of. Proof. Suppose that every proper centralizer of is abelian and let X = {x 1,..., x r } be a set of pairwise non-commuting elements having maximal size. Consider a proper centralizer C (x) of. Then there exists an i {1,..., r} such that x C (x i ) by the maximality of X. Since C (x i ) is abelian, C (x i ) C (x). On the other hand x i C (x) gives C (x) C (x i ). Therefore C (x) = C (x i ) and so cent() = {, C (x i ) 1 i r}. This completes the proof in one direction, since for any two distinct i and j, x i x j x j x i. Conversely suppose that cent() = r + 1, where r is the maximal size of a set of pairwise non-commuting elements of. Assume that X = {x 1,..., x r } is a set of pairwise non-commuting elements of. Then Proposition 2.5(a)(1) implies that for each i {1,..., r}, C (x i ) is abelian, which completes the proof. Lemma 2.7. Let be a finite 7-centralizer group. Then Z() is not a 2-group. Proof. Suppose, on the contrary, that Z() is a 2-group. Suppose that {x 1,..., x r } is a set of pairwise non-commuting elements of having maximal size such that : C (x i ) = α i with α 1 α r. Then {C (x i ) i = 1,..., r} is an irredundant cover for with intersection Z(). It follows from Lemma 2.4 that r = 5 or 6. Suppose that r = 5. Then α 2 4 by Lemma 2.1. It follows from Proposition 2.5(a)(2) that α 2 = 4 and Lemma 2.1 yields that α i = 4 for i 2. Now Proposition 2.5(b) implies that there exists a proper centralizer C (x) which is not abelian and contains at least three C (x i )s, say for i = 2, 3, 4. Therefore

7 ROUPS WITH SPECIFIC NUMBER OF CENTRALIZERS 49 = C (x 1 ) C (x 5 ) C (x) and so by Lemma 2.1, α 5 = 2 which is a contradiction. Now suppose that r = 6. Then { C (x i ) Z() i = 1,..., 6} is a partition with trivial kernel by Lemma 2.6 and Remark 2.1(4). Assume that C (x i ) Z() = n i for 1 i 6. Then 6 i=1 (n i 1) = Z() 1 and so 6 i=1 n i is an odd integer which is a contradiction, since all n i s are even. We recall that a group is called capable if there exists a group H such that = H Z(H). Lemma 2.8. The groups x, y x 5 = y 4 = 1, y 1 xy = x 1, C 3 D 10, S 3 C 5 and x, y x 6 = 1, y 2 = x 3, y 1 xy = x 1 are not capable. Proof. Suppose, for a contradiction, that there exists a group H such that H Z(H) = L := x, y x 5 = y 4 = 1, y 1 xy = x 1. Then there exist elements h 1, h 2 H such that h 1 Z(H) = 4 and h 2 Z(H) = 10 with h 2 1Z(H) = h 5 2Z(H). Therefore C H (h 1 ) and C H (h 2 ) are subgroups of C H (h 2 1) and so the lcm(4, 10) divides C H (h 2 1 ) Z(H). This forces h2 1 to be in Z(H), which is a contradiction. Now assume, on the contrary, that there exists a group such that Z() = D 10 C 3. Then there exist two elements x and y in such that xz() = 15 and yz() = 6 with x 5 Z() = y 2 Z(). It follows that C (x 5 ) contains both of C (x) and C (y). Thus C (x 5 ) =, so that x 5 Z(), which is not possible. K Next suppose, for a contradiction, that there exists a group K such that Z(K) = S 3 C 5. Thus there exist elements x, y K such that xz(k) = 15 and yz(k) = 10 with x 3 Z(K) = y 2 Z(K). Therefore C (x 3 ) contains both C (x) and C (y). It follows that C (x 3 ) =, a contradiction. Lastly suppose, on the contrary, that there exists a group L such that L Z(L) = x, y x 6 = 1, y 2 = x 3, y 1 xy = x 1. Then there exist elements l 1, l 2 L such that l 1 Z(L) = 4 and l 2 Z(L) = 6 with l1z(l) 2 = l2z(l). 3 Therefore C L (l 1 ) and C L (l 2 ) are subgroups of C L (l1) 2 and so lcm(4, 6) divides C L(l 2 1 ) Z(L). This implies that l2 1 Z(L), which cannot happen. Remark 2.2. Let A be a finite abelian group. Let p be a prime number and i > 0 be an integer. Suppose that r(a, p i ) is the number of cyclic direct summands of order p i in the decomposition of A into cyclic groups of prime power orders.

8 50 A. ABDOLLAHI, S.M. JAFARIAN AMIRI, AND A.M. HASSANABADI Baer in [4] showed that A is capable if and only if for every prime number p, r(a, p i ) = 1 implies that A contains elements of order p i+1. Lemma 2.9. (1) The only capable groups of order 12 are D 12 and A 4. (2) The only capablegroups oforder 20 are D 20 and x, y x 5 = y 4 = 1, y 1 xy = x 3. (3) The only capable group of order 30 is D 30. Proof. Firstly we note that, if n 2 is an integer, then D 2n is a capable group D (since 4n Z(D 4n ) = D 2n ). Also clearly every centreless group is capable. (1) There are exactly four non-cyclic groups of order 12, namely C 2 C 2 C 3, D 12, A 4, and x, y x 6 = 1, y 2 = x 3, y 1 xy = x 1. It follows from Remark 2.2 and Lemma 2.8 that D 12 and A 4 are the only non-cyclic capable groups of order 12. (2) There are exactly four non-cyclic groups of order 20, namely C 2 C 2 C 5, D 20, T := x, y x 5 = y 4 = 1, y 1 xy = x 3 and x, y x 5 = y 4 = 1, y 1 xy = x 1. Since Z(T ) = 1, it follows from Remark 2.2 and Lemma 2.8 that D 20 and x, y x 5 = y 4 = 1, y 1 xy = x 3 are the only non-cyclic capable groups of order 20. (3) There are exactly three non-cyclic groups of order 30, namely D 30, C 3 D 10 and S 3 C 5. It follows from Lemma 2.8 that D 30 is the only capable group of order 30. Proof of Theorem A. Suppose that is a 7-centralizer group and {x 1,..., x r } is a set of pairwise non-commuting elements of having maximal size. Let : C (x i ) = α i such that α 1 α r. Then {C (x i ) i = 1,..., r} is an irredundant r-cover for with intersection Z() by Remark 2.1(1). It follows from Lemma 2.4 that r = 5 or 6. If r = 5, then α 2 4 by Lemma 2.1. If α 2 3, then cent() = 4 or 5 by Proposition 2.5(a)(2). Therefore α 2 = 4 and so α i = 4 for i 2 by Lemma 2.1. Now Proposition 2.5(b) implies that there exists a proper centralizer C (x) which is not abelian and contains i S C (x i ) for some S {1,..., 5} with S = 3. Thus = ( i T C (x i )) C (x), where T = {1,..., 5}\S. Now Lemma 2.1 implies that α i = 2 for some 2 i 5, which is a contradiction. Therefore r = 6. Then since cent() = 7, by Proposition 2.5(a)(1) all proper centralizers of are abelian. It follows from Remark 2.1(4) that {C (x i ) : 1 i 6} is a partition with kernel Z() for. Now it follows from Lemma 2.1 that α 2 5. Also Proposition 2.5(a)(2) implies that α 2 2, 3. If α 2 = 4, then

9 ROUPS WITH SPECIFIC NUMBER OF CENTRALIZERS 51 4 divides Z() and Z() 16. Now Lemma 2.7 implies that Z() = 12 and so Z() = D 12 or A 4 by Lemma 2.9. But this is impossible, since Proposition 2.2 and Theorem 3.6 of [2] imply that cent() = 6 or 8. Therefore α 2 = 5. If α 1 = 2, then Z() = D 10, as required. If α 1 = 3, then Z() is of order 15 and so it is cyclic. Thus, in this case cent() = 1, which is not possible. If α 1 = 4, then Z() = 20 and Lemma 2.9 implies that If Z() = x, y x 5 = y 4 = 1, y 1 xy = x 3 or D 20. Z() = D 20, then cent() = 12, a contradiction. Hence Z() = x, y x 5 = y 4 = 1, y 1 xy = x 3, as required. Now assume that α 1 = 5. Since α 2 = 5, by Lemma 2.1 we have α 1 = α 2 = = α 6 = 5. It follows that Z() = 25 and so Z() = C 5 C 5, as required. This completes the only if part of the theorem. If Z() = D 10 or C 5 C 5, then cent() = 7 by Theorem 5 of [5]. Suppose that Z() = x, y x 5 = y 4 = 1, y 1 xy = x 3. Then gz() {1, 2, 4, 5} for each g. If gz() = 4 or 5, then C (g) = g Z() and if gz() = 2, then there exists g 1 such that g 1 Z() = 4 and g = g1z 2 for some z Z(). So C (g) = C (g1) 2 = C (g 1 ). On the other hand, has five Sylow 2-subgroups and one Sylow 5-subgroup which are all cyclic. Thus cent() = 7 and the proof is complete centralizer groups In this section we prove Theorems B and C. Lemma 3.1. Let be a finite 8-centralizer group. Then Z() is a {2, 3}-group. Proof. Let {x 1,..., x r } be a set of pairwise non-commuting elements of having maximal size. Then {C (x i ) i = 1,..., r} is an irredundant r-cover with intersection Z() by Remark 2.1(1). It follows from Lemma 2.4 that r = 5, 6 or

10 52 A. ABDOLLAHI, S.M. JAFARIAN AMIRI, AND A.M. HASSANABADI 7. Assume that : C (x i ) = α i, where α 1 α r and consider a p-element x such that p is a prime number greater than 5. Then x Z() by Lemma 2.1 of [6] and so Z() is a {2, 3, 5}-group. By Lemma 2.1, we have α 2 6. If r 6, then Z() 36 by Remark 2.1(3). If r = 7 then, by Remark 2.1(4), C (x 1 ) C (x 2 ) = Z() which implies that Z() 36. If 5 is a divisor of Z(), then Z() {10, 15, 20, 25, 30, 35}. Since every group of order 15 or 35 is cyclic and Z() is not cyclic, Z() {10, 20, 25, 30}. If Z() = 10 or 25, then Z() = D 10 or C 5 C 5 and so cent() = 7 by Theorem A, a contradiction. If = 20 or 30, then Lemma 2.9 implies that Z() Z() = x, y x 5 = y 4 = 1, y 1 xy = x 3, D 20 or D 30. Therefore cent() = 7, 12 or 17 by Theorem A and Proposition 2.2, which is impossible. This completes the proof. Lemma 3.2. Let be a finite 8-centralizer group. If Z() = C 2 C 2 C 2. Z() is a 2-group, then Proof. Suppose that X = {x 1,..., x r } is a set of pairwise non-commuting elements of having maximal size. Assume that : C (x i ) = α i such that α 1 α r. Then {C (x i ) i = 1,..., r} is an irredundant cover for with intersection Z() by Remark 2.1(1). It follows from Lemma 2.4 that r 3 or 4. If r = 5, then α 2 4. But α 2 2 or 3 by Proposition 2.5(a)(2) and so α i = 4 for i 2 by Lemma 2.1. Therefore by Proposition 2.5(c), there exists a proper centralizer K := C (x) which is not abelian and contains at least two C (x i )s, say for i = 1, 2. Thus = C (x 3 ) C (x 4 ) C (x 5 ) C (x) which implies that α i 3 for some i {3, 4, 5} by Lemma 2.1, a contradiction. If r = 6, then α 2 = 4 by Proposition 2.5(a)(2) and Lemma 3.1. Thus there exists a proper centralizer C (x) which is not abelian and by Proposition 2.5 (b) it contains at least three C (x i )s, say for i = 1, 2, 3. Therefore = C (x 4 ) C (x 5 ) C (x 6 ) C (x) which implies that α i 3 for at least one i {4, 5, 6}. If α 1 = 2, then Proposition 2.5(a)(2) implies that Z() 8. This implies that C (x) Z() 4 and so C (x) is abelian, a contradiction. Thus α i 4 for every i {1,..., 6}, which is impossible. Finally suppose that r = 7. Then C (x i ) is abelian for each i {1,..., 7} by Proposition 2.5(a)(1) and so C (x i ) C (x j ) = Z() by Remark 2.1 (4) for any two distinct i, j {1,..., 7}. It follows from Lemma 2.1 that α 2 = 4 and so Z() α 1α If Z() = 8, then Z() = C 2 C 2 C 2, since other groups

11 ROUPS WITH SPECIFIC NUMBER OF CENTRALIZERS 53 of order 8 contain an element of order 4 and so they cannot have an irredundant 7-cover. If Z() = 16, then α 1 = 4 by Proposition 2.5(a)(2). If C (x i ) is not a normal subgroup of for i = 1 or 2, then Z() = core (C (x i )) and so Z() divides 24, which is wrong. Hence C (x i ) is a normal subgroup of for i = 1, 2. It follows that Z() is abelian and so Z() = C 4 C 4, C 2 C 2 C 2 C 2 or C 2 C 2 C 4. However the group C 2 C 2 C 4 is not capable by Remark 2.2. The group C 4 C 4 cannot have such a partition. For, if C (x i ) Z() = n i, then 7 i=1 n i = 22 and so 7 i=3 n i = 14. It yields that n 3 = n 4 = 4 and n 5 = n 6 = n 7 = 2 (if r = 7, then α 1 = α 2 = α 3 = α 4 = 4 and α 5 = α 6 = α 7 = 8). This is impossible, since C 4 C 4 has only three elements of order 2. Finally we show that Z() C 2 C 2 C 2 C 2. Consider P = { C (x i ) Z() i = 1,..., 7} and A i = C (x i ) Z() for 1 i 7. For each x, let x = xz(), and let A 1 = {1, a, b, ab} and A 2 = {1, c, d, cd} for some a, b, c, d \Z(). Now using the fact that P is a partition of Z() with trivial kernel, having exactly three members of size 2 and four members of size 4; one can show that P is equal to one of the following: (1) { A 1, A 2, {1, ac}, {1, bc, ad, abcd}, {1, bd, abc, acd}, {1, abd}, {1, bcd} }. (2) { A 1, A 2, {1, ac}, {1, bc, abd, acd}, {1, ad, abc, bcd}, {1, abcd}, {1, bd} }. (3) { A 1, A 2, {1, ac, bd, abcd}, {1, bc}, {1, ad, abc, bcd}, {1, abd}, {1, acd} }. (4) { A 1, A 2, {1, ac, bd, abcd}, {1, bc, abd, acd}, {1, ad}, {1, bcd}, {1, abc} }. (5) { A 1, A 2, {1, ac, abd, bcd}, {1, bd, abc, acd}, {1, bc}, {1, ad}, {abcd} }. (6) { A 1, A 2, {1, ac, abd, bcd}, {1, bc, ad, abcd}, {1, bd}, {1, acd}, {1, abc} }. For any elements x, y, we denote x 1 y 1 xy by [x, y]. Note that if x, y A i, then [x, y] = 1, and if 1 x A i, 1 y A j and i j, then [x, y] 1. Now we reach to a contradiction in each of the above cases. For example in the case (1), we have 1 = [bc, ad] = [b, d][c, a] and 1 = [bd, abc] = [b, c][d, a][d, b] which yield that [abd, bcd] = 1, a contradiction (note that in writing the latter equalities we are using the facts that [x, y] 2 = 1 and [x, y] Z() for all x, y ). Lemma 3.3. Let be a finite 8-centralizer group. Then Z() 24, 36. Proof. Let {x 1,..., x r } be a set of pairwise non-commuting elements of having maximal size. Then {C (x i ) i = 1,..., r} is an irredundant r-cover with intersection Z(). Assume that : C (x i ) = α i, where α 1 α r. Suppose, for a contradiction, that Z() = 36. Then r = 6 or 7 by Remark 2.1(2,3). If r = 6, then α 2 4 by Lemma 2.1 and Lemma 3.1. Now Proposition 2.5(a)(2) implies that α 2 2 or 3 and so α 2 = 4. If α 1 3, then Z() 12 by

12 54 A. ABDOLLAHI, S.M. JAFARIAN AMIRI, AND A.M. HASSANABADI Proposition 2.5(a)(2), which is not so. Thus α 1 = 4 and so α i 4 for each i 1. On the other hand by Proposition 2.5(b), there exists a proper centralizer C (x) which is not abelian and contains three subgroups C (x i ) for i {1,..., 6}. Thus = i T C (x i ) C (x) for some T {1,..., 6} such that T = 3. Therefore by Lemma 2.1, there exists at least one of C (x i )s whose index is smaller than 4, which is a contradiction. Hence r = 7 and {C (x i ) i = 1,..., 7} forms a partition with kernel Z() by Remark 2.1(4). Since Z() = 36, by Lemma 2.1 and Proposition 2.5(a)(2), we have α 1 = = α 7 = 6. This gives that C (x i ) is Z() Z() a cyclic subgroup of order 6 for every i {1,..., 7} and so has 14 elements of order 3, contradicting the fact that every group of order 36 has a unique Sylow 3-subgroup. Now suppose that If r = 6, then α 2 = 4 by Lemma 2.1 and Z() = 24. Proposition 2.5(a)(2). If α 1 3, then Z() = 24 by Proposition 2.5(a)(2). Therefore α i 4 for each i {1,..., 6}. On the other hand by Proposition 2.5(b), contains a proper centralizer C (x) which is not abelian and contains C (x i1 ), C (x i2 ) and C (x i3 ) for some distinct i 1, i 2, i 3 {1,..., r}. Again it follows that there exists an index i such that α i is smaller than 4 for some i {1,..., 6}, which is not possible. If r = 7, then α 2 6 by Lemma 2.1. Now Proposition 2.5(a)(2) implies that α 2 2 or 3. If α 2 = 4, then Z() α 1α 2 16, which is wrong. Therefore α i = 6 for 2 i 7 by Lemma 2.1. If α 1 = 6, then Z() = 36 by inclusion-exclusion principal. Thus α 1 = 4 and so C (x 1 ) Z() is the unique cyclic subgroup of order 6. Therefore = Z() has the unique Sylow C (x i ) Z(), 3-subgroup U = w, where w = wz() for some w. Since = 7 i=1 C (x i ) C (x j ) Z() Z() = 1 for distinct i and j, α 1 = 4 and α i = 6 for 2 i 7, has 20 elements of order 2 or 4. It follows that the number of Sylow 2-subgroups of is three, namely P, P w and P w2. Also P P w P w 2 = 20, so that P P w w P 2 = 2. Thus core (P ) = 2. Next since = U P, we have C (U) = Ucore (P ). So that C (U) = 6. But embeds into Aut(U) = C C (U) 2 and so 12, giving the final contradiction. Proof of Theorem B. Let {x 1,..., x r } be a set of pairwise non-commuting elements of having maximal size. Then {C (x i ) i = 1,..., r} is an irredundant r-cover with intersection Z(). Assume that : C (x i ) = α i, where α 1 α r. By Lemma 2.4, r = 5, 6 or 7. Now suppose that r = 5. Then Z() f(5) = 16. If α 2 3, then cent() = 4 or 5 by Proposition 2.5(a)(2), a contradiction. Therefore α 2 = = α 5 = 4

13 ROUPS WITH SPECIFIC NUMBER OF CENTRALIZERS 55 by Lemma 2.1. If α 1 = 2, then Z() 8 by Proposition 2.5(a)(2). Now, using Theorems 2 and 4 of [5] it is clear that 4 or 6. It follows from Lemma 3.2 Z() that Z() = C 2 C 2 C 2. If α 1 = 3, then Z() = 12 and so Z() = A 4 or D 12 by Lemma 2.9, and since α i = 4 for i = 2, 3, 4, 5, we have Z() = A 4. Assume that α 1 = 4. Proposition 2.5(c) implies that there exists a proper centralizer C (x) which is not abelian and contains at least two of C (x i )s for 1 i 5. Thus = i T C (x i ) C (x) for some T {1, 2, 3, 4, 5} such that T 3. It follows that α i 3 for some i T, which is impossible. Suppose that r = 6. Then α 2 5 and so α 2 = 4 by Lemma 3.1 and Proposition 2.5(a)(2). If α 1 = 2, then Z() C 2 C 2, a contradiction. If α 1 = 3, then Z() = 8 from which it follows that Z() = C 2 = 12. It follows from Lemma 2.9 that Z() = A 4 or D 12, which is impossible since α 1 = 3. Thus we have an irredundant 7-cover in this case. Therefore α 1 = 4 and α i 4 for i {1,..., 6}. On the other hand there exists a proper centralizer C (x) which is not abelian and contains at least three of C (x i )s for 1 i 6 by Proposition 2.5(b). Therefore = i T C (x i ) C (x) for some T {1,..., 6} with T 3 which implies that there exists at least an index i T such that α i 3, giving a contradiction. Finally suppose that r = 7. Then {C (x i ) : 1 i 7} is a partition with kernel Z() for by Remark 2.1 and Proposition 2.6. By Lemma 2.1, we have α 2 6. Now by Proposition 2.5(a)(2) we have that α 2 4 and so α 2 = 4 or 6 by Lemma 3.1. Now assume that α 2 = 4. If α 1 = 2, then Z() 8 and so Z() = 8. Thus by Lemma 3.2, Z() = C 2 C 2 C 2 which is not possible, since C 2 C 2 C 2 does not have an irredundant 7-cover with α 1 = 2. If α 1 = 3, then Z() = 12 by Proposition 2.5(a)(2). It follows from Lemma 2.9 that Z() = A 4 or D 12. This is impossible since every irredundant cover of A 4 with α 1 = 3 has five members and D 12 has an element of order 6 which implies that α 1 = 2. Thus α 1 = 4 and so Z() α 1α It follows that Z() = 8, 12 or 16. Since α 1 = α 2 = 4, Z() D 12. Hence by Lemma 3.2 and Lemma 2.9, Z() = C 2 C 2 C 2 or A 4, as required. Now suppose that α 2 = 6. Then Lemma 2.1 implies that α i = 6 for i {2,..., 7}. Since Z() = 7 C (x i ) i=1 Z() and C (x i ) C (x j ) Z() Z() = 1 for any two distinct i, j {1,..., 7}, we have Z() = 6α 1. If α 1 = 2, then Z() = 12 and so Z() = D 12 or A 4. Now since A 4 does not contain a subgroup of index 2, we

14 56 A. ABDOLLAHI, S.M. JAFARIAN AMIRI, AND A.M. HASSANABADI have Z() = D 12, as required If α 1 = 3, then Z() = 18 and K := C (x 1 ) Z() is a unique subgroup of = Z() of order 6. Therefore there exists an element y = yz() K such that y = 2 and so y is a subgroup of Z( Z() ). Thus = y P, where P is a normal Sylow 3-subgroup of. Hence is abelian and we have = C 2 C 3 C 3 which is not possible, since by Remark 2.2, C 2 C 3 C 3 is not capable. Finally if α 1 = 4 or 6, then Z() = 24 or 36 which is impossible by Lemma 3.3. This completes the proof. Proof of Theorem C. Suppose that is a finite 8-centralizer group. Then by Theorem B, Z() is isomorphic to A 4, D 12, or C 2 C 2 C 2. Now the result follows since cent(a 4 ) = 6 and cent(d 12 ) = 5. We know that for a finite group with Z() = A 4 or D 12, we have cent() = 6 or 8 (see [2, Theorem 3.6] and Proposition 2.2). For the case Z() = C 2 C 2 C 2, we also have: Proposition 3.4. Let be a finite group. If Z() cent() = 6 or 8. = C 2 C 2 C 2, then Proof. Suppose that {x 1,..., x r } is a set of pairwise non-commuting elements of having maximal size. Then Remark 2.1 yields that { C (x i ) Z() 1 i r} Z() forms an irredundant r-cover with trivial intersection for. Now the group C 2 C 2 C 2 is covered by at most seven proper subgroups and so r 7. Now we claim that every proper centralizer of is abelian. Since Z() = C 2 C 2 C 2, C (x) Z() = C 2 or C 2 C 2 for every x \Z() and so C (x) is abelian. Thus Lemma 2.6 implies that cent() = r+1 and so cent() 8. Now by Theorems 2 and 4 of [5] and Theorem A, we have cent() 4, 5 or 7, which completes the proof. Finally we note that there exist groups i for i = 1, 2 of order 64 such that i Z( i ) = C 2 C 2 C 2, cent( 1 ) = 6 and cent( 2 ) = 8 (for example, 1 = Smallroup(64, 60) and 2 := Smallroup(64, 73) in [8]). Also for = Smallroup(24, 3) in [8] we have Z() = A 4 and cent() = 8 (this is also pointed out in [2, Remark 3.7]).

15 ROUPS WITH SPECIFIC NUMBER OF CENTRALIZERS 57 References [1] A. Abdollahi, M. J. Ataei, S. M. Jafarian Amiri and A. Mohammadi Hassanabadi, On groups with a maximal irredundant 6-cover, Comm. Algebra 33 (2005), [2] A. R. Ashrafi, On finite groups with a given number of centralizers, Algebra Colloq. 7 (2000), No. 2, [3] A. R. Ashrafi, Counting the centralizers of some finite groups, Korean J. Comput. Appl. Math. 7 (2000), No.1, [4] R. Baer, roups with preassigned central and central quotient group, Trans. Amer. Math. Soc. 44 (1938), [5] S. M. Belcastro and. J. Sherman, Counting centralizers in finite groups, Math. Mag. 5 (1994), [6] R. A. Bryce, V. Fedri and L. Serena, Covering groups with subgroups, Bull. Austral. Math. Soc. 55 (1997), [7] J. H. E. Cohn, On n-sum groups, Math. Scand. 75 (1994), [8] The AP roup, AP roups, Algorithms, and Programming, Version 4.3; 2002, ( [9] D. reco, Sui gruppi che sono somma di quattro o cinque sottogruppi, Rend. Accad. delle Scienze di Napoli (4) 23 (1956), [10] N. Ito, On finite groups with given conjugate types I, Nagoya Math. J. 6 (1953), [11] B. H. Neumann, roups covered by finitely many cosets, Publ. Math. Debrecen 3 (1954), [12]. Scorza, ruppi che possono pensarsi come somma di tre sottogruppi, Boll. Un. Mat. Ital. 5 (1926), [13] M. J. Tomkinson, roups covered by finitely many cosets or subgroups, Comm. Algebra 15 (1987), Received November 5, 2004 Revised version received August 16, 2005 (A. Abdollahi) Department of Mathematics, University of Isfahan, Isfahan , and, Institute for Studies in Theoretical Physics and Mathematics (IPM), Tehran, Iran address: a.abdollahi@math.ui.ac.ir (S.M. Jafarian Amiri) Department of Mathematics, University of Isfahan, Isfahan , and, Institute for Studies in Theoretical Physics and Mathematics (IPM), Tehran, Iran address: sm.jafarian@sci.ui.ac.ir (A. Mohammadi Hassanabadi) Department of Mathematics, University of Isfahan, Isfahan , and, Institute for Studies in Theoretical Physics and Mathematics (IPM), Tehran, Iran address: aamohaha@yahoo.com