A new characterization for some extensions of PSL(2,q) for some q by some character degrees

Transcription

1 Proc. Indian Acad. Sci. (Math. Sci.) Vol. 126, No. 1, February 2016, pp c Indian Academy of Sciences A new characterization for some extensions of PSL(2,q) for some q by some character degrees BEHROOZ KHOSRAVI 1,, BEHNAM KHOSRAVI 2 and BAHMAN KHOSRAVI 3 1 Department of Pure Math., Faculty of Math. and Computer Sci., Amirkabir University of Technology (Tehran Polytechnic), 424, Hafez Ave., Tehran 15914, Iran 2 Department of Mathematics, Institute for Advanced Studies in Basic Sciences, Zanjan , Iran 3 Faculty of Mathematics, Department of Sciences, Qom University of Technology, Qom, Iran * Correspondence author. khosravibbb@yahoo.com MS received 5 December 2013 Abstract. In [22] (Tong-Viet H P, Simple classical groups of Lie type are determined by their character degrees, J. Algebra, 357 (2012) 61 68), the following question arose: Which groups can be uniquely determined by the structure of their complex group algebras? The authors in [12] (Khosravi B et al., Some extensions of PSL(2,p 2 ) are uniquely determined by their complex group algebras, Comm. Algebra, 43(8) (2015) ) proved that each extension of PSL(2,p 2 ) of order 2 PSL(2,p 2 ) is uniquely determined by its complex group algebra. In this paper we continue this work. Let p be an odd prime number and q = p or q = p 3.LetM be a finite group such that M =h PSL(2,q),whereh is a divisor of Out(PSL(2,q)). Also suppose that M has an irreducible character of degree q and 2p does not divide the degree of any irreducible character of M. As the main result of this paper we prove that M has a unique nonabelian composition factor which is isomorphic to PSL(2,q). As a consequence of our result we prove that M is uniquely determined by its order and some information on its character degrees which implies that M is uniquely determined by the structure of its complex group algebra. Keywords. Characterization; character degrees; order; complex group algebra Mathematics Subject Classification. 20C15, 20D05, 20D Introduction and preliminary results If n is an integer, then we denote by π(n) the set of all prime divisors of n.letg be a finite group. Then π( G ) is denoted by π(g). Let Irr(G) be the set of irreducible characters of G and denote by cd(g), the set of irreducible character degrees of G. The degree pattern of G, which is denoted by X 1 (G) is the set of all irreducible complex character degrees of G counting multiplicities. The character degree graph of a finite group G is the graph whose vertices are the prime divisors of the irreducible character degrees of G and two distinct vertices p 1 and p 2 are joined by an edge if p 1 p 2 divides some irreducible character degree of G. 49

2 50 Behrooz Khosravi et al. A finite group G is called a K 3 -group if G has exactly three distinct prime divisors. Recently Xu et al. in [25] and [26] proved that all simple K 3 -groups and the Mathieu groups are uniquely determined by their orders and one or both of their largest and second largest irreducible character degrees. In [9] it is proved that the projective special linear group PSL(2,q)is uniquely determined by its group order and its largest irreducible character degree when q is a prime or when q = 2 a for an integer a 2 such that 2 a 1or2 a + 1 is a prime. Let p be an odd prime number. In [11] the authors proved that the simple group PSL(2,p) is uniquely determined by its order and some information on its irreducible character degrees. In [14] it is proved that if p is an odd prime number and G is a finite group where G = PSL(2,p 2 ), p 2 cd(g) and there does not exist any θ Irr(G) such that 2p θ(1), then G = PSL(2,p 2 ). In [13] it is proved that the simple group PSL(2,p 2 ) is uniquely determined by its character degree graph and its order (see also [10, 15, 16]). In Problem 2 of [1], Brauer asked whether two groups G and H are isomorphic given that two group algebras FG and FH are isomorphic for all fields F. Thisisfalse,in general. In fact, Dade [4] constructed two non-isomorphic metabelian groups G and H such that FG = FH for all fields F. In [5], Hertweck showed that this is not true even for the integral group rings. Note that if ZG = ZH, then FG = FH, for all fields F, where Z is the ring of integer numbers. For nonabelian simple groups, Kimmerle obtained a positive answer in [17]. He outlined the proof asserting that if G is a group and H is a nonabelian simple group such that FG = FH for all fields F, then G = H. By Molien s theorem, knowing the structure of the complex group algebra is equivalent to knowing the first column of the ordinary character table. In [22] with the only assumption that their complex group algebras are isomorphic, Tong-Viet proved that each classical simple group is uniquely determined by its complex group algebra. In [22], Tong-Viet posed the following question: Question 1.1. Which groups can be uniquely determined by the structure of their complex group algebras? It was shown in [19] that the symmetric groups are uniquely determined by the structure of their complex group algebras. It was conjectured that all nonabelian simple groups are uniquely determined by the structure of their complex group algebras. This conjecture was verified in [17, 20, 23] for the alternating groups, the sporadic simple groups, the Tits group and the simple exceptional groups of Lie type. In [12], the authors proved that if G is a finite group such that G =2 PSL(2,p 2 ), p 2 cd(g) and there does not exist any θ Irr(G) such that 2p θ(1), then G has a unique nonabelian composition factor isomorphic to PSL(2,p 2 ). Using these results we get that each extension of PSL(2,p 2 ) of order 2 PSL(2,p 2 ) is uniquely determined by its complex group algebra. In this paper we continue this work. Let p 5 be an odd prime number and q = p or q = p 3.LetM be a finite group such that M =h PSL(2,q), where h is a divisor of Out(PSL(2,q)). Also suppose that M has an irreducible character of degree q and 2p does not divide the degree of any irreducible character of M. As the main result of this paper we prove that M has a unique nonabelian composition factor which is isomorphic to PSL(2,q). As a consequence of our result we give new characterizations for some extensions of PSL(2,q)which implies that these groups are uniquely determined by the structure of their complex group algebras.

3 A new characterization for some extensions of PSL(2,q) 51 If N G and θ Irr(N), then the inertia group of θ in G is I G (θ) ={g G θ g = θ}. If the character χ = k i=1 e i χ i, where for each 1 i k, χ i Irr(G) and e i is a natural number, then each χ i is called an irreducible constituent of χ. Lemma 1.2 (Gallagher s theorem) (Corollary 6.17 of [7]). Let N G and let χ Irr(G) be such that χ N = θ Irr(N). Then the characters βχ for β Irr(G/N) are irreducible distinct for distinct β and all of the irreducible constituents of θ G. Lemma 1.3 (Itô s theorem) (Theorem 6.15 of [7]). Let A G be abelian. Then χ(1) divides G : A, for all χ Irr(G). Lemma 1.4 (Theorems 6.2, 6.8, of [7]). Let N G and let χ Irr(G). Let θ be an irreducible constituent of χ N and suppose θ 1 = θ,...,θ t are the distinct conjugates of θ in G. Then χ N = e t i=1 θ i, where e =[χ N,θ] and t = G : I G (θ). Alsoθ(1) χ(1) and χ(1)/θ(1) G : N. Lemma 1.5 (Itô Michler theorem) [6]. Let ρ(g) be the set of all prime divisors of the elements of cd(g). Then p ρ(g) if and only if G has a normal abelian Sylow p-subgroup. Lemma 1.6 (Lemma of [25]). Let G be a nonsolvable group. Then G has a normal series 1 H K G such that K/H is a direct product of isomorphic nonabelian simple groups and G/K Out(K/H). Let q = p f be a prime power. The outer automorphism group of PSL(2,q)is generated by a field automorphism ϕ of order f and, if q is odd, we get a diagonal automorphism δ of order 2 (see [2]). Also we note that PGL(2,q)= PSL(2,q) δ. Lemma 1.7 (Theorem A of [24]). Let S = PSL(2,q), where q = p f > 3 for a prime p, A = Aut(S), and let S H A. Set G = PGL(2,q)if δ H and G = S if δ H, and let H : G =d = 2 a m, m is odd. If p is odd, let ε = ( 1) (q 1)/2. The set of irreducible character degrees of H is cd(h) ={1,q,(q+ ε)/2} {(q 1)2 a i : i m} {(q + 1)j : j d}, with the following exceptions: (1) If p is odd with H S ϕ or if p = 2, then (q + ε)/2 is not a degree of H. (2) If f is odd, p = 3, and H = S ϕ, then i = 1. (3) If f is odd, p = 3, and H = A, then j = 1. (4) If f is odd, p = 2, 3 or 5, and H = S ϕ, then j = 1. (5) If f 2(mod 4), p = 2 or 3, and H = S ϕ or H = S δϕ, then j = 2. Lemma 1.8 (Lemma 2 of [26]). Let G be a finite solvable group of order p α 1 1 pα pα n n, where p 1,p 2,...,p n are distinct primes. If (kp n + 1) p α i i, for each i n 1 and k>0, then the Sylow p n -subgroup is normal in G. Lemma 1.9 (Remark 1 of [3]). The equation p m q n = 1, where p and q are primes and m,n > 1 has only one solution, namely = 1.

4 52 Behrooz Khosravi et al. Nagell and Ljunggren proved the following result: Lemma 1.10 (Page 182 of [21]). Ifm 2, then the only non-zero solutions of X 2 + X + 1 = Y m are ( 1, 1), when m is odd, and ( 1,±1) when m is even. If n is an integer and r is a prime number, then we write r α n, when r α n but r α+1 n. Also if r is a prime number we denote by Syl r (G), the set of Sylow r-subgroups of G and we denote by n r (G), the number of elements of Syl r (G). All groups considered are finite and all characters are complex characters. We write H ch G if H is a characteristic subgroup of G. All other notations are standard and we refer to [2]. 2. The main results Lemma 2.1. Let M be a finite group such that M p = p j, where 1 j 3. If there exists χ Irr(G), where χ(1) = M p, then O p (M) = 1. Proof. If j = 1orj = 2, then the result obtained by Itô s theorem (see Lemma 2.2 in [14]). So let j = 3 and L = O p (M). Ifη Irr(L) such that [χ L,η] = 0, then p 3 /η(1) M : L, by Lemma 1.4. If L =p 3, then η(1) = p 3, which is impossible. If L =por p 2, then L is abelian and so p 3 M : L, which is a contradiction. Therefore L = 1. Theorem 2.2. Let p 5 be a prime number and q = p or q = p 3.IfG is a finite group such that (i) PSL(2,q) G and G Aut(PSL(2,q)), (ii) there exists χ Irr(G) such that χ(1) = q, (iii) there does not exist any θ Irr(G) such that 2p θ(1), (iv) If q = 2 α 1 is a prime number, then there exists an irreducible character degree of G which is divisible by 4, then G is a nonsolvable group. Proof. On the contrary, let G be a solvable group. First let q = p. If p = 5, then G Since there exists no prime power r α of G such that r α = 5k + 1, by Lemma 1.8 we get that Sylow 5-subgroup of G is a normal subgroup of G. Now we get a contradiction by Itô Michler theorem. Therefore p 7. We know that G p(p+1)(p 1).If G = PSL(2,q), then we get the result by [11]. So let G =p(p+1)(p 1) and G be a solvable group. If p = 2 n 1, for each n>0, then using Lemma 1.8, we get that O p (G) = 1, which is a contradiction by Itô Michler theorem. Therefore let p = 2 n 1, for some n>0. By considering a Hall subgroup H of G such that G : H =(p 1)/2 we get a normal subgroup N of G such that N =2 a p, where 0 a n + 1. Using Lemma 1.4 we get that p cd(n) and so p + 1 = 2 n N. Hence N =p(p + 1) or N = H.If N =p(p + 1) = 2 n p, then

5 A new characterization for some extensions of PSL(2,q) 53 as we mentioned above n p (N) = p + 1. So if R Syl 2 (N), then R G. Letθ Irr(R) such that e =[χ R,θ] = 0. Then p = χ(1) = etθ(1), where t = G : I G (θ). This implies that et = p.ife = p and t = 1, then [χ R,χ R ]=e 2 t = p 2 G : R, which is a contradiction. Therefore e = 1 and t = p. This shows that R has p linear characters and R = p + 1, which implies that R is abelian. By assumptions there exists an irreducible character degree which is divisible by 4 and so we get a contradiction by Lemma 1.3. Therefore N = H and so N = 2p(p + 1) = 2 n+1 p. By Lemma 2.1, we get that O p (N) = 1 and so O 2 (H) = 1. Let O 2 (N) = 2 m. We know that O 2 (N/O 2 (N)) = 1. Therefore L/O 2 (N) = O p (N/O 2 (N)) = 1. This implies that we get a normal subgroup L of N such that L =2 m p.alsol has an irreducible character of degree p. This implies that m = n or m = n+1. If m = n, then L =p(p+1) and similarly to the above we get a contradiction. Hence R, the Sylow 2-subgroup of N is a normal subgroup of N.LetZ be the center of R which is nontrivial and Z N.Let Z =2 i, where i>0. By assumptions and Lemma 1.3, 1 < Z 2 n 1.AlsoN/Z G/Z. Nowif Z > 2, then by the above discussions, O p (N/Z) = 1 which implies that O p (G) = 1, a contradiction. Therefore Z =2 and N/Z =p(p + 1) = 2 n p. By Itô s theorem we get that n p (N/Z) = p + 1. We know that R/Z is a normal subgroup of N/Z. Let Z 2 /Z be the center of R/Z and let Q/Z 2 Syl p (G/Z 2 ). Obviously Z 2 4, which implies that n p (G/Z 2 ) = 1 and so Q G. LetP Syl p (Q). (i) If Z 2 (p + 1)/2 = 2 n 1, then O p (Q) = 1, which implies that O p (G) = 1, a contradiction. (ii) If Z 2 =p + 1 = 2 n, then Q =p(p + 1). Similar to the previous cases we get that n p (Q) = 1 and so n p (Q) = p + 1. Therefore n 2 (Q) = 1 and so S G, where S Syl 2 (Q) and S =p + 1 = 2 n.letϕ Irr(S) such that e =[χ S,ϕ] = 0. Then p = e t, where t = G : I G (ϕ). Since [χ S,χ S ] G : S =p(p 1), it follows that e = 1 and t = p. Hence S is abelian, which is impossible, since there exists an irreducible character degree divisible by 4. (iii) If Z 2 =2(p + 1), then R/Z is abelian and Z(R) =2. Hence R is an extraspecial group. Now using [8, 4A.4], we get that R : Z(R) =2 2e, which is a contradiction since n is an odd prime. Now suppose that q = p 3. We know that if S = PSL(2,p 3 ), then S =p 3 (p+1)(p 1)(p 2 p+1)(p 2 +p+1)/2. Hence G 3p 3 (p + 1)(p 1)(p 2 p + 1)(p 2 + p + 1) and S G. In the sequel we discuss about prime powers, say r α, such that r α G and r α = kp + 1, for some k>0. Now we consider the following cases: Case 1. Let p = 2 n + 1 and p = 2 n 1, for each n N. Hence p 11. Let m = G 2. Then m 2(p 1) or m 2(p + 1). By assumptions, m 2(p + 1)/3. Let H be a Hall subgroup of G such that G : H =m. Then G/H G S m and so if N 1 = H G, we get that p 3 N 1 and N 1 is odd. So p 3 cd(n 1 ), by Lemma 1.4. Now we prove that the only divisor of p 2 + εp + 1 of the form 1 + kp, where ε =±1 and k> 0, is p 2 + εp + 1. Let p 2 + εp + 1 = t(1 + kp), forsomet> 0. Then p (t 1) and so t = 1ort p + 1. If t = 1, then 1 + kp = p 2 + εp + 1. Otherwise p 2 + εp + 1 = t(kp + 1) (p + 1) 2, which is a contradiction. Also we

6 54 Behrooz Khosravi et al. know that 3 (p 2 ± p + 1). Let3 (p 2 + εp + 1). Now we consider the following two subcases: (a) First suppose that p 2 εp + 1 is not a prime power and so p 2 εp + 1 = rs, where r, s > 1 and (r,s) = 1. If r and s are greater than p, then rs (p + 1) 2 >p 2 + εp + 1, which is impossible. So without loss of generality let 1 <r<p. Now by considering a Hall subgroup H 1 of N 1 such that N 1 : H 1 =r, we get a normal subgroup N 2 of N 1 such that p 3 N 2, p 3 cd(n 2 ) and N 2 9p 3 (p + 1)(p 1) (p2 + εp + 1) 3 (p 2 εp + 1). r We note that (p 1,p+ 1) = 2, (p 1,p 2 + p + 1) 3 and (p + 1,p 2 p + 1) 3. So if t α N 2, where t is a prime number and t α = kp + 1, for some k> 0, then t α 9(p 1)/2 ort α 9(p + 1)/2, and by easy computation, we see that these cases are impossible. Therefore using Lemma 1.8 we get that the Sylow p-subgroup of N 2 is a normal subgroup of N 2, and so O p (N 2 ) = 1, which is a contradiction by Lemma 2.1. (b) Let p 2 εp+1 be a prime power. Now using Lemma 1.10 we get that r = p 2 εp+1 is a prime number. Let H 1 be a Hall subgroup of N 1 such that N 1 : H 1 =r. Similarly to the above we get that there exists a normal subgroup N 2 of N 1 such that N 1 /N 2 S r and N 1 : N 2 is a divisor of r(r 1) = p(p ε), since N 1 /N 2 is a solvable group and its order is divisible by r. Hence p 2 N 2. Now similarly to the above using Lemma 1.8 we get that the Sylow p-subgroup of N 2 is a normal subgroup of N 2, which is a contradiction by Lemma 2.1. Case 2. Let p = 2 n + 1orp = 2 n 1. Let p = 2 n + 1. If p = 5, then G By considering a Hall subgroup H of G such that G : H =7, we get a normal subgroup N 1 of H such that 5 3 cd(n 1 ) and N Again by considering a Hall subgroup K of N 1 such that N 1 : K =31, we get a normal subgroup N 2 of N 1 such that 5 2 N 2 and N Nowusing Lemma 1.8 and Itô Michler theorem we get a contradiction. So in the sequel suppose that p 7. By assumptions, 3 (p 2 p + 1) and G 2 2(p 1). Now similarly to the previous case by considering a Hall subgroup H of G such that G : H = G 2 we get a normal subgroup N of G such that p 2 N, N is odd and N 9p 3 (p 2 ε + 1) p p 2 + εp If (p + 1)/2 = 3 β, then p = 17. Now G Similarly to the above let H be a Hall subgroup of G such that G : H =307. Then we get a normal subgroup N of G such that G / N Therefore 17 2 N, which implies that 17 2 cd(n) and this is a contradiction since O 17 (N) = 1 by Lemma 1.8. Otherwise G 3 9(p + 1)/10 <p and exactly similar to Case (1) we get a contradiction by Lemmas 1.10 and 2.1. If p = 2 n 1, for some n N, then 3 (p 2 + p + 1) and exactly similar to the above we get a contradiction.

7 A new characterization for some extensions of PSL(2,q) 55 Therefore G is not a solvable group. Theorem 2.3. Let p 5 be a prime number and q = p or q = p 3.IfG is a finite group such that (i) G Aut(PSL(2,q)) and PSL(2,q) G, (ii) q cd(g), (iii) there does not exist any θ Irr(G) such that 2p θ(1), (iv) if q = 2 α 1 is a prime number, then 4 divides an irreducible character degree of G, then G has a unique nonabelian composition factor which is isomorphic to PSL(2,q). Proof. Using Theorem 2.2, we get that G is a nonsolvable group and by Lemma 1.6 we get that G has a normal series 1 H K G such that K/H is a direct product of m copies of a nonabelian simple group S and G/K Out(K/H). Let p G/K. We know that Out(K/H) = Out(S) S m.ifp m, then 60 p 60 m K/H G <p 9 /2, which is a contradiction. Therefore p>mand so p Out(S). Since p 5, we get that S is not isomorphic to a sporadic simple group or an alternating group. Hence S is a nonabelian simple group of Lie type over GF(q ), where q = p α 0. This is a well known result that q (q 2 1)/2 S. Ifp divides the order of a field automorphism of S, then p α. Therefore 2 p 2 α q and so 2 3p 2 q 3 /4 S p 9 /2, which is impossible. If p divides the order of a diagonal automorphism of S, then S = PSL(n,q ), where p (n,q 1); ors = PSU(n,q ), where p (n,q + 1). Let S = PSL(n,q ), where p (n,q 1). Then p n and p q 1. Then p p(p+1)/2 q n(n+1)/2 S p 9 p(p + 1)/2 9, which is a contradiction for p 5. Similarly we get that S PSU(n,q ).Soitfollows that p G/K. Therefore p 3 K, which implies that p 3 cd(k). Letq = p 3,χ Irr(K) and χ(1) = p 3. Now we consider the following cases: (1) p 3 H ; (2) p 2 H and p K/H ; (3) p H and p 2 K/H ; (4) p 3 K/H (1) Let p 3 H. As we mentioned above we get that p 3 cd(h) and since K/H is the direct product of nonabelian simple groups, we get that there exists θ Irr(K) such that 2p 3 θ(1) by Gallagher s theorem and this is a contradiction. (2) Let p 2 H and p K/H. In this case K/H is a nonabelian simple group and K/H = S p(p 6 1)/2. Also p S. Consider θ Irr(H) such that [χ H,θ] = 0. Then χ(1)/θ(1) G : H and so θ(1) = p 2. Hence H p 4. Therefore K/H = S G/H p 5. Since 2p does not divide the degree of any irreducible character of G and 2p K/H, we get that the character degree graph of K/H is not complete. Also finite simple groups with non-complete character degree graphs are known (see [14, 18]) and so K/H {J 1,M 11,M 23,A 8, 2 B 2 (q ), PSL(3,q ), PSU(3,q ), PSL(2,q )}.

8 56 Behrooz Khosravi et al. If K/H = J 1, then p = 7, 11 or 19. But in each case easily we can see that J 1 does not divide 6 PSL(2,p 3 ). Similarly we get that K/H M 11, M 23, A 8. If K/H = 2 B 2 (q ), where q = 2 2m+1 (m 1), then p (q 2 + 1), since 2 p in the character degree graph of G. On the other hand, q 2 2(p 1) or q 2 2(p + 1), which implies that p = q 2 + 1, but this is a contradiction since 5 (q 2 + 1). If K/H = PSL(3,q ) or PSU(3,q ), then 2 is adjacent to each prime divisor of K/H and specially p, which is a contradiction. Therefore K/H = PSL(2,q ).IfK/H = PSL(2, 2 n ), then p (2 n 1) or p (2 n + 1), sop 1 2 n. On the other hand, 2 n 2(p 1) or 2 n 2(p + 1). Therefore p = 2 n + 1orp = 2 n 1. If p = 2 n 1, then K/H =p(p + 1)(p + 2) and so (p + 2) G, which implies that p = 7, since (p + 2, G ) Therefore K/H = PSL(2, 8) and so H Also 7 2 cd(h). Now by considering Hall subgroups of index 17 and (6, H ), we get a contradiction. If p = 2 n + 1, then K/H =2 n (2 n 1)(2 n + 1) = (p 2)(p 1)p. Easily we can check that (p 2, G ) Hence (p 2) 3 2 7, which implies that p = 5, i.e. K/H = PSL(2, 5). If K/H = PSL(2,q ), then K/H = PSL(2,p), since 2 p in the character degree graph of G. Hence H =dp 2 (p 2 p + 1)(p 2 + p + 1), where d 6, and p 2 cd(h). Also H is solvable. Let 3 (p 2 + εp + 1), where ε =±1 and K be a Hall subgroup of H such that H : K = H 3. Then H/K H S m, where m = 3orm = 9, and so K H is a normal subgroup of H such that p 2 cd(k H ). Now exactly similarly to the proof of Theorem 2.2 by considering p 2 εp + 1 = rs, where r, s > 1 and (r,s) = 1, and p 2 εp + 1 = r, a prime number, we get a normal subgroup N of K H such that its Sylow p-subgroup is a normal subgroup of N and this is a contradiction by Itô Michler theorem. (3) Let p H and p 2 K/H. Since 2 p in the character degree graph of G, we get that K/H is a nonabelian simple group since we know that K/H is the direct product of m copies of a nonabelian simple group. Also K/H has a non-complete character degree graph. Now similarly to Case (2) we get that K/H {J 1,M 11,M 23,A 8, 2 B 2 (q ), PSL(3,q ), PSU(3,q ), PSL(2,q )}. If K/H = 2 B 2 (q ), where q = 2 2m+1 (m 1), then similarly to the previous case we get that p 2 (q 2 + 1) and q 2 2(p 1) or q 2 2(p + 1), which are impossible. Similarly it follows that K/H = PSL(2, 2 n ). Therefore K/H = PSL(2,p 2 ). Then K/H = p 2 (p 2 + 1)(p 2 1)/2 and so (p 2 + 1) G. Easily we can check that (p 2 + 1, G ) 2, which is a contradiction. (4) Let p 3 K/H. As we mentioned above we get that K/H is a nonabelian simple group with non-complete character degree graph. Similarly to the previous cases we get that K/H = PSL(2,p 3 ). If q = p, then we can proceed exactly similar to the above and since p G, we get that K/H = PSL(2,p), because 2 p in the character degree graph of G. Weomitthe details of the proof for convenience. Theorem 2.4. Let p 5 be a prime number and ε = ( 1) (p 1)/2. Let G be a finite group such that p cd(g), and there do not exist any θ Irr(G) such that 2p θ(1). Also

9 A new characterization for some extensions of PSL(2,q) 57 if p = 2 α 1 is a prime number, then 4 divides an irreducible character degree of G. Then (1) if G = PSL(2,p), then G = PSL(2,p); (2) if G =2 PSL(2,p) and (p + ε)/2 cd(g), then G = PGL(2,p); (3) if G = 2 PSL(2,p), (p + ε)/2 cd(g) and (p ε)/2 cd(g), then G = Z 2 PSL(2,p); (4) if G =2 PSL(2,p) and (p + ε)/2,(p ε)/2 cd(g), then G = SL(2,p). So we get new characterizations for PSL(2,p),PGL(2,p),Z 2 PSL(2,p)and SL(2,p). Proof. Using Theorem 2.3 it follows that PSL(2,p)is uniquely determined by the stated conditions of Theorem 2.3. Also we see that if G is a finite group of order PGL(2,p), then G has a normal series 1 H K G such that K/H = PSL(2,p)and H G/K = 2. If G/K =2, then G = PGL(2,p)and if H =2, then H Z(G). Now the Schur multiplier of PSL(2,p)implies that G = Z 2 PSL(2,p)or G = SL(2,p). Using [18] by comparing the character degrees of these groups we get the result. Theorem 2.5. Let p 5 be a prime number. Let q = p 3 and ε = ( 1) (q 1)/2. Let G be a finite group such that q cd(g), and there do not exist any θ Irr(G) such that 2p θ(1). Then (1) if G = PSL(2,p 3 ), then G = PSL(2,p 3 ); (2) if G =2 PSL(2,p 3 ) and (q + ε)/2 cd(g), then G = PGL(2,p 3 ); (3) if G =2 PSL(2,p 3 ) and (q + ε)/2 cd(g) and (q ε)/2 cd(g), then G = Z 2 PSL(2,p 3 ); (4) if G =2 PSL(2,p 3 ) and (q + ε)/2,(q ε)/2 cd(g), then G = SL(2,p 3 ); (5) if G =3 PSL(2,p 3 ) and 3(q 1) cd(g), then G = Z 3 PSL(2,p 3 ), otherwise G = PSL(2,p 3 ).3. Proof. Similarly to the above if G is a finite group such that G =h PSL(2,q), where h 6, then G has a normal series 1 H K G such that K/H = PSL(2,q) and H G/K =h.if H =3, then H is a central subgroup of G and by the Schur multiplier of PSL(2,p 3 ), we get that K = Z 3 PSL(2,p 3 ).If H =2, then K = Z 2 PSL(2,p 3 ) or K = SL(2,p 3 ). Finally if H =1 and G/K =2 or 3, then G = PSL(2,q).2 = PGL(2,q)or G = PSL(2,q).3. Therefore using [18] by comparing the character degrees of these groups we get the result. Remark 2.6. In the above theorems we get new characterizations for PSL(2,p), PGL(2,p), Z 2 PSL(2,p),SL(2,p), PSL(2,p 3 ),PGL(2,p 3 ), Z 2 PSL(2,p 3 ), Z 3 PSL(2,p 3 ), PSL(2,p 3 ).3 and SL(2,p 3 ). As a consequence of our results we get the following result: COROLLARY 2.7 Let p be a prime number and M be a finite group which stated in Remark 2.6. If G is a finite group such that X 1 (G) = X 1 (M), then G = M.

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