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ISSN (print): 2251-7650, ISSN (on-line): 2251-7669 Vol. 5 No. 2 (2016), pp. 61-74.
c
⃝2016 University of Isfahan
www.ui.ac.ir
CHARACTERIZATION OF SOME SIMPLE K4-GROUPS BY SOME
IRREDUCIBLE COMPLEX CHARACTER DEGREES
SOMAYEH HEYDARI AND NEDA AHANJIDEH∗
Communicated by Mark L. Lewis
Abstract. In this paper, we examine that some finite simpleK4-groups can be determined uniquely by their orders and one or two irreducible complex character degrees.
1. Introduction
Throughout this paper, letGbe a finite group and let all characters be complex characters. Also, let
l(G) be the largest irreducible character degree ofG,s(G) be the second largest irreducible character degree of Gand t(G) be the third largest irreducible character degree of G. The set of all irreducible characters of G is shown by Irr(G) and the set of all irreducible character degrees ofG is shown by cd(G). In [4], B. Huppert conjectured that if Gis a finite group andS is a finite non-abelian simple
group such that cd(G) = cd(S), thenG∼=S×A, whereA is an abelian group. In [7], [11] and [12], it is shown that L2(p), simple K3-groups and Mathieu simple groups are determined uniquely by their orders and one or two irreducible character degrees. In [6], it is proved that if 2a+ 1 or 2a−1 is
prime, then L2(2a) is determined uniquely by its order and the largest irreducible character degree. Also, in [3], the finite groups with the same order and the same largest and second largest irreducible
character degrees as P GL(2, p) have been determined. In this paper, we are going to prove that:
Main Theorem. Let Gbe a finite group.
(I)Let M beA7,L2(25),L2(81) or Sz(8). ThenG∼=M if and only if |G|=|M|and l(G) =l(M). (II)LetM beU3(7),Sz(32),U3(8) orU3(9). ThenG∼=M if and only if|G|=|M|ands(G) =s(M). MSC(2010): Primary 20C15; Secondary 20E99.
Keywords: Irreducible complex character degree, Finite simpleK4-group. Received: 10 October 2014, Accepted: 24 February 2015.
∗Corresponding author.
(III)LetM beS6(2),L3(8),L3(5), O8+(2),L3(7), L2(49),G2(3), U3(5) orS4(7). ThenG∼=M if and only if |G|=|M|,l(G) =l(M) andt(G) =t(M).
(IV) Let M beL3(4), U4(3), 2F4(2)′ orU5(2). ThenG ∼=M if and only if |G|=|M|, l(G) =l(M) and s(G) =s(M).
(V) Let M be U3(4), 3D4(2), S4(4) or S4(5). Then G ∼=M if and only if |G|= |M|, s(G) = s(M) and t(G) =t(M).
(VI)LetL2(3m) be a simpleK4-group as Lemma 1.6(ii)(4), which appears below. ThenG∼=L2(3m) if and only if |G|=|L2(3m)|,l(G) =l(L2(3m)) ands(G) =s(L2(3m)).
Note that by [7] and [6], the finite simple K4-groups L2(p) and L2(2a) are determined uniquely by their orders and one or two irreducible character degrees. Also, the statements in the main
theo-rem do not hold for the theo-remaining finite simple K4-groups that are not covered in the main theorem. So they have been studied in the separate work.
Throughout this paper, we use the following notations: For two natural numbersband na natural number n,π(n) is the set of all prime divisors of n. The set of all prime divisors of π(|G|) is denoted
by π(G) and also, if|π(G)|=n, then we say thatGis a Kn-group. For two natural numbersb andn
and a prime a, we write |b|a =an when an ∥b i.e., an |b while an+1 ∤b. For p∈π(G), the set of all
p-Sylow subgroups ofGis denoted by Sylp(G) and we writenp =|Sylp(G)|. For the subgroupH ofG,
we setHG =∩g∈GHg. IfH is a characteristic subgroup ofG, then we writeHchG. Ifχ=
∑N
i=1niχi,
where for every 1 ≤i≤N, χi ∈Irr(G), then those χi withni >0 are called irreducible constituents
of χ.
In the following, we bring some lemmas, which are used in the proof of the main theorem. Also,
we mention that throughout this paper, for the orders, character degrees, the orders of the outer
automorphism groups of finite simple groups, we refer the reader to [1].
Lemma 1.1. (Ito’s theorem) [5, Theorem 6.15] Let A ⊴ G be abelian. Then χ(1)|[G:A], for all
χ∈Irr(G).
Lemma 1.2. (Clifford’s theorem)[5, Theorem 6.2 and Corollary 11.29] Let N ⊴G and χ ∈ Irr(G).
Let θ be an irreducible constituent ofχN and suppose thatθ1 =θ, . . . , θt are the distinct conjugates of θ in G. Then χN =e
∑t
i=1θi, where e= [χN, θ]. Also, χ(1)/θ(1)|[G:N].
Lemma 1.3. [12] Let G be a non-solvable group. Then G has a normal series 1⊴H ⊴K ⊴G such that K/H is a direct product of isomorphic non-abelian simple groups and |G/K| | |Out(K/H)|.
Lemma 1.4. [12] Let Gbe a finite solvable group of order ∏ni=1piai, where p1, p2, . . . , pn are distinct
primes. If kpn+ 1∤piai for each i≤n−1 and k >0, then the pn-Sylow subgroup of G is normal in
G.
Proof. Since N ≤Z(CG(N)) andN ∈Sylt(CG(N)), lemma follows. □
Lemma 1.6. (i) [2] If G is a simple K3-group, then G is isomorphic to one of the following groups:
A5,A6, L2(7), L2(8), L2(17), L3(3), U3(3) or U4(2).
(ii) [8, 10] Let G be a simple K4-group. Then G is isomorphic to one of the following groups:
(1) A7, A8, A9, A10, M11, M12, J2, L2(16), L2(25), L2(49), L2(81), L3(4), L3(5), L3(7), L3(8),
L3(17), L4(3), S4(4), S4(5), S4(7), S4(9), S6(2), O8+(2), G2(3), U3(4), U3(5), U3(7), U3(8),
U3(9), U4(3), U5(2), Sz(8), Sz(32), 3D4(2), 2F4(2)′;
(2) L2(r), wherer is a prime, 17̸=r ≥11, r2−1 = 2a3bvc, v >3 is a prime, a, b∈N, and c is either1 or 2 for r∈ {97,577};
(3) L2(2m), where m≥5, u= 2m−1, andt= (2m+ 1)/3 are primes;
(4) L2(3m), where m andu= (3m+ 1)/4 are odd primes and (3m−1)/2 is either a prime or 112 (for m= 5).
Lemma 1.7. For n∈ {3,4}, let G be a finite Kn-group. If there is not any finite simple group L in
Lemma 1.6 such that π(L)⊆π(G), then G is solvable.
Proof. It follows immediately from Lemmas 1.3 and 1.6. □
Lemma 1.8. If G is one of the groups in the main theorem, thenG is non-solvable.
Proof. On the contrary, let Gbe solvable. We are going to complete the proof in the following cases: Case a. LetM be one of the groups mentioned in the main theorem (I) or (II).
Note that l(A7) = 5·7, l(L2(25)) = 2·13, l(L2(81)) = 2·41, l(Sz(8)) = 7·13, s(U3(7)) = 23·43,
s(Sz(32)) = 52 ·41, s(U3(8)) = 33 ·19 and s(U3(9)) = 2·5 ·73. Let χ ∈ Irr(G) such that for
M =A7,L2(25), L2(81) and Sz(8), χ(1) = l(G) =l(M) and forM =U3(7), Sz(32), U3(8), U3(9),
χ(1) = s(G) = s(M). For M = A7, set p = 5, for M = L2(25) and M = Sz(8), set p = 13, for
M =L2(81), set p= 41, for M =U3(7), set p= 43, forM =Sz(32), setp= 41, for M =U3(8), set
p= 19 and forM =U3(9), setp= 73. LetP ∈Sylp(G). Then since|M|=|G|,P is a cyclic group of orderp. SinceGis solvable and, for everyt∈π(M) and every natural numberk,pk+ 1∤|M|t=|G|t, Lemma 1.4 shows that the p-Sylow subgroupP of Gis normal in G. Thus Ito’s theorem implies that
χ(1)|[G:P], which is impossible. So,Gis non-solvable.
Case b. Let M be one of the groups mentioned in the main theorem (III), (IV) or (V).
Note that|S6(2)|= 29·34·5·7,|L3(5)|= 25·3·53·31,|O8+(2)|= 212·35·52·7,|L2(49)|= 24·3·52·72,
|G2(3)|= 26·36·7·13, |U3(5)| = 24·32 ·53·7, |L3(7)| = 25·32 ·73·19, |S4(7)| = 28 ·32·52·74,
|L3(8)| = 29 ·32 ·72·73, |L3(4)| = 26·32·5·7, |U5(2)| = 210·35 ·5·11, |U4(3)| = 27 ·36·5·7,
|2F4(2)′|= 211·33·52·13,|S
4(4)|= 28·32·52·17,|U3(4)|= 26·3·52·13,|3D4(2)|= 212·34·72·13,
|S4(5)|= 26·32·54·13,l(S6(2)) = 29,t(S6(2)) = 34·5,l(L3(5)) = 2·3·31,t(L3(5)) = 53,l(O8+(2)) = 35 ·52, t(O+
8(2)) = 212, l(L3(7)) = 23 ·3·19, t(L3(7)) = 73, l(L2(49)) = 2·52, t(L2(49)) = 24·3,
l(S4(7)) = 27 ·52, t(S4(7)) = 74 (see [9]), l(G2(3)) = 26 ·13, t(G2(3)) = 36, l(U3(5)) = 24 ·32,
s(L3(4)) = 32·7, l(U4(3)) = 27·7,s(U4(3)) = 36, l(U5(2)) = 35·5, s(U5(2)) = 210, s(3D4(2)) = 212,
t(3D4(2)) = 34·72,s(U3(4)) = 13·5, t(U3(4)) = 26,s(S4(4)) = 28,t(S4(4)) = 3·5·17,s(S4(5)) = 54 and t(S4(5)) = 24·3·13.
Letχ, β∈Irr(G) such that forM =S6(2), L3(8), L3(5), O8+(2), L3(7), L2(49), G2(3), U3(5) andS4(7),
χ(1) =l(G) =l(M) andβ(1) =t(G) =t(M), forM =L3(4), U4(3),2F4(2)′ andU5(2), χ(1) =l(G) =
l(M) andβ(1) =s(G) =s(M) and forM =U3(4),3D4(2),S4(4) andS4(5), χ(1) =s(G) =s(M) and
β(1) = t(G) =t(M). First, let M ̸∈ {L3(8), S4(4), U3(4)}. Let N be a normal minimal subgroup of
G. Then for somet∈π(G), N is a t-elementary abelian subgroup ofG. Thus applying Ito’s theorem toN,χand β shows that ifM =S6(2), O8+(2), G2(3) orU3(5), then t= 7 =|N|, if M =L3(5), then
t= 2 and|N| ≤24, ifM =L2(49), thent= 7 and|N| |72, ifM =L3(4) orU4(3), then|N|=t= 5, if
M =U5(2), then|N|=t= 11, ifM =3D4(2), thent= 13 =|N|, ifM =2F4(2)′, thent∈ {5,13}and ifM ∈ {L3(7), S4(5), S4(7)}, then t∈ {2,3}. Let M =S6(2), O8+(2), G2(3), U3(5), L3(4), U4(3), U5(2) or 3D4(2). Then since G/CG(N) ,→Aut(N) =∼Z6, Z4, Z10 orZ12, we deduce thatN < CG(N). But t∥|G|and hence by Lemma 1.5, CG(N) = N ×C, where C is a (π(CG(N))− {t})-Hall subgroup of CG(N), which is a normal subgroup ofG. Therefore, there exists a normal minimal subgroup N1 of
G such that N1 ≤ C and hence, |N1| | |G|/t = |M|/t, which is a contradiction with the fact that every normal minimal subgroup of G is a t-group. Now, let M = L3(5), t = 2 and 1 ̸= |N| ≤ 24. Then applying Lemma 1.4 to G/N shows that for P ∈Syl31(G),P N/N⊴G/N and hence, P N⊴G. But |P N| |24·31 and hence, applying Lemma 1.4 toP N shows that P ch P N. ThusP is a normal minimal subgroup ofG, which is a contradiction with the fact that every normal minimal subgroup of
Gis a 2-group. LetM =L3(7) and letL1be a maximal normal 2-subgroup ofGandL2 be a maximal normal 3-subgroup ofG. Set L=L1×L2 and letθ∈Irr(L1) such that [χL1, θ]̸= 0. Then by Lemma 1.2, |χ(1)|2/[G:L1]2 |θ(1). Also,θ2(1) + 1≤ |L1|. Thus we can check at once that either |L1| ≤4 or
|L1|= 8 and θ(1) = 2, and hence, in the latter caseL1 =D8 orQ8. The same argument guarantees that |L2| ≤3. ThusG/CG(L),→Aut(L1)×Aut(L2), where Aut(L1)∈ {D8, S4, GL2(2), Z2,{1}} and Aut(L2) ∈ {Z2,{1}}. This forces 73·19 | |CG(L)|. If|L1|= 8, then |CG(L)L1/L|2 ≤ 22 and hence,
P L/L chCG(L)L1/L⊴G/L, whereP ∈Syl7(CG(L)). Thus P×L⊴Gand hence, P⊴G, which is a
contradiction with the above statements. Now, let|L1| ≤4. ThenLis abelian and hence,L≤CG(L).
Also, by the above statements, |CG(L)/L| ̸= 1. Now, let S/L be a normal minimal subgroup of G/L
such that S/L ≤ CG(L)/L. Then for some s ∈ π(CG(L)), S/L is s-elementary abelian and hence,
there exists a s-subgroup S1 of G such that S = S1×L⊴G. Thus S1 ⊴G. But by assumptions on L,s̸∈ {2,3}, which is a contradiction with the fact that every normal minimal subgroup ofG is a 2-group or a 3-group. Let M = L2(49) and let L be a maximal normal 7-subgroup of G. Then
|L| | 72. On the other hand, G/CG(L) ,→ Aut(L) and hence, |G/CG(L)| | 1,6,|GL2(7)| or 6·7. Thus L < CG(L). Now, applying the argument given for M =L3(7) leads us to get a contradiction.
Also, if M ∈ {S4(7),2F4(2)′, S4(5)}, then repeating the argument given for L3(7) leads us to get a contradiction. Let M =L3(8). Suppose that H is a Hall-subgroup of G of order 29·32·73. Then
N is a t-elementary abelian subgroup of G. On the other hand, by Ito’s theorem, χ(1) = 32·73 and
β(1) = 29 divide |G|/|N|, which is impossible. Now, let M = S4(4) and let H be a (|M|/9)-Hall subgroup of G. Then G/HG ,→ S9 and hence, 5·17| |HG|. LetL be a normal minimal subgroup of
Gsuch thatL≤HG. Then Ito’s theorem shows thatL is a 5-elementary abelian group. Note thatL
is a maximal normal 5-subgroup of Gand |L|= 5. Also, HG/CHG(L),→ Aut(L) and hence, we can
see at once that 5·17| |CHG(L)|. IfE/L is a normal minimal subgroup ofCHG(L)/L, then for some e∈π(CHG(L)),E/L ise-elementary abelian and hence, there exists ane-subgroupE1 of Gsuch that
E =E1×L⊴G. Thus E1⊴G. But by assumption onL, e̸= 5, which is a contradiction with the fact that every normal minimal subgroup of G which is a subgroup of HG is a 5-group. Finally let M = U3(4). Then since for every natural number k, 13k+ 1 ∤ 26,3 or 52, Lemma 1.4 implies that
P ⊴G, where P ∈Syl13(G). On the other hand, P is abelian and hence, Ito’s theorem shows that
χ(1) = 13·5|[G:P], which is impossible.
Case c. LetM =L2(3m) be a simpleK4-group as Lemma 1.6(ii)(4).
Let χ, β ∈Irr(G) such that χ(1) = l(G) = l(L2(3m)) = 3m+ 1 and β(1) = s(G) = s(L2(3m)) = 3m. Note that by Lemma 1.6(ii)(4), mandu= (3m+ 1)/4 are odd primes and (3m−1)/2 is either a prime
or 112 (for m = 5). Thus π((3m−1)/2) = {t}. Also, |G| = |L2(3m)| = (3m·(3m−1)(3m+ 1))/2. Let P ∈Sylu(G). Then since for every natural numberk,ku+ 1∤|G|2,|G|3 or|G|t, we conclude by Lemma 1.4 that P ⊴G. Moreover, |P|=u. Thus P is abelian and hence, Ito’s theorem shows that
χ(1) = 3m+ 1 = 4u|[G:P], which is impossible.
These contradictions show that Gis non-solvable. □
2. Proof of the Main Theorem
We only need to prove the sufficiency of the main theorem. Note that since |G|=|M|and M is a
K4-group, Gis aK4-group.
Proof of (I). Let χ ∈ Irr(G) such that χ(1) = l(G) = l(M). Then Lemma 1.8 shows that G is
non-solvable. Therefore by Lemma 1.3, there exists a normal series 1⊴H ⊴K ⊴G such that K/H
is a direct product of isomorphic non-abelian simpleK3 orK4-groups and|G/K| | |Out(K/H)|. Now, by comparing the order of G and the orders of the non-abelian simple groups mentioned in Lemma 1.6, we can see that K/H is a simple group, which is isomorphic to one of the following groups:
i. LetM =A7. Then|G|=|A7|= 23·32·5·7, χ(1) =l(G) =l(A7) = 5·7 and K/H is isomorphic to A5, A6, L2(7), L2(8) or A7. If K/H ∼= A5, then |G/K||H|= 2·3·7 and |G/K| | |Out(A5)|= 2. Thus |H|= 2t·3·7, where 0 ≤t≤1. Let P ∈Syl7(H). Then we can see that P chH ⊴G. Thus
P is a normal abelian subgroup of G and hence, Ito’s theorem shows that χ(1) = 5·7 | [G : P],
which is impossible. If K/H ∼= A6, then |G/K||H| = 7 and |G/K| | |Out(A6)| = 4. Thus |H| = 7 and hence, Ito’s theorem implies that χ(1) = 5·7 |[G:H], which is impossible. IfK/H ∼=L2(7) or
L2(8), then since |Out(L2(7))| = 2 and |Out(L2(8))| = 3, Lemma 1.3 implies that |H| = 3·5 or 5. Let P ∈ Syl5(H). Then we can see that P ch H ⊴ G. Thus P is a normal abelian subgroup of G
and hence, Ito’s theorem shows that χ(1) = 5·7 |[G:P], which is impossible. If K/H ∼=A7, then
ii. Let M = L2(25). Then |G| = |L2(25)| = 23 ·3·52·13, χ(1) = l(G) = l(L2(25)) = 2·13 and
K/H ∼= A5 or L2(25). If K/H ∼=A5, then |H||G/K| = 2·5·13 and |G/K| | |Out(A5)| = 2. Thus
|H|= 2t·5·13, where 0≤t≤1 and hence, we can see at once that forP ∈Syl
13(H),P chH ⊴G. Moreover,P is abelian and hence, Ito’s theorem shows thatχ(1) = 2·13|[G:P], which is impossible. Thus K/H ∼=L2(25) and hence,G∼=L2(25).
iii. Let M = L2(81). Then |G| = |L2(81)| = 24 ·34 ·5·41, χ(1) = l(G) = l(L2(81)) = 2·41 and K/H ∼= A5, A6 or L2(81). Suppose that K/H ∼= A5. Then |H||G/K| = 22 ·33 ·41 and
|G/K| | |Out(A5)| = 2. Thus |H| = 2t·33 ·41, where 1 ≤ t ≤ 2. Therefore we can see at once that a P ∈Syl41(H) is normal in H. Thus P chH ⊴G. Moreover, P is abelian. So, Ito’s theorem implies thatχ(1) = 2·41|[G:P], which is impossible. LetK/H ∼=A6. Then |H||G/K|= 2·32·41 and |G/K| | |Out(A6)|= 4. Thus|H|= 2·32·41 or 32·41 and hence, a 41-Sylow subgroup P of H is normal in H. Therefore repeating the argument used in the case when K/H ∼=A5 leads us to get a contradiction. ThusK/H ∼=L2(81) and hence,G=K∼=L2(81).
iv. Let M = Sz(8). Then |G| = |Sz(8)| = 26 ·5·7·13, χ(1) = l(G) = l(Sz(8)) = 7·13 and
K/H ∼=Sz(8). It follows thatG=K ∼=Sz(8), as desired.
Proof of (II). Let χ ∈ Irr(G) such that χ(1) = s(G) = s(M). Then Lemma 1.8 shows that G
is non-solvable and so, by Lemma 1.3, there exists a normal series 1⊴ H ⊴K ⊴ Gsuch that K/H
is a direct product of isomorphic non-abelian simpleK3 orK4-groups and|G/K| | |Out(K/H)|. Now by comparing the order of G and the orders of the non-abelian simple groups mentioned in Lemma 1.6, we can see that K/H is isomorphic to one of the following groups:
i. Let M = U3(7). Then |G| = |U3(7)| = 27 ·3·73 ·43, χ(1) = s(G) = s(U3(7)) = 23 ·43 and
K/H ∼=L2(7) orU3(7). First, suppose thatK/H ∼=L2(7). Then|G/K| | |Out(L2(7))|= 2 and so, for some 3≤t≤4,|H|= 2t·72·43. Hence, we can see at once that a 43-Sylow subgroupP ofH is normal in H. ThusP ch H ⊴G and since P is abelian, Ito’s theorem implies that χ(1) = 23·43 |[G:P], which is impossible. Therefore K/H ∼=U3(7). ThusH ={1}=G/K and hence, G=K∼=U3(7). ii. LetM =Sz(32). Then|G|=|Sz(32)|= 210·52·31·41 andχ(1) =s(G) =s(Sz(32)) = 52·41. Also,Sz(32) is the only non-abelian simple group mentioned in Lemma 1.6 such that the set of prime divisors of its order is a subset of π(G). It follows that K/H ∼=Sz(32) and hence, G=K ∼=Sz(32), as claimed.
iii. Let M = U3(8). Then |G| = |U3(8)| = 29 ·34 ·7·19, χ(1) = s(G) = s(U3(8)) = 33 ·19 and K/H ∼= L2(7), L2(8), U3(3) or U3(8). Let K/H =∼ L2(7). Then |G/K||H| = 26 ·33 ·19 and
|G/K| | |Out(L2(7))|= 2. Thus |H|= 26·33·19 or 25·33·19. Hence, we can see at once by Lemma 1.7 that H is solvable and a P ∈ Syl19(H) is normal in H. Therefore, P ch H ⊴ G. Moreover, since |P|= 19,P is abelian. Now, Ito’s theorem leads us to get a contradiction. Applying the same argument rules out the cases when K/H ∼=L2(8) or U3(3). Thus K/H ∼= U3(8), which implies that
G=K ∼=U3(8).
iv. LetM =U3(9). Then|G|=|U3(9)|= 25·36·52·73,χ(1) =s(G) =s(U3(9)) = 2 ˙5·73 and K/H∼=
|H|= 2t·35 ·5·73, where 2 ≤t ≤3. Let H be solvable and also, let P be a 73-Sylow subgroup of
H. Then an easy calculation and Lemma 1.4 showP chH ⊴G. Now, sinceP is cyclic, Ito’s theorem leads us to get a contradiction. Therefore H is a non-solvable group of order 2t·35·5·73, where 2 ≤t ≤3. So, considering the order of H, and Lemmas 1.3 and 1.6 show that there exists a normal series as 1⊴H1 ⊴K1 ⊴H such that for t= 3, K1/H1 =∼A5 orA6 and for t= 2,K1/H1 ∼=A5. Thus
π(H1)⊆ {2,3,73}and hence, Lemma 1.7 guarantees thatH1is solvable. Suppose thatP ∈Syl73(H1). Then by Lemma 1.4,P chH1 ⊴H. On the other hand,P is a 73-Sylow subgroup ofH. ThusP ⊴G. Now, Ito’s theorem leads us to get a contradiction. Also, applying the same argument rules out the
cases when K/H ∼=A5×A5 orA6. ThusK/H ∼=U3(9). It follows that G∼=U3(9), as claimed.
Proof of (III). Let χ, β ∈ Irr(G) such that χ(1) = l(G) = l(M) and β(1) = t(G) = t(M). Then
Lemma 1.8 shows that G is non-solvable and hence, Lemma 1.3 implies that G has a normal series as Lemma 1.3 such that K/H is a direct product of isomorphic non-abelian simpleK3 orK4-groups and |G/K| | |Out(K/H)|. Now by comparing the order ofGand the orders of the non-abelian simple groups mentioned in Lemma 1.6, we have the following cases:
i. Let M =S6(2). Then |G|= |S6(2)| = 29·34·5·7, χ(1) = l(G) = l(S6(2)) = 29, β(1) = t(G) =
t(S6(2)) = 34·5. Let N be a normal minimal solvable subgroup of G. Then applying Ito’s theorem to χ, β and N shows that N = Z7 or N = 1. Since G/CG(N) ≲ Aut(Z7) ∼= Z6, |CG(N)/N| ̸= 1.
Now, we claim that CG(N)/N is non-solvable. On the contrary, let CG(N)/N be solvable. Then
we conclude that CG(N) is solvable and so, Lemma 1.5 shows that CG(N) = N ×C, where C is
a (π(CG(N))− {7})-Hall subgroup of CG(N). Thus C ch CG(N) ⊴ G. So, C is a normal solvable
subgroup of G, which is a contradiction with the fact that every normal minimal solvable subgroup of G is a 7-group. This contradiction shows that CG(N)/N is non-solvable. Let L/N be a normal
minimal subgroup of G/N such that L/N ≤CG(N)/N. Then the above statement shows that L/N
is a direct product of t-copies of isomorphic non-abelian simple K3 or K4-groups. Thus considering Lemma 1.6 shows that
L/N ∼=A5, A6, L2(7), L2(8), U3(3), U4(2), A7, A8, A9, L3(4) or S6(2). (2.1)
SinceLis a central extension ofN byL/N, considering the schur multiplier ofL/N shows thatLis a central product ofN andL/N. Also,L⊴Gand hence, Lemma 1.2 applied toχ andβ , respectively, allows us to assume that there exists α, η∈cd(L) such that |L|2 |α and |L|3· |L|5 |η. On the other hand, cd(L) = cd(L/N), becauseLis a central product ofN andL/N andN is abelian. Therefore, we
can assume thatα, η ∈cd(L/N) and hence, considering the character degrees of the groups mentioned in 2.1 forces
L/N∼=L2(7), L2(8), U3(3), A8, L3(4) orS6(2). (2.2)
Thus 7 | |L/N| and hence, N = 1. Now, let C be a normal minimal subgroup of G such that
CG(L) = 1 and hence, G≤Aut(L). Thus considering the orders of the automorphism groups of the
groups in 2.2 guarantees that L∼=S6(2) and so,G=L∼=S6(2), as desired.
ii. Let M = L3(8). Then |G| = |L3(8)| = 29 ·32 ·72 ·73, χ(1) = l(G) = l(L3(8)) = 32 ·73,
β(1) =t(G) =t(L3(8)) = 29 and K/H ∼=L2(7), L2(7)×L2(7), L2(8) orL3(8). IfK/H ∼=L2(7), then
|G/K| | |Out(L2(7))|= 2 and hence,|H|= 26−t·3·7·73, where 0≤t≤1. LetP ∈Syl73(H). Then an easy calculation shows thatP chH⊴G. Now, sinceP is abelian, Ito’s theorem forces χ(1) = 32·73 to divide |G|/73 = |M|/73, which is a contradiction. The same reasoning rules out the cases when
K/H ∼=L2(7)×L2(7) orL2(8). Therefore K/H ∼=L3(8) and hence,G=K ∼=L3(8).
iii. Let M = L3(5). Then |G| = |L3(5)| = 25 ·3·53 ·31, χ(1) = l(G) = l(L3(5)) = 2·3 ·31,
β(1) = t(G) = t(L3(5)) = 53 and K/H ∼= A5, L2(31) or L3(5). Let K/H ∼= A5. Then since
|Out(A5)| = 2, |H|= 2t·52·31, where 2≤ t ≤3 and hence, Lemma 1.7 shows that H is solvable. Let P ∈ Syl31(H). Then since for every natural number k, 31k+ 1 ∤ 52 or 2t for t = 2 or 3, Lemma 1.4 guarantees that P ch H ⊴ G. But P is cyclic and hence, Ito’s theorem shows that 2 ·3·31 = χ(1) | [G : P], which is impossible. Also, if K/H ∼= L2(31), then by Lemma 1.3,
|G/K| | |Out(L2(31))|= 2 and so, |H|= 52. Now, applying Ito’s theorem toH and β leads us to get a contradiction. Therefore K/H ∼=L3(5) and hence, G=K ∼=L3(5).
iv. Let M = O8+(2). Then |G| = |O8+(2)| = 212·35 ·52·7, χ(1) = l(G) = l(O+8(2)) = 35·52 and
β(1) = t(G) = t(O+8(2)) = 212. Let N be a normal minimal solvable subgroup ofG. Then applying Ito’s theorem toχ,β and N shows that N =Z7 orN = 1. Let L/N be a normal minimal subgroup of G/N such that L/N ≤ CG(N)/N. Then applying the same reasoning as that used for the
non-solvability of CG(N)/N in (i) shows thatCG(N)/N is non-solvable and so,L/N is a direct product of t-copies of isomorphic non-abelian simple K3 or K4-groups. Thus considering Lemma 1.6 shows that
L/N ∼=A5, A5×A5, A6, A6×A6, L2(7), L2(8), U3(3), (2.3)
U4(2), A7, A8, L3(4), S6(2), A9, A10, J2 orO8+(2). Now, by the similar argument as that used in (i), we get N = 1 and
L/N ∼=A8, L3(4), S6(2), L2(7), L2(8), U3(3) or O8+(2). (2.4)
and CG(L) = 1. Thus G ≤ Aut(L). So, considering the order of the automorphism groups in 2.4
forcesL∼=O+8(2) and henceG=L∼=O+8(2), as desired.
v. Let M = L3(7). Then |G| = |L3(7)| = 25 ·32 ·73 ·19, χ(1) = l(G) = l(L3(7)) = 23·3·19,
β(1) = t(G) = t(L3(7)) = 73 and K/H ∼= L2(7), L2(8) or L3(7). If K/H ∼= L2(7) or L2(8), then
|G/K| | |Out(K/H)| = 2 or 3. Thus |H| | 22·3·72 ·19 or |H| | 22 ·72·19 and hence, Lemma 1.7 guarantees that H is solvable. Therefore, an easy calculation and Lemma 1.4 show that a 19-Sylow subgroup of H is normal in H and so, it is normal in G. Thus G has a normal abelian 19-Sylow subgroup and hence, Ito’s theorem implies that χ(1) = 23 ·3 ·19 | [G : P], which is impossible. ThereforeK/H ∼=L3(7) and hence,G=K∼=L3(7), as desired.
vi. Let M = L2(49). Then |G| = |L2(49)| = 24 ·3·52 ·72, χ(1) = l(G) = l(L2(49)) = 2 ·52,
|G/K| | |Out(K/H)|= 2 and|G/K||H|= 22·5·72 or 2·52·7. Therefore,|H| |22·52·72 and hence, Lemma 1.7 guarantees that H is solvable. Let P ∈ Syl5(H). Then since there is not any natural number k such that 5k+ 1|7t or 2t, where 1≤t≤2, Lemma 1.4 shows thatP chH ⊴G. ThusP
is an abelian normal subgroup ofG and hence, Ito’s theorem shows thatχ(1) = 2·52 divides [G:P], which is impossible. So, K/H ∼=L2(49) and hence, G=K ∼=L2(49), as desired.
vii. Let M = G2(3). Then |G| = |G2(3)| = 26 ·36 ·7·13, χ(1) = l(G) = l(G2(3)) = 26·13 and
β(1) = t(G) = t(G2(3)) = 36. Let N be a normal minimal solvable subgroup of G. Then applying Ito’s theorem toχ,β and N shows that N =Z7 orN = 1. Let L/N be a normal minimal subgroup of G/N such thatL/N ≤CG(N)/N. Then the same reasoning as that used for the non-solvability of CG(N)/N in (i) shows that CG(N)/N is non-solvable and hence, L/N is a direct product oft-copies
of isomorphic non-abelian simpleK3 orK4-groups. Thus by considering Lemma 1.6, we can see that
L/N∼=L2(7), L2(8), L3(3), U3(3), L2(27), L2(13), orG2(3) Now, repeating the argument given for (i) shows that N = 1,
L∼=L2(7), L2(8), U3(3) or G2(3) (2.5)
and CG(L) = 1. Therefore,G≤Aut(L). Thus considering the orders of the automorphism groups of
the groups in (2.5) guarantees thatL∼=G2(3) and hence, L=G∼=G2(3), as desired.
viii. Let M = U3(5). Then |G| = |U3(5)| = 24 ·32 ·53 ·7, χ(1) = l(G) = l(U3(5)) = 24 ·32,
β(1) =t(G) =t(U3(5)) = 53 andK/H∼=A5, A6, A7, L2(7), L2(8) orU3(5). Suppose thatK/H ∼=A5. Then by Lemma 1.3, |G/K| | |Out(A5)|= 2 and hence, |H| = 22·3·52·7 or 2·3·52·7. If H is solvable, then an easy calculation and Lemma 1.4 show that P ch H⊴G, whereP ∈Syl5(H). Thus by Ito’s theorem, we get β(1) = 53 |[G: P], which is impossible. Hence, H is a non-solvable group of order 22·3·52·7, by considering Lemma 1.7. Thus Lemma 1.3 shows thatH has a normal series as 1 ⊴ H1 ⊴ K1 ⊴ H such that K1/H1 ∼= A5. Now, since |Out(A5)| = 2, by Lemma 1.3, we get
|H1| = 5·7. Let Q ∈ Syl5(H1). Then Q ch H1 ⊴ H. Let θ ∈ Irr(H) such that [βH, θ] ̸= 0. Then
Lemma 1.2 forces θ(1) = 52. Now, Ito’s theorem implies thatθ(1) = 52 |[H :Q], which is impossible. The same argument rules out the cases when K/H ∼= A6, A7, L2(7) or L2(8). Thus K/H ∼=U3(5), which implies that G∼=U3(5), as claimed.
ix. Let M = S4(7). Then |G| = |S4(7)| = 28 ·32 ·52 ·74, χ(1) = l(G) = l(S4(7)) = 27 ·52,
β(1) = t(G) =t(S4(7)) = 74 and K/H ∼=A5,A5 ×A5,A6, A7, L2(49), L2(7), L2(7)×L2(7), L2(8),
L3(4), L4(2) or S4(7). Suppose that η, θ ∈ Irr(H) such that [χH, η] ̸= 0 and [βH, θ] ̸= 0. Then by
Lemma 1.2, χ(1)/η(1) | [G : H] and β(1)/θ(1) | [G : H]. If K/H ̸∼= L2(7) and S4(7), then since
|G/K| | |Out(K/H)|, we can check at once that η2(1) > |H|, θ2(1) > |H| or η2(1) +θ2(1) > |H|, which is a contradiction. Now, letK/H∼=L2(7). Then either|H|= 25·3·52·73 or|H|= 24·3·52·73. IfH is solvable, then letN be a normal minimal subgroup ofGsuch thatN ≤H. Thus Ito’s theorem forces |N| | 2 or |N| | 3. Let L1 be a maximal normal 2-subgroup of G and let L2 be a maximal normal 3-subgroup of G such that L1, L2 ≤ H. Then applying Lemma 1.2 to L1 and η shows that
|CH(L)/L| ̸= 1. Now, the same reasoning as that used for the non-solvability of L3(7) leads us to
get a contradiction. Thus H is non-solvable. Therefore by Lemma 1.3, there exists a normal series as 1⊴H1 ⊴K1 ⊴H such that K1/H1 is a direct product of isomorphic non-abelian simple K3 or K4 -groups and|H/K1| | |Out(K1/H1)|. Now comparing the order ofH and the orders of the non-abelian simple groups mentioned in Lemma 1.6 shows thatK1/H1 ∼=L2(7) orA5. Therefore, 5·72 | |H1|and
|H1| | 23·52·73. Thus H1 is solvable and Lemma 1.4 shows that P ∈Syl5(H1) is normal in H1 and hence, P ⊴H. So, Ito’s theorem forces η(1)|[H :P], which is impossible. Therefore K/H ∼=S4(7) and hence, G=K∼=S4(7), as desired.
Proof of (IV) Let χ, β ∈ Irr(G) such that χ(1) = l(G) = l(M) and β(1) = s(G) = s(M). Then by Lemma 1.8, G is non-solvable and hence, Lemma 1.3 shows that there exists a normal series 1 ⊴ H ⊴ K ⊴ G such that K/H is a direct product of isomorphic non-abelian simple K3 or K4 -groups and|G/K| | |Out(K/H)|. Now by comparing the order ofGand the orders of the non-abelian simple groups mentioned in Lemma 1.6, we have the following cases:
i. Let G =L3(4). Then |G|= |L3(4)|= 26 ·32·5·7, χ(1) = l(G) = l(L3(4)) = 26, β(1) = s(G) =
s(L3(4)) = 32·7 and K/H ∼=A5, A6, A7, L2(8), A8, L2(7) or L3(4). Suppose that η, θ∈Irr(H) such that [χH, η] ̸= 0 and [βH, θ]̸= 0. Then by Lemma 1.2, χ(1)/η(1) |[G :H] and β(1)/θ(1)| [G :H].
Therefore, |H|2 | η(1) and |H|3 · |H|7 | θ(1). Also, |G/K| | |Out(K/H)| and hence, if K/H ∼= A5, then |H| = 2t·3·7, where 3 ≤ t ≤ 4. So, θ2(1) = (3·7)2 > |H|, which is a contradiction. If
K/H ∼=A6, then|G/K| | |Out(A6)|= 22and hence,|H|= 2t·7, where 1≤t≤3. So, eithert≤2 and
θ2(1) = 72 >|H|orη2(1) = 26>|H|, which is a contradiction. IfK/H∼=A7, then|H|= 22 or 23 and
η(1) =|H|, which is a contradiction. Let K/H ∼=L2(8). Then |H|= 23·5. Thus η2(1) = 26 >|H|, which is a contradiction. If K/H ∼=A8, then G∼=A8. Thus 32·7∈cd(A8), which is a contradiction. Finally, let K/H ∼= L2(7). Then |G/K| | |Out(L2(7))| = 2 and hence, |H| = 23·3·5 or 22 ·3·5. If H is solvable, then an easy calculation and Lemma 1.4 show that P ∈ Syl5(G) is normal in G. Also, H/CH(P) ,→ Aut(P) ∼= Z4 and hence, 3,5 ∈ π(CH(P))⊆ π(H) and CH(P) =P ×C, where C ̸= 1 is a (π(CH(P))− {5})-Hall subgroup of CH(P). Thus C is a normal solvable subgroup of
G such that |C| | 23·3. Thus C contains a 2 or 3-elementary abelian normal subgroup of G which is impossible by considering Ito’s theorem. Therefore H is a non-solvable group of order 23·3·5 or 22·3·5 and hence, Lemma 1.3 implies that H has a normal series as 1 ⊴ H1 ⊴ K1 ⊴ H such that K1/H1 ∼= A5. Also, by Lemma 1.3, |H/K1| | |Out(A5)| = 2 and hence, |H1| = 1 or 2. If
|H1| = 2, then by Ito’s theorem, we get η(1) = |H|2 = 23 | [H : H1], which is impossible. Thus
|H1| = 1 and hence, H = A5·2 or A5. Therefore, considering the fact that G/CG(H) ,→ Aut(H)
guarantees thatL2(7)≤CG(H) and hence,K =L2(7)×S5 orK =L2(7)×A5. So, Lemma 1.2 shows that β(1) = 32·7 ∈ cd(K) = cd(L2(7)×cd(A5)) or β(1) = 32 ·7 ∈ cd(L2(7)×cd(S5)), which is a contradiction. These contradictions show that K/H∼=L3(4) and hence,G=K∼=L3(4).
ii. Let M = U4(3). Then |G| = |U4(3)| = 27 ·36 ·5·7, χ(1) = l(G) = l(U4(3)) = 27 ·7 and
β(1) = s(G) = s(U4(3)) = 36. Let N be a normal minimal solvable subgroup of G. Then applying Ito’s theorem forces N =Z5 or N = 1. Let L/N be a normal minimal subgroup of G/N such that
β show that L/N is a direct product of t-copies of isomorphic non-abelian simple K3 or K4-groups. Thus considering Lemma 1.6 shows thatL/Nis isomorphic toA5, L2(7), A6, L2(8), A7, A8, A9, L3(4),
U3(3), U4(2) orU4(3). Now repeating the argument given for (i) in the proof of (III) shows thatN = 1,
L∼=A5, A6, U4(2) or U4(3) (2.6)
and CG(L) = 1. Therefore,G≤Aut(L). Thus considering the orders of the automorphism groups of
the groups in (2.6) guarantees thatL∼=U4(3) and hence,L=G∼=U4(3), as desired.
iii. Let M = 2F4(2)′. Then |G| = |2F4(2)′| = 211·33 ·52 ·13, χ(1) = l(G) = l(2F4(2)′) = 211,
β(1) =s(G) =s(2F4(2)′) = 26·33 and K/H ∼=A5, A5×A5, A6, L3(3), L2(25), U3(4) or2F4(2)′. Sup-pose thatθ, η∈Irr(H) such that [χH, θ]̸= 0 and [βH, η]̸= 0. Then Lemma 1.2 implies thatθ(1) =|H|2 and |η(1)|3=|H|3. LetK/H ∼=A5. Then since |G/K| | |Out(K/H)|, we have,|H|= 29·32·5·13 or 28·32·5·13. LetHbe solvable. ThenGhas a normal minimal subgroup, namelyN, such thatN ≤H. Thus for some t∈ π(H), N is a t-elementary abelian group and so, applying Ito’s theorem to N, χ
and β shows thatt∈ {5,13}. Now, by the similar argument as that used for the non-solvability of G, we get a contradiction. Hence,H is non-solvable. Thus Lemma 1.3 shows thatH has a normal series as 1⊴ H1 ⊴K1 ⊴H such that K1/H1 ∼=A5 or A6. Now, since |Out(A5)|= 2 and|Out(A6)|= 22,
|H1| = 2s·3·13, where 5 ≤ s ≤ 7 or 2s·13, where 3 ≤ s ≤ 6. So, Lemma 1.7 shows that H1 is solvable and hence, H1 has a normal minimal subgroup namely L, such that L≤H1. Thus for some
t ∈ π(H1), L is a t-elementary abelian group. Again, applying Ito’s theorem to L, θ and η shows that |L| = t = 13 and by a similar argument as that used for the non-solvability of G , we get a contradiction. Also, the similar reasoning as the case when K/H ∼=A5, rules out K/H ∼= A5 ×A5 or A6. If K/H ∼= L3(3), then for t ∈ {6,7}, |H| = 2t·52, if K/H ∼= L2(25), then for t ∈ {6,7,8},
|H|= 2t·32 and ifK/H ∼=U3(4), then for t∈ {3,4,5},|H|= 2t·32. Thus by considering the order of H in the above cases, we can see that η2(1)>|H|orθ2(1)>|H|, which is a contradiction. Thus
K/H ∼=2F4(2)′, which implies that G=K ∼=2F4(2)′, as claimed.
iv. Let M = U5(2). Then |G| = |U5(2)| = 210 ·35 ·5 ·11, χ(1) = l(G) = l(U5(2)) = 35 ·5,
β(1) =s(G) =s(U5(2)) = 210 and K/H ∼=A5, A6, U4(2), M11, M12, L2(11) or U5(2). IfK/H ∼=A5, then |H| |28·34·11 and 27·34·11 | |H|, if K/H =∼A6, then |H| |27·33·11 and 25·33·11 | |H|, if K/H ∼= U4(2), then |H| | 24 ·3·11 and 23 ·3·11 | |H|, if K/H ∼= M11, then |H| = 26 ·33, if
K/H ∼=L2(11), then|H| |28·34 and 27·34| |H|and ifK/H∼=M12, then|H| |24·32 and 23·32| |H|. Hence, Lemma 1.7 shows that H is solvable. We note that in the above cases, the order of H has at least two prime divisors. It follows that H has a s-elementary abelian subgroup, namely N, which is normal in G. If s∈ {2,3,5}, then applying Ito’s theorem toN,β and χ leads us to get a
contradic-tion. Thus s = 11. Since|N|=|CH(N)|11 and H/CH(N) ,→ Aut(N) = Z10, the order of H in the
above cases and Lemma 1.5 guarantee that π(CH(N))− {11} ̸= ∅ and CH(N) = N ×C, where C
is a (π(CH(N))− {11})-Hall subgroup of CH(N). Thus we can see at once that C ⊴ G and hence,
for some u∈π(CH(N))− {11},C contains anu-elementary abelian normal subgroup ofG, which is
contradiction with the fact that the only normal minimal subgroups of G are 11-groups. Therefore
Proof of (V). Let χ, β ∈ Irr(G) such that χ(1) = s(G) = s(M) and β(1) = t(G) = t(M). Then Lemma 1.8 shows that G is non-solvable and hence, Lemma 1.3 implies that there exists a normal
series 1 ⊴ H ⊴ K ⊴ G such that K/H is a direct product of isomorphic non-abelian simple K3 or K4-groups and |G/K| | |Out(K/H)|. Now, by comparing the order of G and the orders of the non-abelian simple groups mentioned in Lemma 1.6, we have the following cases:
i. Let M = U3(4). Then |G| = |U3(4)| = 26 ·3 ·52 ·13, χ(1) = s(G) = s(U3(4)) = 13·5,
β(1) = t(G) =t(U3(4)) = 26 and K/H =∼A5, L2(25) or U3(4). Suppose that K/H ∼= A5 or L2(25). Then since |G/K| | |Out(K/H)|,|H|= 2t·5·13, where 3≤t≤4 or |H|= 2t, where 1≤t≤3. Now, letθ, η∈Irr(H) such that [χH, η]̸= 0 and [βH, θ]̸= 0. Then Lemma 1.2 implies that|H|5·|H|13|η(1) and|H|2 |θ(1). Thusη2(1)>|H|orθ2(1)>|H|, a contradiction. ThereforeK/H ∼=U3(4). It follows that G=K ∼=U3(4), as claimed.
ii. Let M = 3D4(2). Then |G| = |3D4(2)|= 212·34·72 ·13, χ(1) = s(G) = s(3D4(2)) = 212 and
β(1) = t(G) = t(3D4(2)) = 34·72. Let N be a normal minimal solvable subgroup of G. Then ap-plying Ito’s theorem to χ, β and N shows that N =Z13 or N = 1. Let L/N be a normal minimal subgroup ofG/N such thatL/N ≤CG(N)/N. Then applying the same reasoning as that used for the
non-solvability of Gshows that L/N is a direct product of t-copies of isomorphic non-abelian simple
K3 orK4-groups. Thus considering Lemma 1.6 shows that
L/N ∼=L2(7), L2(8), L3(3), U3(3), L2(13), L2(27), L2(7)×L2(7), L2(8)×L2(8) or 3D4(2) (2.7)
SinceLis a central extension ofN byL/N, considering the schur multiplier ofL/N shows thatLis a central product ofN andL/N. Suppose that η, θ∈Irr(L) = Irr(L/N)×Irr(N) = Irr(L/N) such that [χL, η]̸= 0 and [βL, θ]̸= 0. Then by Lemma 1.2,χ(1)/η(1)|[G:L] andβ(1)/θ(1)|[G:L]. Therefore, |L|2 | η(1) and |L|3 · |L|7 |θ(1). Thus considering the character degrees of the groups mentioned in (2.7) shows that
L/N ∼=L3(3) or3D4(2) (2.8)
This forces N = 1, because 13∥|G|. Now, let C be a normal minimal subgroup of G such that
C ≤ CG(L). Obviously, by repeating the argument given for L/N, C = 1 or C is isomorphic to
one of the groups mentioned in (2.8). If C ̸= 1, then C×L ⊴ G and |L|3 · |C|3 > 34, which is a contradiction. This forces C = 1 and hence, CG(L) = 1. If L ∼= L3(3), then G ,→ Aut(L3(3)). It
follows that |G| | |Aut(L3(3))|= 25·33·13, which is a contradiction. Therefore L/N ∼=3D4(2) and hence,G=L∼=3D4(2), as claimed.
iii. Let M = S4(4). Then |G| = |S4(4)| = 28 ·32 ·52 ·17, χ(1) = s(G) = s(S4(4)) = 28 and
β(1) =t(G) =t(S4(4)) = 3·5·17. Applying Ito’s theorem to the normal minimal solvable subgroup of
G,s(G) andt(G) shows that the normal minimal solvable subgroup ofGis a 3-group or 5-group. Let
N1 be a maximal normal 3-subgroup ofG,N2be a maximal normal 5-subgroup ofGandN =N1×N2. Then|N| |3·5. LetL/N be a normal minimal subgroup ofG/N such thatL≤CG(N). Then applying
non-abelian simple K3 orK4-groups. Thus considering Lemma 1.6 shows that
L/N ∼=A5, A6, A5×A5, L2(16), L2(17) orS4(4). (2.9)
Now, let C/N be a normal minimal subgroup of G/N such that C ≤ CG(L). Obviously, N ≤ C
and by our assumption on N and repeating the argument given for L/N, C = N or C/N is iso-morphic to one of the groups mentioned in (2.9). Let C ̸= N. If L/N ∼= A5 ×A5, then con-sidering the order of G/N shows that |C/N| | 24 ·17 and hence, C/N is solvable. Thus our as-sumption on N forces C = N, which ia a contradiction. Now, let L/N ∼= A5. Then repeating the above argument shows that C/N ∼= A5 or L2(16). Thus L/N ×C/N ∼= A5 ×A5 ⊴ G/N or
L/N×C/N ∼=A5×L2(16) ⊴ G/N. IfL/N×C/N ∼=A5×A5, then repeating the argument given for the case when L/N ∼=A5×A5 leads us to get a contradiction. Now, letL/N×C/N ∼=A5×L2(16). Then since |A5×L2(16)|5|A5×L2(16)|3=|G|5|G|3,N = 1. Hence,A5×L2(16) ⊴ G. Now, Lemma 1.2 forces β(1) = 3·5·17 ∈ cd(A5 ×L2(16)), which is a contradiction. Also a similar argument shows that if L/N ∼=L2(16), then C/N ∼=A5 and as above, we get a contradiction. If L/N ∼=A6 or
L/N ∼=L2(17), then|C/N| |25·5·17 or |C/N| |24·52 and hence, considering the order of the finite simpleK3-groups shows that C/N is solvable. Therefore, our assumption on N forcesC=N, which is a contradiction. Thus in the mentioned groups, C/N = 1, and so G/N ,→ Aut(L/N), which is a
contradiction by considering the order ofGand the order of Aut(L/N). Therefore,L/N ∼=S4(4) and hence,N = 1 andL=G∼=S4(4).
iv. Let M = S4(5). Then |G| = |S4(5)| = 26 ·32 ·54 ·13, χ(1) = s(G) = s(S4(5)) = 54 and
β(1) = t(G) = t(S4(5)) = 24 ·3·13. Let N be a normal minimal solvable subgroup of G. Then applying Ito’s theorem toχ,β andN shows that|N| |23·3. LetL/N be a normal minimal subgroup of G/N such thatL≤CG(N). Then applying the same reasoning as that used for the non-solvability
of G shows L/N is a direct product of t-copies of isomorphic non-abelian simple K3 or K4-groups. Thus Lemma 1.6 shows that L/N ∼= A5, A5 ×A5, A6, L2(25), U3(4) or S4(5). Then repeating the argument given for (iii) completes the proof.
Proof of (VI). Let χ, β ∈Irr(G) such that χ(1) = l(G) =l(L2(3m)) = 3m+ 1 and β(1) = s(G) =
s(L2(3m)) = 3m. Then Lemma 1.8 shows thatG is non-solvable. Now, since L2(3m) is a simple K4 -group,|π(G)|= 4. Also, 4∥|G|and u, t∈π(G). Thus we can see by Lemma 1.3 thatGhas a normal series 1 ⊴ H ⊴ K ⊴ G such that K/H ∼= L2(2e), L2(r) or L2(3e), where L2(2e), L2(r) and L2(3e) are simple K4-groups. If K/H ∼= L2(2e), then since 4∥|L2(3m)| = |G|, we deduce that 2e ≤ 22 and hence,|K/H|= 22·3·5, which is a contradiction. IfK/H ∼=L2(r), thenr =toru. First letm̸= 5. Let r =t= (3m−1)/2. Then since |L2(r)|=r(r2−1)/2 | |G|, we deduce that 3(3m−1−1)/8|3m, which is impossible. Thus r=u= (3m+ 1)/4. Now, since t= (3m−1)/2| |K/H|=r(r2−1)/2, we deduce that t|(u−1)/2 or (u+ 1)/2, which is impossible. Also, ifm = 5, then an easy calculation shows that |L2(r)|∤|G|. Therefore K/H =∼L2(3e), wheree and u′ = (3e+ 1)/4 are odd primes and (3e−1)/2 is either a prime or 112 (fore= 5). Hence,u=u′ ort′. But (3m+ 1)/4∤3e−1 and hence,
Acknowledgments
The authors wish to thank the referee for the valuable comments which helped to improve the
manuscript.
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Somayeh Heydari
Department of Mathematics, University of Shahrekord, P. O. Box 115, Shahrekord, Iran
Email: heydari.somayeh@stu.sku.ac.ir
Neda Ahanjideh
Department of Mathematics, University of Shahrekord, P.O.Box 115, Shahrekord, Iran
