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

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.

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 cd(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 GM 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 1HKG 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/KOut(M/N) and K/HM/N is a direct product of isomorphic non-abelian simple groups. Thus 1HKG 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 in1 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 PG. Let Q be a Sylow p n -subgroup of P. Then P=NQ. By Sylow’s theorem, |P: N P (Q)|= p i l (lm 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 QP and QG. □

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 2 511 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 χIrr(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 1HKG 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 2 511, 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 311, where t=1 or 2. Let χIrr(G) such that χ(1)=L(G)=55 and θIrr(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 311, we have H is solvable. Let N be a Sylow 11-subgroup of H. Then NH 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 θIrr(H) such that e=[ χ H ,θ]0 and let t=|G: I G (θ)|. Then θ(1)=1 and et=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 3511, where 2b3, 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 571123 and L(G)= 2 3 1123. Then O 23 (G)=1. If not, then | O 23 (G)|=23 and O 23 (G) is abelian. Hence L(G)= 2 3 1123|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 1HKG 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 571123, 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 371123, where m=4 or 5. Suppose that H is non-solvable. By Lemma 1, H has a normal series 1ABH such that B/A is a direct product of isomorphic non-abelian simple groups and |H/B||Out(B/A)|. Since |H|= 2 m 371123, we have B/A L 2 (7) and |H/B|2. Thus |A|= 2 a 1123, where 0a2. Let N be a Sylow 23-subgroup of A. Then NA by Lemma 2. Hence we get a subnormal series of G, NcharABHKG, which implies that NG. 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 571123, we have that |H|=1 and G=K M 23 .

Case 1.3 M= M 12

In this case, |G|= 2 6 3 3 511 and L(G)= 2 4 11. Since 11L(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 1HKG 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 3 511, 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 χIrr(G) such that χ(1)=L(G)= 2 4 11 and θIrr(H) such that e=[ χ H ,θ]0. Then χ(1)=etθ(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 et=44. Since |H|= 2 4 3 2 , we have that H has at most eight irreducible characters of degree 4. Hence t4. 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 HG, we have that O 3 (H)G. Let λIrr( 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 SG and G/S S 8 . By the Jordan-Hölder theorem, S has a normal series 1 O 3 (H)CDS such that D/C L 2 (11) and |C/ O 3 (H)|=1,2 or 4. Let αIrr(S) such that [ χ S ,α]0. Since χ(1)/α(1)|G/S|, we have that 22α(1). Since λ g is invariant in S, for each gG 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 44α(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 3511, 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 GH× 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 3 511, 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 571123 and S(G)= 2 2 3 2 723. Let χIrr(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 1HKG 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 571123, 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 71123, where t=7 or 8. Let θIrr(H) such that [ χ H ,θ]0. Since χ(1)/θ(1)|G/H|, it implies that 23θ(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: 1NMH 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 71123, 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 GH× 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 2 723 in G, a contradiction.

If K/H M 24 , since |G|= 2 10 3 3 571123, 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 5711 and L(G)=385. Let χIrr(G) such that χ(1)=L(G)=5711. 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 1HKG 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 5711, 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 3711, 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 1NMH such that M/N is a direct product of isomorphic non-abelian simple groups and |H/M||Out(M/N)|. As |H|= 2 t 3711, one has that M/N L 3 (2) and |N|= 2 s 11, where 0s2. 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 PG, 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 37, 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 SZ(H), and then SG. Let θIrr(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 λIrr(H/ H ) if β is invariant in G. Since |H|= 2 α 3711, where α=4 or 5, H has at most 12 irreducible characters of degree 7. Let t=|G: I G (β)|. Then t12. 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 λIrr(H/ H ), which forces t10, 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 1m3 and 0n1. 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 AH such that |H/A|=7, then H/AZ(G/A). Since Mult( L 2 (11))=2, we have G/AH/A× L 2 (11). Hence G has a normal series 1NMG 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 GH× M 11 , where |H|= 2 3 7. Let θIrr(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 5711, we have that |H|=1 and G=K M 22 , which completes the proof of Theorem C. □

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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).

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Correspondence to Guiyun Chen.

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Authors’ contributions

HX carried out the study of the Mathieu groups M 11 , M 12 and M 23 . YY carried out the study of the Mathieu group M 24 . GC carried out the study of the Mathieu group M 22 . All authors read and approved the final manuscript.

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Xu, H., Yan, Y. & Chen, G. A new characterization of Mathieu-groups by the order and one irreducible character degree. J Inequal Appl 2013, 209 (2013). https://doi.org/10.1186/1029-242X-2013-209

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Keywords

  • finite group
  • simple group
  • character degree