Proc. Indian Acad. Sci. (Math. Sci.) Vol. 126, No. 1, February 2016, pp. 49–59.
c Indian Academy of Sciences
A new characterization for some extensions of PSL(2, q)
for some q by some character degrees
BEHROOZ KHOSRAVI1,∗ , BEHNAM KHOSRAVI2
and BAHMAN KHOSRAVI3
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 45137-66731, Iran
3 Faculty of Mathematics, Department of Sciences, Qom University of Technology,
Qom, Iran
* Correspondence author. E-mail: [email protected]
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) 3330–3341) 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 . Let M 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 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.
2000 Mathematics Subject Classification.
20C15, 20D05, 20D60.
1. Introduction and preliminary results
If n is an integer, then we denote by π(n) the set of all prime divisors of n. Let G 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 X1 (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 p1 and p2 are joined by an edge if p1 p2 divides some irreducible
character degree of G.
49
50
Behrooz Khosravi et al.
A finite group G is called a K3 -group if |G| has exactly three distinct prime divisors.
Recently Xu et al. in [25] and [26] proved that all simple K3 -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 = 2a for an integer a ≥ 2 such that 2a − 1 or 2a + 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, p2 )|, p2 ∈ cd(G) and there does not exist any θ ∈ Irr(G) such that
2p | θ (1), then G ∼
= PSL(2, p2 ). In [13] it is proved that the simple group PSL(2, p2 ) 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. This is false, 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, p2 )|,
2
p ∈ 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, p2 ). Using these results we
get that each extension of PSL(2, p2 ) of order 2|PSL(2, p2 )| 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 = p3 . Let M 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.
A new characterization for some extensions of PSL(2,q)
51
g
If N G and θ ∈ Irr(N),
k then the inertia group of θ in G is IG (θ ) = {g ∈ G | θ = θ }.
If the character χ = i=1 ei χi , where for each 1 ≤ i ≤ k, χi ∈ Irr(G) and ei is a natural
number, then each χi 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 χN = 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, 11.29 of [7]). Let N G and let χ ∈ Irr(G). Let θ be an
irreducible constituent
of χN and suppose θ1 = θ, . . . , θt are the distinct conjugates of θ
in G. Then χN = e ti=1 θi , where e = [χN , θ ] and t = |G : IG (θ )|. Also θ (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 = pf 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 δ̄
of order 2 (see [2]). Also we note that PGL(2, q) = PSL(2, q)δ̄
.
Lemma 1.7 (Theorem A of [24]). Let S = PSL(2, q), where q = pf > 3 for a prime p,
A = Aut(S), and let S ≤ H ≤ A. Set G = PGL(2, q) if δ̄ ∈ H and G = S if δ̄ ∈ H, and
let |H : G| = d = 2a 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)2a i : i | m} ∪ {(q + 1)j : j | d},
with the following exceptions:
(1)
(2)
(3)
(4)
(5)
If p is odd with H ≤ Sϕ
or if p = 2, then (q + ε)/2 is not a degree of H .
If f is odd, p = 3, and H = Sϕ
, then i = 1.
If f is odd, p = 3, and H = A, then j = 1.
If f is odd, p = 2, 3 or 5, and H = Sϕ
, then j = 1.
If f ≡ 2(mod 4), p = 2 or 3, and H = Sϕ
or H = Sδ̄ϕ
, then j = 2.
Lemma 1.8 (Lemma 2 of [26]). Let G be a finite solvable group of order p1α1 p2α2 . . . pnαn ,
where p1 , p2 , . . . ,pn are distinct primes. If (kpn + 1) ∤ piαi , for each i ≤ n − 1 and k > 0,
then the Sylow pn -subgroup is normal in G.
Lemma 1.9 (Remark 1 of [3]). The equation pm − q n = 1, where p and q are primes and
m, n > 1 has only one solution, namely 32 − 23 = 1.
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Behrooz Khosravi et al.
Nagell and Ljunggren proved the following result:
Lemma 1.10 (Page 182 of [21]). If m ≥ 2, then the only non-zero solutions of X2 + 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 Sylr (G), the set of Sylow r-subgroups of G
and we denote by nr (G), the number of elements of Sylr (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 = pj , where 1 ≤ j ≤ 3. If there
exists χ ∈ Irr(G), where χ (1) = |M|p , then Op (M) = 1.
Proof. If j = 1 or j = 2, then the result obtained by Itô’s theorem (see Lemma 2.2
in [14]). So let j = 3 and L = Op (M). If η ∈ Irr(L) such that [χL , η] = 0, then
p3 /η(1) | |M : L|, by Lemma 1.4. If |L| = p3 , then η(1) = p3 , which is impossible. If
|L| = p or p2 , then L is abelian and so p3 | |M : L|, which is a contradiction. Therefore
L = 1.
Theorem 2.2. Let p ≥ 5 be a prime number and q = p or q = p3 . If G is a finite group
such that
(i)
(ii)
(iii)
(iv)
|PSL(2, q)| | |G| and |G| | |Aut(PSL(2, q))|,
there exists χ ∈ Irr(G) such that χ (1) = q,
there does not exist any θ ∈ Irr(G) such that 2p | θ (1),
If q = 2α − 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| | 23 × 3 × 5. Since there exists no prime power r α of |G| such that
= 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 = 2n − 1,
for each n > 0, then using Lemma 1.8, we get that Op (G) = 1, which is a contradiction
by Itô–Michler theorem. Therefore let p = 2n − 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 | = 2a p, where 0 ≤ a ≤ n + 1. Using Lemma 1.4 we get that p ∈ cd(N) and so
p + 1 = 2n | |N|. Hence |N | = p(p + 1) or N = H . If |N| = p(p + 1) = 2n p, then
rα
A new characterization for some extensions of PSL(2,q)
53
as we mentioned above np (N) = p + 1. So if R ∈ Syl2 (N), then R ⊳ G. Let θ ∈ Irr(R)
such that e = [χR , θ ] = 0. Then p = χ (1) = etθ (1), where t = |G : IG (θ )|. This
implies that et = p. If e = p and t = 1, then [χR , χR ] = e2 t = p2 ≤ |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) = 2n+1 p. By Lemma 2.1, we get that
Op (N) = 1 and so O2 (H ) = 1. Let O2 (N) = 2m . We know that O2 (N/O2 (N)) = 1.
Therefore L/O2 (N) = Op (N/O2 (N)) = 1. This implies that we get a normal subgroup
L of N such that |L| = 2m p. Also L 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. Let Z be
the center of R which is nontrivial and Z ⊳ N. Let |Z| = 2i , where i > 0. By assumptions
and Lemma 1.3, 1 < |Z| ≤ 2n−1 . Also N/Z ⊳ G/Z. Now if |Z| > 2, then by the above
discussions, Op (N/Z) = 1 which implies that Op (G) = 1, a contradiction. Therefore
|Z| = 2 and |N/Z| = p(p + 1) = 2n p. By Itô’s theorem we get that np (N/Z) = p + 1.
We know that R/Z is a normal subgroup of N/Z.
Let Z2 /Z be the center of R/Z and let Q/Z2 ∈ Sylp (G/Z2 ). Obviously |Z2 | ≥ 4,
which implies that np (G/Z2 ) = 1 and so Q ⊳ G. Let P ∈ Sylp (Q).
(i) If |Z2 | ≤ (p + 1)/2 = 2n−1 , then Op (Q) = 1, which implies that Op (G) = 1, a
contradiction.
(ii) If |Z2 | = p + 1 = 2n , then |Q| = p(p + 1). Similar to the previous cases we get
that np (Q) = 1 and so np (Q) = p + 1. Therefore n2 (Q) = 1 and so S ⊳ G, where
S ∈ Syl2 (Q) and |S| = p + 1 = 2n . Let ϕ ∈ Irr(S) such that e′ = [χS , ϕ] = 0. Then
p = e′ t ′ , where t ′ = |G : IG (ϕ)|. Since [χS , χS ] ≤ |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 |Z2 | = 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)| = 22e , which is a contradiction
since n is an odd prime.
• Now suppose that q = p3 .
We know that if S = PSL(2, p3 ), then |S| = p3 (p+1)(p−1)(p2 −p+1)(p2 +p+1)/2.
Hence |G| | 3p3 (p + 1)(p − 1)(p2 − p + 1)(p2 + 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 = 2n + 1 and p = 2n − 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/HG ֒→ Sm and so if
N1 = HG , we get that p3 | |N1 | and |N1 | is odd. So p3 ∈ cd(N1 ), by Lemma 1.4.
Now we prove that the only divisor of p2 + εp + 1 of the form 1 + kp, where ε = ±1
and k > 0, is p2 + εp + 1. Let p2 + εp + 1 = t (1 + kp), for some t > 0. Then
p | (t − 1) and so t = 1 or t ≥ p + 1. If t = 1, then 1 + kp = p2 + εp + 1.
Otherwise p2 + εp + 1 = t (kp + 1) ≥ (p + 1)2 , which is a contradiction. Also we
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Behrooz Khosravi et al.
know that 3(p2 ± p + 1). Let 3 | (p2 + εp + 1). Now we consider the following two
subcases:
(a) First suppose that p2 − εp + 1 is not a prime power and so p2 − εp + 1 = rs, where
r, s > 1 and (r, s) = 1. If r and s are greater than p, then rs ≥ (p + 1)2 > p2 + εp + 1,
which is impossible. So without loss of generality let 1 < r < p. Now by considering a
Hall subgroup H1 of N1 such that |N1 : H1 | = r, we get a normal subgroup N2 of N1
such that p3 | |N2 |, p3 ∈ cd(N2 ) and
|N2 | | 9p3 (p + 1)(p − 1)
(p2 + εp + 1) (p2 − εp + 1)
.
3
r
We note that (p − 1, p + 1) = 2, (p − 1, p2 + p + 1) | 3 and (p + 1, p2 − p + 1) | 3.
So if t α | |N2 |, where t is a prime number and t α = kp + 1, for some k > 0, then
t α | 9(p − 1)/2 or t 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 N2 is a
normal subgroup of N2 , and so Op (N2 ) = 1, which is a contradiction by Lemma 2.1.
(b) Let p2 −εp +1 be a prime power. Now using Lemma 1.10 we get that r = p2 −εp +1
is a prime number. Let H1 be a Hall subgroup of N1 such that |N1 : H1 | = r. Similarly
to the above we get that there exists a normal subgroup N2 of N1 such that N1 /N2 ֒→ Sr
and |N1 : N2 | is a divisor of r(r − 1) = p(p − ε), since N1 /N2 is a solvable group and its
order is divisible by r. Hence p2 | |N2 |. Now similarly to the above using Lemma 1.8 we
get that the Sylow p-subgroup of N2 is a normal subgroup of N2 , which is a contradiction
by Lemma 2.1.
Case 2. Let p = 2n + 1 or p = 2n − 1.
Let p = 2n + 1. If p = 5, then |G| | 23 33 53 7 × 31. By considering a Hall subgroup H
of G such that |G : H | = 7, we get a normal subgroup N1 of H such that 53 ∈ cd(N1 ) and
|N1 | | 23 33 53 31. Again by considering a Hall subgroup K of N1 such that |N1 : K| = 31,
we get a normal subgroup N2 of N1 such that 52 | |N2 | and |N2 | | 23 33 53 . Now using
Lemma 1.8 and Itô–Michler theorem we get a contradiction. So in the sequel suppose that
p ≥ 7. By assumptions, 3 | (p2 − 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 p2 | |N |, |N| is odd and
|N | | 9p3 (p2 − ε + 1)
p + 1 p2 + εp + 1
.
2
3
If (p + 1)/2 = 3β , then p = 17. Now |G| | 173 25 34 × 7 × 13 × 307. 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 | | 307 × 306. Therefore 172 | |N|, which implies that
172 ∈ cd(N) and this is a contradiction since O17 (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 = 2n − 1, for some n ∈ N, then 3 | (p2 + p + 1) and exactly similar to the above
we get a contradiction.
A new characterization for some extensions of PSL(2,q)
Therefore G is not a solvable group.
55
Theorem 2.3. Let p ≥ 5 be a prime number and q = p or q = p3 . If G is a finite group
such that
(i)
(ii)
(iii)
(iv)
|G| | |Aut(PSL(2, q))| and |PSL(2, q)| | |G|,
q ∈ cd(G),
there does not exist any θ ∈ Irr(G) such that 2p | θ (1),
if q = 2α − 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) ≀ Sm . If p ≤ m, then 60p ≤ 60m ≤
9
|K/H | ≤ |G| < p /2, which is a contradiction. Therefore p > m and 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 ′ = p0α .
This is a well known result that q ′ (q ′2 − 1)/2 ≤ |S|. If p divides the order of a field
automorphism of S, then p | α. Therefore 2p ≤ 2α ≤ q ′ and so 23p−2 ≤ q ′3 /4 ≤ |S| ≤
p9 /2, which is impossible. If p divides the order of a diagonal automorphism of S, then
S ∼
= PSU(n, q ′ ), where p | (n, q ′ + 1). Let
= PSL(n, q ′ ), where p | (n, q ′ − 1); or S ∼
′
′
∼
S = PSL(n, q ), where p | (n, q − 1). Then p ≤ n and p ≤ q ′ − 1. Then
pp(p+1)/2 ≤ q ′n(n+1)/2 ≤ |S| ≤ p9 ⇒ p(p + 1)/2 ≤ 9,
which is a contradiction for p ≥ 5. Similarly we get that S ≇ PSU(n, q ′ ). So it follows
that p ∤ |G/K|. Therefore p3 | |K|, which implies that p3 ∈ cd(K). Let q = p3 , χ ∈
Irr(K) and χ (1) = p3 . Now we consider the following cases:
(1)
(2)
(3)
(4)
p3 | |H |;
p2 | |H | and p|K/H |;
p|H | and p2 | |K/H |;
p3 | |K/H |
(1) Let p3 | |H |. As we mentioned above we get that p3 ∈ cd(H ) and since K/H is the
direct product of nonabelian simple groups, we get that there exists θ ∈ Irr(K) such that
2p3 | θ (1) by Gallagher’s theorem and this is a contradiction.
(2) Let p2 | |H | and p|K/H |. In this case K/H is a nonabelian simple group and
|K/H | = |S| | p(p6 − 1)/2. Also p | |S|. Consider θ ∈ Irr(H ) such that [χH , θ ] = 0.
Then χ (1)/θ (1) | |G : H | and so θ (1) = p2 . Hence |H | ≥ p4 . Therefore |K/H | =
|S| ≤ |G/H | ≤ p5 . 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 ∈ {J1 , M11 , M23 , A8 , 2 B2 (q ′ ), PSL(3, q ′ ), PSU(3, q ′ ), PSL(2, q ′ )}.
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Behrooz Khosravi et al.
If K/H ∼
= J1 , then p = 7, 11 or 19. But in each case easily we can see that |J1 | does not
divide 6|PSL(2, p3 )|. Similarly we get that K/H ≇ M11 , M23 , A8 .
2
If K/H ∼
= 2 B2 (q ′ ), where q ′ = 22m+1 (m ≥ 1), then p | (q ′ + 1), since 2 ≁ p in the
2
′
character degree graph of G. On the other hand, q | 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 ′ ). If K/H ∼
=
n
n
n
PSL(2, 2 ), then p | (2 − 1) or p | (2 + 1), so p − 1 ≤ 2n . On the other hand,
2n | 2(p − 1) or 2n | 2(p + 1). Therefore p = 2n + 1 or p = 2n − 1. If p = 2n − 1,
then |K/H | = p(p + 1)(p + 2) and so (p + 2) | |G|, which implies that p = 7, since
(p + 2, |G|) | 32 7. Therefore K/H ∼
= PSL(2, 8) and so |H | | 6 × 72 17 × 43. Also
2
7 ∈ cd(H ). Now by considering Hall subgroups of index 17 and (6, |H |), we get a
contradiction. If p = 2n + 1, then |K/H | = 2n (2n − 1)(2n + 1) = (p − 2)(p − 1)p.
Easily we can check that (p − 2, |G|) | 32 7. Hence (p − 2) | 32 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 | = dp2 (p2 − p + 1)(p2 + p + 1), where d | 6, and p2 ∈ cd(H ).
Also H is solvable. Let 3 | (p2 + εp + 1), where ε = ±1 and K be a Hall subgroup of
H such that |H : K| = |H |3 . Then H /KH ֒→ Sm , where m = 3 or m = 9, and so KH
is a normal subgroup of H such that p2 ∈ cd(KH ). Now exactly similarly to the proof
of Theorem 2.2 by considering p2 − 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 KH 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 p2 |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 ∈ {J1 , M11 , M23 , A8 , 2 B2 (q ′ ), PSL(3, q ′ ), PSU(3, q ′ ), PSL(2, q ′ )}.
If K/H ∼
= 2 B2 (q ′ ), where q ′ = 22m+1 (m ≥ 1), then similarly to the previous case we get
2
that p | (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, p2 ). Then |K/H | =
= PSL(2, 2n ). Therefore K/H ∼
2
2
2
2
p (p + 1)(p − 1)/2 and so (p + 1) | |G|. Easily we can check that (p2 + 1, |G|) | 2,
which is a contradiction.
(4) Let p3 | |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, p3 ).
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. We omit the
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
A new characterization for some extensions of PSL(2,q)
57
if p = 2α − 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 ∼
=
Z2 × 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), Z2 × 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 1H 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 ∼
= Z2 × 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 = p3 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, p3 )|, then G ∼
= PSL(2, p3 );
3
(2) if |G| = 2|PSL(2, p )| and (q + ε)/2 ∈ cd(G), then G ∼
= PGL(2, p3 );
3
(3) if |G| = 2|PSL(2, p )| and (q + ε)/2 ∈ cd(G) and (q − ε)/2 ∈ cd(G), then G ∼
=
Z2 × PSL(2, p3 );
(4) if |G| = 2|PSL(2, p3 )| and (q + ε)/2, (q − ε)/2 ∈ cd(G), then G ∼
= SL(2, p3 );
3
∼
(5) if |G| = 3|PSL(2, p )| and 3(q − 1) ∈ cd(G), then G = Z3 × PSL(2, p3 ), otherwise
G∼
= PSL(2, p3 ).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, p3 ), we get that K ∼
= Z3 × PSL(2, p3 ). If |H | = 2, then K ∼
= Z2 × PSL(2, p3 )
3 ). Finally if |H | = 1 and |G/K| = 2 or 3, then G ∼ PSL(2, q).2 =
or K ∼
SL(2,
p
=
=
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), Z2 × PSL(2, p), SL(2, p), PSL(2, p3 ), PGL(2, p3 ), Z2 × PSL(2, p3 ), Z3 ×
PSL(2, p3 ), PSL(2, p3 ).3 and SL(2, p3 ).
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 X1 (G) = X1 (M), then G ∼
= M.
58
Behrooz Khosravi et al.
The following result is an answer to Question 1.1.
COROLLARY 2.8
Let p be a prime number and M be a finite group which stated in Remark 2.6. If G is a
group such that CG ∼
= CM, then G ∼
= M. Thus M is uniquely determined by the structure
of its complex group algebra.
Acknowledgements
The authors express their gratitude to the referee and Professor Derek Holt for very useful
comments.
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C OMMUNICATING E DITOR: B Sury