 Research
 Open Access
 Published:
A new characterization of Mathieugroups by the order and one irreducible character degree
Journal of Inequalities and Applications volume 2013, Article number: 209 (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 Mathieugroup 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 nonabelian simple group G is characterized by the set cd(G) of degrees of its complex irreducible characters. In [1–4], it was shown that many nonabelian 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\cong M if and only if the following conditions are fulfilled:

(1)
G=M;

(2)
L(G)=L(M).
Theorem B Let G be a finite group. Then G\cong {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\times {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 nonsolvable group. Then G has a normal series 1\u22b4H\u22b4K\u22b4G such that K/H is a direct product of isomorphic nonabelian simple groups and G/K\mid Out(K/H).
Proof Let G be a nonsolvable group. Then G has a chief factor M/N such that M/N is a direct product of isomorphic nonabelian simple groups. Hence {C}_{G/N}(M/N)\cap M/N=Z(M/N)=1, and so
Let K/N={C}_{G/N}(M/N)\times M/N and H/N={C}_{G/N}(M/N). Then G/K\le Out(M/N) and K/H\cong M/N is a direct product of isomorphic nonabelian simple groups. Thus 1\u22b4H\u22b4K\u22b4G 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}}\cdots {p}_{n}^{{a}_{n}}, where {p}_{1},{p}_{2},\dots ,{p}_{n} are distinct primes. If k{p}_{n}+1\nmid {p}_{i}^{{a}_{i}} for each i\le n1 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\u22b4G. Let Q be a Sylow {p}_{n}subgroup of P. Then P=NQ. By Sylow’s theorem, P:{N}_{P}(Q)={p}_{i}^{l} (l\le m\le {a}_{i}) and {p}_{n}\mid {p}_{i}^{l}1. But this means that k{p}_{n}+1\mid {p}^{{a}_{i}}, and then k=0 by assumption. Hence Q\u22b4P and Q\u22b4G. □
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}\cdot {3}^{2}\cdot 5\cdot 11 and L(G)=55. We first show that G is nonsolvable. Assume the contrary. By Lemma 2, we know that the Sylow 11subgroup of G is normal in G. Let N be the 11Sylow subgroup of G. Since N is abelian, we have \chi (1)\mid G/N for all \chi \in Irr(G). But L(G)=55 and 55\nmid G/N, a contradiction. Therefore, G is nonsolvable.
Since G is nonsolvable, by Lemma 1, we get that G has a normal series 1\u22b4H\u22b4K\u22b4G such that K/H is a direct product of isomorphic nonabelian simple groups and G/K\mid Out(K/H). As G={2}^{4}\cdot {3}^{2}\cdot 5\cdot 11, we have K/H\cong {A}_{5},{A}_{6},{L}_{2}(11)\text{or}{M}_{11}.
We first assume that K/H\cong {A}_{5}. Since Out({A}_{5})=2, we have G/K\mid 2 and H={2}^{t}\cdot 3\cdot 11, where t=1 or 2. Let \chi \in Irr(G) such that \chi (1)=L(G)=55 and \theta \in Irr(H) such that [{\chi}_{H},\theta ]\ne 0. Then \theta (1)=11 by the Clifford theorem (see Theorem 6.2 in [5]). On the other hand, since H={2}^{t}\cdot 3\cdot 11, we have H is solvable. Let N be a Sylow 11subgroup of H. Then N\u22b4H by Lemma 2. Hence \theta (1)\mid H/N={2}^{t}\cdot 3, a contradiction.
By the same reason as above, one has that K/H\ncong {A}_{6}.
Suppose that K/H\cong {L}_{2}(11). Since Out({L}_{2}(11))=2, we have G/K\mid 2 and so H={2}^{a}\cdot 3, where a=1 or 2. Let \theta \in Irr(H) such that e=[{\chi}_{H},\theta ]\ne 0 and let t=G:{I}_{G}(\theta ). Then \theta (1)=1 and et=\chi (1)/\theta (1)=55. Since H={2}^{t}\cdot 3, where a=1 or 2, we have that 55\nmid Aut(H/{H}^{\mathrm{\prime}}). Hence t=1, e=55. But {(55)}^{2}={e}^{2}t=[{\chi}_{H},{\chi}_{H}]>G:H={2}^{b}\cdot 3\cdot 5\cdot 11, where 2\le b\le 3, a contradiction.
If K/H\cong {M}_{11}, by comparing the orders of G and {M}_{11}, we have H=1. Therefore G=K\cong {M}_{11}.
Case 1.2 M={M}_{23}
In this case, we have G={2}^{7}\cdot {3}^{2}\cdot 5\cdot 7\cdot 11\cdot 23 and L(G)={2}^{3}\cdot 11\cdot 23. Then {O}_{23}(G)=1. If not, then {O}_{23}(G)=23 and {O}_{23}(G) is abelian. Hence L(G)={2}^{3}\cdot 11\cdot 23\mid G/{O}_{23}(G), a contradiction.
If G is solvable, then the Sylow 23subgroup of G is normal in G by Lemma 2, which leads to a contradiction as above. Therefore G is nonsolvable.
Since G is nonsolvable, by Lemma 1, we get that G has a normal series 1\u22b4H\u22b4K\u22b4G such that K/H is a direct product of isomorphic nonabelian simple groups and G/K\mid Out(K/H). As G={2}^{7}\cdot {3}^{2}\cdot 5\cdot 7\cdot 11\cdot 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\cong {A}_{5}. Since Out({A}_{5})=2, we have G/K\mid 2 and H={2}^{m}\cdot 3\cdot 7\cdot 11\cdot 23, where m=4 or 5. Suppose that H is nonsolvable. By Lemma 1, H has a normal series 1\u22b4A\u22b4B\u22b4H such that B/A is a direct product of isomorphic nonabelian simple groups and H/B\mid Out(B/A). Since H={2}^{m}\cdot 3\cdot 7\cdot 11\cdot 23, we have B/A\cong {L}_{2}(7) and H/B\mid 2. Thus A={2}^{a}\cdot 11\cdot 23, where 0\le a\le 2. Let N be a Sylow 23subgroup of A. Then N\u22b4A by Lemma 2. Hence we get a subnormal series of G, N\mathit{char}A\u22b4B\u22b4H\u22b4K\u22b4G, which implies that N\u22b4G. But {O}_{23}(G)=1, a contradiction. If H is solvable, then the Sylow 23subgroup 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\cong {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\cong {M}_{23}, since G={2}^{7}\cdot {3}^{2}\cdot 5\cdot 7\cdot 11\cdot 23, we have that H=1 and G=K\cong {M}_{23}.
Case 1.3 M={M}_{12}
In this case, G={2}^{6}\cdot {3}^{3}\cdot 5\cdot 11 and L(G)={2}^{4}\cdot 11. Since 11\mid 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 nonsolvable. If G is solvable, then the Sylow 11subgroup of G is normal in G by Lemma 2, a contradiction. Therefore, G is nonsolvable.
By Lemma 1, we get that G has a normal series 1\u22b4H\u22b4K\u22b4G such that K/H is a direct product of isomorphic nonabelian simple groups and G/K\mid Out(K/H). As G={2}^{6}\cdot {3}^{3}\cdot 5\cdot 11, we have K/H\cong {A}_{5},{A}_{6},{L}_{2}(11),{M}_{11}\text{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\cong {L}_{2}(11). Since Out({L}_{2}(11))=2, we have G/K\mid 2 and H={2}^{a}\cdot 3, where a=3 or 4. Suppose that G/K=1. Then H={2}^{4}\cdot {3}^{2}. Let \chi \in Irr(G) such that \chi (1)=L(G)={2}^{4}\cdot 11 and \theta \in Irr(H) such that e=[{\chi}_{H},\theta ]\ne 0. Then \chi (1)=et\theta (1)=176, where t=G:{I}_{G}(\theta ). Since \chi (1)/\theta (1)\mid G/H, we have that \theta (1)=4 or 8. If \theta (1)=4, then et=44. Since H={2}^{4}\cdot {3}^{2}, we have that H has at most eight irreducible characters of degree 4. Hence t\le 4. We assert that {I}_{G}(\theta )=G. If not, then {I}_{G}(\theta )<G. Let U containing {I}_{G}(\theta ) be a maximal subgroup of G. Then 1\le G:U\mid G:{I}_{G}(\theta )=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}(\theta )=G, and so t=1 and e=44. But {e}^{2}\cdot t=[{\chi}_{H},{\chi}_{H}]>G:H, a contradiction. If \theta (1)=8, then {O}_{3}(H)=9 and {I}_{G}(\theta )=G. Since H\u22b4G, we have that {O}_{3}(H)\u22b4G. Let \lambda \in Irr({O}_{3}(H)) such that [{\theta}_{{O}_{3}(H)},\lambda ]\ne 0. Since \theta (1)=8, we have 4\le H:{I}_{H}(\lambda )\le 8. But {I}_{G}(\theta )=G, which implies that 4\le G:{I}_{G}(\lambda )=H:{I}_{H}(\lambda )\le 8. Let S={\bigcap}_{g\in G}{I}_{G}{(\lambda )}^{g}. Then S\u22b4G and G/S\lesssim {S}_{8}. By the JordanHölder theorem, S has a normal series 1\u22b4{O}_{3}(H)\u22b4C\u22b4D\u22b4S such that D/C\cong {L}_{2}(11) and C/{O}_{3}(H)=1,2\text{or}4. Let \alpha \in Irr(S) such that [{\chi}_{S},\alpha ]\ne 0. Since \chi (1)/\alpha (1)\mid G/S, we have that 22\mid \alpha (1). Since {\lambda}^{g} is invariant in S, for each g\in G and 4\le G:{I}_{G}(\lambda ), we have that each irreducible character is invariant in S and {O}_{3}(H)\le Z(S). Therefore, the following conclusions hold:

(a)
S\cong {L}_{2}(11)\times {O}_{3}(H) if C/{O}_{3}(H)=1;

(b)
S\cong (2\cdot {L}_{2}(11))\times {O}_{3}(H) or ({Z}_{2}\times {L}_{2}(11))\times {O}_{3}(H) if C/{O}_{3}(H)=2.
By checking the character table of 2\cdot {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\mid \alpha (1). Since {O}_{3}(H)\le Z(S), one has that C\cong {O}_{3}(H)\times B, where B is a group of order 4. Let β be an irreducible component of {\alpha}_{C} and {t}_{1}=S:{I}_{S}(\beta ). Then \beta (1)=1 and {t}_{1}\mid \alpha (1)/\beta (1)\mid 44. Since the indexes of the maximal subgroups of S containing {I}_{S}(\beta ) divide {t}_{1} and {t}_{1}\mid Aut(C), we have that {t}_{1}=1. Hence [{\alpha}_{C},{\alpha}_{C}]>S:C={2}^{2}\cdot 3\cdot 5\cdot 11, a contradiction.
Similarly, we can show that G/K\ne 2.
Suppose that K/H\cong {M}_{11}. Since Out({M}_{11})=Mult({M}_{11})=1, we have G\cong H\times {M}_{11}, where H={2}^{2}\cdot 3. By checking the character table of {M}_{11}, we see that G has no irreducible character of degree L(G)={2}^{4}\cdot 11, a contradiction.
If K/H\cong {M}_{12}, since G={2}^{6}\cdot {3}^{3}\cdot 5\cdot 11, we conclude that H=1 and G=K\cong {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}\cdot {3}^{3}\cdot 5\cdot 7\cdot 11\cdot 23 and S(G)={2}^{2}\cdot {3}^{2}\cdot 7\cdot 23. Let \chi \in Irr(G) such that \chi (1)=S(G). If {O}_{23}(G)\ne 1, then {O}_{23}(G)=23, which implies that \chi (1)\mid G:N, a contradiction. Hence {O}_{23}(G)=1.
We have to show that G is nonsolvable. Assume the contrary, by Lemma 2, we have that the Sylow 23subgroup is normal in G, a contradiction. Therefore, G is nonsolvable.
Since G is nonsolvable, by Lemma 1, one has that G has a normal series 1\u22b4H\u22b4K\u22b4G such that K/H is a direct product of isomorphic nonabelian simple groups and G/K\mid Out(K/H). As G={2}^{10}\cdot {3}^{3}\cdot 5\cdot 7\cdot 11\cdot 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\cong {A}_{5}. Since Out({A}_{5})=2, we have G/K\mid 2 and H={2}^{t}\cdot {3}^{2}\cdot 7\cdot 11\cdot 23, where t=7 or 8. Let \theta \in Irr(H) such that [{\chi}_{H},\theta ]\ne 0. Since \chi (1)/\theta (1)\mid G/H, it implies that 23\mid \theta (1). If H is solvable, then {O}_{23}(H)\ne 1 by Lemma 2, which implies that {O}_{23}(H)={O}_{23}(G)\ne 1, a contradiction. Thus H is nonsolvable. Then there exists a normal series of H: 1\u22b4N\u22b4M\u22b4H such that M/N is a direct product of isomorphic nonabelian simple groups and H/M\mid Out(M/N). As H={2}^{t}\cdot {3}^{2}\cdot 7\cdot 11\cdot 23, we have M/N\cong {L}_{2}(7) or {L}_{2}(8), which implies that 23\mid N. Hence {O}_{23}(N)\ne 1 by Lemma 2, which implies that {O}_{23}(N)={O}_{23}(G)\ne 1, a contradiction.
By the same arguments as the proof of K/H\cong {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\cong {M}_{23}. Since Out({M}_{23})=Mult({M}_{23})=1, we have that G\cong H\times {M}_{23}, where H={2}^{3}\cdot 3. By checking the character table of {M}_{23}, it is easy to see that there exists no irreducible character of degree {2}^{2}\cdot {3}^{2}\cdot 7\cdot 23 in G, a contradiction.
If K/H\cong {M}_{24}, since G={2}^{10}\cdot {3}^{3}\cdot 5\cdot 7\cdot 11\cdot 23, one has that H=1 and G=K\cong {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}\cdot {3}^{2}\cdot 5\cdot 7\cdot 11 and L(G)=385. Let \chi \in Irr(G) such that \chi (1)=L(G)=5\cdot 7\cdot 11. We assert that {O}_{11}(G)=1. Otherwise, we have that {O}_{11}(G)=11 and {O}_{11}(G) is abelian. Hence \chi (1)\mid G/{O}_{11}(G), a contradiction. Similarly, {O}_{5}(G)={O}_{7}(G)=1.
If G is solvable, then the Sylow 11subgroup of G is normal in G by Lemma 2. But {O}_{11}(G)=1, a contradiction. Therefore, G is nonsolvable.
Since G is nonsolvable, by Lemma 1, we get that G has a normal series 1\u22b4H\u22b4K\u22b4G such that K/H is a direct product of isomorphic nonabelian simple groups and G/K\mid Out(K/H). As G={2}^{7}\cdot {3}^{2}\cdot 5\cdot 7\cdot 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\cong {A}_{5}. Since Out({A}_{5})=2, we have G/K\mid 2 and H={2}^{t}\cdot 3\cdot 7\cdot 11, where t=4 or 5. If H is solvable, then {O}_{11}(H)={O}_{11}(G)\ne 1 by Lemma 2, a contradiction. Hence H is nonsolvable and H has a normal series 1\u22b4N\u22b4M\u22b4H such that M/N is a direct product of isomorphic nonabelian simple groups and H/M\mid Out(M/N). As H={2}^{t}\cdot 3\cdot 7\cdot 11, one has that M/N\cong {L}_{3}(2) and N={2}^{s}\cdot 11, where 0\le s\le 2. Let P be the Sylow 11subgroup of N. Then P is normal in N by Sylow theorem. Since P is also a Sylow 11subgroup in G and N is subnormal in G, we have P\u22b4G, 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\cong {L}_{2}(11). Since Out({L}_{2}(11))=2, we have G/K\mid 2 and H={2}^{\alpha}\cdot 3\cdot 7, where \alpha =4 or 5. Suppose that H={H}^{\mathrm{\prime}}. Then H has a normal subgroup S such that H/S\cong {L}_{2}(7), where S=2 or 4. Obviously, we know that S\le Z(H), and then S\u22b4G. Let \theta \in Irr(S) such that [{\chi}_{S},\theta ]\ne 0. Then \theta (1)=1 since S is abelian. Let e=[{\chi}_{S},\theta ] and t=G:{I}_{G}(\theta ). Then t=1 and e=\chi (1)=385 by the Clifford theorem (see Theorem 6.2 in [5]). But {e}^{2}\cdot t=[{\chi}_{H},{\chi}_{H}]>G:H, a contradiction. Hence {H}^{\mathrm{\prime}}<H. Suppose that H/{H}^{\mathrm{\prime}}=2. Then H/{H}^{\mathrm{\prime}} is central in G/H. Let β be an irreducible component of {\chi}_{H}, and let θ be an irreducible component of {\beta}_{{H}^{\mathrm{\prime}}}. Then \theta (1)=\beta (1)=7 and θ is extendible to β. Hence λβ is invariant in G for every \lambda \in Irr(H/{H}^{\mathrm{\prime}}) if β is invariant in G. Since H={2}^{\alpha}\cdot 3\cdot 7\cdot 11, where \alpha =4 or 5, H has at most 12 irreducible characters of degree 7. Let t=G:{I}_{G}(\beta ). Then t\le 12. Since the index of the maximal subgroup of U containing {I}_{G}(\theta ) 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 \lambda \in Irr(H/{H}^{\mathrm{\prime}}), which forces t\le 10, a contradiction. Therefore t=1 and e=55. But {(55)}^{2}=[{\chi}_{H},{\chi}_{H}]>G:H, a contradiction. By the same reasoning as before, we can prove that H/{H}^{\mathrm{\prime}}\ne {2}^{m}\cdot {3}^{n}, where 1\le m\le 3 and 0\le n\le 1. If H/{H}^{\mathrm{\prime}}={2}^{m}\cdot {3}^{n}, where m=4 or 5, then the Sylow 7 subgroup of {H}^{\mathrm{\prime}} is normal in {H}^{\mathrm{\prime}}, and so it is normal in G, a contradiction. Now we assume that 7\mid H/{H}^{\mathrm{\prime}}. Let {H}^{\mathrm{\prime}}\le A\u22b4H such that H/A=7, then H/A\le Z(G/A). Since Mult({L}_{2}(11))=2, we have G/A\cong H/A\times {L}_{2}(11). Hence G has a normal series 1\u22b4N\u22b4M\u22b4G such that M/N\cong {L}_{2}(11) and N={2}^{\alpha}\cdot 3, where \alpha =4 or 5. Let φ be an irreducible component of {\chi}_{M}, and let η be an irreducible component of {\phi}_{N}. Then \phi (1)=55 and \eta (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}(\eta )=1. Therefore {(55)}^{2}=[{\phi}_{N},{\phi}_{N}]>M:N, a contradiction.
Suppose that K/H\cong {M}_{11}. Since Out({M}_{11})=Mult({M}_{11})=1, we have G\cong H\times {M}_{11}, where H={2}^{3}\cdot 7. Let \theta \in Irr(H) such that [{\chi}_{H},\theta ]\ne 0. Since \theta (1)\mid \chi (1) and \theta (1)\mid H, one has that \theta (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\cong {M}_{22}. Since G={2}^{7}\cdot {3}^{2}\cdot 5\cdot 7\cdot 11, we have that H=1 and G=K\cong {M}_{22}, which completes the proof of Theorem C. □
References
Huppert B: Some simple groups which are determined by the set of their character degrees, I. Ill. J. Math. 2000, 44(4):828–842.
Huppert B: Some simple groups which are determined by the set of their character degrees, II. Rend. Semin. Mat. Univ. Padova 2006, 115: 1–13.
Wakefield TP:Verifying Huppert’s conjecture for PS{L}_{3}(q) and PS{U}_{3}({q}^{2}). Commun. Algebra 2009, 37: 2887–2906. 10.1080/00927870802625661
TongViet HP, Wakefield TP:On Huppert’s conjecture for {G}^{2}(q), q\ge 7. J. Pure Appl. Algebra 2012, 216: 2720–2729. 10.1016/j.jpaa.2012.03.028
Isaacs IM: Character Theory of Finite Groups. Academic Press, New York; 1976.
Conway JH, Nortion SP, Parker RA, Wilson RA: Atlas of Finite Groups. Oxford University Press, Oxford; 1985.
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), GraduateInnovation Funds of Science of SWU (Grant No. ky2009013) and Natural Science Foundation Project of CQ CSTC (Grant No. cstc2011jjA00020).
Author information
Authors and Affiliations
Corresponding author
Additional information
Competing interests
The authors declare that they have no competing interests.
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.
Rights and permissions
Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License ( https://creativecommons.org/licenses/by/2.0 ), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
About this article
Cite this article
Xu, H., Yan, Y. & Chen, G. A new characterization of Mathieugroups by the order and one irreducible character degree. J Inequal Appl 2013, 209 (2013). https://doi.org/10.1186/1029242X2013209
Received:
Accepted:
Published:
DOI: https://doi.org/10.1186/1029242X2013209
Keywords
 finite group
 simple group
 character degree