Open Access

A new characterization of Mathieu-groups by the order and one irreducible character degree

Journal of Inequalities and Applications20132013:209

https://doi.org/10.1186/1029-242X-2013-209

Received: 29 December 2012

Accepted: 15 April 2013

Published: 26 April 2013

Abstract

The main aim of this article is to characterize the finite simple groups by less character quantity. In fact, we show that each Mathieu-group G can be determined by their largest and second largest irreducible character degrees.

MSC:20C15.

Keywords

finite groupsimple groupcharacter degree

1 Introduction and preliminary results

Classifying finite groups by the properties of their characters is an interesting problem in group theory. In 2000, Huppert conjectured that each finite non-abelian simple group G is characterized by the set c d ( G ) of degrees of its complex irreducible characters. In [14], it was shown that many non-abelian simple groups such as L 2 ( q ) and S z ( q ) satisfy the conjecture. In this paper, we manage to characterize the finite simple groups by less character quantity. Let G be a finite group; L ( G ) denotes the largest irreducible character degree of G and S ( G ) denotes the second largest irreducible character degree of G. We characterize the five Mathieu groups G by the order of G and its largest and second largest irreducible character degrees. Our main results are the following theorems.

Theorem A Let G be a finite group and let M be one of the following Mathieu groups: M 11 , M 12 and M 23 . Then G M if and only if the following conditions are fulfilled:
  1. (1)

    | G | = | M | ;

     
  2. (2)

    L ( G ) = L ( M ) .

     

Theorem B Let G be a finite group. Then G M 24 if and only if | G | = | M 24 | and S ( G ) = S ( M 24 ) .

Theorem C Let G be a finite group. If | G | = | M 22 | and L ( G ) = L ( M 22 ) , then either G is isomorphic to M 22 or H × M 11 , where H is a Frobenius group with an elementary kernel of order 8 and a cyclic complement of order 7.

We need the following lemmas.

Lemma 1 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 ) | .

Proof Let G be a non-solvable group. Then G has a chief factor M / N such that M / N is a direct product of isomorphic non-abelian simple groups. Hence C G / N ( M / N ) M / N = Z ( M / N ) = 1 , and so
M / N C G / N ( M / N ) × M / N C G / N ( M / N ) G / N C G / N ( M / N ) Aut ( M / N ) .

Let K / N = C G / N ( M / N ) × M / N and H / N = C G / N ( M / N ) . Then G / K Out ( M / N ) and K / H M / N is a direct product of isomorphic non-abelian simple groups. Thus 1 H K G is a normal series, as desired. □

Lemma 2 Let G be a finite solvable group of order p 1 a 1 p 2 a 2 p n a n , where p 1 , p 2 , , p n are distinct primes. If k p n + 1 p i a i for each i n 1 and k > 0 , then the Sylow p n -subgroup is normal in G.

Proof Let N be a minimal normal subgroup of G. Then | N | = p m for G is solvable. If p = p n , by induction on G / N , we see that normality of the Sylow p n -subgroup in G. Now suppose that p = p i for some i < n . Now consider G / N . By induction, the Sylow p n -subgroup P / N of G / N is normal in G / N . Thus P G . Let Q be a Sylow p n -subgroup of P. Then P = N Q . By Sylow’s theorem, | P : N P ( Q ) | = p i l ( l m a i ) and p n p i l 1 . But this means that k p n + 1 p a i , and then k = 0 by assumption. Hence Q P and Q G . □

2 Proof of theorems

Proof of Theorem A We only need to prove the sufficiency. We divide the proof into three cases.

Case 1.1 M = M 11

In this case, we have | G | = 2 4 3 2 5 11 and L ( G ) = 55 . We first show that G is non-solvable. Assume the contrary. By Lemma 2, we know that the Sylow 11-subgroup of G is normal in G. Let N be the 11-Sylow subgroup of G. Since N is abelian, we have χ ( 1 ) | G / N | for all χ Irr ( G ) . But L ( G ) = 55 and 55 | G / N | , a contradiction. Therefore, G is non-solvable.

Since G is non-solvable, by Lemma 1, we get that 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 ) | . As | G | = 2 4 3 2 5 11 , we have K / H A 5 , A 6 , L 2 ( 11 )  or  M 11 .

We first assume that K / H A 5 . Since | Out ( A 5 ) | = 2 , we have | G / K | 2 and | H | = 2 t 3 11 , where t = 1 or 2. Let χ Irr ( G ) such that χ ( 1 ) = L ( G ) = 55 and θ Irr ( H ) such that [ χ H , θ ] 0 . Then θ ( 1 ) = 11 by the Clifford theorem (see Theorem 6.2 in [5]). On the other hand, since | H | = 2 t 3 11 , we have H is solvable. Let N be a Sylow 11-subgroup of H. Then N H by Lemma 2. Hence θ ( 1 ) | H / N | = 2 t 3 , a contradiction.

By the same reason as above, one has that K / H A 6 .

Suppose that K / H L 2 ( 11 ) . Since | Out ( L 2 ( 11 ) ) | = 2 , we have | G / K | 2 and so | H | = 2 a 3 , where a = 1 or 2. Let θ Irr ( H ) such that e = [ χ H , θ ] 0 and let t = | G : I G ( θ ) | . Then θ ( 1 ) = 1 and e t = χ ( 1 ) / θ ( 1 ) = 55 . Since | H | = 2 t 3 , where a = 1 or 2, we have that 55 | Aut ( H / H ) | . Hence t = 1 , e = 55 . But ( 55 ) 2 = e 2 t = [ χ H , χ H ] > | G : H | = 2 b 3 5 11 , where 2 b 3 , a contradiction.

If K / H M 11 , by comparing the orders of G and M 11 , we have | H | = 1 . Therefore G = K M 11 .

Case 1.2 M = M 23

In this case, we have | G | = 2 7 3 2 5 7 11 23 and L ( G ) = 2 3 11 23 . Then O 23 ( G ) = 1 . If not, then | O 23 ( G ) | = 23 and O 23 ( G ) is abelian. Hence L ( G ) = 2 3 11 23 | G / O 23 ( G ) | , a contradiction.

If G is solvable, then the Sylow 23-subgroup of G is normal in G by Lemma 2, which leads to a contradiction as above. Therefore G is non-solvable.

Since G is non-solvable, by Lemma 1, we get that 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 ) | . As | G | = 2 7 3 2 5 7 11 23 , we have that K / H can be isomorphic to one of the simple groups: A 5 , L 2 ( 7 ) , A 6 , L 2 ( 8 ) , L 2 ( 11 ) , A 7 , M 11 , L 3 ( 4 ) , A 8 , M 22 and M 23 .

We first assume that K / H A 5 . Since | Out ( A 5 ) | = 2 , we have | G / K | 2 and | H | = 2 m 3 7 11 23 , where m = 4 or 5. Suppose that H is non-solvable. By Lemma 1, H has a normal series 1 A B H such that B / A is a direct product of isomorphic non-abelian simple groups and | H / B | | Out ( B / A ) | . Since | H | = 2 m 3 7 11 23 , we have B / A L 2 ( 7 ) and | H / B | 2 . Thus | A | = 2 a 11 23 , where 0 a 2 . Let N be a Sylow 23-subgroup of A. Then N A by Lemma 2. Hence we get a subnormal series of G, N char A B H K G , which implies that N G . But O 23 ( G ) = 1 , a contradiction. If H is solvable, then the Sylow 23-subgroup of H is normal in H by Lemma 2, which leads to a contradiction as before.

By the same arguments as the proofs of K / H A 5 , we show that K / H cannot be isomorphic to one of the simple groups: A 6 , L 2 ( 7 ) , L 2 ( 8 ) , L 2 ( 11 ) , A 7 , M 11 , L 3 ( 4 ) , A 8 and M 22 .

If K / H M 23 , since | G | = 2 7 3 2 5 7 11 23 , we have that | H | = 1 and G = K M 23 .

Case 1.3 M = M 12

In this case, | G | = 2 6 3 3 5 11 and L ( G ) = 2 4 11 . Since 11 L ( G ) , by the same arguments as the proofs of Case 1.2, we have that O 11 ( G ) = 1 .

We will show that G is non-solvable. If G is solvable, then the Sylow 11-subgroup of G is normal in G by Lemma 2, a contradiction. Therefore, G is non-solvable.

By Lemma 1, we get that 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 ) | . As | G | = 2 6 3 3 5 11 , we have K / H A 5 , A 6 , L 2 ( 11 ) , M 11  or  M 12 .

By the same arguments as the proofs of Case 1.2, we can prove that K / H cannot be isomorphic to A 5 or A 6 .

Assume that K / H L 2 ( 11 ) . Since | Out ( L 2 ( 11 ) ) | = 2 , we have | G / K | 2 and | H | = 2 a 3 , where a = 3 or 4. Suppose that | G / K | = 1 . Then | H | = 2 4 3 2 . Let χ Irr ( G ) such that χ ( 1 ) = L ( G ) = 2 4 11 and θ Irr ( H ) such that e = [ χ H , θ ] 0 . Then χ ( 1 ) = e t θ ( 1 ) = 176 , where t = | G : I G ( θ ) | . Since χ ( 1 ) / θ ( 1 ) | G / H | , we have that θ ( 1 ) = 4 or 8. If θ ( 1 ) = 4 , then e t = 44 . Since | H | = 2 4 3 2 , we have that H has at most eight irreducible characters of degree 4. Hence t 4 . We assert that I G ( θ ) = G . If not, then I G ( θ ) < G . Let U containing I G ( θ ) be a maximal subgroup of G. Then 1 | G : U | | G : I G ( θ ) | = 4 . By checking the maximal subgroups of L 2 ( 11 ) (see ATLAS table in [6]), it is easy to get a contradiction. Hence I G ( θ ) = G , and so t = 1 and e = 44 . But e 2 t = [ χ H , χ H ] > | G : H | , a contradiction. If θ ( 1 ) = 8 , then | O 3 ( H ) | = 9 and I G ( θ ) = G . Since H G , we have that O 3 ( H ) G . Let λ Irr ( O 3 ( H ) ) such that [ θ O 3 ( H ) , λ ] 0 . Since θ ( 1 ) = 8 , we have 4 | H : I H ( λ ) | 8 . But I G ( θ ) = G , which implies that 4 | G : I G ( λ ) | = | H : I H ( λ ) | 8 . Let S = g G I G ( λ ) g . Then S G and G / S S 8 . By the Jordan-Hölder theorem, S has a normal series 1 O 3 ( H ) C D S such that D / C L 2 ( 11 ) and | C / O 3 ( H ) | = 1 , 2  or  4 . Let α Irr ( S ) such that [ χ S , α ] 0 . Since χ ( 1 ) / α ( 1 ) | G / S | , we have that 22 α ( 1 ) . Since λ g is invariant in S, for each g G and 4 | G : I G ( λ ) | , we have that each irreducible character is invariant in S and O 3 ( H ) Z ( S ) . Therefore, the following conclusions hold:
  1. (a)

    S L 2 ( 11 ) × O 3 ( H ) if | C / O 3 ( H ) | = 1 ;

     
  2. (b)

    S ( 2 L 2 ( 11 ) ) × O 3 ( H ) or ( Z 2 × L 2 ( 11 ) ) × O 3 ( H ) if | C / O 3 ( H ) | = 2 .

     

By checking the character table of 2 L 2 ( 11 ) and L 2 ( 11 ) , we see that both conclusions (a) and (b) are not satisfied with the above conditions. Now, we suppose that | C / O 3 ( H ) | = 4 . Then 44 α ( 1 ) . Since O 3 ( H ) Z ( S ) , one has that C O 3 ( H ) × B , where B is a group of order 4. Let β be an irreducible component of α C and t 1 = | S : I S ( β ) | . Then β ( 1 ) = 1 and t 1 α ( 1 ) / β ( 1 ) 44 . Since the indexes of the maximal subgroups of S containing I S ( β ) divide t 1 and t 1 | Aut ( C ) | , we have that t 1 = 1 . Hence [ α C , α C ] > | S : C | = 2 2 3 5 11 , a contradiction.

Similarly, we can show that | G / K | 2 .

Suppose that K / H M 11 . Since | Out ( M 11 ) | = Mult ( M 11 ) = 1 , we have G H × M 11 , where | H | = 2 2 3 . By checking the character table of M 11 , we see that G has no irreducible character of degree L ( G ) = 2 4 11 , a contradiction.

If K / H M 12 , since | G | = 2 6 3 3 5 11 , we conclude that | H | = 1 and G = K M 12 , which completes the proof of Theorem A. □

Proof of Theorem B We only need to prove the sufficiency.

In this case, we have | G | = 2 10 3 3 5 7 11 23 and S ( G ) = 2 2 3 2 7 23 . Let χ Irr ( G ) such that χ ( 1 ) = S ( G ) . If O 23 ( G ) 1 , then | O 23 ( G ) | = 23 , which implies that χ ( 1 ) | G : N | , a contradiction. Hence O 23 ( G ) = 1 .

We have to show that G is non-solvable. Assume the contrary, by Lemma 2, we have that the Sylow 23-subgroup is normal in G, a contradiction. Therefore, G is non-solvable.

Since G is non-solvable, by Lemma 1, one has that 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 ) | . As | G | = 2 10 3 3 5 7 11 23 , then K / H can be isomorphic to one of the following simple groups: A 5 , L 2 ( 7 ) , A 6 , L 2 ( 8 ) , L 2 ( 11 ) , A 7 , U 3 ( 3 ) , M 11 , L 3 ( 4 ) , A 8 , M 12 , M 22 , M 23 and M 24 .

We first assume that K / H A 5 . Since | Out ( A 5 ) | = 2 , we have | G / K | 2 and | H | = 2 t 3 2 7 11 23 , where t = 7 or 8. Let θ Irr ( H ) such that [ χ H , θ ] 0 . Since χ ( 1 ) / θ ( 1 ) | G / H | , it implies that 23 θ ( 1 ) . If H is solvable, then O 23 ( H ) 1 by Lemma 2, which implies that O 23 ( H ) = O 23 ( G ) 1 , a contradiction. Thus H is non-solvable. Then there exists a normal series of H: 1 N M H such that M / N is a direct product of isomorphic non-abelian simple groups and | H / M | | Out ( M / N ) | . As | H | = 2 t 3 2 7 11 23 , we have M / N L 2 ( 7 ) or L 2 ( 8 ) , which implies that 23 | N | . Hence O 23 ( N ) 1 by Lemma 2, which implies that O 23 ( N ) = O 23 ( G ) 1 , a contradiction.

By the same arguments as the proof of K / H A 5 , we show that K / H cannot be isomorphic to one of the simple groups: L 2 ( 7 ) , A 6 , L 2 ( 8 ) , L 2 ( 11 ) , A 7 , U 3 ( 3 ) , M 11 , L 3 ( 4 ) , A 8 , M 12 and M 22 .

Suppose that K / H M 23 . Since | Out ( M 23 ) | = Mult ( M 23 ) = 1 , we have that G H × M 23 , where | H | = 2 3 3 . By checking the character table of M 23 , it is easy to see that there exists no irreducible character of degree 2 2 3 2 7 23 in G, a contradiction.

If K / H M 24 , since | G | = 2 10 3 3 5 7 11 23 , one has that | H | = 1 and G = K M 24 , which completes the proof of Theorem B. □

Proof of Theorem C We only need to prove the sufficiency.

In this case, | G | = 2 7 3 2 5 7 11 and L ( G ) = 385 . Let χ Irr ( G ) such that χ ( 1 ) = L ( G ) = 5 7 11 . We assert that O 11 ( G ) = 1 . Otherwise, we have that | O 11 ( G ) | = 11 and O 11 ( G ) is abelian. Hence χ ( 1 ) | G / O 11 ( G ) | , a contradiction. Similarly, O 5 ( G ) = O 7 ( G ) = 1 .

If G is solvable, then the Sylow 11-subgroup of G is normal in G by Lemma 2. But O 11 ( G ) = 1 , a contradiction. Therefore, G is non-solvable.

Since G is non-solvable, by Lemma 1, we get that 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 ) | . As | G | = 2 7 3 2 5 7 11 , we see that K / H is isomorphic to one of the simple groups: A 5 , L 2 ( 7 ) , A 6 , L 2 ( 8 ) , L 2 ( 11 ) , A 7 , M 11 , L 3 ( 4 ) , A 8 and M 22 .

We first assume that K / H A 5 . Since | Out ( A 5 ) | = 2 , we have | G / K | 2 and | H | = 2 t 3 7 11 , where t = 4 or 5. If H is solvable, then O 11 ( H ) = O 11 ( G ) 1 by Lemma 2, a contradiction. Hence H is non-solvable and H has a normal series 1 N M H such that M / N is a direct product of isomorphic non-abelian simple groups and | H / M | | Out ( M / N ) | . As | H | = 2 t 3 7 11 , one has that M / N L 3 ( 2 ) and | N | = 2 s 11 , where 0 s 2 . Let P be the Sylow 11-subgroup of N. Then P is normal in N by Sylow theorem. Since P is also a Sylow 11-subgroup in G and N is subnormal in G, we have P G , a contradiction.

Similarly, K / H cannot be isomorphic to the simple groups: L 2 ( 7 ) , A 6 , L 2 ( 8 ) , L 2 ( 11 ) , A 7 , L 3 ( 4 ) or A 8 .

Assume that K / H L 2 ( 11 ) . Since | Out ( L 2 ( 11 ) ) | = 2 , we have | G / K | 2 and | H | = 2 α 3 7 , where α = 4 or 5. Suppose that H = H . Then H has a normal subgroup S such that H / S L 2 ( 7 ) , where | S | = 2 or 4. Obviously, we know that S Z ( H ) , and then S G . Let θ Irr ( S ) such that [ χ S , θ ] 0 . Then θ ( 1 ) = 1 since S is abelian. Let e = [ χ S , θ ] and t = | G : I G ( θ ) | . Then t = 1 and e = χ ( 1 ) = 385 by the Clifford theorem (see Theorem 6.2 in [5]). But e 2 t = [ χ H , χ H ] > | G : H | , a contradiction. Hence H < H . Suppose that | H / H | = 2 . Then H / H is central in G / H . Let β be an irreducible component of χ H , and let θ be an irreducible component of β H . Then θ ( 1 ) = β ( 1 ) = 7 and θ is extendible to β. Hence λβ is invariant in G for every λ Irr ( H / H ) if β is invariant in G. Since | H | = 2 α 3 7 11 , where α = 4 or 5, H has at most 12 irreducible characters of degree 7. Let t = | G : I G ( β ) | . Then t 12 . Since the index of the maximal subgroup of U containing I G ( θ ) divides t, we have that t = 1 or 11 by checking maximal subgroups of L 2 ( 11 ) (see ATLAS table in [6]). If t = 11 , then H has exactly 12 irreducible characters of degree 7, and one of them, say δ, is invariant in G. Hence, λδ is also invariant in G for λ Irr ( H / H ) , which forces t 10 , a contradiction. Therefore t = 1 and e = 55 . But ( 55 ) 2 = [ χ H , χ H ] > | G : H | , a contradiction. By the same reasoning as before, we can prove that | H / H | 2 m 3 n , where 1 m 3 and 0 n 1 . If | H / H | = 2 m 3 n , where m = 4 or 5, then the Sylow 7 subgroup of H is normal in H , and so it is normal in G, a contradiction. Now we assume that 7 | H / H | . Let H A H such that | H / A | = 7 , then H / A Z ( G / A ) . Since Mult ( L 2 ( 11 ) ) = 2 , we have G / A H / A × L 2 ( 11 ) . Hence G has a normal series 1 N M G such that M / N L 2 ( 11 ) and | N | = 2 α 3 , where α = 4 or 5. Let φ be an irreducible component of χ M , and let η be an irreducible component of φ N . Then φ ( 1 ) = 55 and η ( 1 ) = 1 by the Clifford theorem (see Theorem 6.2 in [5]). Since | Aut ( N ) | is not divided by 5 and 11, one has that t = | M : I M ( η ) | = 1 . Therefore ( 55 ) 2 = [ φ N , φ N ] > | M : N | , a contradiction.

Suppose that K / H M 11 . Since | Out ( M 11 ) | = Mult ( M 11 ) = 1 , we have G H × M 11 , where | H | = 2 3 7 . Let θ Irr ( H ) such that [ χ H , θ ] 0 . Since θ ( 1 ) χ ( 1 ) and θ ( 1 ) | H | , one has that θ ( 1 ) = 7 , which implies that H is a Frobenius group with an elementary kernel of order 8 and a cyclic complement of order 7.

Suppose that K / H M 22 . Since | G | = 2 7 3 2 5 7 11 , we have that | H | = 1 and G = K M 22 , which completes the proof of Theorem C. □

Declarations

Acknowledgements

Article is supported by the Natural Science Foundation of China (Grant No. 11271301; 11171364), The Fundamental Research Funds for the Central Universities (Grant No. XDJK2009C074), Graduate-Innovation Funds of Science of SWU (Grant No. ky2009013) and Natural Science Foundation Project of CQ CSTC (Grant No. cstc2011jjA00020).

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Southwest University
(2)
Department of Mathematics and Information Engineering, Chongqing University of Education

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