ON FINITE GROUPS WITH A CERTAIN NUMBER OF ...

J. Appl. Math. & Computing Vol. 17(2005), No. 1 - 2, pp. 217 - 227

ON FINITE GROUPS WITH A CERTAIN NUMBER OF CENTRALIZERS ALI REZA ASHRAFI∗ AND BIJAN TAERI

Abstract. Let G be a finite group and #Cent(G) denote the number of centralizers of its elements. G is called n-centralizer if
#Cent(G) = n, and G = n. primitive n-centralizer if #Cent(G) = #Cent Z(G) In this paper we investigate the structure of finite groups with at most 21 element centralizers. We prove that such a group is solvable and if G is a finite group such that G/Z(G) ' A5 , then #Cent(G) = 22 or 32. Moreover, we prove that A5 is the only finite simple group with 22 centralizers. Therefore we obtain a characterization of A5 in terms of the number of centralizers AMS Mathematics Subject Classification : 20D99, 20E07. Key Words and phrases : Finite group, n-centralizer group, primitive ncentralizer group, simple group.

1. Introduction Throughout this paper all groups mentioned are assumed to be finite. Z(G) denotes the center of G. Most notation is standard and it is taken mainly from [1] and [14]. Following Belcastro and Sherman [4] we shall use the notation #Cent(G) to denote the number of distinct centralizers in the group G. A group G is called n-centralizer  if #Cent(G) = n. Also, we say that G is primitive n-

G centralizer if #Cent Z(G) = n. It is obvious that G is 1-centralizer if and only if G is abelian and there is no n-centralizer group, for n = 2, 3.

Received November 7, 2003. Revised December 29, 2003. ∗ Corresponding author. c 2005 Korean Society for Computational & Applied Mathematics and Korean SIGCAM.

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Belcastro and Sherman [4] proved that G is 4-centralizer if and only if G ' Z2 × Z2 . Z(G) Furthermore, they proved that G is 5-centralizer if and only if G ' Z3 × Z3 or S3 , Z(G) the symmetric group on three letters. By these results one can see that there is no primitive 4-centralizer group and that G is a primitive 5-centralizer group if G and only if Z(G) ' S3 . Ashrafi [2] and [3] proved that if G is 6-centralizer, then G ' D8 , A4 , Z2 × Z2 × Z2 or Z2 × Z2 × Z2 × Z2 . Z(G) Ashrafi showed [2] that if G/Z(G) ' A4 , then #Cent(G) = 6 or 8. We consider the case G/Z(G) ' A5 and prove that Theorem 1. If G is a finite group and G/Z(G) ' A5 , then #Cent(G) = 22 or 32. Note that A5 has 21 Sylow subgroup and if x is an element of a Sylow subgroup P , then CG (x) = P . Thus #Cent(A5 ) = 22. In fact we prove that A5 is the only finite simple group with 22 centralizers. Therefore we obtain a characterization of A5 in terms of the number of centralizers. Theorem 2. If G is a finite simple group and #Cent(G) = 22, then G ' A5 . Therefore we can prove that if G is a finite group and #Cent(G) ≤ 21, then G is solvable. 2. Proof of the Theorem 1 Let x be an element of a group G. In this section we denote by C(x) := CG (x), the centralizer of x in G. In order to prove Theorem 1 firstly we prove the following Lemma. Lemma 1. Let x be an element of a finite group G such that |C(xZ)| = p, p a prime, where Z = Z(G). Then for all y ∈ G with C(xZ) = C(yZ) we have C(x) = C(y).

On finite groups with a certain number of centralizers

Proof. Clearly

C(x) Z

≤ C(xZ). So

C(x) Z

219

= C(xZ). Now

C(y) C(x) ≤ C(yZ) = = C(xZ), Z Z C(x) and so C(x) = C(y), as Z = p.



Lemma 2. Let x be an element of a finite group G such that |C(xZ)| = pq, C(g) C(x) where p, q are primes not necessary distinct and C(x) Z = C(xZ). If Z ≤ Z for some g ∈ G, then C(g) = C(x).

< C(x) and Proof. Suppose on the contrary that C(g) < C(x). Then C(g) Z Z C(g) C(g) = p or q. Suppose = p. Then C(g) is abelian and Z(C(g)) = C(g). Z Z Note that Z(C(x)) ⊆ Z(C(g)) = C(g). For, if u ∈ Z(C(x)), then [u, v] = 1, for all v ∈ C(x). In particular, when v = g, we have u ∈ C(g). Also for all v ∈ C(g), [u, v] = 1, so u ∈ Z(C(g)). Now Z(G) ⊆ Z(C(x)) ⊆ C(g) and |C(g) : Z(G)| = p. Hence either Z(C(x)) = C(g) or Z(G) = Z(C(x)). If Z(C(x)) = C(g), then for all u ∈ C(x) we have [g, u] = 1, and so u ∈ C(g). Therefore C(x) = C(g), a contradiction. If Z(G) = Z(C(x)), then x ∈ Z(G), which is a contradiction. This completes the proof.  Lemma 3. Let G be a finite group, G/Z(G) ' A5 , and let x ∈ G such that |C(xZ)| = 4, where Z = Z(G). If C(x) = C(xZ), then for all y ∈ G with Z C(y) |C(yZ)| = 4, we have Z = C(yZ). Proof. First of all, we can see that C(xZ) is a Sylow 2-subgroup of G/Z and, for every yZ ∈ C(xZ), we have C(xZ) = C(yZ). If yZ ∈ C(xZ), then C(y) C(x) ≤ C(yZ) = C(xZ) = Z Z and, by Lemma 2, C(x) = C(y). Therefore C(x) C(y) = = C(xZ) = C(yZ). Z Z Now suppose yZ 6∈ C(xZ). Then C(yZ) is a Sylow 2-subgroup of G/Z different from C(xZ). Since A5 acts transitively, by conjugation, on the set of its Sylow

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2-subgroups, there exists u ∈ G such that u−1 xuZ ∈ C(yZ). Hence C(yZ) = C(u−1 xuZ) and u  C(y) C(xu ) C(x) ≤ C(yZ) = C(aZ)u = . = Z Z Z Thus, by Lemma 1,

C(y) Z

=

C(xu ) Z ,

and so

C(y) Z

= C(yZ).



Proof of Theorem 1. Let np denote the number of Sylow p-subgroups of A5 . Then n2 = 5, n3 = 10, n5 = 6. Let C(xi Z), i = 1, 2, . . . , 21, be proper and distinct centralizer of G/Z, where |C(xi Z)| = 4, i = 1, 2, . . . , 5, |C(xi Z)| = 3, i = 6, 7, . . . , 15, |C(xi Z)| = 5, i = 16, 17, . . . , 21. By Lemma 3, C(xi ) = C(xi Z), i = 6, 7, . . . , 21. Z i) Thus C(x Z , i = 6, 7, . . . , 21 are all of the Sylow subgroups of G/Z of order 3, 5. Note that if y ∈ C(xi )\Z for some 6 ≤ i ≤ 21, then C(xi Z) = C(yZ) and C(xi ) = C(y) > . Therefore among the set of elements of G which have order 3 or 5 modulo Z, we have 16 distinct centralizers C(xi ), i = 6, 7, . . . , 21. i) i) Now if C(x = C(xi Z) for some 1 ≤ i ≤ 5, then, by Lemma 3, C(x = Z Z C(xi Z), for all 1 ≤ i ≤ 5. Therefore, in this case, among the set of elements of G which have order 2 modulo Z, we have 5 distinct centralizers C(xi ), i = i) 1, 2, . . . , 5. Hence #Cent(G) = 22. So suppose C(x < C(xi Z) for all i for Z 1 ≤ i ≤ 5. Let C(xi Z) = {Z, xi Z, yi Z, ti Z}, i = 1, 2, . . . , 5. Then C(xi ), C(yi ), C(ti ), i = 1, 2, . . . , 5 are all distinct, and among the set of elements of G which have order 2 modulo Z, we have 15 distinct centralizers. Hence #Cent(G) = 32.  In what follows, we show that there exists a finite group G such that G/Z(G) ' A5 and #Cent(G) = 32. To do this, we assume that G is a group generated by two permutations (1, 20, 17, 5, 12)(2, 3, 9, 19, 10)(4, 14, 22, 11, 6)(7, 8, 15, 13, 16) and (2, 18)(5, 1)(6, 21)(7, 24)(9, 17)(10, 16)(12, 23)(13, 20)(14, 19)(15, 22),

On finite groups with a certain number of centralizers

221

then |G| = 240, G/Z(G) ' A5 and #Cent(G) = 32. 3. Proof of the Theorem 2 Let G be finite group, p a prime divisor of the order of G. We denote by vp (G), the number of Sylow p-subgroups of G which pairwise intersect trivially. Lemma 4. Let G be finite group, and p a prime divisor of the order of G. Then #Cent(G) ≥ vp (G) + 1. Proof. Let P and Q be Sylow p-subgroups of G such that P ∩ Q = 1. Let 1 6= x ∈ Z(P ) and 1 6= y ∈ Z(Q). If CG (x) = CG (y), then P, Q ≤ CG (x) = CG (y), as x ∈ Z(P ). Thus [P, y] = 1 and P hyi is a subgroup of G. Now |P hyi | =

|P || hyi | = |P || hyi | > |P |, |P ∩ hyi |

which is impossible. Therefore CG (x) 6= CG (y), and the result follows.



Lemma 5. Let G be a finite group, and p a prime divisor of the order of G. If p2 6 | |G|, and G has more than one Sylow p-subgroup, then #Cent(G) ≥ 2 + kp, where k is the least positive integer such that 1 + kp divides |G|. Proof. Since vp (G) = 1 + rp, for some r > 1, and vp (G) is the number of Sylow p-subgroups of G, we have by Lemma 4, #Cent(G) ≥ vp (G) + 1 = 1 + rp + 1 ≥ 2 + kp, for k ≤ r.



Lemma 6. Let K be a subgroup of a finite group G. Then #Cent(K) ≤ #Cent(G). Proof. Let Xi = CK (ki ), i = 1, 2, . . . , m be distinct centralizers in K. Then Xi = K ∩ CG (ki ), and so CG (ki ) 6= CG (kj ), for all i 6= j.  To prove Theorem 2 we use the classification of finite simple groups. First we prove the result for projective special linear groups. Lemma 7. If n ≥ 3 or n = 2 and q > 5, then #Cent(P SL(n, q)) > 22.

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Proof. Let q = pm , where p is prime and m ≥ 1. Suppose that #Cent(P SL(n, q)) ≤ 22. If n = 2, by [11, Theorem 8.2, p. 191], we have 22 ≥ #Cent(P SL(2, pm )) ≥ vp (P SL(2, pm )) + 1 = pm + 2. Thus we have the following cases. (i) p = 2, m = 3 or p = 2, m = 4. (resp., pm + 1), then vr (P SL(2, pm )) =

If r is an odd prime which divides pm − 1

pm (pm + 1) pm (pm − 1) (resp., vr (P SL(2, pm)) = ), 2 2

by [11, II, §8]. For q = 23 , we put r = 7, and obtain that 21 ≥ v7 (P SL(2, 23)) =

23 (23 + 1) = 36, 2

a contradiction. For q = 24 , we put r = 17, and obtain that 21 ≥ v17 (P SL(2, 24 )) = a contradiction. (ii) q = 7, 11, 13, 17, or 19.

24 (24 − 1) = 120, 2

Put r = 3, then

7(7 + 1) = 28, 2 11(11 − 1) v3 (P SL(2, 11)) = = 55, 2 17(17 − 1) = 136, and v3 (P SL(2, 17)) = 2 19(19 + 1) v3 (P SL(2, 19)) = = 190. 2 v3 (P SL(2, 7)) =

Thus in these cases we also obtain a contradiction. Now suppose n ≥ 3. By checking the order of P SL(n, pm ) it follows that there is a prime r such that r divides |P SL(n, pm )| and r2 does not divide |P SL(n, pm )| and r ≥ 23. Therefore, by Lemma 4, #Cent(P SL(2, pm )) ≥ 23, unless in the following cases: n = 3 and pm = 2, 4, 3, 9, 7, 11 or n = 4 and pm = 2, 4, 3, 7, or n = 5 and pm = 3, or n = 6 and pm = 3.

On finite groups with a certain number of centralizers

223

By an easy program which is written in GAP [15] we find the following data G P SL(3, 2) P SL(3, 4) P SL(3, 3) P SL(4, 2) P SL(4, 7) P SL(5, 3)

vr (G) v3 = 22 · 7 v3 = 26 · 3 · 5 v3 = 22 · 7 v19 = 22 · 3 · 7 v19 = 29 · 3 · 52 · 76 v13 = 27 · 35 · 5

G P SL(3, 7) P SL(3, 11) P SL(3, 7) P SL(4, 3) P SL(4, 4) P SL(6, 3)

vr (G) v19 = 22 · 3 · 7 v19 = 24 · 52 · 113 v13 = 27 · 35 · 5 v13 = 27 · 35 · 5 v17 = 210 · 34 · 5 · 7 v5 = 23 · 314 · 7 · 11 · 132

and in each case we obtain a contradiction.



Note that P SL(2, 2) ' S3 , P SL(2, 3) ' A4 , and P SL(2, 4) ' P SL(2, 5) ' A5 . Therefore we have #Cent(P SL(2, 2)) = 5, #Cent(P SL(2, 3)) = 6, and #Cent(P SL(2, 4)) = #Cent(P SL(2, 5)) = 22. Now we consider sporadic simple groups Lemma 8. If S is a sporadic simple group, then #Cent(S) > 22. Proof. If S is one of the sporadic simple group M23 , M24 , Co1 , Co2 , Co3 , F i23 , F i024 , Ru, T h, O0 N, B, J4 , Ly, M, then |S| has a prime divisor p ≥ 23 in which p2 6 ||S| (see Atlas [6, p. viii]). Thus, by Lemma 4, #Cent(S) ≥ 23. If S is one of the sporadic group, M11 , M12 , J1 , M22 . By the Atlas [6, p. 18, 33, 36,39], P SL(2, 11) < S. Since #Cent(P SL(2, 11)) > 22, by Lemma 7, we have #Cent(S) > 22. By the Atlas [6], P SL(2, 7) < U3 (3) < J2 (p.14, 42), M11 < HS(p.80), P SL(2, 19) < J3 (p.82), M11


22, for all sporadic simple group S. 

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Lemma 9. If L is a finite simple group of Lie type not isomorphic to A5 , then #Cent(L) > 22. Proof. If L = An (q) ' P SLn+1 (q), then by Lemma 7, #Cent(L) > 22. If L = Bn (q) ' P Ω2n+1 (q), n > 1, then by checking the order of L, it follows that there is a prime r such that r divides |L| and r2 does not divide |L| and r ≥ 23. So, by Lemma 4, #Cent(L) ≥ 23, unless in the following cases: B2 (3), B2 (5), B2 (7), B3 (3). But using the fact that Bn (q) ' P Ω2n+1 (q) and GAP, we obtain that v5 (B2 (3))

=

24 · 34 , v13 (B2 (5)) = 24 · 32 · 54 , and

v13 (B3 (3))

=

28 · 38 · 5 · 7.

Also we have |B2 (7)| = |P Ω4 (7)| = 28 · 32 · 52 · 74 . If P, Q are Sylow subgroups of B2 (7) of order 52 and 28 respectively, then using GAP, P x ∩ P y = 1 for all x, y ∈ Q, with x 6= y. Thus v5 (B2 (7)) ≥ 28 . If L = Cn (q) ' P Sp2n (q), n > 2, then, as before, by checking the order of L, it follows that we must consider the following cases: C3 (2), C3 (4), C3 (3). Using the fact that Cn (q) ' P Sp2n (q) and GAP, we obtain that v7 (C3 (2)) = 28 · 38 · 5 · 7, and v17 (C3 (4)) = 214 · 33 · 52 · 7 · 13. Also we have |C2 (7)| = |P Sp4 (7)| = 28 · 32 · 52 · 74 . If P, Q are Sylow subgroups of C2 (7) of order 32 and 52 respectively, then using GAP, P x ∩ P y = 1 for all x, y ∈ Q, withx 6= y. Thus v5 (B2 (7)) ≥ 52 . Suppose that L = Dn (q) ' P Ω+ 2n (q), n > 3. As before, by checking the order of L, it follows that we must consider the following cases: D4 (2), D4 (4), D4 (4), D4 (3). Using the fact that Dn (q) ' P Ω+ 2n (q) and GAP, we obtain that v7 (D4 (2)) = 211 · 34 · 52 , v13 (D4 (4)) = 223 · 34 · 52 · 7 · 17, and v13 (D4 (3)) = 211 · 311 · 52 · 17. Thus in any case #Cent(Dn (q)) > 22. If L = E6 (q), E7 (q), E8 (q),2 Gn (q), we obtain a prime r such that r divides |L|, and r2 does not divides |L|, and r ≥ 23. So, by Lemma, 4, #Cent(L) ≥ 23. If L = G2 (q), then we must consider G2 (4) and G2 (3). Since J2 < G2 (4) (see [12, p. 317]) and, by Lemma 8, #Cent(J2 ) > 22, we have #Cent(G2 (4)) ≥ #Cent(J2 ) > 22.

On finite groups with a certain number of centralizers

225

Now, by the Atlas [6, p. 61], P SL(2, 13) < G2 (3). So, by Lemma 7, #Cent(G2 (3)) ≥ #Cent(P SL(2, 13)) > 22. If L = F4 (q), then we must consider F4 (2). We know (see [12, p. 318]) that G2 (2) < F4 (2), and also G2 (2)0 ' U3 (3) (see [1, p. 253]). Now P SL(2, 7) < U3 (3), and so #Cent(F4 (2)) ≥ #Cent(G2 (2)) ≥ #Cent(U3 (3) ≥ P SL(2, 7) > 22. Suppose that L = 2An (q) ' Un+1 (q), n > 1. We must consider 2

A2 (4),

2

2

A2 (8),

A2 (3),

2

A3 (2),

2

A3 (4),

2

A3 (8),

2

A3 (3),

2

A3 (5).

Using the fact that L = 2An (q) ' Un+1 (q) and GAP, we find that v7 (2A2 (3)) = 25 · 32, v7 (2A2 (5)) = 24 · 3 · 5, v5 (2A3 (2)) = 24 · 34 , v7 (2A3 (3)) = 27 · 35 · 5. Now we have |2A2 (4)| = v13 (2A2 (4))

=

2

| A2 (8)| = v19 (2A2 (8))

=

|2A3 (4)| = v13 (2A2 (4))

=

26 · 3 · 52 · 13, and so 1 + 13k ≥ 40, 29 · 34 · 7 · 19, and so 1 + 19k ≥ 96, 212 · 32 · 52 · 13 · 17, and so 1 + 13k ≥ 40.

2

Also for A3 (5), let P, Q be Sylow subgroups of order 13 and 34 respectively. Then, using GAP, we find that P x ∩ P y = 1, for all distinct x, y ∈ Q, and so v13 (2A2 (5)) ≥ 34 . For 2A3 (8), let P, Q be Sylow subgroups of order 19 and 72 respectively. Then, using GAP, we find that P x ∩ P y = 1, for all distinct x, y ∈ Q, and so v19 (2A3 (8)) ≥ 72 . If L = 2B2 (q) ' Sz(q), where q = 22m+1 , m ≥ 1, then we must consider 2 B2 (8). But v2 (2B2 (8)) = v2 (Sz(8)) = 65 and so #Cent(2B2 (8)) ≥ 65. Suppose L = 2Dn (q) ' P Ω− 2n (q), n > 3. We must consider 2D4 (2) and 2D4 (2). Using the fact that 2Dn (q) ' P Ω− 2n (q) and GAP we obtain that v17 (2D4 (2)) = 210 · 34 · 5 · 7 and v17 (2D5 (2)) = 217 · 36 · 52 · 7 · 11. 3

If L = 3D4 (q), then we must consider 3 D4 (2). By [12, p. 318], G2 (2) < D4 (2). Since G2 (2)0 ' U3 (3) and v7 (U3 (3)) = 288, we have #Cent(3 D4 (2)) ≥ #Cent(G2 (2)) ≥ 288.

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Finally suppose that L = 2F4 (q),2 E6 (q). By T a result of Tomkinson [16] if G n is covered by n subgroup X1 , . . . , Xn , then |G/ i=1 Xi | ≤ f2 (n), where f2 (n) = max{(n − 1)2 , (n − 2)3 }(n − 3)!. So if G is simple and #Cent(G) = 22, then |G| ≤ f2 (21). Since |L| > f2 (21), we have #Cent(L) > 22.  Proof of Theorem 2. The proof follows from Lemmas 7, 8 and 9.



It may seem that if G and H are finite simple groups and |G| ≤ |H|, then #Cent(G) ≤ #Cent(H). But this is not true. For example #Cent(P SL(2, 7)) = 79 > 74 = #Cent(P SL(2, 8)). The following question arises naturally Question. Let G and H be finite simple groups. Is it true that if #Cent(G) = #Cent(H), then G is isomorphic to H?

Acknowledgement The authors would like to thank the referee for his/her helpful remarks.

References 1. M. Aschbacher, Finite group theory, Cambridge Univ. Press, (1986). 2. A. R. Ashrafi, On finite groups with a given number of centralizers, Algebra Colloquium, 7(2) (2000), 139-146. 3. A. R. Ashrafi, Counting the centralizers of some finite groups, Korean J. Comput. & Appl. Math. 7(1) (2000), 115-124. 4. S. M. Belcastro and G. J. Sherman, Counting centralizers in Finite Groups, Math. Mag. 5 (1994), 366-374. 5. J. H. E. Cohn, On n-sum groups, Math. Scand. 75 (1994), 44-58. 6. J. H. Conway, R. T. Curtis, S. P. Norton and R. A. Wilson, Atlas of finite groups, Oxford Univ. Press (Clarendon), Oxford, (1985). 7. M. R. Darafsheh and Z. Mostaghim, Computation of the complex characters of the group AU T (GL7 (2)), Korean J. Comput. & Appl. Math. 4 (1997), 193-210. 8. M. R. Darafsheh and Nowroozi Larki, F., The character table of the group GL2 (q) when extended by a certain group of order two, Korean J. Comput. & Appl. Math. 7 (2000), 643-654. 9. D. Gorenstien, Finite simple groups, Plenum Press, New York, (1982). 10. P. B. Howlett, L. J. Rylands and D. E. Taylor, Matrix generators for exceptional groups of Lie type, J. Symbolic Computation 31 (2001), 429-445.

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11. B. Huppert, Endliche Gruppen I, Springer - Verlag, Berlin (1967). 12. P. B. Kleidman and R. A. Wilson, Sporadic simple subgroups of finite exceptional groups of Lie type, J. Algebra 157 (1993), 316-330. 13. B. H. Neumann, Groups covered by permutable subsets, J. London Math. Soc. 29 (1954), 236-248. 14. D. J. S. Robinson, A course in the theory of groups, 2nd ed., Springer - Verlag, Berlin (1996). 15. M. Schonert et al., GAP: Groups, Algorithms and Programming, Version 4.3, Aachen, St Andrews, 2003. (http://www-gap.dcs.st-andrews.ac.uk/∼ gap) 16. M. J. Tomkinson, Groups covered by finitely many cosets or subgroups, Comm. Alg. 15 (1987), 845-859. Ali Reza Ashrafi received his BSc from Teacher Training University of Tehran, MSc from Shahid Beheshti University and Ph. D at the University of Tehran under the direction of Mohammad Reza Darafsheh. Since 1996 he has been at the University of Kashan. Now he is an associate professor of mathematics and supervised ten MSc students in the field of computational group theory and hyperstructures. His research interests focus on computational group theory, finite lattice, hyperstructures and mathematical chemistry. Department of Mathematics, Faculty of Science, University of Kashan, Kashan, Iran e-mail: [email protected] Bijan Taeri received his BSc from Isfahan University of Technology, MSc from Isfahan University and Ph. D at the Isfahan University under the direction of Aliakbar Mohammadi Hassanabadi. Since 1990 he has been at the Isfahan University of Technology. Now he is an associate professor of mathematics. His main intrests include finite and infinite group theory. Department of Mathematics, Isfahan University of Technology, Isfahan, Iran e-mail: [email protected]